Handbook_on_Material_and_Energy_Balance_Calculat1

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 225

EXAMPLE 4.11 — Catalyst Reactivation.

A granular catalyst is reactivated by oxidation in a fluid-bed furnace. Tests showed that the
catalyst was properly reactivated if it remained in the furnace between 7 and 13 minutes. Less than
7 minutes, the impurities were incompletely oxidized. Longer than 13 minutes, the catalyst pores
became overly sintered and it lost some effectiveness. Determine the optimum bed capacity Mfor
a feed (and withdrawal) rate of 100 lb/min for a one-stage and two-stage system

Solution. Since the midpoint of the time span is 10 min, the retention time Θ for a single reactor
should be about 10 min, so Mwill be about 1000 lb. Thus the exponential factor in Equation [4.5]
will be -100^/M For M= 1000 lb, the fraction of catalyst particles remaining in the bed after 7
minutes is 0.4966, while the fraction remaining after 13 minutes is 0.2725. Thus only 22.41 % of
the catalyst particles will reside in the bed for the specified 6 minute time span. Figure 4.59 shows
results of calculations carried out for a range of M values. The optimum bed mass is about 970 lb,
but still only 22.42 % of the catalyst particles will be properly reactivated. Practically there is very
little difference in the per cent retained for a bed mass between 900 and 1050 lb.

Figure 4.59 Fraction of catalyst particles that remain in reactor bed between 7 and 13 minutes at a
feed rate of 100 lb/min. Text box shows Trendline equation for the relationship.

For two reactors connected in series, a calculation can be made using Equation [4.5]. The
results were plotted for a range of M as shown in Figure 4.60. With two reactors, the optimum
total bed mass is about 730 lb (365 lb for each reactor) and about 36.1 % of the catalyst can be
satisfactorily reactivated. This is clearly a major improvement over a single reactor.

In this example, we fixed the feed rate at 100 lb/min and varied the reactor capacity.
However the critical variable is really the mean retention time 0, which is probably a better way of
expressing system characteristics in this type of reactor.

Assignment. Suppose the oxidizing conditions and reactor temperature could be adjusted so that
satisfactory catalyst reactivation could occur over a longer time span with no danger of sintering
the catalyst material. Laboratory tests indicate that the fraction of a mass of catalysts that is
properly reactivated conforms to the following equation, where fa = fraction reactivated, t =
residence time in minutes, and τ is an equation coefficient in minutes.

fa/(l-fa) = t/x

A single batch test indicated that fa = 0.5 after 10 minutes of exposure to the reactor
conditions. Calculate the bed mass for a single and a 2-stage reactor such that 75% of the catalyst
in the exit stream would be reactivated at a feed rate of 100 lb/min.

226 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

Figure 4.60 Fraction of catalyst particles that remain in two-reactor system between 7 and 13
minutes at a feed rate is 100 lb/min. M refers to the total bed mass for the system. Text box shows
Trendline equation for the relationship.

4.14.2 Unsteady-State Systems

If a CSTR operating steadily experiences a change in composition of a substance in its feed,
there will be a gradual change in the composition of the exit stream. For example, if a step change
in powder composition in the feed is made from a mass fraction of 0.013 to 0.018 at / = 0, the
value of nt at / = oo would be 0.018. Thus, the overall change in mass fraction with time is 0.018 -
0.013 = 0.005 and hence is equivalent to n0 in Equation [4.5]. The change in mass fraction at any
time t is the concentration at / - 0.013, and hence is equivalent to nt in Equation [4.5]. This gives:

cx-c2~e [4-8]

where с is the composition of a solute species (or a second phase) at any time t, c\ is the
composition at t = 0, and c2 is the composition at infinite time (providing there are no further
changes in the system).

An example of the application of retention time calculations can be shown using Equation
[4.6] for three different retention times, as shown in Figure 4.61. Suppose at / = 0 a change is
made in an input stream. At the retention time 0 for each curve, the value of/ = 0.368, which
indicates that the response to the initial change is 63.2 % complete. The response is 86.5 %
complete after 20. Another characteristic of the mixing curve is that the most probable retention
time is zero; that is, a larger fraction of the input stream species leaves in the first second after
mixing than in any other one-second interval.

Figure 4.61 illustrates quantitatively our earlier observation that a substantial part of the new
feed material is retained in the tank for a very short time, while another fraction remains much
longer than the mean retention time. We saw earlier that one way to mitigate this situation is to
employ multiple smaller devices in series. Another way is to design the CSTR to have a much
different width than length, so that some degree of plug flow is attained.

The retention time is therefore an indicator of the rate of response to a change in feed or
reaction rate. If well-mixed, the change in composition of the exit stream will be immediate, but
there will be a considerable lag before the change is 70 or 80 % complete. An appreciation of the
nature and magnitude of the time lag of a system in responding to changes is important in
attempting to control or adjust the process.

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 227

ί 0.8 Unsteady-State Mixing
Я RT = 100 min
S 0.6 RT = 150 min

- - - RT = 200 min
О RT values

£E£я 0.4
0.2

100 200 300 400 500
minutes after change

Figure 4.61 Fraction of newly-added stream remaining in tank vs. time, for three different
retention times (RT). The symbol О represents the fraction at the retention time Θ.

Going back to the case where powder is being fed with water to a mixing tank, suppose the
flow of water into a 25 000-gallon tank was 150 gal/min, and the flow of powder was 3 lb/min. At
steady state, the effluent slurry will contain 0.02 lb of powder/gal of solution*. If the flowrate of
powder is increased to 5 lb/min, the concentration of powder in the effluent slurry will increase
with time. How fast will this occur?

The retention time 0 for the mixing tank is 25 000/150 =167 min, c\ = 0.02 lb/gal, and c2 =
0.0333 lb/gal. A series of calculations using Equation [4.8] from 0 to 500 min gives the results
shown in Figure 4.62. The concentration reaches 63.2 % of the final concentration at 167 minutes
(the retention time).

0.034 -1 Powder Concentration in Tank
_ 0.032
5> 0.030 ^00Ju—— "
| 0.028
§ 0.026 100 200 300 400 5()0
n 0.024 time, min

0.022
0.020

()

Figure 4.62 Concentration of powder in tank and effluent slurry as a function of time after
changing the input flowrate of powder from 3 to 5 lb/min. Concentration asymptotically
approaches the new inlet concentration of 0.0333 lb/gal. The point labeled D indicates the
concentration at the retention time of 167 minutes (0.0284 lb/gal).

Two other factors affecting retention time are short-circuiting and stagnation. Both
conditions cause concentration gradients in the tank. In the first condition, the position of the feed
entrance and product exit may be too close. Baffles can be added to mixing tanks to minimize
short-circuiting. Stagnation occurs when part of the reactor contents are static, or accumulate on
walls or the bottom as accretions. A very large effort has been made to understand flow and

The assumption is made here thatflowsand volume are effectively constant whatever the powder content.

228 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

mixing during the refining of steel and aluminum to minimize short-circuiting and stagnant zones.
The results have been very effective in changing the way refining and casting vessels are designed.

The foregoing discussion of concentration and retention time also apply to temperature. At
steady state, the feed stream and tank contents are at the same temperature. If the feed stream
temperature changes, the response is similar to that of a change in composition. The tank will
reach about 63 % of the way to the new temperature after Θ minutes.

EXAMPLE 4.12 — Dissolution ofZnCl2.

A zinc plating process produces 3000 kg/h of residue containing wZnCl2 between 30 and 40
%. A new batch of residue is brought to the leach plant every eight hours; each new batch usually
has a different ZnCl2 content. The ZnCl2 is leached from the residue by water at 40 °C in an 8000
L tank. The dissolution rate of ZnCl2 is very fast. The aim is to produce an effluent stream having
wZnCl2 of 15 % in the liquid portion of the effluent stream, but the operator tolerates between 13
% and 17 % ZnCl2. If the % ZnCl2 goes out of this range, the flow rate of water is changed.
Calculate the mean retention time for operation at 15.0 % ZnCl2. Then, given the data in the table
below (where at / = 0 a new batch of residue arrived at the leach plant), determine if any corrective
action should be taken to avoid having the process go out of tolerance. The flow rate of water at /
= 0 was 96.9 kg/min.

Data. The density of ZnCl2 is 2.9 kg/L, and the density of water is 0.992 kg/L. The density of
ZnCl2 solutions is given by: pZnC\2 = 0.985 + 0.0096(%ZnCl2).

time, min 0 5 10 15 20 25 30
% ZnCl2 15.0 15.2 15.5 15.8 15.9 16.1 16.3

Solution. Assuming steady state up to t = 0, the amount of ZnCl2 leached has been 17.1 kg/min.
Neglecting the insoluble phase, the mass flowrate of the feed streams (water + ZnCl2) and effluent
solution has been 114 kg/min. The density of the effluent stream is 1.13 kg/L. The table below
shows the one-minute volumetric and mass flow balance for the system at t = 0.

ZnCl2 IN V, L Solution OUT V, L
H20 mass, kg 5.9 mass, kg 100.9
97.7
17.1 114

96.9

The retention time based on volumetric flow is 8000/100.9 or 79.3 min, which is the same
value obtained by a mass flow calculation.

The table data indicate a steady increase in ZnCl2 content because of an increase in % ZnCl2
in the residue composition at / = 0. We can estimate the final % ZnCl2 by rearranging the terms in
Equation [4.8] to linearize the composition function:

c = (15-c2/+c2 [4.9]

where/= e~t/e. Figure 4.63 shows the results. Excel's Trendline tool was used to make a statistical

analysis of the data.

The straight line has an intercept at the composition at infinite time, and a slope of c\ - c2.
The results indicate that the final composition (at infinite time, when ел1в = zero) to be 19.15 %
ZnCl2, and the value of c2 - c\ is 4.15, as it should be for a linear relationship. The results indicate
that if there are no further changes in residue composition, the upper limit of 17 % ZnCl2 will be
reached in about 50 min. Therefore, either the feed rate of residue must be decreased or the feed

rate of water increased.

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 229
Concentration Function for Leach Tank

0.95 0.9 0.85 .„,0.8 0.75 0.7 0.65
e

Figure 4.63 Composition of ZnCl2 vs. e_t/ö. Relationship should be linear for a step function
change in composition at t = 0. The heavy line is a statistically determined straight line through
the data points as determined by Excel's Trendline tool. The text box shows the linear equation.

Assignment. Suppose the feed rate and wZnCl2 in the feed remain the same. Calculate the feed
rate of water at the 30-minute point that will bring the % ZnCl2 to the aim composition in 30
minutes.

4.14.3 Inert Gas Flushing

Once gases like hydrogen or nitrogen get into a metal they are difficult to remove. When
removal is necessary, one of the following three methods can be used.

1. Vacuum treatment, whereby the steel is exposed to a pressure below 0.04 atm.

2. Flushing of the liquid metal by injection of an inert and insoluble gas like argon*.

3. Adding a reactive element to "tie up" the dissolved gas as a liquid or solid impurity that
can be subsequently removed. This method is generally inapplicable to hydrogen because
there are no hydrogen compounds of adequate chemical stability.

The method discussed in this section is the second method, in which the dissolved impurity
gas is removed by bubbling an inert gas up through a ladle of molten metal. The dissolved gas
equilibrates with the rising inert gas bubbles to give the equilibrium partial pressures. The inert
gas thereby carries with it a certain amount of the dissolved gas, depending on Sieverts' law for the
dissolved gas. Please review Section 2.7.3 for an explanation of Sieverts' law for gases dissolved
in metals.

At a steady flow of flushing gas, the change in dissolved gas composition varies with time, as
does the partial pressure of gas in equilibrium with the rising inert gas bubbles. The concentration
of dissolved gas thus decreases more slowly with time. The change in dissolved gas concentration
can be expressed as a differential equation, but an integrated version is more useful for practical
application. For an ambient pressure of one atm:

/ 7 1 5 0 * f (c0-c)(Ks 2 Л
-cxc0)
cxctо [4.10]

The symbols have the following meaning:
V= volume of flush gas introduced into the metal (ft3 at STP per short ton of metal);

* In steel, CO is an effective flushing gas, and is generated by the reaction between dissolved carbon and
injected oxygen.

230 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

c0 = original composition of dissolved gas (wt %);
с = composition of dissolved gas at end of flushing operation (wt %);
M= molecular weight of the diatomic gas molecule dissolved; and
Ks = Sieverts' law constant at temperature, as defined by Equation [2.26] (wt %/atm/2).

Equation [4.10] gives the maximum rate of dissolved gas removal because at every moment it
assumes that the rising inert gas bubbles come to equilibrium with the dissolved gas before both
gases leave the system. Such a condition is seldom met in practice. It requires a deep bath of
metal, dispersion of inert gas into tiny bubbles that don't rise too fast, good circulation of metal in
the ladle, and a low concentration of interstitial solutes in the steel that could interfere with gas-
metal transport at the bubble-metal interface. It has been found that small amounts of sulfur and
oxygen in liquid steel do slow down the interfacial mass transfer, so actual rates of dissolved gas
removal will be less than predicted by Equation [4.10]. One way to enhance the flushing process is
to include a small amount of a reactive substance like Ca with the Ar. The Ca vapor effectively
removes interfacial "poisons" like S and O, thus minimizing the interference of these elements on
the interfacial mass transfer of gas from metal to bubble.

Equation [4.10] is based on insert gas flushing of an open ladle of metal where the ambient
pressure is one atm. If the flushing takes place in a vacuum container, the value of с will be less.
Although Equation [4.10] can be modified to include an ambient pressure term for reduced
pressure flushing, we must recognize that the rising gas bubbles don't attain that (reduced) pressure
until they are approaching the surface of the melt. Their residence time is so short that the
dissolved gas wouldn't have time to approach the lowered pressure, so the addition of an ambient
pressure term in Equation [4.10] is probably not justified.

EXAMPLE 4.13 — Removal of Hydrogen from Steel.

A ladle containing 100 tons of steel at 1600 °C contains initially wH = 1.6 x 10~3 % and is to
be lowered to 1.0 x 10"4 % by argon flushing. Calculate the required volume of argon.

Data. The value of Ks for H in liquid iron at 1600 °C is given by Equation [2.27] as 27.5 when the
composition unit for H is weight parts per million. K$ for mass % units as required by Equation
[4.10] is 2.75 x 10"3.

Solution. Insertion of the variables into Equation [4.10] gives V Ar = 250 ft3/ton, or 25 000 ft3 for
the ladle. This volume of Ar is the minimum required. In practice, an even larger amount would
be required since equilibrium might not be reached for reasons mentioned earlier. Therefore, a
vacuum degassing process would be more efficient in going to such a low H content.

Assignment. Calculate the volume composition of the first gas bubble that leaves the ladle of steel,
and the mass of H removed by the first 10 ft3 of Ar.

So far, we have looked at unsteady-state processes that can be described mathematically. This
allows integration of the differential equation and calculation of the stream flow and/or
composition vs. time. However, some processes are so complex that a mathematical description is
nearly impossible. For example, some processes undergo composition changes that are difficult to
describe by a time-variable differential equation. One way to handle this situation is to make a
series of calculations at small time intervals, such that the overall change can be approximated by
summing the small changes. This technique is called iterative stagewise approximation. The
smaller the incremental change, the better is the approximation.

EXAMPLE 4.14 — Vacuum Refining of a Cd-Zn Alloy.

A still is separating cadmium and zinc at 1000 К (please refer to Example 4.1 for orientation).
The still initially contains 500 kg of a Cd-Zn alloy containing wCd = 0.600. The vaporization rate
is 2.00 kg/min and additional alloy of wCd = 0.600 is fed at the same rate. Calculate the

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 231

composition of the alloy as a function of time until it reaches wCd = 0.400, and calculate the
composition of the alloy at infinite time.

Data. The vapor pressure in atm of pure liquid Cd and Zn near 1000 К is given by the following
equations:

Log(p°Zn) = -6130/7 + 5.20 [4.11a] Log(p°Cd) = -5250/7 + 5.05 [4.11b]

Cd-Zn liquid alloys can be represented by the regular solution approximation, with a = 1800
over a range of xCd between 0.25 and 0.55. (The value of a will be significantly different outside
this range of alloy composition). Please review section 2.72, and Equations [2.23] and [2.24],
which define the activity (and hence the vapor pressure) of a constituent of a regular solution, and
Example 2.11.

Solution. The vapor pressure of pure Cd at 1000 К is 0.631 atm, while that of pure Zn is 0.117, so
the alloy vapor will initially be higher in Cd than Zn. The actual vapor pressure of Cd and Zn
depends on the mole fraction of each species in the alloy. Rewriting the regular solution equation
in terms of common instead of natural logarithms gives:

Γ(1ο§γι) = 393(^)2 [4.12]

where i and j are the two constituents of a regular solution. Substituting in log(p°Cd) and
log(/7°Zn) at 1000 К into the regular solution equation and collecting terms gives an expression for
log(pCd) and log(pZn) as a function of alloy composition at 1000 K:

Log(pZn) = -0.930 + log(xZn) + 0.393(1 -xZn)2 [4.13]

Log(pCd) = -0.200 + log(xCd) + 0.393(1 - Cd)2 [4.14]

The initial alloy contains 300 kg of Cd and 200 kg of Zn, so xCd = 0.4660, and xZn = 0.5340.
Therefore, the initial vapor phase haspCd = 0.3806 atm, andpZn = 0.0762 atm. The gas contains
xCd = 0.8332 and xZn = 0.1668. On a mass fraction basis, wCd in the gas = 0.8957. The
evaporation process clearly diminishes Cd in the alloy.

The continuous process is simulated by a series of discreet vaporization and addition stages.
Each stage consists of vaporization for a time interval, followed by the addition of the new alloy
(of wCd = 0.600). The vapor pressures of Cd and Zn are assumed to remain constant during each
time interval. A time interval of 10 minutes was selected as being small enough to adequately
simulate the process, while being large enough to keep the calculational effort to a reasonable
minimum. 20.00 kg of metal vaporize during each 10-minute interval, followed by the addition of
20.00 kg of the original alloy at the end of the interval (which is the same as at the beginning of the
next interval). The mass of alloy in the still thus fluctuates between 480 and 500 kg. A mass
balance is made at the end and beginning of each interval to track the change in composition of the
alloy. The calculations were carried out to more significant figures than justified to avoid
accumulating round-off errors.

