Handbook_on_Material_and_Energy_Balance_Calculat1

Chapter 5 Stoichiometry and the Chemical Equation 275

5.9.1 Calculations in Gas-Condensed Phase Processes

Many material production and processing operations employ gas-condensed phase reactions,
in particular, gas-solid reactions. A very common situation is when a metal oxide is reduced by a
gas to form solid metal. Even though the reaction may look unfavorable because of a small value
of Кщ, an excess of reducing gas can be used to enhance the amount of solid oxide reduced.

EXAMPLE 5.7 — Reduction of Molybdenum Oxide with Hydrogen.

Powdered molybdenum can be prepared by hydrogen reduction of molybdenum dioxide. The
reaction device is a continuously operated fluidized bed. Figure 5.5 shows a sketch of the
flowsheet. Calculate the STP volume of H2 required for equilibrium production of 1 tonne of Mo
at reduction temperatures between 800 and 1000 °C. Assume that the spent gas is in equilibrium
with Mo and M0O2, but a stoichiometrically insignificant amount of Mo02 is present in the solid
product.

M0O2 Spent
*gas
Fluid bed
reduction -► Mo
furnace
H2

Figure 5.5 Flowsheet for the reduction of Mo02 with H2.
Data. The table below shows values of Кщ for M0O2 reduction.

Temperature, °C 800 850 900 950 1000
Кщ (рЯ20/рЯ2) 0.392 0.480 0.576 0.678 0.786

Solution. There is only one independent chemical reaction for the process. For the reduction of Vi
mole of M0O2 with one mole of H2:

H2(g) + V2Mo02{c) -► H20(g) + V2Mo{c) [5.59]

The basis of the problem is 1000 kg of Mo, or 1000/95.94 = 10.42 kmol of Mo (and Mo02).
The stoichiometric mole ratio of H2/Mo is two, so the consumption of H2 must be 2(10.42)(22.4) =
467 m3 of H2. Similarly, the mass of water produced must be 2(10.42)(18.01) = 375 kg. These
values are independent of temperature or pressure in the system, and depend only on the

coefficients of Equation [5.59].

Once the reaction attains equilibrium, it is deemed reversible. The reaction stops once the

product gas reaches the equilibrium position. Consider a situation where one mole of H2 comes in
contact with a large amount of Mo02 at 900 °C. The reaction represented by Equation [5.59]
proceeds until/>H20//?H2 = 0.576. A material balance on the system is:

Overall: moles H2 + moles H20 = 1 [5.60]
Кщ: moles H20/moles H2 = 0.576 [5.61]·

After the reaction stops, the gas contains 0.635 moles of H2 and 0.365 moles of H20, so 36.5
% of the H2 was used to reduce Mo02, while 63.5 % remained unreacted. The reduction of one
mole of M0O2 requires the entry of 2/0.365 = 5.48 moles of H2 to the reactor. Two moles are

Notice that the Кщ expression can use mole ratios because they are equivalent to partial pressure ratios in
this case.

276 Chapter 5 Stoichiometry and the Chemical Equation

consumed by reduction, while 3.48 pass through unreacted. For steady-state reduction with
products at equilibrium, the spent gas would contain 63.5 % H2 and 36.5 % H20. Thus, 2.74 times
more H2 must flow into the reactor than would be the case if the reaction was spontaneous and
complete. Moreover, since the process will only approach, rather than actually reach equilibrium,
in practice more than the equilibrium amount would be added to assure complete reduction of the
Mo02. The equilibrium-limited case thus serves to indicate the upper limit on how much Mo02
could be reduced by a certain amount of H2.

Extrapolating these results to the problem basis, the amount of H2 required for entry to the
reactor for the production of 1000 kg of Mo is 2.74(467) = 1280 m3. 467 m3 is consumed (i.e.,
converted to H20), and 813 m3 exit as part of the spent gas. Figure 5.6 shows the results for a 200°
range of temperatures. Less H2 is required as temperature increases because Кщ increases with
temperature.

This example shows how to use the value of Кщ to determine the maximum progress of a
thermodynamically limited reaction. The stoichiometric equation indicates the relative amount of
substances participating in the reaction, while Ä^eq is used to calculate how much of a particular
reactant must be added to accomplish a specific aim. In making material balances where there is a
thermodynamic limit to the progress of a reaction, the Кщ relationship is used as a subsidiary
relation, similar to vapor pressure, solubility, and the like. In the above equilibrium-constrained
process, XRMo02 reduction approaches one as the amount of H2 added approaches 1280 m3.

Reduction of Mo02 by H2

25%

800 850 900 950 1000
temperature, °C

Figure 5.6 Mass balance results for the reduction of 1000 kg of molybdenum under equilibrium
conditions.

Assignment. A furnace at 950 °C contains 1330 g of Mo02. 1050 STP liters of H2 are passed
through the furnace. Assuming gas/solid equilibrium, calculate R-R and Xi?H2, and the maximum
amount of Mo that can form in the furnace.

EXAMPLE 5.8 — Carbothermic Reduction of Zinc Oxide.
Zinc oxide is mixed with 115 % of the stoichiometric amount of С for reduction to Zn(g) plus

some amount of CO and C02. The process is carried out in a retort above atmospheric pressure.
Assuming that the process is equilibrium-constrained, calculate the temperature and the amount of
С required to reduce one mole of ZnO to produce a gas at 1.2 atm total pressure. The independent
reaction set consists of two reactions, which were presented earlier as Equations [5.3} and [5.4].
Data: Values of ^eq for the two relevant reactions is shown below.

t,°C 890 910 930 950 960
Keq[53]; CO formation 0.159 0.297 0.543 0.973 1.