At the end of the first interval, the alloy lost 17.909 kg of Cd and 2.091 kg of Zn, and has wCd
= 0.5877. At the beginning of the second step (and every step thereafter), the alloy gains 12 kg of
Cd and 8 kg of Zn. The alloy then has wCd = 0.5882 (xCd = 0.4539). The pCd has dropped to
0.3751 atm while the/>Zn has increased to 0.0773 atm. The second interval brings a Cd loss of
17.858 kg and a Zn loss of 2.142 kg, and so forth. A series of calculations was performed using
Excel, and the results of 200 min of vaporization are shown in Figure 4.64. The wCd line shows a
slight curvature over the 200 min time interval, which is well represented by an empirical quadratic
fit to the data, as shown on the figure. Application of this equation indicates that the alloy requires
185 minutes to reach wCd = 0.40.

As time extends to infinity, the alloy approaches a steady-state composition. This occurs
when the composition of the vapor asymptotically approaches that of the incoming alloy (wCd =
0.600; xCd = 0.4660). The steady-state alloy composition was calculated by setting up Equations
[4.13] and [4.14] to calculate thepCd/pZu. At infinite time, this ratio is equal to the mole fraction

232 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems
ratio of the incoming alloy (i.e., at xCd/xZn = 0.466/0.531, pCd/pZn = 0.8727). Goal Seek was
used to calculate the steady-state alloy composition in the still as wCd = 0.1022 (xCd = 0.0621).

з 0.6 f^^ Stagewise Evaporation from a Cd-Zn Alloy at 1000 К
wCd = 5.8x10 " ( Г ) ■ 0.00119* + 0.600
°·55

о 0.5
'<&
о
2 0.45

8 0.4

CO

E 0.35

Figure 4.64 Composition of alloy in still during addition of 2.0 kg/min of alloy of wCd = 0.600.
Calculations based on stagewise additions of 20 kg (10-minute intervals). The solid line shows
wCd of the alloy, while the dashed line indicates a linear extrapolation of wCd based on the first
10-minute interval. The dashed line was added to show more clearly the changing slope of the
actual line with time. Excel's Trendline feature was used to fit a quadratic equation to the data.

Assignment. Repeat the example at 1050 К to see if a higher temperature would shorten or
lengthen the time needed to get wCd = 0.400.

4.15 Graphical Representation of Material Balances

So far, we've represented the results of mass balance calculations in the form of tables,
graphs, or labels on flowsheets. Sometimes a different kind of representation is useful, in the form
of a diagram where the mass of a particular stream is represented by the width of the line that
represents the stream. In this way, the relative amounts of materials entering and leaving a system
are easily visualized. Figure 4.65 shows such a diagram for Example 4.6.

New Feed — Evaporator Crystallizer feed Crystallizer Filter cake (2000 kg)
and 1500kgofFeSO477H2O
(3000 kg of solution) (2800 kg of solution) Filter
30 % FeS04 37.84 % FeS04 500 kg of solution
66 % H20 51.71 %H20
4% H2S04

Recycle (800 kg)
80 kg of FeS04?7H20

720 kg of solution

Figure 4.65 Mass flow diagram for the crystallization and recovery of FeS04-7H20 from acidic
solution. The width of each flow indicator is proportional to the mass of material in the stream.

4.16 Measures of Performance

All processes operate as a compromise between an optimum in one aspect or another. As we
have seen, features such as recycle, counter-current flow, and larger numbers of reactors can have
a beneficial effect on the recovery of a desired species, or an improvement in its purity. But these

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 233

features may have drawbacks, such as allowing unwanted reactions to occur. Usually there are one
or two characteristics of a process that are most important, and it would be nice to have some
measure of performance to indicate how well a process is performing.

Let's look at some performance measures in connection with a common process for disposing
of scrap automobiles. After the more valuable components have been removed, the hulk is sent to
a shredder, which converts it to fist-size or smaller pieces. The most valuable substances in the
shredded mass (the shred) are the metals, mainly steel, aluminum, and copper. Steel is removed
magnetically and aluminum by eddy current, as shown in Figure 4.66. The remainder is processed
by a variety of ways to recover copper and other valuable alloys, and dispose of the remainder.
The objective in the separation of the steel and aluminum is to recover all of this material
uncontaminated by anything else. This objective is never reached, mainly because each piece of
shred is unavoidably contaminated with more or less of everything else that was in the hulk —
glass, upholstery, rubber, other metals, etc.

The main objective of the shredder is liberation, or the release of valuable substances from
themselves and all other substances. Complete liberation is never attained, but is improved by
decreasing the upper size limit on the shred. However, this can adversely affect other aspects of
the shred, in addition to raising the operating cost. Therefore, we are faced with separating metals
from shred containing pieces of all sizes and degrees of contamination.

Fe-rich mB=1180lb Al-rich mD = 160lb
fraction wBFe = 0.95 fraction ^ wDFe = 0.075

В w A I = 0.01 wDAI ·· 0.814

Shred A Magnetic m c = 820lb Eddy- E Residue
separator wcFe = 0.194 current
mA = 2000lb wcAI = 0.181 separator m — 660 lb
wAFe = 0.64 wEFe = 0.228
wAAI = 0.08 wEAI = 0.027

Figure 4.66 Flowsheet for separation of metal fractions from shredded auto hulks. Basis is 1 ton
of shred. The mass fraction of iron-rich material (stream B) is 59.0 % of the shred and the mass
fraction of the aluminum-rich material (stream D) is 8.0 % of the shred.

One measure of the metal separation process is recovery, expressed as a percentage of the
total metal contained in the raw material. The steel recovery is 87.6 %. The purpose of calculating
the recovery is to determine the distribution of a metal (or any other valuable species) between the
various products of a separation process. A second measure of performance is the grade, or purity,
of the recovered product. The aluminum-rich product has a grade of 81.4 %.

There is an inverse relationship between recovery and grade of product for all separation
processes. The recovery of steel can be increased (for example by increasing the magnetic field
intensity over the shred), but something else also happens: The magnetic fraction will contain
more particles that contain less steel, thus the grade will be lowered. A separation process is
always a compromise between attaining a high measure of performance for recovery and grade.
The compromise is usually reached by economic considerations.

Another measure of performance is yield, which is the amount of some specified output
stream divided by the amount of a specified input stream. In the example above, the yield of
magnetic fraction is 59.0 %. The inverse of yield is called the consumption.

Since the objective of processing the auto hulk is to separate the metals from everything else,
we may define a separation measure of performance: distribution coefficient. This measures the
ability of a process to separate a specific component from everything else in the system. The

234 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

distribution coefficient may be defined between input and output streams, or between two different
output streams. For example, the distribution coefficient of steel between the magnetic and eddy
current streams is 93.4.

In other cases, separating one component from a specific second component is more important
than the distribution coefficient. To quantify how well a process performs this task, its selectivity,
or separation factor, is calculated, using the formula:

or = (% M in M - rich fraction)(% N in N - rich fraction) [4.151

(% M in N - rich fraction)(% N in M - rich fraction)

where M is the desired product in its stream, and N is the undesired product in its stream. In the
case of iron separation from aluminum, SF = 1030. The selectivity index (SI) is the square root of
SF, or in the above case, SI = 32.1.

4.17 Controllers

Every process requires a method for keeping it under control. In the simplest sense, an
operator observes the flow of materials in and out of a system, and then adjusts a valve or switch to
maintain a desired output. More likely, the system is fitted with devices called controllers, which
monitor the properties of streams and continually adjust the stream properties according to some
predetermined specification. Whatever the method, controllers have three features: a set-point, a
sensor, and an actuator. A setpoint is any stream specification chosen by the operator as a
specified value. Typical set-points are temperature, pressure, composition, and flowrate. A sensor
is a device that detects some physical or chemical property of a stream or device, and converts the
property into an electrical or mechanical signal. The controller measures the difference or error
between the setpoint value and the sensor signal. An actuator is a device that controls some stream
feature that in turn affects the signal. Actuators are usually electrically operated devices like
valves to control mass flow, electrical switches to control current flow, and motors to control
pressure and agitation. The controller varies the position of the actuator until the stream variable
attains the setpoint. Any portion of the system that is controlled is defined as having a constraint
on the system. The constraint is located on the stream containing the actuator.

There are three main types of flow controllers: flowrate, feed-forward, and feedback. The
simplest type is a flowrate controller, which has the sensor and actuator on the same stream. It sets
the flow of any stream in the system. The flowrate controller was implicit in all of the examples
discussed in this Chapter where a stream flow was specified. The feed-forward controller has the
sensor on one stream and the actuator on an adjacent stream, and is used to maintain a fixed flow
ratio or flow difference between two streams or stream components. The flowrate and feed-
forward controllers are often used in conjunction with each other. When the setpoint on the
flowrate controller for stream A is changed, the feed-forward controller senses the new flow on
stream A, and adjusts the setpoint on stream В to maintain the same flow ratio or difference. A
similar logic applies to temperature controllers where either the same temperature ratio or
temperature difference must be maintained between two streams. A feed-forward controller may
also contain a logic circuit programmed to calculate the effect of a changing upstream variable on a
downstream property, and adjust accordingly.

The feedback controller has a sensor at a downstream point, and adjusts an actuator on an
upstream point to control the properties of the downstream point so that it conforms to the setpoint.
Owing to the often-complex relationship between the upstream variable and the downstream
setpoint, the feedback controller often includes a logic system that simulates the degree of response
expected from a change in a stream variable. Figure 4.67 shows a sketch of a flowsheet showing
the three different types of controllers.

FlowBal accepts control logic in the form of equations relating stream flow, composition, and
temperature. Please see the User's Guide for control logic examples.

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 235

The subject of process controllers is much more involved than the preceding discussion, but
the basic concepts are important in making material and energy balances on multi-device
processes. Example 4.15 illustrates this point.

Figure 4.67 Flowsheet showing the application of three different types of controllers. The symbol
® represents an actuator, and the symbol 0 represents a sensor. The fiowrate (FRC) controller on
stream A maintains the set-point flow of stream A. The feed-forward (FFC) controller senses the
fiowrate of stream A, and maintains the flow of stream В according to a specified function. The
feedback (FBC) control senses some property of stream D, such as composition of a species,
temperature, or fiowrate, and controls some property of stream E to maintain stream D at the
desired setpoint.

EXAMPLE 4.15 — Control Strategyfor Upgrading Spent Reducing Gas.

The gas-phase reduction of iron ore generates a "spent" gas containing varying (but known)
amounts of CO, C02, H2, and H20. This gas can be reused by removing much of the H20 and
C02. The H20 is removed first by cooling to 25 °C to condense most of the water, followed by
C02 removal with an amine solution that brings the/>C02 to 0.014 atm. The objective of the gas
treatment process is to remove enough H20 and C02 such that the product gas has a volume
fraction of (C02 + H20) of 0.080. This is termed the gas quality, designated Q. If the H20 and
C02 removal sections remove more than is necessary to attain the specified gas quality, some of
the spent gas can bypass these sections, and be blended into the final gas stream. This gas, with
added natural gas, will be recycled to the iron oxide reduction device. Figure 4.68 shows the
flowsheet for the process. Set up a balance program that will calculate the relationship between the
fraction of spent gas bypassing the processing devices, and gas quality Q for any spent gas
composition.

Iron

oxide Iron Gas У\#9+АГ 1r Gas
splitter mixer
^^^™^^ \r

oxide "■^™™""™"Ψ H20 яшштттттттшΨ co2
reduction removal
furnace removal
Iron CH4 J1
ii i1
4 1

__

Figure 4.68 Flowsheet for iron oxide reduction device and spent gas processing.

Solution. The problem implies that we must develop a generic solution that will be valid for any
spent gas composition. Figure 4.69 shows the flowsheet for the spent gas treatment part of the
process. We are interested only in the numbered (gas) streams involving H20 and C02 removal,
which means we neglect the amount of amine.

236 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

Spent gas & r' / FBCT
F1 = 100 n/min
Amine 0.014
pCO = 0.14
H20 F4 = ? C02
0>CO2 = O.21 removal pH20 = removal
φ Η2 = 0.26 0.0308
φ H20 = 0.39
8

Figure 4.69 Flowsheet for processing spent gas from iron ore reduction process. A feedback
controller on stream 6 measures the gas composition and controls a valve on stream 2. The
controller circuit is shown as a dotted line.

No basis is given, so we are free to select one. Since the only phase entering in stream 1 is a
gas, it is convenient to select 100 moles of spent gas (stream 1) as a basis. The example will be
worked for a selected arbitrary spent gas composition to illustrate the method, but a spreadsheet
solution will be developed to make a balance for any spent gas composition. The system pressure
is deemed to be 1.00 atm. We begin by noting that the composition of streams 2 and 3 is the same,
and defining the stream 2 splitter fraction R as the fraction of gas in stream 1 that reports to stream
2. The objective is to calculate the relationship between R and Q for any spent gas composition.

The procedure is demonstrated with the spent gas composition from the flowsheet:

<pCO = 0.14, фСО2 = 0.21, φΗ2 = 0.26, φΗ2Ο = 0.39

with an arbitrary R value of 0.09. Streams 1, 2, and 3 are thereby completely specified. A DOF
analysis around the splitter indicates that SV = 12, IB = 4, 1С = 3 and F = 1. The subsidiary
relations for a splitter require some additional comment. First, each specified split ratio constitutes
a SR. Second, owing to the relationships between the compositions of the split streams, the
number of special splitter restrictions is (N - 1)(S - 1), where N is the number of branches from
the splitter, and S is the number of species. This gives SR = 4 (R was set at 0.09, and the number
of special splitter restrictions is three). The DOF around the splitter is therefore 0, so it is correctly
specified.

The mass balance around the splitter is expressed by a series of equations for each component.
For CO for example:

л3СО = (л1 CO)(l - R) - 14(1 - 0.09) - 12.74

и2СО = (nlCO) - n3CO = 14.00 - 12.74 - 1.26

Similar balances can be written for the other species. For the H20 removal device, the only
change is the moles of H20. This is calculated from the expression for/?H20; the/?H20 in stream 4
is obtained from Equation [2.15] as 0.0308 atm.

pU20 = 0.0308 = л4Н20/(л4Н20 + 12.74 + 19.11 + 23.66); n4R20 = 1.76

A similar procedure was used to calculate the moles of C02 in stream 5:

pC02 = 0.014 = л5С02/(и5С02 + 12.74 + 23.66 + 1.79); л5С02 - 0.54.

The stream 5 value of Q (remember, Q = <pC02 + φΗ20) is 0.060, which will increase with
the admixture of stream 2. In this example, the split fraction was set as 0.09 so that stream 2
contains 9 moles of gas having the same composition as stream 1. After mixing with stream 5, the
composition of stream 6 is:

<pCO = 0.293, фС02 = 0.051, φΗ2 = 0.545, φΗ20 = 0.111

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 237

The gas quality in stream 6 is 0.162; therefore the split fraction of 0.09 is too large. The
correct split fraction can be calculated two ways. First, a graphical technique. Successive
calculations were made at different splitter ratios, and the results plotted as shown in Figure 4.70.
From the text-box equation, the desired Q of 0.08 is obtained at R of about 0.0163. In order to get
Q = 0.08, over 98 % of the spent gas must pass through the water and C02 removal devices. If
none of the spent gas bypassed the devices (i.e., R = 0), Q would be about 0.06.

Alternatively, the balance can be set up in Excel to use the Goal Seek tool. Table 4.15 shows
the results, where the numbers represent the moles of gas in each stream. Goal Seek gave R as
0.0164 to get Q = 0.08, very close to the graphical technique. The benefit of the Goal Seek
solution is that any spent gas composition can be entered into the cells for that gas, and Goal Seek
can be used to solve for R and the stream 6 composition. In addition, the values used for H20 and
C02 removal, and the gas quality parameter can be changed for different process conditions
without any additional arithmetic.

0.11 Gas Processing Results

CO Q = 1.20R +0.0605
R2 = 0.9998
E 0.10
0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04
CO split fraction R to stream 2
CD

ъ 0.09

(Л

ä 0.08

05 0.07

3
ö"

CO 0.06
w
О) 0.05

0

Figure 4.70 Relationship between the gas quality in stream 6 and the fraction of spent gas
bypassed. ExcePs Trendline tool was used to obtain a linear equation expressing the functions.
The trendline equation is displayed in the text box. The value of R2 = 0.9998 in the textbox
indicates the "goodness-of-fit" of the equation, as defined in Chapter 3.

Table 4.15 Amount of species in streams in order to obtain a gas quality Q of 0.08.

Species CO co2 H2 H20 total

Stream 1 14.00 21.00 26.00 39.00 100.00

Stream 2 0.23 0.35 0.43 0.64 1.64 Goal Seek varied
Stream 3 13.77 20.66 25.57 38.36 98.36 the Splitter
Stream 4 13.77 20.66 25.57 1.91 61.90
Stream 5 13.77 0.59 25.57 1.91 41.84 fraction R until
Stream 6 14.00 0.93 26.00 2.55 43.48 the gas quality
Stream 7 36.45 36.45 reached 0.08
Stream 8 — — — 20.07
Сотр. of stream 6 — — — 100.0%
20.07
32.2% 2.1% 59.8% 5.9%

Splitter fraction R = 0.0164 Gas quality = 0.0800

In practice, the FBC sensor on stream 6 would determine the gas quality and compare it with
the set-point value. As the spent gas composition changes, the gas quality would change, and the
FBC would adjust the stream 2 flow up or down until the value of Q returned to the setpoint. The

238 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

effectiveness of the FBC in adjusting the flow of stream 2 would be enhanced by incorporating
some logic into the FBC that would use some of the relationships determined above.

Assignment. Change the position of the H20 and C02 removal devices, and set up a program to
calculate the value of R to produce a gas quality of 0.08.

If you've tried FlowBal, you may have noticed that it does not have a controller device.
However, inasmuch as a controller simply forces a certain relationship between one stream
variable and another, it's possible to simulate a controller device by inserting an appropriate
controller equation in FlowBal. The flowsheet was also set up as an example for FlowBal. Please
see the FlowBal User's Guide worksheet Spentgas to see how this was done.

We conclude this section by expanding a bit on the principles of control logic. Early in
Chapter 3, we discussed the control of temperature of a furnace. We said that when the
temperature went a few degrees above the set point, the controller turned the power off, and when
it cooled a few degrees below the set point, the controller turned the power on. The result is a
cycling furnace temperature that can be deleterious to the process being carried out in the furnace.
This is an extremely crude control system, and is almost never used in practice. More commonly,
a proportional-integral-derivative controller (PID controller) is used, which employs three types of
control action to prevent the cycling behavior of a simple on-off controller.

The PID can adjust process outputs based on the history and rate of change of the error signal,
which gives more accurate and stable control. "PID" is named after its three correcting functions,
whose sum constitutes the output of the PID controller.

Proportional - To handle the immediate error, the error is multiplied by a constant P (for
"proportional"), and added to the controlled quantity. P is only valid in the band over which a
controller's output is proportional to the error of the system. For example, for a furnace, a
controller with a proportional band of 10 °C and a setpoint of 900 °C would have an output of 100
% on at 890 °C, 50 % on at 895 °C and 10% on at 899 °C. Note that when the error is zero, a
proportional controller's output is zero. In the simplest case, the furnace temperature tends to
"settle down" a bit below the set point. The proportional band width can be adjusted as desired.

Integral - To learn from the past, the error is integrated (added up) over a period of time, and then
multiplied by a constant / (making an average), and added to the controlled quantity. A simple
proportional system either oscillates, moving back and forth around the setpoint because there's
nothing to remove the error when it overshoots, or oscillates and/or stabilizes at a too low or too
high value. By adding a proportion of the average error to the process input, the average difference
between the process output and the setpoint is continually reduced. In essence, the integral
function moves the proportional band up or down. Therefore, eventually, a well-tuned PID loop's
process output will settle down at the setpoint.

Derivative - To handle rapidly-changing processes, the first derivative (the slope of the error) over
time is calculated, multiplied by another constant 2), and also added to the controlled quantity. The
derivative term controls the response to a rate of change in the system. The larger the derivative
term, the more rapidly the controller responds to changes in the process's output. Its 3) term is the
reason a PID loop is also sometimes called a "predictive controller." The 3) term is reduced when
trying to dampen a controller's response to short term changes. In essence, the derivative function
senses how fast the furnace temperature is approaching the proportional band, and if the approach
is deemed to be so fast that temperature overshoot is likely, it can move the proportional band
(temporarily) to a lower range. Practical controllers for slow processes can do without the
derivative function.

There are several methods for tuning a process controller to improve its control action. Some
controllers have logic circuits that can "learn" from experience, while others have programmable
circuits that accept a mathematical model of the material and energy balance of the system.
Devices that interface with (or replace) a PID controller are called programmable logic controllers.

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 239

Controllers and data recorders have become increasingly sophisticated and affordable, and are
pervasive throughout the materials processing world. Statistical techniques are available to
analyze the enormous amount of data collected during the operation of a complex process for the
purpose of developing predictive process models. All such flow-based models, empirical or
phenomenological, contain a conservation of mass constraint.

4.18 Summary

Every process analysis involves writing and solving material balance equations. The chapter
began with a description of different device types and their purposes, and introduced the use of
flowsheet diagrams. Certain conventions were introduced for designating the flows and
composition of streams. A systematic method was developed for making a material balance on a
system. The essence of this method is summarized below.

Identifying the nature of the system. For systems that operate continuously (no change in system
conditions with time), the accumulation is zero. A differential balance applies over an instant of
time, while an integral balance over a specified time interval. A batch process is one in which no
material flows in or out during operation. A semi-batch process is one in which material can be
added or removed at intervals until the final condition is reached.

Writing and labeling a flowsheet. The devices for carrying out the processes of a system should be
characterized according to their main function, and each device labeled according to its main
function. Each stream flowing between devices should be labeled, and the stated variables noted
on the flowsheet. A stream is completely specified if a) the stream flow and composition are
defined, or b) the mass flow or amount flow of each stream constituent is specified. A basis of
calculation can be chosen from a stated stream or component flow rate, or a mathematically
convenient basis can be chosen arbitrarily and the result scaled to the requested basis. For complex
systems, start with a simplified version of the flowsheet, and add details later.

Constructing a material balance ledger. The ledger was introduced as a valuable technique for
presenting the flowsheet data and displaying interim and final mass balance results. A spreadsheet
is a particularly useful way to display a ledger because cells can be incorporated in formulae for
calculating other ledger entries.

Determining the degree of freedom. A DOF analysis is a way of determining if enough (or too much)
information is available to calculate balances for all species. If the DOF is zero, the system is
correctly specified. Even if the DOF is not zero, a material balance can be performed under special
circumstances that will give useful information. A DOF calculation for multi-unit systems requires
special care to avoid over-counting the stream variables in connecting streams.

Writing the material balance equations. The strategy is to first write a component balance equation
around each device. Next, write equations that define the sum of compositions for a stream.
Finally, write special restriction equations, such as relationships between stream flows or
compositions, the partial pressure of a substance, the mass fraction of a slurry, etc. Finally,
determine exactly how many unknowns the system has. You will need as many independent
equations as there are unknowns.

Solving the equations. For simple systems it's often possible to use a hand calculator or Excel to do the
necessary arithmetic. But in most cases, the number and character of equations is such that Goal
Seek or Solver is useful, or even necessary. Super Goal Seek and Super Solver are useful when
multiple or repetitive calculations are required.

Process design improvements using recycle, bypass, and counter-current flow streams. The efficiency
of a process may be enhanced by recycling part of a product stream back to an earlier device
operation, while other efficiencies are attained when part of a stream bypasses a device. Counter-
current flow systems are employed in the extraction of selected constituents from one stream to
another as illustrated with leaching and solvent extraction examples. These situations increase the

240 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

mathematical complexity of a material balance equation set, and may require the use of Excel's
Goal-Seek or Solver tool.

Calculating the retention time in a device. The extent of a reaction depends partly on the contact time
of the reactants. The retention time of a substance was introduced first in connection with the
desire to attain a certain product characteristic that depended on time. Connecting devices in series
increases the fraction of material retained in a certain time interval. The retention time concept
was extended to unsteady-state processes. The retention time can be used to calculate the
concentration or composition of a component versus time.

Measuring the performance. Criteria are described for assessing the ability of a process to attain a
specified goal. A common technique is to calculate one or two measures of performance versus
one of the main process variables to seek optimum conditions.

Controlling the process. Every process must be kept under control. A sensor is required for detecting a
stream variable, and an actuator for using sensor information to keep the stream variable at its set
point. The basic concepts of flowrate, feed-forward, and feedback controllers were described, and
examples presented of their use.

All of the basic concepts for making a material balance were introduced, and techniques
described for applying these concepts. The methods presented in this chapter are used universally
in the materials processing industry. The next two chapters will extend these concepts to systems
where chemical reactions occur.

Computational tools are indispensable in complex processes. The MMV-C program is useful
for converting stream compositions and amounts from one set of units to another. FlowBal (which
uses Solver) was introduced as a special computational tool that may simplify making material
balances. FlowBal may require a change in the way the flowsheet streams are drawn so that each
stream has only one phase (or one phase assemblage). These tools will be used extensively in later
chapters, especially when making simultaneous material and energy balances.

Special attention was paid to a technique for simulating a complex continuous unsteady-state
process by dividing it into a series of small steps. This technique is called iterative stagewise
approximation.

References and Further Reading

ChemicaLogic Corporation, SteamTab Companion, [Online].
http://www.chemicalogic.com/steamtab/companion/default.htm. November 2009.

Green, William, Jr., Course Materials for 10.37 Chemical and Biological Reaction Engineering,
spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology.
Downloaded on [12/11/09].

U. S. Environmental Protection Agency on-line course, Basic Concepts in Environmental Sciences,
Module 1 : Basic Concepts - Material Balance

http://www.epa.gov/apti/bces/modulel/material/material.htm Module 5: Flowcharts and
Ventilation Systems.

Wikipedia contributors, "Mass balance", "Conservation of mass", "Closed system", "Isolated
system", "Open system", "Process flow diagram", "Bayer process", "Chemical reactor", "Unit
operation", "Unit processing", "Process design", "List of chemical process simulators", "Liquid-
liquid extraction", "Leaching", "Process control", "Control system", "Van der Waals equation",
Wikipedia, thefree encyclopedia, December 2010. http://en.wikipedia.org/wiki/Main_Page.

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 241

Exercises

4.1. Magnesium-calcium scrap is being processed by a distillation/condensation process. The feed
to the condenser is 6000 kg/h of vapor containing Mg, Ca, and Ar. The condensate liquid flowrate
(stream C) is 2400 kg/h. A flowsheet for the process is shown below. Make a mass balance on the
system, and then use MMV-C to calculate the stream volume in streams A and В at operating
temperature, and the volume fraction of species for P = 0.97 atm.

A(1450°C) Condenser B(1230°C) FB = ? kg/h
Vapor wBCa = ?
Feed wBMg = ?
FA = 6000 kg/h C(1215°C) wBAr=?
wAMg = 0.58 Liquid
wACa = 0.21 F c = 2400 kg/h
wAAr = 0.21 wcCa: 0.42
wcMg = 0.58

4.2. The surface of steel castings contains a film of iron oxide (magnetite, Fe304). The film (and
some of the iron) is removed by sandblasting with Si02. The product powder is to be treated in a
magnetic separator to separate the Si02 from the iron/Fe304 mixture. A flowsheet is sketched
below. Preliminary design testing of the magnetic separator indicates that 90 % of the iron/Fe304
mixture is recovered in the magnetic stream. Make a mass balance to determine the expected

flowrate and composition of stream C, and the flowrate of stream B.

A Magnetic В ^ FB = ?
separator (magnetic wBS«02 = 0.20
Feed
F A = 1000 kg/h stream) wBFe/Fe304= 0.80
wASi02 = 0.70
wAFe/Fe304= 0.30 " wcSi02 = ?
wcFe/Fe304=?

4.3. The separation process described above is put into operation, and its performance is to be
evaluated according to one of the measures of performance described in Section 4.16. Each stream
is sampled, and the following results obtained:

Mass Fraction of Com jonents in Streams

Stream A В С

Fe/Fe304 0.282 0.788 0.051
Si02 0.718 0.212 0.949

Make a mass balance and calculate the separation efficiency of the process.

4.4. Foundry sand is prepared by mixing clay and sand, and then adding water, such that the mass
fraction of moisture in the product is 9.3%. The moisture content of the original sand and clay is
1.3 % and 7.7 % respectively. How much water should be added to a mixture of 1870 lb of sand
and 77 lb of clay? Calculate the mass fraction of clay in a dried sample of the moist foundry sand.

4.5. A plant making powdered silver for specialized solder consists of a melting device, an
atomizing device, and screens to separate the powder into three fractions: oversize, correct size,
and undersize. The over- and undersize fractions are recycled to the melting device. A flowsheet
for the process is shown below. The oversize material (stream C) is 10 % of the atomizer product,
and 20 % of the material passing the 40 mesh screen also passes through the 100 mesh screen
(stream E). Make a mass balance to determine the flowrate of correctly-sized powder (stream D),
and the amount of silver melted per day (sum of streams A and F).

242 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

A *- Melter and кVВ 40 mesh
FA = 375kg/h atomizer screen
С, ^ —
——► 100 mesh

F

й _1 screen

4.6. Develop a charge calculation worksheet in Excel for use by stainless steel melt shop
personnel. Set up the worksheet so that the operator enters the desired composition of the stainless
steel into cells, the desired mass of the heat of steel, and the worksheet calculates the amount of
each charge material. Assume no significant loss of alloying elements to the slag. The plant
produces 304-type stainless steel that has a mass fraction of each alloy element in the following
range:

wCr: 16.3-18.2%. wNi: 7.8-8.7%. wMn: 0 . 4 - 0 . 9 % .

The shop has the following materials available for making up the charge to the furnace.

Mass Fraction of Elements in Charge Materials

Element Cr Ni Fe Mn

Alloy scrap 0.68 0.2 0.1 0.02
Ferrochromium 0.75 — 0.25 —
Electrolytic Ni — 1.00 — —
Electrolytic Fe — — 1.00 —

4.7. The desired composition of a copper-rich bronze alloy is as follows:

wSn: 5.0%. wBi: 10.0%. wZn: 5.0%.

The charge materials available have the following composition:

Mass Fraction of Elements in Charge Materials

Element Cu Sn Zn Bi
Yellow brass 0.700 — 0.300 —
0.835
Bronze A 0.690 0.150 — 0.015
Cu-Bi 1.00 0.010 — 0.300
—
Electrolytic Cu — —

During melting, there is a loss of zinc of between 5 and 10 % of the mass of zinc charged.
Calculate the amount of each charge material required as a function of the loss of zinc, and plot the
results.

4.8. A feed slurry containing BaS04, H2S04 and water is washed with a dilute solution of H2S04.
The washing takes place in a washer/settler tank such that the washed BaS04 slurry has wDBaS04
= 0.48. A sketch of the wash system is shown on the next page. The washed BaS04 is
subsequently heated in a dryer to remove 90 % of the mass of the water (none of the H2S04 is
removed). Make a mass balance for the system to determine the effect of flow of wash solution on

the mass fraction of H2S04 in the washed and nearly-dry BaS04.

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 243

FB = ? Wash solution

wBH20 = 0.956

BaS04 wBH2S04 = 0.044 1 Decanted Fc = ?

slurry feed solution wc/LH20 = ?

F A = 1880 Ib/h wc/LH2S04 = ?
wNLH20 = 0.46 wcBaS04 = 0.02

w ^ S C ^ = 0.54 Washed FD = ?
wABaS04 = 0.35 BaS04 ivD/LH20 = ?

(Stream D to drier) wD/LH2S04 = ?
w°BaSOa
= 0AS

4.9. A feed slurry containing BaS04, H2S04 and water is washed with water in three stages. The
washing takes place in three washer/settler tanks such that the washed slurry moves from tank to

tank, while the decanted solution (free of BaS04) is withdrawn and collected in a common stream.
80 % of the H2S04 is removed in each tank. The slurry has a mass fraction of BaS04 of 0.56. The
washed BaS04 is dried to remove 90 % of the mass of the water (none of the H2S04 is removed).
A sketch of the wash system is shown below. Make a mass balance for the system to determine

the flow of water to each tank and the mass fraction of H2S04 in the washed and dried BaS04.

BaS04 slurry Water Water Water Washed
i* BaS04
F = 1880 Ib/h 1r 1r
ivH2O = 0.30 Wash/settling F =?
wBaSO4 = 0.35 tank I Wash/settling Wash/settling
ννΗ2ο04 = 0.35 tank II tank III ivH20 = ?
ivH2S04 =?
ivBaS04 — 0 56

ir Decanted
solution
Decanted solution mixer
F = ? wH20 = ?

ivH2S04 = ?

4.10. A three-stage cooling system is used to cool a hot annealing furnace offgas with a cold water
spray. The hot gas contains CO, C02, H2, H20, and N2 at a flowrate of 125 ft3/s, measured at 1300
°F. A sketch of the system is shown below. Calculate the volumetric flow and composition of all
streams. P= 1.00 atm. Use VLE data for water from Chapter 2.

Water Water Water
F = 240 gal/h F = 320 gal/h F = 400 gal/h I

Hot gas Spray tank I Spray tank II Spray tank III Cold gas
(1300 °F) 365 °F 160 °F 95 °F (95 °F)

F = 125ft3/s Water Water
<pCO = 0.20 F =? F =?
φΟΟ2 = 0.15
φΗ2 = 0.30
φΗ2Ο = 0.20
φΝ2 = 0.15

244 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

4.11. A spent reducing gas is processed to produce two gas streams; one for use as combustion
fuel, the other for recycling back to an iron ore reducing furnace. The flowsheet is shown below.
The process operates at 1 atm pressure.

Water To combustion Amine
2 I F 2 = 500 kg 7 I F7 = 300 kg
—-1 -t
S(p1e0n7t° gCa)s 1 H2O Condenser CO2 Absorber Recycle gas
► 4 Splitter 6 (7°C)
(4°C) p (15° C) to reducer
F1 = 1000 m3 ^"^^^^ F8 = 290 m3

φ Η2 = 47% Water I 3 Loaded amine 9
pH20 = 28% to C02 strip 1
pCO = 16%

φ C02 = 9 %

The (warm) spent gas is quenched from 107 °C to 4 °C in the condenser, which removes most
of the water vapor. The portion to be recycled is stripped of 80 % of its C02 in the absorber by an
amine (CH3NH2). The objective is to produce 290 m3 gas (volume measured at 7 °C) that can be
recycled back to the reduction furnace. Use FlowBal to make a material balance on the system.
Select volume percent for gas stream units, mass for the two water streams and the incoming
amine, and mass % for the loaded amine.

4.12. Residue from a coating operation consists of ZnS04, CuS04 and an insoluble residue. The
wCuS04 and wZnS04 in the residue remain close to 0.43 and 0.39, respectively. The first step in
reclaiming the zinc and copper is a hot water leach. The flowrate of hot water is set equal to the
flowrate of residue. The slurry is filtered and the insoluble filter cake (wsolid = 0.625) is disposed
of in a hazardous waste facility. The filtrate is cooled to precipitate as much copper sulfate
pentahydrate (CuS04-5H20, called CSPH) as possible. CSPH has the highest value since it has the
most copper. The CSPH precipitate has wsolid = 0.50. The current facility sends the supernatant
liquid from the precipitator to another part of the plant for recovery of ZnS04 and the unrecovered
CuS04. The plant engineer wonders if recycling part of the supernatant liquid stream back to the
stream leaving the dissolution tank might increase the recovery of copper as the dry CSPH
precipitate, and possibly its purity. The objective is to maximize the recovery of copper without
unduly overloading the system by an excessive amount of recycle. Make a mass balance for the
process to evaluate this idea, and determine the relationship between the flow of stream J and I, the
recovery of copper, and the zinc content of the dry CSPH cake. The flowsheet is shown next page.
Double-line arrows indicate multiphase streams.

Water Proposed To Cu & Zn Supernatant liquid
recycle recovery
FB = 1000kg/h stream J
wBH2O = 1.0

Residue Dissolution Mixer
reactor
FA=1000kg/h 0.625 Filter
wEliquid = 0.375 c a k e CSPH
ivACuS04 = 0.43 cake
wAZnS04 = 0.39
wAinsol = 0.18 nrCuS04-5H20 = 0.50

wG/LCuSO4 = 0.13

The current flowsheet has no recycle (i.e., stream J does not exist). At saturation, the wCuS04
solubility = 0.13. The solubility of ZnS04 is much higher than CuS04 and is not expected to
precipitate anywhere in the system. CSPH has wCuS04 of 0.6392.

4.13. An aqueous starting solution of wBa(C2H302)2 = 0.185 and wC2H402 = 0.075 is used to
prepare a solution of wBa(C2H302)2 = 0.564 by vacuum extraction of water and acetic acid in a

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 245

flash evaporator. The evaporator removes 88 % of the entering acetic acid. In practice, the
process operates more controllably by removing excessive vapor in the evaporator, analyzing the
composition of stream E, and attaining the correct composition by using an evaporator bypass
stream of the original feed. A flowsheet for the process is sketched below. The term ВAc refers to
barium acetate and the term HAc refers to acetic acid. Make a mass balance on the system to
determine the flow and composition of all streams. Water (liquid or vapor) is present in all
streams, but is not labeled on the flowsheet.

Vapor
(stream D)

Feed (stream A) Bypass (stream B) Product
FB = ? (stream F)
FA=1080kg/h FF = ?
wABAc = 0.185
wAHAc = 0.075 wFBAc = 0.564

v/HAc = ?

4.14. Petroleum refining requires the use of ceramic catalysts impregnated with nickel. When
these catalysts become contaminated, they are roasted (often with other Ni-containing-scrap) to
convert the nickel to NiO. The NiO is then leached in sulfuric acid to recover the nickel. The
roasted mixture contains wNiO = 0.288. Leaching is done counter-currently in three leach/settle
tanks (see sketch below), with the insoluble material passing in a direction opposite to the acid
leach solution. 75 % of the NiO in the insoluble fraction is leached in each tank. For every
kilogram of NiO leached, 1.313 kg of H2S04 is consumed, producing 2.138 kg of NiS04 and 0.241
kg of H20. The primary leach reagent is a dilute sulfuric acid solution proportioned so that the
product leach solution has wH2S04 = 0.038. The slurry stream has wsolid = 0.556. Calculate the
tonnes of acid leach solution per tonne of oxide leached, and the percentage loss of nickel to the
waste slurry.

Oxide feed
F = ? ivNiO = 0.288

Acid leach Product
solution solution

F =? Leach tank Leach tank Leach tank F =?
wH2S04 = I II wH2S04 0.038
ivH20 = ? Slurry III wH20 = ?
Slurry II F =? wNiS04 = ?
Waste slurry F =? winsol = 0.556
winSOl : 0.556
F =?
winsol = 0.556

4.15. A copper solvent extraction plant receives 1000 m3/d of PLS, which varies in copper content
from 2.0 to 3.0 g/L. The operator wishes to keep the PLS flowrate constant, and to recover 90 %
of the contained copper during two-stage extraction by varying the flow of organic reagent. The
flowsheet for the process is similar to that depicted in Figure 4.48 and 4.49. Laboratory data for
extraction of copper give the following relationship:

g/L Cu in Organic / . -0.883^1

— - = 2.52lg/L Cu in Aqueous I

g/L Cu in Aqueous

Data for acid stripping showed that the above ratio = 0.0244. Make a material balance on the

system to determine the flow of organic phase to attain the desired 90 % copper extraction for the

specified variation in PLS copper.

246 Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems

4.16. An ammonium hydroxide (NH4OH) solution is prepared by injecting water and gaseous
ammonia (NH3) into a 1000-gallon mixing tank on a continuous basis. The flow of water is 45
gal/min, and the flow of ammonia is 32 lb/min. The temperature of the ammonium hydroxide
solution is kept at 77 °F. Although the tank is lidded, it is possible for fumes to escape the tank
through the feedholes. To prevent undue worker exposure to ammonia, an ammonia sensor is
located at one of the feedholes. It is set to go off when the <pNH3 > 0.09. At a certain time, an
obstruction in the pipe cut the water flow to 25 gal/min. How long will it take to set off the alarm?
The partial pressure of NH3 in equilibrium with ammonium hydroxide solutions at typical tank
compositions is given by: /?NH3 (atm) = 1.08(wNH3) - 0.0080.

4.17. Ceramic insulators are prepared by slip casting, drying, and then firing to give an object of
precise shape. The slip cast objects contain wH20 between 10.4 and 11.6 %, depending on the
type of ceramic. They are dried under very carefully controlled conditions to prevent cracking.
They move down a tunnel dryer counter-current to a flow of warm air, such that they exit at 110 °C
completely dry. The engineer wishes to develop a model of the drying process by simulating it as
a series of discrete steps, each step at an airflow such that the same mass of water will be removed
in each step. A sketch of the simulated model involving five stages is shown below. The path of
the ceramic objects is shown as solid arrows, while the air streams are shown as dashed lines. Dry
air can be added to the entering humid air stream for the first three stages, and humid air can be
withdrawn from the exit stream from the middle three stages, but the amounts of air additions must
be kept to a minimum to conserve heat. The humid air leaving each stage (except the last one) is at
100 % relative humidity. P = 0.93 atm throughout.

Dry air
1^ 1^

Humid 1r 1r 1r
air
4 40 °C 60 °C * . «. . 4- - - Dry air
110 °C 4- - -
75 °C 85 °C

Wet Dry
ceramic
ceramic

_Jt _ J f 2L _
Humid air

Make a mass balance around each stage to determine how much air should be added, and how
much (if any) withdrawn. Plot the results for three different moisture levels in the incoming
ceramic parts.

4.18. A dust leaching process uses water to dissolve CuS04 from 25 t/h of baghouse dust. There
are no chemical reactions, just simple dissolutions. The Fe203 does not dissolve. The flowsheet is
shown below.

Baghouse dust Wash water
wFe203 = 0.68, wCuS04 = ?

Rich leach sol'n ^
wH20 = ? wCuS04 = ?

Filtrate I t Filter cake II

иЖ20 = ?
wCuS04 = ?

Filtrate II
ivH20 = ? vvCuS04 = ?

The leach tank slurry is passed to a filter to produce a filter cake and a clear liquid (Filtrate I)
that is returned to the leach tank. Filter cake I is washed with water, producing a slurry (Slurry II)

Chapter 4 Fundamentals of Material Balances in Non-Reacting Systems 247

that is also filtered to produce a cake and a clear liquid. All of the CuS04 dissolves in the leach
tank. Slurry I has a liquid mass fraction of 50%, Filter cake I has a liquid mass fraction of 13 %,
and Filter cake II has a liquid mass fraction of 11 %. Liquid streams are designated with a dashed
line, and solid phase streams with a solid line.

a) Make a mass balance to calculate the mass of wash water required to attain a mass fraction
of CuS04 in the liquid portion of Filter cake II of 0.0025.

b) Make a mass balance to calculate the mass of wash water required to attain an overall
recovery of CuS04 in the Rich leach solution of 99.9 %.

c) Suppose the dust feed rate was raised to 50 t/h with the same equipment size, which causes
the Rich leach stream to contain a solid fraction of 3 %. This outstream is filtered, and the cake,
containing 13 % water, is recycled to the leach tank. Repeat the calculations of parts a and b.

4.19. The extraction of titania from ore by a chloride process produces an intermediate titania/salt
mixture which must be washed to remove the salt. Washing is accomplished by a three-stage
cross-current process as shown in the flowsheet below. The slurry leaving each mixer/settler has
wTi02 = 50 %.

Solid feed Wash 3 Wash ivß6Ti02 = 0.96 Wash
(1000 lb) water " T i O 2 - 0 . 9 0
water watef
w1TiO2 = 0.60
w1NaCI = 0.30 a- Щ-(1000 lb) w3NaCI = bai (Ю00 lb) w6NaCI = bai,i(Ю00 lb)
w1H20 = bai. ■0 ,
T■ 8
T - w10TiO2 = 1.00

M/S1 M/S 2 Г^г M/S3 ! 10 Washed
- · - £ * · slurry

Wash 11. 12; (?lb)
solution 14 Solution mixer 13 i

The slurry leaving each stage is shown as two streams: the solid portion of the slurry (solid
line) and the liquid portion (dashed line). The salt is partially leached in M/S 1 and M/S 2, and
completely leached in M/S 3, as shown by the composition labeling for the solid portion of each
slurry. Use FlowBal to make a material balance on the process. Then use the results to calculate
the composition of the solid obtained by drying the washed slurry (i.e., mixed streams 9 and 10).

4.20. The solubility of H in aluminum was measured with the following results:

725 °С,рН2 = 1.0 atm, <pH = 0.90 wppm. 820 °C,pH2 = 0.70 atm, <pH = 1.30 wppm.

The H content of liquid Al at 670 °C is to be lowered from 0.50 to 0.010 wppm by inert gas
flushing with Ar. Calculate the volume of flush gas required per ton of Al.

CHAPTER 5

Stoichiometry and the Chemical Equation

Most material balance problems involve chemically reacting species. This requires the
establishment of molecular formulas for chemical compounds, the understanding of atom
rearrangement during chemical reactions, and the representation of chemical reactions by
stoichiometric equations. Based on the work of Dalton in 1803, a large body of chemical
information was rationalized by assuming that chemical substances consist of fixed ratios of
different atoms. This understanding, together with the work of Avogadro (1811) on the molecule,
led to the now-familiar method of expressing what happens in a chemical reaction in terms of a
stoichiometric equation. This Chapter deals with techniques used to write and use stoichiometric
equations, and shows how stoichiometry imposes restrictions on a material balance. We will also
illustrate the use of stoichiometric principles to introduce material balance calculations in reactive
system.

5.1 Atomic and Molecular Mass

The mole was defined in Section 1.4:
1. The mole is the amount of substance of a system that contains as many elementary entities
as there are atoms in 0.012 kg of carbon 12.
2. When the mole is used, the elementary entities must be specified and may be atoms,
molecules, ions, electrons, other particles, or other specified groups of such particles.
The amount-of-substance unit mole was developed as a unit of measure for the number of
atoms in 0.012 kg (exactly) of the isotope carbon 12, which is 6.02214 x 1023. This number is
designated as Avogadro's number NA, which for most purposes is rounded off to 6.022 x 1023
entities. The mass of one mole of an element in its atomic form (6.022 x 1023 atoms) is equal to
the atomic mass in grams. For example, one mole of atomic nitrogen has a mass of 14.0067 g, and
a mole of N2 molecules has a mass of 28.0134 g*. The molar mass of any chemical entity is
obtained by adding the molar masses of the constituent atoms. For example, the molar mass of
S02 is 32.06 + 2(15.9994) = 64.0588 g.
It's important to distinguish between actual and empirical molecular formulae. The
molecular structure of a gas is easily determined, so we know for example that benzene actually
consists of a molecule of six atoms of carbon and six of hydrogen. The actual formula for benzene
is thus C6H6. For solids and liquids, especially ionic species, the molecular nature is largely
unknown, so an empiricalformula of small numbers is used. The molar mass of NaCl(c) is based
on an empirical molecule of one atom of Na plus one atom of Cl, so the empirical formula for solid
sodium chloride is NaCl. There are no NaCl molecules in table salt, just an array of Na+ and СГ
ions. The empirical formula for benzene is CH, as is that for acetylene (otherwise C2H2).
However, not all ionic solids have simple integral amounts of elements. The iron oxide phase
wustite, when in equilibrium with hot iron, has xFe = 0.4864. It can be given either the empirical
formula Feo.9470 or FeOi.o56.

* The preferred SI molar mass unit (symbol M) is kg/mol, such that M(N2) is 0.0280134 kg. However,
g/mol and its equivalent kg/kmol are acceptable. A kmol has 1000 times the mass of a mol.

Chapter 5 Stoichiometry and the Chemical Equation 249

Sometimes it's handy to have an empirical formula for a mixture of two or more species. For
example we often use air as a reactant, and where high accuracy is not required, the composition of
dry air is set as φΝ2 = 79 % and φ02 = 21 %. The empirical formula for one mole of air is then
Ni.5sO.42. However, creating an empirical formula for a mole of a more complex formulation of air
requires a little more work than shown above. Suppose a special formulation of air contained xN2
= 0.7400, x02 = 0.2100, xAr = 0.0200, and xH20 = 0.0300. One mole of this gas contains 1.4800
mole of N, 0.4500 mole of O, 0.0200 mole of Ar, and 0.0600 mole of H, for a total of 2.0100
moles of elements. The empirical formula for one mole is then N1.48O.45Ar.02H.06, and the gas
contains xN = 0.73632, JCO = 0.22389, xAr = 0.00995, and xH - 0.02985. The relative molecular
mass of the gas is 28.7891. The program MMV-C will do this type of arithmetic for you.

5.2 Composition of Compounds and the Gravimetric Factor

It's necessary to be able to express the composition of a compound in both atomic and mass
fractions. Consider calcium carbonate, which has the empirical formula CaC03. A mole of CaC03
consists of one mole of Ca, one mole of C, and three moles of O. Since the empirical СаСОз
molecule consists of five atoms, the amount-of-substance (mole) fraction of each element is
calculated by dividing the number of atoms of each element by five. Thus, xCa is 0.2, xC is 0.2,
and xO is 0.6. However, it is also correct to consider the CaC03 molecule as made up of one
molecule of CaO plus one of C02. In that case, xCaO = 0.5 and xC02 = 0.5.

The mass fraction of each element in a compound is obtained by comparing the mass of each
element to the molar mass of the compound. For CaC03:

1 Ca: 40.08 g

1С: 12.011 g

ЗО: 3(15.999)-47.997 g

M(CaC03) - 40.08 + 12.011 + 47.997 = 100.09 g/mol.

The mass fraction of each element is then:

wCa = 40.08/100.09 = 0.4004, or 40.04 %.

wC = 12.011/100.09 - 0.1200, or 12.00 %.

wO = 47.997/100.09 = 0.4795, or 47.95 %.

Similarly, the mass fraction of CaO and C02 in CaC03 is:

wCaO - 56.08/100.09 = 0.5603, or 56.03 %.

WC02 = 44.01/100.09 = 0.4397, or 43.97 %.

It is often useful to relate the amount of an element entering as a species to the amounts of
other elements brought in by the same species. When CaC03 is added to a system, we see that
every mole of О brought in by CaC03 also brings in Уз mole of Ca and Уз mole of С Thus, the
molar ratio of C/O entering as CaC03 is 0.3333 . . . A similar calculation can be made for the ratio
of mass by calculating the mass ratio of two entities. For CaC03, 40.08 g of calcium enter for
every 47.997 g of O, so the mass ratio of Ca/O entering is 0.8351. This is called a gravimetric
factor, and is used extensively in analytical chemistry and by some material balance texts. The
gravimetric factor is most useful in making repeated calculations for a system involving the same
substances. A gravimetric factor can be calculated for any material, even if it is not a compound,
so long as its composition is known. The program MMV-C is useful in making calculations
involving the gravimetric factor.

250 Chapter 5 Stoichiometry and the Chemical Equation

EXAMPLE 5.1 — Use of the Gravimetric Factorfor Silicon.

Silicon is a raw material used in the semiconductor industry, but the crude metallurgical grade
must be purified for that use. The first step is chlorinating the metallic components of the material.
A partial analysis of a ferrosilicon feed material gave the following results:

W¥Q = 0.098; wC = 0.012; wSi = 0.869

The difference between the sum of the listed mass fractions and 1 turns out to be oxygen,
which enters as Si02; silica does not react with chlorine during the process. Use the gravimetric
factor to determine the mass of Si02 entering the chlorination reactor per 1000 kg of crude alloy,
and the mass of Si chlorinated.
Solution. By difference, wO = 0.021. The gravimetric factor for converting the amount of О to the
amount of Si02 in the crude alloy is obtained by calculating the mass ratio of Si02/0 in Si02.
Recognizing that every mole of О entering in the alloy brings in Vi mole of Si02, the mass ratio is
calculated as follows:

m(Si02) _ m(O)
(28.086 + 32.00)" 216.00

m(Si02)/m(0) = Y2(2S.0S6 + 32.00)/16.00 = 1.8777

The mass of Si02 entering the chlorination reactor is obtained by multiplying the mass of О
entering the reactor by the gravimetric factor:

m(Si02) = 0.021(1000)(1.8777) - 39.43 kg.

The gravimetric factor for converting the amount of О to the amount of Si in the alloy as Si02
is calculated in a similar way. Here again, one mole of О entering brings Vi mole of Si as Si02:

/w(Si as Si02)//w(0) = Уг(2%.Ш)1\6.№ = 0.8777

The mass of Si entering as Si02 is obtained by multiplying the mass of О entering by the
gravimetric factor:

m(Si as Si02) = 0.021(1000)(0.8777) = 18.43 kg.

The mass of Si that is potentially chlorinated is obtained by the difference between the total
mass of Si in the alloy and the mass of Si as Si02 in the alloy:

Mass of Si chlorinated = 0.869(1000) - 18.43 = 850.6 kg
Assignment. An x-ray diffraction analysis of the ferrosilicon showed that the С in the alloy is
present as silicon carbide, SiC. Use the gravimetric factor to calculate the mass of SiC present in
1000 kg of alloy.

EXAMPLE 5.2 — Mineralogical Constituents of a Concentrate.

A copper concentrate consists of chalcopyrite (CuFeS2), pyrite (FeS2) and silica (Si02). The
laboratory reported the chemical analysis as 29.8 %Cu and 30.2 %Fe. Calculate the mass fraction
of each mineral in the concentrate.

Solution. Although not stated explicitly, the context of the example indicates that the concentrate
analysis is mass fraction, expressed as parts per 100. The gravimetric factors involved are:

CuFeS2 from Cu = 183.56/63.55 = 2.888

FeS2 from Fe - 120.01/55.85 - 2.149

Fe from CuFeS2 = 55.85/183.56 = 0.3043

Chapter 5 Stoichiometry and the Chemical Equation 251

The mass fraction of CuFeS2 is obtained directly from the gravimetric factor, while the mass
fraction of FeS2 is obtained after subtracting the Fe present as CuFeS2:

xCuFeS2 = 2.888(0.298) = 0.8606 = 86.1 %.

xFeS2 - 2.149(0.302 - 0.3043 x 0.8606) = 0.0862 - 8.6 %.

The wSi02 is obtained by difference as 1 - 0.8606 - 0.0862 = 0.0532 = 5.3 %.
Assignment. Calculate the mass fraction of sulfur in the concentrate.

5.3 Writing and Balancing Chemical Equations

A chemical equation is a statement describing the reorganization of atoms in one or more
substances. Chemical bonds are broken and new ones formed. The reactants are placed on the left
side of the equation, and the products on the right side. The requirement for conservation of mass
requires that there be the same number of atoms on both sides of the equation. Furthermore, a
useful chemical reaction must be balanceable in only one unique way.

It's important to differentiate between a chemical reaction and a physical transformation. In
the latter case, a substance undergoes a phase change, like melting or vaporization. Chemical
bonds are intact, but there is a different physical structure as a result of the transformation.
Sometimes, in a casual way, the term "reaction" is used in this Handbook to refer to a physical
transformation. This is usually in connection with the use of FlowBal, where one must designate
physical transformation as "reactions" in order to tell FlowBal that two different forms of a species
are involved. Please be alert to this terminology.

Correctly expressing the amount of each species on each side of a chemical reaction is called
balancing a chemical equation for a reaction. In most cases, balancing a reaction is
straightforward, even trivial. Consider a reaction for the formation of a compound from the
elements; here sulfur trioxide from diatomic sulfur gas and oxygen. Start by writing the reaction
with letters а, й, etc to represent the amount of each substance:

aS2(g) + M)2(g)^cS03(g) [5.1]

Next, pick a basis of a small integer for one of the constituents. Here, we select a value of с =
one. The reactants must then have one atom of S and three atoms of O, so the equation is easily
balanced at a, = Vi and b = 1 lA.

On the other hand, some reaction balancing can take a bit of algebra, and may require an
element balance. Consider the oxidation of methane with carbon dioxide to form carbon monoxide
and water vapor:

aCH4(g) + Z>C02(g) -> cCO(g) + dH20(g) [5.2]

If we set a = 1, then d must = 2. An oxygen atom balance shows that 2b = с + 2 and a carbon
atom balance shows that b + 1 = c. Clearly, 6 = 3 and с = 4. Later in the Chapter, we'll look at
reaction-writing and balancing in more detail, but for now, please review the coverage of this
subject in a typical engineering chemistry textbook.

Theoretically, a chemical reaction can proceed in either direction. Sometimes the direction is
obvious, especially to a person with experience in chemical processes. But even a novice knows
that S02 is a very stable substance, so S2(g) and 02(g) tend to react until one or the other is
consumed. This indicates that Equation [5.1] goes only in the direction written. But if CH4(g),
C02(g), CO(g), and H20(g) are mixed, there is no way to tell a priori which direction Equation
[5.2] tends to go. The directional tendency may be to the right or the reverse, so the reactants in
one circumstance are the products in another. You must then be told the direction of a reaction, or
given information that you can use to calculate the direction. Some of the factors that favor one
reaction direction over another are discussed later in Section 5..

252 Chapter 5 Stoichiometry and the Chemical Equation

5.3.1 Chemical Reaction Concepts

Correctly-written chemical reactions are crucial in understanding what goes on in a process.
Some major points are:

• The chemical formula of the reactant and product species tells us the amount of each
element making up the compound.

• A chemical reaction gives information on the relative amount of each species taking part.
The relative amounts of reactants and products are given by the coefficients in the
balanced equation. For example, Equation [5.2] states that the oxidation of one mole of
CH4(g) by three moles of C02(g) produces four moles of CO(g) and two moles of H20(g).
The number of atoms (and therefore the mass) on both sides of the equation is the same,
but not the number of molecules. When no coefficients are shown, the number 1 (one) is
implied as the molar amount of each species.

• Some heterogeneous reactions produce intrinsic stoichiometric ratios for the products.
Consider two simultaneous reactions that occur during the reduction of zinc oxide by
carbon:

ZnO(c) + C(c) -> Zn(g) + CO(g) [5.3]

2ZnO(c) + C(c) -* 2Zn(g) + C02(g) [5.4]

When ZnO reduction is carried out by heating a mixture of ZnO and С in an evacuated
chamber or in the presence of an inert gas, an intrinsic stoichiometric ratio exists between the
amount of zinc vapor and the amount of CO(g) + C02(g).

nZn = nCO + 2nC02

This occurs because the stoichiometric atom ratio of O/Zn in ZnO = 1, therefore there must be
as many moles of О as Zn in the products. This relationship holds regardless of the ratio of solid
ZnO to solid carbon in the reactants, the extent of reaction, the system volume, or the temperature*.
Since there are no other reactive species present, the intrinsic stoichiometric ratio causes the
following relation to hold:

pZn=pCO + 2pC02 [5.5]

• The relative masses of reactants and products can be calculated from the molar masses and

the equation coefficients. Equation [5.1] states that 32.06 g of S2(g) combines
stoichiometrically with 48.00 g of 02(g). Since mass is conserved, the reaction produces
80.06 g of S03(g).

• The physical states of reactants and products should be added to the formulas to indicate
the form of each substance. We'll use the following terminology to indicate the physical
form of a substance:

State Symbol
Solid (crystalline)
Liquid (c )
Gas
Dissolved in water (aqueous) (/)
Dissolved in non-aqueous solvent
(g)
(aq )
(d)

If a solid species exists in more than one crystalline form, the symbol will so indicate. For
example, СаСОз exists in two crystalline forms: calcite (rhombohedral crystalline form), and
aragonite (orthorhombic form).

* The intrinsic stoichiometric relation is not observed if the ZnO/C mixture is heated in the presence of a
reactive gas, such as CO.

Chapter 5 Stoichiometry and the Chemical Equation 253

• Writing an equation does not mean that it will take place, can take place, or how far it goes
if it does take place, or if it tends to go in the direction written or the opposite. Equation
[5.1] does not say that if one mole of S2(g) is mixed with three moles of 02(g), two moles
of S03(g) will form. It says that if a mole of S2 reacts, three moles of 0 2 reacts with it to
form two moles of S03. If a process states that one mole of S2(g) is mixed withfour moles
of 02(g), you might be tempted to write a reaction with these amounts, and include a mole
of unreacted 02(g) on the right-hand side of the equation. This is incorrect, and gives rise
to a rule not to include the same species on both sides of a chemical equation.

• A chemical reaction is not a chemical process. Other species might form as a result of
mixing the specified reactants. For example, under some circumstances, S02(g) might be
a product of the Equation [5.1] reactants. Also, sometimes a process includes an inert
species that helps a reaction take place but does not participate chemically. For example,
air might be mixed with S2(g) instead of pure 02(g), but since N2 is not a reactant or
product in the chemical reaction, it does not (and should not) appear in Equation [5.1].
This gives rise to a rule not to include inert substances in chemical equations.

5.3.2 Writing and Balancing Chemical Reactions for Simple Processes

One of the first steps in making a material balance in chemically reactive systems is to decide
if a single chemical reactions is adequate to represent the process. In an existing process, there is
usually enough data to suggest if a single reactions dominates to an extent that all others can be
considered irrelevant or inconsequential. For process modification or a new process, some
guesswork may be required, based on experience with similar process chemistry. The material
balance calculations are greatly simplified when a single chemical reaction adequately describes a
process. In some cases, two or more chemical reactions are required to describe what's taking
place in a reactor. This Section shows examples of reaction stoichiometry applied to simple
systems with one, or at most two chemical reactions.

EXAMPLE 5.3 — Reduction of Wustite by CO.

Wustite of composition 76.10 %Fe is reduced by carbon monoxide to form iron and carbon
dioxide. Write a balanced chemical reaction for the production of one mole of iron.

Solution. Wustite is an oxide of iron that has a variable ratio of О to Fe, but the ratio is close to
one. This tells us that although the composition unit is not explicitly stated, it must be as wFe. A
convenient basis is 100 g of wustite. The table of atomic mass gives MFe = 55.845 and MO =
16.00. Therefore, 100 g of the above wustite contains 1.3627 moles of Fe and 1.4938 moles of O.
The O/Fe ratio is thus 1.096, so we can express the empirical chemical formula for wustite as
FeOi.o96*· Thus, we must remove 1.096 moles of О to produce one mole of Fe. On a basis of one
mole of Fe, our preliminary reaction is:

FeOi.o96 + aCO -» Fe + eC02

By inspection we see that a carbon balance requires that a = b. An oxygen balance tells us
that a + 1.096 = 2b. We can solve for a and b in our head to write:

F e C W O + 1.096CO(g) -* Fe(c) + 1.096CO2(g)

Assignment. Write a balanced chemical reaction for the reduction of wustite to produce one mole
ofC02.

The wustite reduction equation (like the other reactions shown early in the Chapter) was easy
to balance. However, it is sometimes difficult to balance an equation containing several species.
For example, the production of tungsten carbide coatings on cutting tools involves a reaction

We could also have expressed the empirical composition as Fe0.9i2O.

254 Chapter 5 Stoichiometry and the Chemical Equation

between tungsten hexafluoride and methane. The (unbalanced) chemical reaction is written first
with letter symbols indicating the (unknown) equation coefficients:

*WF6(g) + j;CH4(g) -> aW2C(c) + Z>HF(g) [5.6]

First, we check to see if the equation can be balanced. This means checking to see if all of the
reactant elements appear as product elements (they do). Next, set up element balance equations

W balance: x = 2a [5.7]

С balance: y = a [5.8]

H balance: 4y = b [5.9]

F balance:6x = b [5.10]

There are four equations and four unknowns, so it should be easy to solve for each unknown.

However, after a little effort, you will find that there is no feasible solution to this equation set, so
the equation can't be balanced as written. There appears to be a mismatch between the amount of

H and F. Four possibilities are: F2 is a product, С forms, WC is a product, or H2 must be added as
a reactant. Since we are interested in conversion of all the WF6 without forming C, the best option
is to try adding H2 to the reactants to see if this allows the equation to be balanced. The revised
reaction and four element balance equations are:

*WF6(g) + yCU4(g) + zH2(g) - aW2C(c) + Mfffe) [5.11]
W balance: x = 2a [5.12]

С balance: y = a [5.13]

H balance: 4y + 2z =~-b [5.14]

F balance: 6x = b [5.15]

With five unknowns and four equations, one of the variables must be chosen as a basis. For a
basis of a = 1 (i.e., one mole of W2C formed), x = 2 and у = 1, so b = 12. Therefore, z = 4. The
assumption of H2 being required as a reactant appears valid, at least from a reaction-balancing
standpoint.* The balanced reaction is:

2WF6(g) + CH4(g) + 4H2(g) - W2C(c) + 12HF(g) [5.16]

The above situation shows what happens when reaction writing omits a required species.
Another type of balancing difficulty arises when we include too many species in a reaction.

Consider the combustion of one mole of carbon with less than one mole of oxygen (as 02). A look
at possible reactants shows that CO and C02 are common products of combustion, so our reaction
and two element balance equations are:

dgraph) + x02(g) -> aCO(g) + £C02(g) [5.17]

С balance: l = a + b [5.18]

О balance: 2x = a + 2b [5.19]

Clearly, the equation set cannot be solved because there are too many unknowns. The
equation can be balanced by choosing a value of x (something between 0.5 and 1), but there are an
infinite set of values between these limits. Therefore, Equation [5.17] was inadvertently written
with too many products. A little thought tells us that the С - О system is better described by two
separate reactions. Two possibilities are:

C(graph) + 02(g)^C02(g) [5.20]
Cigraph) + V202(g) - CO(g) [5.21 ]

Try writing a balanced reaction without H2 as a reactant, and forming CF4(g).

Chapter 5 Stoichiometry and the Chemical Equation 255

Alternatively:

C(graph) + 02(g) - C02(g) [5.22]

C(graph) + C02(g) - 2CO(g) [5.23]

At elevated temperatures, we know that carbon and oxygen react vigorously. Since carbon is
present in excess, oxygen will be completely consumed whichever reaction set we decide is most
valid. (However, if oxygen had been present in excess, carbon would have been completely
consumed). Picking one reaction set out of two or more possible sets to simulate a process is a
matter of arbitrary choice unless we have some knowledge of the process mechanism. Suppose
there's evidence that when carbon is in excess, oxygen is completely consumed to form C02 via
the first reaction in each of the above sets. The second set (involving [5.23] to form CO) is thus
probably a better representation of the carbon oxidation process.

As we apply the writing and balancing of chemical reactions to making material balances, we
often find that the process requirements set the basis rather than the reaction writer. It's important
to be able to take any given process basis, and write chemical reactions involving the reactants and
products. Since a process may involve dozens (or more) species, it is often very difficult to
determine which reactions take place. More difficult yet is to determine which reactions have a
dominant effect on the process vs. those that have little or no effect, and may therefore be omitted
from the material and heat balance calculations.

EXAMPLE 5.4 — Production of Molybdenum Carbide.

Molybdenum carbide is an important ingredient in drill bits. It can be prepared at 1800 °F by
the reduction of molybdenum dioxide with carbon monoxide according to:

яМо02 + bCO -► xMo2C + j;C02 [5.24]

Calculate the volume of CO consumed at 1800 °F for every ton of dioxide converted. Assume the
pressure is 1.00 atm.

Solution. The first step is to select a basis, then balance the equation. On the basis of one mole of
Mo02 (a = 1), the material balances equations are:

Mo balance: 1=2JC [5.25]

О balance: 2 + b = 2y [5.26]

С balance: b = x+y [5.27]

The solution to this equation set yields x = 0.5, b = 3, and у = 2.5. Rewriting the reaction to
include physical state designators and integers for coefficients gives:

2Mo02(c) + 6CO(g) -+ Mo2C(c) + 5C02(g) [5.28]

The basis of the problem is one ton (2000 lb) of oxide reacting. The amount of oxide reacting
is obtained by dividing 2000 by its molar mass of 127.94, or 15.632 lb-mol. The coefficients in
Equation [5.28] must be multiplied by 15.632/2, or 7.816 to convert to the process basis. So
46.896 lb-mol of CO are consumed for reaction with 15.632 lb-mol of Mo02 to form 7.816 lb-mol
of Mo2C and 39.08 lb-mol of C02.

The mass of Mo2C produced is obtained by multiplying the amount by the molar mass (7.816
x 203.89). The volumes of CO and C02 are obtained by multiplying the number of moles of each
by the molar volume of a lb-mol (359 ft3 at STP). The results are shown in the next-page ledger.

This example is an illustration of the need first to write reactions in terms of simple integer
molar coefficients before converting to a process basis. Reaction equations are always written in
amount units (moles) that can be uniquely balanced.

256 Chapter 5 Stoichiometry and the Chemical Equation

Reactants Pro(iucts co2

Mo02 CO Mo2C

Moles (reaction) 26 15

lb-mols (process) 15.632 46.896 7.816 39.08

Mass (lb) 2000 1313.6 1593.6 1719.9
Volume (ft3, STP)
lume (ft3, 1800 °F) 4.716E+05 6.175E+05

2.168E+06 2.839E+06

Assignment. Suppose the oxide feed was molybdenum trioxide (M0O3). Calculate the volume of
CO consumed (at 1800 °F) per ton of Mo2C produced.

5.4 Calculations Involving Excess and Limiting Reactants

A balanced chemical equation expresses the relative amount of substances participating in the
reaction. But remember: a chemical reaction is not a process. Processes are seldom carried out by
mixing reactants together in stoichiometric quantities as indicated by the balanced chemical
equation. For example, in a process based on Equation [5.28], the molar feed ratio of CO to Mo02
would always be more than three. Using more than stoichiometric CO means that when all of the
M0O2 has been converted, there is still some CO remaining. The CO is thus present in excess. The
amount of Mo2C formed is then determined by the quantity of Mo02 present. Once the Mo02 is
consumed, no more Mo2C can form because the Mo02 limits the reaction, hence Mo02 is called
the limiting reactant. From a chemical reaction standpoint, the limiting reactant is the one that
would be consumed first if the reaction were to go to completion as written.

A system may have more than one limiting reactant. If we expose a large amount of iron to a
small amount of gas containing F2, Cl2, and Br2, the iron will react with each halogen in turn,
starting with F2 until all the F2 is consumed, then with Cl2, and then Br2. So each of the halogens is
a limiting reactant, and all will be completely consumed when iron is present in excess.

There are a number of reasons for the use of an excess amount of a reactant.

• There are thermodynamic limitations on the extent to which a reaction may proceed. The
extent of a reaction (sometimes called the degree of completion) improves if one of the
reactants is added in excess. The thermodynamic limit on reaction extent is determined by
the numerical value of the equilibrium constant for the balanced chemical reaction. The
concepts of degree of completion and thermodynamic limitations will be explained in
more detail later in Section 5.5.5.

• There are kinetic limitations on how fast a reaction will proceed. Adding an excess
reactant is one way to increase the reaction rate, thus minimizing the time required to
produce a certain amount of product. Where a gas-is reducing a metal oxide, adding
excess gas will make it diffuse more rapidly to the location where reduction occurs.
Where a solid (like lime) is used to desulfurize a gas, adding excess lime increases the
surface area for reaction and speeds up the process.

The use of an excess reactant does not always mean a loss of that substance in the product
stream. It's common to use an excess amount of reducing gas in the production of direct reduced
iron, and then send the spent product gas to a treatment plant for recovery of the unused CO and
H2. The gas is recycled to the reduction furnace (with more reducing gas). A flowsheet for spent
reduction gas reprocessing was given in Example 4.15. Economic factors also play a role in the
use of excess reactants. The cost of the excess reactant and its reclamation is balanced against the
value of increased productivity of the process. The optimum amount of the excess reactant is
reached when the cost/benefit ratio is a minimum.

The degree of excess reactant added to or required for a process is defined as a function of the
stoichiometric amount according to:

Chapter 5 Stoichiometry and the Chemical Equation 257

% excess = 100 " i n " " s t o i c h [5.29]

^stoich

where nm is the amount or mass of reactant fed to the reactor and «stoich is the amount or mass
required by the stoichiometric coefficients. Therefore, the designation of "excess" reactant must
be defined in reference to a specific reaction. The term "% of stoichiometric" or "% of theoretical"
is sometimes used, and is simply the % excess + 100. Example 5.5 illustrates the concept of using
an excess reactant. The amount of excess reactant can also be expressed as fractional excess,
which is just the % excess/100.

EXAMPLE 5.5 — Production of Titanium by the Kroll Process.

The reduction of a metal by another is called metallothermic reduction. Metallic titanium is
produced by reduction of titanium tetrachloride (TiCl4) with magnesium. In one plant, a steel
reactor vessel is charged with 320 lb of Mg bars, and heated in an inert atmosphere to about 850 °C
to form a pool of molten Mg on the bottom of the furnace. Then 1070 lb of liquid TiCl4 is run in
through a pipe at the top, while a small amount of inert gas is added to maintain a positive pressure
in the reactor. Figure 5.1 shows a sketch of the Kroll reactor vessel. Determine the excess reactant
and its % excess.

TiCL Inert gas

Figure 5.1 Schematic of a Kroll reactor for the production of titanium sponge. Diagram shows
molten Mg at the point of adding the first TiCl4.

Solution. The TiCl4 vaporizes and reacts with the Mg according to:

2Mg(/) + TiCl4(g) -+ 2MgCl2(/) + Ti(c) [5.30]

The stoichiometric mass ratio of Mg/TiCl4 - 2(24.305)/189.712 = 0.2562. The actual mass

ratio put in for the process is 320/1070, or 0.2991, which shows that Mg is an excess reagent. The

stoichiometric requirement for Mg is 1070(0.2562) = 274 lb, so the % excess Mg is calculated

using Equation [5.31]:

%excessMg = 100 320 - 274 - 1 6 . 8 % [5.3i]

Since the price of titanium is several times that of magnesium, it makes sense to add excess
magnesium even if (as in this case) the excess magnesium cannot be recovered economically from
the mass of titanium sponge.

Assignment. A variation of the Kroll process uses sodium as a reducing agent. Calculate the mass
of TiCl4 to be put in if the reactor was initially charged with 600 lb of sodium, and the aim is to use
15 % excess sodium.

258 Chapter 5 Stoichiometry and the Chemical Equation

The concept of an excess reactant can be applied to systems where more than one reaction
equation occurs, such as fuel oil combustion. Fuel oil is a mixture of several hydrocarbons of
variable composition plus a small amount of sulfur, so oil has no specific formula. The
composition of fuel oil (as with most liquids and solids) is given as elemental mass fractions. A
typical fuel oil composition is 87 %C and 13 %H. When excess air is used to assure complete
combustion, only two products form: C02 and H20. In addition, the excess 0 2 and all of the N2
are in the product gas, although they are not reaction products per se. The chemical equations for
the combustion of the С and H portions of the fuel are:

H(/) + %02(g) -+ V2tì20(g) [5.32]

C(/) + 0 2 ( g ) ^ C 0 2 ( g ) [5.33]

where (/) indicates the liquid state of fuel oil.* The chemical equation shows that one mole of 0 2 is
required per mole of С in the oil, and % mole of 0 2 per mole of H.

One kg of the above oil has 128.97 moles of H and 72.43 moles of С According to the

equation coefficients, oxidation of the H requires %(128.97) = 32.24 moles of 0 2 and oxidation of
the С requires 72.43 moles of 02. Assuming air to have a volume fraction of 0 2 of 0.21, the
stoichiometric amount of air required to combust one kg of fuel oil is:

32 24 + 72 43
«air, stoich. = —:
0.21 — = 498 [5.34]

LJ

If we wanted 15 % excess air, the amount of air is 1.15(498) = 573 moles, or 12.8 m3 air

(STP)/kg oil.

5.5 Progress of a Reaction

There is no a priori way to know how far a reaction goes in a process. It may continue until
one of the reactants is consumed, or it may hardly go at all. It may go to equilibrium, but it cannot
go further. Here we introduce some general terms for defining how far a reaction can go.

There are three main sources of information about reaction progress. First, in an existing
process, the net amount of products formed can be measured by sampling the in- and outstreams.
The information so gained is entirely empirical, and is valid only for the conditions of the process
being sampled. Second, if the reaction is limited by chemical kinetics, we might be able to apply
kinetic principles by using rate constants to calculate the amount of products formed. Third, for
any process, thermodynamic considerations limit how far a reaction goes; its equilibrium position
is limited by the value of the equilibrium constant (Keq) for the reaction. Equilibrium calculations
do not tell us howfast a reaction goes, but instead howfar it can go if kinetics are fast.

5.5.1 Extent of Species Reaction and Rate of Reaction Terminology

To be useful, reactions must proceed at reasonable rates. The simplest case occurs when a
reaction is so fast that it closely approaches the equilibrium position before leaving the system.
Quite often, this does not happen. This section introduces two terms that we will use to describe
how far a reaction proceeds, whether it reaches equilibrium or not. Consider a process for the
oxidation of sulfur dioxide to form sulfur trioxide. Equation [5.35] states that one mole of oxygen
is required to oxidize two moles of sulfur dioxide. The time basis is one second. We now define
the stoichiometric reaction coefficient с as negative for reactants and positive for products. For the
reaction described by Equation [5.35], cS03 = +2, cS02 = -2, and c02 = - l .

2S02(g) + 02(g)-2S03(g) [5.35]

There is an advantage in mass balance arithmetic in writing the equation coefficients in terms of one mole
of the atomic reactant when the reactant composition is listed in elemental mass fraction units.

Chapter 5 Stoichiometry and the Chemical Equation 259

Fin = 10.0/7/s Reactor F°ut=7.5A7/s
F°utSO2 = 1.0/7/s
F*nS02 = 6.0 nls *F°utS03 = 5.0 nls
Fin02 = 4.0 /7/s F°ut02 = 1.5/7/s

There are two types of flows for the process described by the flowsheet: stream (or process)
flow and species flow. We designate R as the net molar rate offormation of a species (the rate of
consumption is defined as the negative of the rate of formation). We see that 7?S03 = 5.0 - 0.0 =
5.0,7?S02 = 1 - 6 = - 5 , and R02 = 1.5 - 4.0 = -2.5. The overall net Äpr0Cess = 7.5 - 10 - -2.5. The
units of R are amount per unit time, but once we explicitly state the time unit in the process setup,
it is permissible to simplify the problem discussion by omitting the time unit from R.

The rate of reaction R-R is defined as the net molar rate of change of a species divided by the
stoichiometric reaction coefficient c, where s refers to the species:

R-R = ^ [5.36]

Cs

This definition gives a unique value for the rate of any reaction, with no distinction for any
given species. When using Equation [5.35] for the process, R-R based on S03 is 5.0/2.0 = 2.5. R-
R based on S02 is -5.0/-2 = 2.5, and R-R based on 0 2 is -2.5/-1 = 2.5. Therefore, we can indicate
the progress of a reaction by specifying its R-R without reference to any of the reaction species.

The word rate is used in the above context to describe the application of the R-R term in
making steady-state material balances. It has a different meaning than the same word used in
describing the kinetics of a chemical reaction. (Chemical reaction kinetics are described in Section
5.5.2.) The R-R term is based on measurements taken from existing processes, laboratory
experiments, or expert knowledge from persons intimately familiar with the process. Among other
things, the value of R-R is flowrate-dependent. If the molar rate of change of a species were
constant, and we were to double the inflow, R-R would also double even if the outstream
composition remained the same. For this reason, R-R is seldom used as an independent variable in
making a material balance.

The second term used in connection with the progress of a reaction is the extent of reaction
XR of a particular reactant species s:

с* in _ c.out

XRs = — [5.37]
Fins

The XRS02 = (6.0 - 1.0)/6.0 = 0.833, while the XR02 = (4.0 - 1.5)/4.0 = 0.625. XR varies
between 0 and 1, and is handy to use because it is always a small positive number. XR is most
often applied to the limiting reactant for a system. Unlike the R-R term, the XR is species-
dependent and is flowrate independent if the outstream composition does not change with process
flow. In this Handbook, the two reaction progress terms are used almost exclusively when
FlowBal is the computational tool.

The R-R and XR terms are related to each other, and if known, may be used as part of the
equation set for solving the material balance. Alternatively, if a process balance is known from
stream sample data, R-R and XR can be calculated from the data. However, it is dangerous to use
XR or R-R values obtained from one process to another one at different conditions.

5.5.2 Chemical Reaction Kinetics

There are many factors that limit the rate of approach to equilibrium. One factor is how fast
the reactant molecules interact with each other to form products. The rate of species-to-species
chemical reaction is part of the subject called chemical kinetics. A detailed description of the

260 Chapter 5 Stoichiometry and the Chemical Equation

subject is beyond the scope of this Handbook. Readers seeking a more exposition should consult a
typical freshman engineering chemistry text. The subject is briefly considered here to provide a
better understanding of reaction rates, and to clarify the XR and R-R terms introduced in the
previous section.

Consider the situation where a single reactant tends to decompose to certain products.
Suppose the reactant is heated to a temperature where Кщ for its decomposition is a large positive
number. The reactant tends to decompose, and if the reaction kinetics are finite, decomposition
occurs until the equilibrium position is reached. Strictly speaking, the equilibrium position is
never reached. Instead, according to the law of mass action, the kinetics of decomposition and
product reformation approach each other as the equilibrium position is approached. Once there is
no discernable change in composition with time, we consider the system as being at equilibrium.

For example, sometimes it's useful to prepare a protective atmosphere free of oxygen-
containing gases. Hydrogen itself is ideal, but is difficult to store and handle. On the other hand,
pure ammonia is easily stored as a liquid under pressure, and tends to decompose to the elements at
elevated temperature according to Equation [5.38], a process known as thermal decomposition. In
the absence of a catalyst, this is a homogeneous gas-phase reaction. Two moles of product are
produced for every mole of ammonia that decomposes.

2NH3(g)-N2(g)+3H2(g) [5.38]

Chemical kinetics can be studied in several ways. Commonly, an isothermal experiment is

devised to measure the progress of a reaction with time. The kinetics of NH3 decomposition can
be studied by heating pure NH3 in a batch reactor and analyzing the gas in the reactor as a function
of time. In one set of data at 700 K, after 35 minutes, 10 % of the NH3 had decomposed. In the
next 35 minutes, 10 % of the remaining NH3 had decomposed, and so on until the experiment was
stopped after 200 minutes. The value of Кщ for Equation [5.38] at 700 К is large enough so that

at equilibrium, over 99 % of the ammonia will decompose. Therefore, the reverse reaction is

unimportant until about 60 % of the NH3 decomposes. Up to that point, the composition of NH3 in
the product gas gives results that are unambiguously defined by the kinetics of NH3 decomposition.
Table 5.1 and Figure 5.2 show the results from a second experiment, with samples taken at

intervals between 40 and 360 minutes

The experiment was carried out in a batch reactor having a volume 57.44 L, which is exactly

the molar volume at 700 К and 1 atm, and the gas was analyzed for xNH3. From a material
«NH3 = 2xNH3/(l + xNH3), ηΐί2 = 1.5(1 -
balance based on one mole of NH3 initially present, by dividing nH2 and nN2 by the total moles
«NH3), and riN2 = ηΆ2β. xH2 and xN2 were obtained

out. The product gas volume increased with time, and since the volume was fixed, the pressure in
the reactor also increased with time. Notice that the increase in number of moles produced is

equivalent to the fractional conversion of NH3 in the reactor. For example, at 360 minutes, 59.3 %

of the incoming NH3 decomposed.

Table 5.1 Products of the isochoric thermal decomposition of 57.44 L (one mole) of ammonia at

700 К and 1 atm in a batch reactor, based on analysis of NH3. Samples were taken until about 60
% of the ammonia had decomposed.

time, min. 40 80 120 180 260 360

xNH3 0.827 0.692 0.588 0.469 0.352 0.255
0.130 0.230 0.309 0.399 0.485 0.559
xH2 0.043 0.077 0.103 0.133 0.162 0.186
xN2 0.905 0.818 0.741 0.639 0.521 0.407
0.142 0.273 0.389 0.542 0.719 0.890
«NH3 0.047 0.091 0.130 0.181 0.240 0.297
1.095 1.182 1.259 1.361 1.479 1.593
«H2
«N2

total moles produced

Chapter 5 Stoichiometry and the Chemical Equation 261

Figure 5.2 Composition of gas produced by the isochoric thermal decomposition of one mole of
NH3 at 700 K. Time refers to the minutes following the introduction of pure NH3 into the reactor.

In the above context, we've defined the reaction rate as the change in amount of a species
divided by the change in time. By this definition, the reaction rate decreases with time; note that
0.087 moles of NH3 decompose between 40 and 80 minutes, but only 0.077 moles decompose
between 80 and 120 minutes. The average ammonia reaction rate between 40 and 120 minutes is
0.00205 mol/min. The instantaneous reaction rate of any of the species is the slope of the tangent
to any of the lines in Figure 5.2 at a particular time period.

Now consider the results for the extent of NH3 reaction for a 260 minute reaction time. One
mole of NH3 entered, and 0.521 mole of NH3 remain, so 0.479 decomposed. Our definition of
species extent of reaction from Equation [5.37] tells us that JLZ?NH3 = 0.479. This value is
independent of the amount of NH3 present initially in the reactor and the balancing numbers used
for Equation [5.38]. Our definition of R-R for the ammonia decomposition reaction can be based
on the rate of formation of N2 or H3, or the rate of decomposition of NH3. Decomposition of one
mole of NH3 produced 0.719 mole of H2, so R-R, based on the stoichiometric balancing numbers
used in Equation [5.38] is 0.719/3 = 0.2397. Notice that this is twice the value of XÄNH3.
However, if two moles of NH3 were initially present, R-R would be 0.479. Also, if one mole of
NH3 was initially present and Equation [5.38] was written with 1 as the balancing number for NH3,
R-R would be 0.479. We must be careful in using and interpreting R-R because it is a function of
the instream flow and the numbers used to balance the controlling reaction equation.

It may be useful to go a bit deeper into the subject of chemical kinetics of a simple
decomposition reaction. We noted that the reaction rate of Equation [5.38] decreased with time;
actually the correct observation is that the reaction rate is a function of the amount of the reactant
present:

Rate = £(ШН3;Г [5.39]

The above expression is called a rate law, the proportionality constant к is the rate constant,
and the exponent n is the order of the reaction. The application of Equation [5.39] to the data in
Table 5.1 applies only to the decomposition of NH3 via Equation [5.38]; a different set of data are
required to study the reverse of Equation [5.38]. Some formulations of Equation [5.39] use
concentration instead of amount, such as mole fraction or mol/L. However, for the case of
increasing-moles reactions, mole fraction gives an incorrect result. Mol/L is acceptable for an
isochoric reactor because the volume remains the same.

The integration of Equation [5.39] gives a useful form of the rate law for the decomposition of
a single species. In terms of the amount of species /:

262 Chapter 5 Stoichiometry and the Chemical Equation

\n(ni)n = -kt + ln(ni)0n [5.40]

where the first term represents the amount of i for any time t, and the second composition term is
the initial value of ni. If a plot of 1п(ш) vs. / (or even better, 1п[(ш)0/(ш)] vs. f) is a straight line, n
= 1, and the decomposition reaction is first-order. Figure 5.3 shows this is indeed the case for
ammonia decomposition during isochoric decomposition in a batch reactor. The unit of к is
reciprocal time, and for this set of data, has the value 0.00250 min-1. For the stated process
conditions, where the reverse reaction is insignificant, ln(wNH3) = -(Л min)/400.

Ammonia Decomposition Kinetics at 700 К and 1 atm

o.o ^ _ 1, 11 1 1!

-0.1 ^^^^ y = -0.00250x
^^^^W>4.
-0.2 ^^^^
^^^N^
Ä -0.3
^4^^
CO
^*^^^
I -0.4
^""v^^
z ^s^.

5, -0.5
■S -0.6

-0.7

Γ\Ι-0.8

-0.9

-1.0 40 80 120 160 200 240 280 320 360|
0 time, minutes

Figure 5.3 Graphical determination of the order of the reaction for the decomposition of pure

ammonia at 700 К in an isochoric batch reactor. Data from Table 5.1. Trendline tool was used to

develop a linear equation which was constrained to pass through the origin. Reaction is first order.

If we had carried out the decomposition in an isobaric batch reactor, some arrangement would
be necessary to allow for the additional amount of product gas. At 260 minutes, for example, the
molar volume of the reactor would need to be 84.95 L. The composition term of mole fraction is
still inappropriate for use in Equations [5.39] and [5.40]. Mol/L is also inappropriate because the
volume varies. Of course, the amount can be obtained by multiplying the molar flow times the
mole fraction.

So far, we've considered reactions where only one reactant is involved. When two reactants
are involved, the rate law requires an additional term for the amount of the second reactant, which
may create a second-order reaction. The experimental determination of к then becomes more
complex. The reaction order n can only be determined by experiment, and it has nothing to do
with the stoichiometric coefficients for the reaction. We've also confined our attention to reactions
taking place in a batch reactor. The situation is more complex when reactions are taking place in a
continuous-flow or plug-flow reactors. This subject will be dealt with in Chapter 6. Finally,
heterogeneous reactions systems may require a term for the area of the solid phase.

Another useful feature of chemical reaction kinetics is that ln(&) is often found to be a linear
function of reciprocal temperature. Where this is true, the reaction rate at two temperatures can be
used to estimate the reaction rate at another temperature. This procedure is valid only over a short
temperature range, and depends on the absence of phase changes between the two temperatures.
Further discussion of this aspect of chemical kinetics is beyond the scope of the Handbook.

Two other factors may control how fast a chemical reaction proceeds: mass transfer and
diffusion. First, when a reaction occurs at a phase boundary, mass transfer may dominate the
reaction rate. Stirring and mixing increase mass transfer to the interface, and speed up the reaction.
Second, when a solid reactant is coated with a porous layer, gaseous reactants may be hindered in

Chapter 5 Stoichiometry and the Chemical Equation 263

reaching the unreacted core. This circumstance often leads to a parabolic rate law for the
oxidation of a metal. Anything that opens the pores to gas diffusion increases the reaction rate.

The main lesson from this discussion is that the rate constant for a simple first-order reaction
can be used to determine the residence time required to attain a certain reaction extent. The rate-
controlling step for more complex reactions (chemical kinetics, mass transfer, diffusion) may be
impossible to discern. If so, we must rely on data from plant measurements or laboratory tests
fitted to empirical equations, without knowing which step (or steps) is rate-controlling. We can
simulate a process by converting this data to XR or R-R values, always being aware of the
approximations introduced when these values are used outside the range of process conditions on
which the data was based. When nothing is known about the kinetics, or where reaction rates are
believed to be fast, the last resort may be to assume reactions go to equilibrium. Using Кщ values
at least allows us to estimate the upper limit on reaction extent, and thus provide a benchmark
against which we can compare actual process efficiency. Fortunately, in the presence of a suitable
catalyst, or at elevated temperatures, many material-related processes do approach equilibrium,
which accounts for the frequent use of Кщ as one of the stream variables in material balances.

5.5.3 Reaction Progress and Кщ

The previous section stated that the progress of a reaction could be limited by thermodynamic
considerations. Under any situation, a spontaneous reaction tends to proceed until equilibrium is
reached between the reactants and products, or until the limiting reactant is consumed. At
equilibrium, the rate of the forward and reverse reactions is the same, and to outward appearances,
the reaction has stopped. This observation is an expression of the law of mass action. The
equilibrium state for a reaction can be calculated from data on the thermodynamic properties of the
reactants and products according to well-known principles of chemical thermodynamics. We will
not go into detail on the thermodynamic laws that cover this material, except to say that the result
of making such calculations give a numerical value for an equilibrium constant (symbol Кщ) for
the reaction. All freshman chemistry texts used in the engineering curricula contain one or more
chapters describing the calculation and use of the equilibrium constant.

We digress briefly from the main subject of this Chapter to review how Кщ is used to
determine the equilibrium position of a system. Product yield will be highest at equilibrium, but
less if the reaction progress is limited by kinetics. It's informative to compare the R-R of a reaction
or the XR of a reactant at the equilibrium position to ones derived from process data to determine
how much more would react if the reaction kinetics could be increased.

While many processes to approach the equilibrium position, the main value in calculating the
equilibrium amount of reaction products is because Кщ allows us to calculate the maximum
possible R-R of a reaction or the XR of a reactant. We will restrict our discussion of the use of Кщ
to gas-phase and gas + condensed phase reactions. In most cases, the format of the Кщ expression
will be given for each example.

The Кщ value is always a finite number. Its numerical value is based entirely on temperature
and the mole numbers used to balance the reaction. At 1000 К for example, Кщ for the formation
of one mole of H20(g) in its standard state from one mole of H2(g) and one-half mole of 02(g) in
their standard states is 1.151 x 1010. If we were to write the formation reaction for two moles of
H20(g), the value of Кщ would be (1.151 x 1010)2, or 1.324 x 1020. Thus for any given reaction
stoichiometry, there is a unique value of Кщ at each temperature. But no matter how we write the
standard reaction, the equilibrium partial pressure relationship is the same.

We must be careful to distinguish between the numerical value of Кщ for a reaction and the
equilibrium position of a reaction, which has an infinite number of possibilities. The Кщ
expression for the formation of one mole of H20(g) from the elements has three terms; the
/>H20(g), the/?H2(g), and the square root of the/?02(g). Any combination of these three terms that
gives the numerical value of Кщ is an equilibrium position for the system. If a reactor is fed with
1 mol/min of H2 and different amounts of 0 2 and reacts to equilibrium at a set temperature, the

264 Chapter 5 Stoichiometry and the Chemical Equation

equilibrium position will vary with amount of 02. However, the partial pressures of each gas will
be such that the numerical value of Кщ is always the same.

It's important to understand the factors that affect the equilibrium position of a reaction when
the temperature is specified (i.e., Кщ is fixed). We can get a qualitative idea of the effect of a
change by applying Le Chateher's principle, which states:

When a change is imposed on a system at equilibrium, the equilibrium position will shift
I in a direction that tends to reduce that change.

We can apply this to the formation reaction of NH3(g) from the elements, which involves a
decrease in volume. In the range of temperatures where the value of Кщ is between 100 and 0.01,
the isothermal equilibrium position will shift with a change in system pressure. Increasing the
pressure decreases the volume of the reactant (II/2H2 + V2N2) species more than the volume of the
NH3, so the equilibrium position will shift to an increased amount of NH3 (i.e., the side with the
lowest volume). If there is no volume change, there is no effect of pressure. For example, the
equilibrium position for the reaction to form HCl(g) from the elements is not affected by system
pressure because one mole of H2 plus one mole of Cl2 forms two moles of HC1.

A more subtle effect of Le Chatelier's principle is what happens when an inert gas is added to
a system where a volume-change reaction occurs. If the system pressure remains constant, the
addition of an inert gas has the effect of lowering the sum of the partial pressures of the reacting
species, and hence causing the equilibrium position to move in the direction of the side with the
larger volume.

Le Chatelier's principle also tells us the effect of temperature on the equilibrium position (i.e.,
the effect of temperature on Кщ). If the reaction is endothermic, an increase in temperature
increases Кщ, and shifts the equilibrium position to the right.

5.5.4 Кщ Values from FREED

For many applications in this Handbook, the necessary values for Кщ are given in the text.
For other calculations, you can find values of Кщ tabulated in reference books or from the
thermodynamic database program FREED, which is on the Handbook CD. FREED is described in
more detail in the Appendix, and has a User's Guide on the CD. FREED can display Кщ data for
the formation of species from the elements. You can use these "formation" values to calculate Кщ
for reactions involving the relevant species. For example, you can calculate Кщ for Equation
[5.23] at 1000 К from FREED's thermodynamic tables that list log(ATeq) for the formation of
species from the elements. For example, Кщ for Equations [5.20] and [5.21] at 1000 К were taken
from FREED tables:

C(c) + 02(g) - C02(g); Кщ = 4.74 х 1020 [5.20]
C(c) + V202(g) -> CO(g); ^eq = 2.88 x 1010 [5.21]

C(c) + C02(g) - 2CO(g); ^eq = ? [5.23]

Equation [5.23] can be obtained by summing the reverse of Equation [5.20] with twice
Equation [5.21]. When the resulting two reactions are summed, the Кщ value of the new reaction
[5.23] is a product of the summed reactions*. Therefore:

zr г r r ^ i ( 2 . 8 8 x l 0 1 0 ) 2 1л / 7п5г LГ5 41J1
tfeqfor 5.23 =■* 4.74 xlO2200

A better way to obtain Кщ for a reaction is to use FREED's Reaction tool. This calculates a
table of thermodynamic properties of a reaction, and makes a chart of the relevant results. FREED
asks the user for a temperature range, energy units, and the temperature interval for each

* Alternatively, we can use log Кщ and sum the various values.

Chapter 5 Stoichiometry and the Chemical Equation 265

calculation, and then calculates four different thermodynamic functions. Table 5.2 shows the
results for the reaction represented by Equation [5.23] and Figure 5.4 shows the chart. Compare
the value of Кщ at 1000 К from the previous text calculation to that in Table 5.2 (bordered cell),
where Кщ is also 1.75. In this Chapter, we are only interested in the log Кщ function column of
the table {Kr in FREED = Кщ in the text). In subsequent chapters, we will be looking at the other
data produced by the Reaction tool.

The Reaction tool plots values for log Kr vs. \IT because log Kr is nearly linear with
reciprocal absolute temperature. You can use the Trendline tool to obtain an equation for the
function. We will use FREED's Reaction tool often in the rest of the Handbook. There are also
on-line sources for reaction properties, such as Fact-Web (see citation in General References).

Table 5.2 Results of FREED's Reaction tool to calculate the thermodynamics of the reaction
between one mole of С and one mole of C02to produce two moles of CO via Equation [5.23].

Table:

Material Balance: System is balanced

Unit: Cai

T(K) 1/T(1/K) dHr dGr LogKr Heat
-21.034 41216
298.15 0.003354 41216 28696 -1.999 48468
-0.751 50012
800 0.00125 41189 7318 0.242 | 51586
53186
900 0.001111 41000 3095 1.050

1000 ] 0.001 40785 -1106

1100 0.000909 40551 -5284

Carbox Reaction

1.5

1.0 ^ * ч * ^ ч ^ y = -8945x + 9.184
0.5 R2 = 1

u 0.0 I I _^^^^^^ ^^$<

ö) -0.5

' -1.0

-1.5 ^^^^J

-2.0

-2.5 0.001 0.00105 0.0011 0.00115 0.0012 0.00125
0.0009 0.00095 1/T(1/K)

Figure 5.4 Chart from FREED's Reaction tool for log(Xeq) for the reaction of one mole of carbon
with carbon dioxide as expressed by Equation [5.23]. The Trendline tool was used to obtain an
equation relating the variables: Log(Xeq) = -8945/Г+ 9.184. The default Reaction chart has been
enhanced for display purposes.

5.5.5 Guidelines for Using Кщ to Determine Maximum Reaction Extent

Some additional terminology will help clarify the extent of a thermodynamically-limited
reaction. Here, for simplicity, we omit the criteria of an increase in entropy as indicating the
degree of spontaneity of a reaction, and instead use the value of the equilibrium constant of a
reaction as a more practical measure of reaction progress.

266 Chapter 5 Stoichiometry and the Chemical Equation

• First, there exists a class of reactions that occur so energetically that they are termed
spontaneous, or irreversible. They proceed in one direction only*. Eventually,
equilibrium is reached or the limiting reactants are consumed. The extent of reaction of
the limiting reactant is (essentially) one. The criterion for a truly spontaneous reaction is a
very large value of Кщ — on the order of 107 or larger. For example, Equations [5.20]
and [5.21] are spontaneous at 1000 K, and C, 0 2 or both can be completely consumed.
Other examples of such reactions are the complete oxidation of hydrocarbons during the
combustion of fuel oil with excess air, and the complete oxidation of sulfur (to S02 and/or
S03) during roasting of sulfide concentrates with excess air. However, considering the
usual accuracy of the stream flow and composition, a reaction may be considered as
essentially complete when Кщ per mole of product is above about 400.

• Second, there exists a class of reactions that can go in either direction. Equilibrium is
reached before the limiting reactant is consumed. These reactions are termed
stoichiometrically reversible, where a significant amount of both reactants and products
are present at the equilibrium position. A change in temperature, pressure or the amount of
a reactant changes the equilibrium position (or the predominant direction) of the reaction,
but reversing the change restores the original (equilibrium) position. Equation [5.23] is a
reversible reaction at 1000 К so long as C(c) is present. Another example is the hydrogen
reduction of wustite to metallic iron. The value of Кщ varies between 0.01 and 100 over
the range of process temperatures commonly used. However, since Кщ for any reaction
varies with temperature, some reactions that are irreversible at low temperatures may
become reversible at high temperatures. For example, the combustion of hydrogen with
oxygen is spontaneous at 1000 К but reversible above about 3000 K. An important point:
once a reversible reaction stops (i.e., reaches equilibrium), some amount of each reactant is
still present. Therefore, a reversible reaction usually does not have a limiting reactant, so
the reactants' extent of reaction is less than one.

• Finally, there exists a class of reactions that hardly take place at all. There is no generally
accepted name for this class of reactions, but we will call them stoichiometrically
negligible reactions. The value of Кщ for these reactions is 0.001 or less. The formation
of nitrogen oxides from nitrogen and oxygen is an example of a negligible reaction. The
concept of a limiting or excess reactant is not relevant in this case, and the extent of
reaction of any reactant of a negligible reaction is « 0 . 0 1 . This means that even if there is
a tiny amount of product formed, for material balance purposes we can consider the initial
amounts of reactants to be unchanged.

These classifications are somewhat arbitrary since the border between one category and
another will depend on process conditions. However, they are a convenient way to guide general
thinking about the maximum extent of species reaction. The use of Кщ may prevent a material
balance error by assuming an XR or an R-R value in excess ofthat to reach equilibrium.

5.5.6 Application of Equilibrium Limitations for Gas-Condensed Phase Reactions

Consider a process for the removal of the water of hydration from calcium hydroxide by

heating it.** The formula for calcium hydroxide is usually given as Ca(OH)2, but a better way of
expressing the formula might be CaOH20, which indicates that it is a hydrated form of lime. The
chemical equation for calcination of the hydroxide is:

Ca(OH)2(c) -+ CaO(c) + H20(g) [5.42]

In theory, no reaction is truly complete because that would imply Кщ = oo. Spontaneous (irreversible)
reactions proceed to the point where the limiting reactant is present in stoichiometrically insignificant
amounts. In the context of this Handbook, a reversible reaction is one which closely approaches an
equilibrium position where stoichiometrically significant amounts of both reactants and products are present.
** The thermal decomposition of hydroxides, carbonates, and sulfates to oxides is called calcination.

Chapter 5 Stoichiometry and the Chemical Equation 267

The equilibrium constant expression for this reaction (both solids present) is given by:

aCaOxoH20 I5"3!
* * ' <,Ca(OH)2

This ^eq expression indicates a relationship between the activities of the two solids and the
partial pressure of water vapor. When both solids are present, their thermodynamic activity = 1, in
which case the value of Кщ is equal to the/?H20. Since Кщ is a function of temperature only, the
pH20(g) is also a function of temperature only. The table shows some representative values for
Кщ for pressure units of atm, so pH20(g) is in atm. The reaction proceeds negligibly below about
500 K, and reversibly between 500 and 800 K.

Temperature, К 400 500 600 700 800

Keq 1.600 x 10-7 9.850 x 10~5 6.589 x 10-3 0.1250 1.085

Suppose one mole of dry air is passed through a large amount of Ca(OH)2 at 700 K. This
wording indicates that Ca(OH)2 will be present throughout the passage of one mole of air. The
extent of Ca(OH)2 decomposition can be calculated by assuming the pR20 remains very close to
the equilibrium value of 0.125 atm during the process. Neither 0 2 nor N2 are involved in the
decomposition chemistry of the hydroxide (i.e., they are inert in this case). They are therefore not
part of Equation [5.42] or the Кщ expression (remember the rule stated in Section 5.3: inert
species are not part of a reaction equation). Although not a reactant, the incoming air causes the
decomposition reaction to proceed by removing a reaction product. As a consequence of the law
of mass action, removing a reaction product causes the rate of the forward reaction to increase until
the rate of the forward and reverse reaction are again the same (i.e., the system always tries to
attain an equilibrium position). If the air flow stopped, the reaction would continue momentarily
until the pU20 in the reaction vessel reached 0.125 atm. If the air flow was started again, each
increment of dry air would cause an increment of Ca(OH)2 to decompose.

According to our definition, if given sufficient time to reach equilibrium, the reaction is
reversible because Кщ is 0.125 at 700 K, which means that we can cause Ca(OH)2 to decompose
or form by controlling the moisture content of the flowing air. Ca(OH)2 decomposes when passing
dry air into the reactor. If we were to pass moist air through the reactor with pH20 in excess of
0.125 atm, Ca(OH)2 would not decompose; instead, CaO (if present) would react to form Ca(OH)2.

A specific calculation to determine the amount of Ca(OH)2 that decomposes as a result of the
passage of one mole of dry air requires use of the relationship between partial pressure and mole
fraction, where the partial pressure of a gas is equal to the mole fraction times the total pressure. If
P = 1.00 atm, and with both solids present, the relationship is:

рЩО= пЯ^ =0.125 [5.44]

гаиг + иН20

For one mole of air, яН20 = 0.143. Therefore, for every mole of dry air (or any dry non-
reactive gas) passed sufficiently slowly (i.e., so equilibrium is maintained) through a bed of

Ca(OH)2 at 700 K, 0.143 moles of CaO and 0.143 moles of H20 will form. Alternatively, if one
mole of Ca(OH)2 was present initially, ХЛСа(ОН)2 = 0.143.

The concept of reversibility is important and so deserves further comment. For Equation
[5.42], reversibility means that passing moist air through a reactor containing both Ca(OH)2 and
CaO at about 700 К could cause the reaction to go in either direction, depending on the /?H20 in
the air. We deem the reaction to be negligible at 500 К because Кщ is such a small number at 500
K. It's practically impossible to maintain the moisture content of air to less than 9.8 x 10"5 atm in
order to coax the reaction to proceed, and even if we could, the amount of hydrate decomposition
would be negligibly small per mole of dry air used. Although the 500 К reaction is theoretically
reversible, it is practically irreversible because of the difficulty of maintaining the equilibrium
amounts of the reactants and products at their stoichiometrically-negligble levels.

268 Chapter 5 Stoichiometry and the Chemical Equation

It's important to emphasize again the correct use of the equilibrium constant expression.
Equation [5.43] states that Кщ = /?H20 only when both solid phases are present. If Ca(OH)2 is
initially present, the first trace of dry air will initiate Equation [5.42], thus causing a trace of CaO
to form. If the Ca(OH)2 decomposition rate is relatively fast compared to the flow of air, /?H20
will be a constant, equal to Кщ. This will be true even if the air contains some moisture, so long
as the/?H20 in the incoming air is less than the equilibriumpU20 for Equation [5.42]. If this is not
the case, no reaction will occur, and no CaO will form. This is a consequence of the law of mass
action, which states that a system tends to change composition until the forward and reverse
reaction rates are the same, which occurs at the equilibrium position.

5.5.7 Application of Equilibrium Limitations to Gas-Phase Reactions

Gas-phase reactions are very common in materials processing, either as a source of thermal
energy or to provide desired conditions for downstream reactions with a condensed phase. The
simplest gas-phase reactions are those with no volume change such as the reaction between one
mole of H2 and one mole of Cl2 to form two moles of HCl. But if there is a change in volume, the
arithmetic becomes more complicated. Consider the reaction between oxygen and sulfur dioxide
that takes place during the manufacture of sulfìiric acid:

S02(g) + V202(g)^S03(g); Кщ= *S°* [5.45]
pS02^p02

This reaction is spontaneous and complete up to about 500 K, proceeds negligibly above
about 3000 K, and is potentially reversible in between. At 1000 K, for example, Кщ = 1.82 in
Equation [5.45]. If S03(g) alone is initially present and heated to 1000 K, it will decompose
spontaneously (but incompletely) by the reverse of Equation [5.45]. Irrespective of which way the
reaction goes, once it reaches equilibrium, the relationship between the partial pressures of the
three gases is expressed by the Кщ expression above, and will be equal to 1.82. For the thermal
decomposition of pure S03, there will also be a unique stoichiometric relationship between the
products: pS02 = 2p02. If we know the total pressure, we have three equations to solve. AtP=l
atm,/?02 = 0.214, pS02 = 0.427, and/?S03 - 0.359.

If we mix one mole of 0 2 with one mole of S02 and heat the mixture to 1000 K, we no longer
have the same unique stoichiometric product relationship as we did for the decomposition of pure

S03. Instead, we must find a different one. Notice that the ratio of S atoms/O atoms in the
equilibrium gas will be the same as for the reactants, so nS/nO = V4. Since the atom ratio is equal
to the partial pressure ratio, the relationship is:

KS _ 1 _ pSQ2 + pSQ3

Ю " 4 " 2pS02 + 3pS03 + 2p02 [5,46]

Again, we have three equations to solve. At P = 1 atm, p02 = 0.422, pS02 = 0.265, and/?S03
= 0.313. The products are n02 = 0.730, nS02 = 0.458, and nS03 = 0.542. At the equilibrium
position, R-R for Equation [5.45] = 0.542, XRS02 = 0.542, andX7?02 = 0.270.

A word of warning on using the atom ratio as a constraint on a system: it is valid only when
the system does not contain any condensed phase reactants that contain S or О as constituent
elements.

We mentioned earlier that the numerical value of Кщ depended only on the temperature, thus
temperature can have a large effect on the extent of reaction of a species in reaching the
equilibrium position. Another factor that may have an effect on the equilibrium position (but not
on Кщ) is pressure P, where P is the sum of the partial pressures of the chemically involved
species. We described this effect earlier in terms of Le Chatelier's principle. For a reversible
reaction, P has an effect if there are a different number of gas molecules on both sides of the
reaction. P therefore has an effect on the R-R for Equation [5.45] and XT? of the reactants, which is
why we needed to use P to solve the equation set.

Chapter 5 Stoichiometry and the Chemical Equation 269

5.6 Practical Indicators of the Progress of Reactions and Processes

**- I U U Productivity vs. Throughput Rate For many industrial processes, mass
transfer and/or chemical kinetics limit
;0£ _ 90
* J1■2υ30 2?50 80 the amount of product. Reactions
70 approach equilibrium if given enough
time, at which point reactor productivity
60 reaches 100 % of theoretical. As the
50 throughput rate increases, material leaves
the reactor before equilibrium is reached,
Ω.

throughput rate so the reactor operates at less than 100%

of theoretical efficiency. This trend is

shown in the sketch for arbitrary

throughput rates. A process engineer often has available a large amount of operating experience to

guide his expectation of the productivity of a certain reactor. Several terms have been used to

describe how far a reaction can or might go. These terms require enhanced definition, with

examples of their application. Please see an earlier discussion of these concepts in Section 4.16.

• Conversion is the fraction of some material in the feed that is converted into products. The
process basis and the nature of the products must, of course, be specified to avoid
confusion. The term may be applied to the overall process or to a specific reaction. When
applied to a single reactant, the term is nearly synonymous with extent of reaction of the
reactant, or in some texts, the degree of completion of a reactant. Both terms may be
expressed as fraction and percentage conversion or completion.

• Selectivity is the ratio of the mass or amount of a particular (usually the desired) product
produced to the mass or amount of another product (usually undesired) produced by a
reaction or process.

• Yield is for a single product and reactant, and is the mass or amount of final product
divided by the mass or amount of the initial reactant fed or consumed. Where sequential
reactions occur in multiple devices, their location must be clearly stated.

• Production rate is a way of indicating the mass or amount of product species per unit time.
The overall production rate refers to a specific product, while the unit production rate is
divided by the amount (not the flowrate) of a specific reactant.

• Extent of species reaction (XR) quantifies how far a reaction goes in terms of a fractional
consumption of a specified reactant (usually the limiting reactant).

Terms such as extent of species reaction and rate of reaction are functions of the system
conditions. They're changed by changing the temperature (which changes the value of Кщ and
the kinetics), the relative amounts of reactants (adding a large excess of a reactant to increase
reaction extent is called swamping the reaction), pressure, and vigorous mixing (this increases
mass transfer between reactants). When using these terms, be aware that they might be defined
differently by others.

These concepts are illustrated by examining a process for reduction of tungsten trioxide

(W03). The reducing agent is a gas having equal amounts of H2 and CO. Reduction is carried out
in a fluidized bed furnace (see Figure 4.3). The feed ratio is 10 moles of reducing gas per mole of

W03. The reduction mechanism is not completely understood but probably takes place in stages.
First, the WO3 is reduced to W02 and then the W02 is reduced to W.

W03(c) + H2(g) -> W02(c) + H20(g) [5.47]

W03(c) + CO(g) - W02(c) + C02(g) [5.48]

W02(c) + 2H2(g) — W(c) + 2H20(g) [5.49]

W02(c) + 2CO(g) - W(c) + 2CQ2(g) [5.50]

270 Chapter 5 Stoichiometry and the Chemical Equation

Without going into details of the values of the respective Кщ, the reactions represented by

Equations [5.47] and [5.48] are irreversible (i.e., they go to completion until one reactant is

completely consumed). Based on the amount of reducing gas in the feed, W03 turns out to be the
limiting reactant. The reactions represented by Equations [5.49] and [5.50] at equilibrium are

reversible, but if CO and H2 are present in sufficient excess, W02 would be the limiting reactant.
This is the case at 900 °C at equilibrium for a molar flow ratio of reducing gas (half CO, half H2)
to W03 often. (The minimum reducing gas/W03 ratio based on equilibrium is about 9.1). At
equilibrium (i.e., slow feed rate of reactants), the product solid should consist entirely of W.

However, at any practical feed rate some trace of W02 (but never any W03) is present in the
solid product because insufficient time is allowed for the reactions to reach equilibrium. The

amount of W02 increases with increased feed rate (while still keeping the flow ratio of reducing
gas to WO3 often) because the corresponding retention time decreases. A small amount of W02 is
not too objectionable because it can be removed in subsequent processing.

The formation of WC can occur by the reaction of CO with freshly-reduced W:

W(c) + 2CO(g) -> WC(c) + C02(g) [5.51]

At high feed rates and the commensurate increase in unreduced W02, the relative amount of
CO vs. C02 in the gas increases, thus favoring the formation of WC, which is very undesirable
even in trace amounts. The following table shows the results of bench-scale tests on W03
reduction at 900 °C at a 10:1 mole ratio of reducing gas to WO3. The amount of W02 at the WO3
feed rate of 1 mol/min is <0.003 mol/min, and is set at zero for this example.

Feed rate (mol/min)

W03 |1 1.5 2 2.5

Production rate (mol/min)

W02 0 0.01 0.02 0.05
WC 0 0 0 0.03

W 1 1.49 1.98 2.42

H2O 1.6 2.39 3.18 4.00

co2 1.4 2.09 2.78 3.43

H2 3.4 5.11 6.82 8.50

CO 3.6 5.41 7.22 9.04

D gas 10 15 20 24.97

The listed indicators of reaction and reactor progress will be calculated for three different
WO3 feed rates, based on the data from the above table at 1.0, 2.0, and 2.5 mol/min. The basis for
calculating each indicator will be clearly stated in each case, as well as the specific reactant or
product involved.

• The conversion of oxide feed to metal is a common indicator of reactor productivity. At
the 1.0 WO3 feed rate, the oxide is completely converted (100 % conversion), and reactions
represented by Equations [5.49] and [5.50] have a 100 % degree of completion. At the 2.0
feed rate, 0.99 moles of W are produced per mole of feed, which is a 99 % conversion (99 %
degree of completion). At the 2.5 feed rate, the oxide reduction is 98 % complete, but the
degree of conversion to metal is 96.8 %*.

• The selectivity of the reaction refers to the production of the desired product to the
undesired one. Here, the desired product is deemed to be W + W02 and the undesired

* If the conversion or degree of completion referred to the amount of W03 converted, the indicators would be
100% inali cases.

Chapter 5 Stoichiometry and the Chemical Equation 111

product is WC. The selectivity at the 1.0 and 2.0 W03 feed rates is infinite, while at the 2.5
feed rate, the selectivity is 82.

The yield refers to the amount of a specified product divided by the amount of some
specified reactant. Here we will define the yield as amount of W per mole of reducing gas.
The yield at 1.0, 2.0, and 2.5 W03 feed rates is 0.1, 0.099, and 0.0968. At feed rates below
1.0, it might be possible to decrease the amount of reducing gas below the feed ratio of 10
and obtain a yield above 0.10. The thermodynamic limit at 900 °C is at a feed ratio of about
9.1, for a yield of 0.11.

The production rate refers to the amount of any product per unit time, and as such is listed
in the production rate columns of the previous table. However, it is sometimes more
instructive to express unit production rates, which is expressed as a ratio of product/feed
(per unit time). The unit production rate of C02 and H20 for the 1.0, 2.0, and 2.5 W03 feed
rates is shown in the table below.

Feed rate (mol/min)

W03 1 2 2.5

Unit production rate (mol gas/mol W03"min)

H20 1.6 1.59 1.6
C02 1.4 1.39 1.372

• The extent of reaction (XR) is a term applied to a single reaction to indicate the
consumption of a particular reactant (usually the limiting reactant). When more than one
reaction is involved in a reactor, the reactions are assumed to occur in the order written,.
The first reaction proceeds to the extent designated, and consumes a certain amount of the
designated reactant. The species XR for the second reaction is then based on the amount
remaining after the first reaction stops. The XR concept is used in later Chapters and in
FlowBal. Consider the reduction of W03 by H2 and CO. Since W03 is limiting, it's all
consumed by the reactions of Equations [5.47] and [5.48], but no information is given as to
the extent of WO3 reduction by H2 vs. that by CO. Therefore, the extent of W02 reduction
by H2 or CO (Equations [5.49] and [5.50]) is ambiguous. For purposes of this example,
suppose 0.54 was the fraction of WO3 reduced by H2 and 0.46 the fraction reduced by CO.
If Equation [5.47] were considered to take place first, the JLRW03 reduction (by H2) would
be 0.54, which on a basis of one mole of WO3 in leaves 0.46 mole of WO3 unreduced. The
reaction of Equation [5.48] occurs next, where all of the remaining W03 is reduced by CO.
Therefore, the XRWO3 reduction by CO is 1 because all of the remaining W03 is completely
reduced by CO. This shows that XR depends on the sequence of reactions chosen to
represent the process. It's essential to clarify the reaction sequence assumptions in order to
apply the XR concept.

With this information, the .A3?W02 reduction by H2 and CO can now be calculated. For a
WO3 feed rate of 1 mol/min, 1.6 - 0.54 = 1.06 mole of H20 is produced during W02 reduction,
therefore, the XRW02 reduction by H2 as shown in Equation [5.49] is 0.53 (i.e., 53 % of the W02
is reduced by H2). In a similar calculation, the XRW02 reduction by CO (Equation [5.50]) is 1
because all of the remaining W02 (0.47 mole) is reduced by the CO. A similar calculation at a
W03 feed rate of 2.0 shows thatXi?W02 reduction by H2 is 0.525, but that by CO is 0.979 because
not all the W02 is reduced (0.02 mole remain). This gives an overall extent of W02 reduction of
0.99. At a W03 feed rate of 2.5, XRW reaction with CO to form WC according to Equation [5.51]
is 0.012, which means that 1.2 % of the W formed by W02 reduction is converted to WC.

All of these indicators of performance are classified as subsidiary relations in the calculation
of the DOF, which must = zero to make possible a solution of the material balance equations.
Many of these indicators have no physical meaning; they are simply a convenient way to designate
a subsidiary relation. Since there is no uniformly accepted definition of these indicators, great care
must be taken to define the species and reactions to which they are applied. Some indicators are

272 Chapter 5 Stoìchiometry and the Chemical Equation

for the overall process while others are for specific reactants in a single reaction. Finally,
remember that these indicators are not constants, but are functions of process conditions. For
example, increasing the flow ratio of reducing gas to W03 from 10 to 12 would likely increase the
progress of all reactions, both desired and undesired.

EXAMPLE 5.6 — The Reaction Between Oxygen and Carbon.

A reducing gas is prepared by passing air through a bed of carbon at 900 °C. At an air

flowrate of 50 mol/min, the product gas had <jpCO = 32.0 % and <jpC02 = 1.6 % (the remainder is
N2). Characterize the process by application of the criteria of % conversion, % yield, unit
production rate, and extent of reaction(s).

Data. The values of Кщ for Equations [5.20] and [5.21] at 900 °C are 3.91 x 109 and 4.29 x 1017
respectively.

Solution. The design of the reactor indicates that carbon is an excess reactant, hence air (really,
oxygen) is the limiting reactant. Nitrogen is assumed inert. As mentioned in Section 5.3, different
equation sets can be written for the reaction of С and 02. The choice of different reaction sets is
discussed in relation to Equations [5.17 - 5.23]. The following four reactions will be used for this
example, all written with the limiting reactant listed first. Since carbon is present in excess,
oxygen is completely consumed (i.e., Equations [5.20/5.52] and [5.21/5.53] are spontaneous and
complete). We know this because Кщ for both of these reactions is a very large number at 900 °C
(see values listed above).

02(g) + C(c)^C02(g) [5.52]

02(g) + 2C(c)^2CO(g) [5.53]

C02(g) + C(c)^2CO(g) [5.54]

02(g) + 2CO(g)-^2C02(g) [5.55]

The % conversion indicator calculation will be based on Equations [5.52] and [5.53]. The

product gas has 20 times as much CO as C02, but the formation of CO requires only half the 0 2
(per mole of C) required by C02. Thus, the consumption of 0 2 for CO production is ten times that
for C02 production. This gives 9.1 % of the 0 2 converted to C02, and 90.9 % to CO. In a
reducing gas, C02 is the undesired species hence the selectivity is 10. If we define the yield as the
amount of CO produced (i.e., the desired species in the product gas) in terms of the amount of С

consumed, the yield is 90.9 %.

The production rate of reducing gas is most easily calculated from the N2 balance. Air has
φΝ2 of 79 %, for a feed rate of 39.5 mol/min. The product gas has φΝ2 = 66.4 %, so the
production rate of reducing gas must be 39.5/0.664, or 59.5 mol/min. The production rates of CO

and C02 are then 19.04 and 0.95 mol/min respectively.

The extent of reaction indicator depends on which reaction set(s) are selected. For a two-
reaction set, the designated species in the first reaction is assumed to react to the extent specified
before the second reaction begins. In other words, the reactions are deemed to occur in the order
written, so when the XR concept is applied to Equations [5.52] and [5.54], Equation [5.52] occurs
first. Since it is common to designate the limiting reactant as the XR species, we specify XR02. 0 2
is completely consumed via Equation [5.52], therefore, Jo?02 for Equation [5.52] is 1. Equation
[5.54] describes the production of CO from C02 and С Based on the requirement that the product
must have 20 times more CO than C02, XRC02 for Equation [5.54] is 0.9091. The amount of C02
consumed by Equation [5.54] is obtained by multiplying XR times the amount of C02 present
(0.9091 x 1.0), leaving 1 - 0.9091 = 0.0909 mole C02 remaining unreacted. 1.9091 mole of С are
consumed to produce 1.8182 mole of CO and 0.0909 mole of C02.

If XR is applied to Equations [5.52] and [5.53], then XR02 via Equation [5.52] is 0.09091.
Based on one mole of 0 2 into the reactor, and XR02 via Equation [5.52], the amount of 0 2

Chapter 5 Stoichiometry and the Chemical Equation 273

available for [5.53] is 0.9091 mole. Since this 0 2 is completely consumed by [5.53], XR02 for
[5.53] = 1, and 1.8182 mole of CO are produced.

Finally, we can use Equations [5.53] and [5.55] and again pick a basis of one mole of 0 2 into
the reactor. In order to produce a product gas having 20 times more CO than C02, XR02 from
[5.53] is 0.9545, leaving 0.04545 mole of 0 2 for [5.55]. Since all of the remaining 0 2 is consumed
by [5.55], XR02 for [5.55] is 1. This is summarized by writing the two reactions on a basis of one

mole of 0 2 in and consumed:

0.9545O2 + 1.9091C -> 1.9091CO

0.04545O2 + 0.0909CO -+ 0.09091C02

The overall result is the reaction of 1 mole of 0 2 with 1.9091 moles of С to form 1.8182
moles of CO and 0.0909 moles of C02, the same as obtained by the other reaction sets.

Clearly, XR of a species depends on the chosen reactions and the way they're sequenced.
Sometimes the reaction order is known from process measurement, but if not, a reasonable
arbitrary choice must be made. It often helps with the material balance arithmetic if the reaction
set and species are chosen so that the limiting reactant is completely consumed by the first reaction
(i.e., the first reaction written has XR for the limiting reactant of 1). The XR and the other
indicators are valid only for the conditions that gave the analyzed gas composition. A change in
temperature, pressure, or air flow rate for oxidation of С would almost certainly change the gas
composition, and hence the process performance indicators calculated from it.

The XR value may be best considered as empirically-derived, based on experimental data and
assumptions of reaction order. Its applicability to real processes depends on the validity of the
assumptions.

Assignment. Calculate the reaction indicators for one mole of air reacting with one mole of carbon
where the product gas composition has reached its thermodynamic limit at 800 °C. Calculate the
amount of carbon unreacted, and the volume fraction of each gas present. The equilibrium
constant for Equation [5.54] when C(c) is present is:

5.7 Parallel, Sequential, and Mixed Reactions

A process may use a reactor in which several chemical reactions take place. There is seldom
an a priori way to decide the order of the reactions, or if all of them occur simultaneously. Our
knowledge of reacting systems is gained mainly by taking samples from inside the reactor or from
the exit streams. The information so obtained can tell us a lot about the nature and progress of the
reactions, and with some additional data on values of Кщ, we can usually select a set of reactions
that best describes what goes on in the reactor.

Before making detailed calculations on a process, it is helpful to distinguish between what are
designated here as parallel and sequential reactions. Parallel reactions require the same reactants,
as with Equations [5.52] and [5.53], where С and 0 2 are reactants in both cases. We may or may
not wish to consider that parallel reactions occur in the sequence written. In sequential reactions,
the products of one reaction are consumed as reactants in a subsequent reaction. Consider the
production of a reducing gas by passing steam across hot carbon. Two sequential reactions can be
written. In the first reaction, carbon is oxidized by steam, and the products of the first reaction are
the reactants in the second reaction.

2H20(g) + C(c) -> 2H2(g) + C02(g) [5.57]

H2(g) + C02(g) -> H20(g) + CO(g) [5.58]

274 Chapter 5 Stoichiometry and the Chemical Equation

The definition of sequential reactions does not mean that the first reaction proceeds entirely
before the second one starts. It simply means that some amount of the first reaction product must
form before the second reaction starts.

Finally in mixed reactions, the reactant of an earlier reaction combines with a product of the
same or an earlier reaction. Equations [5.53] and [5.55] are mixed reactions because in [5.55], 0 2
combines with a product of [5.53].

The choice of reactions and their sequence may affect the ability to properly simulate a
process. Look back at Equations [5.52] and [5.53] where carbon is being oxidized by oxygen.
Both reactions are irreversible, but if the retention time is short, neither may cause complete
consumption of the limiting reactant. It's possible to calculate an XR for alternately-ordered
reactions based on exit gas stream composition. For processes taking place by sequential
reactions, "starvation" logic favors writing the most-favored reaction first. This provides ample
product for the subsequent reactions to take place.

It's generally not necessary to determine a "most likely" reaction set unless the XR factor is
used to define the extent of reactions. So long as the selected reaction set material balance
equations are correctly written, the material balance results will be correct. The main objective is
to determine the nature of all species that are present in amounts that have an effect on the material
balance for all of the other species. Valuable guidance on selecting the important species can be
obtained by the use of thermodynamic software, such as THERBAL, FactSage, or HSC. Please
visit the web site for program descriptions (listed in General References). There are also on-line
programs for calculating an equilibrium position (Fact-Web 2010).

5.8 Independence of Chemical Reactions

The final criteria in selecting a reaction set for a reactor is reaction independence. Consider,
for example, the production of a reducing gas by passing pure oxygen through a bed of carbon, a
process which was discussed earlier in Example 5.6. Four chemical reactions were written as
Equations [5.52 - 5.55]. However, some of these reactions may be obtained by combining others.
For example, the reverse of Equation [5.54] can be obtained by combining [5.52] with the reverse
of [5.53]. Thus the set of four reactions is not independent, in that some of the reactions are
obtainable by manipulation of others. In preparation for making a material balance, it's essential to
identify the fewest number of reactions that are necessary to completely characterize a system. We
designate this as a set of independent reactions.

An independent set of reactions has certain characteristics. First, each reactive species must
appear in at least one reaction equation. Second, none of the stoichiometric reactions can be
obtained from any combination of the others. As a rule of thumb, the number of reactions in an
independent set is the number of reactive species minus the number of reactive elements. In the
case of the carbon - oxygen system, there are two elements and four species, so the number of
reactions in an independent set is two. For example, Equations [5.52] and [5.53] constitute a set of
independent reactions, as do Equations [5.54] and [5.55]. The determination of the number and
nature of the independent reactions for a reactor is essential in calculating the DOF for the device,
and in using FlowBal.

5.9 Practical Examples of Reaction Writing and Stoichiometry

The concepts introduced in this Chapter are now applied to selected examples. The emphasis
is on the selection and use of chemical reactions and stoichiometry in calculating the material
balance. Section 5.9.1 has examples for situations where condensed phases are present, and the
gas comes to equilibrium with the condensed phase assemblage. Section 5.9.2 has an example of a
gas phase system where one reaction reaches equilibrium, but a second reaction does not.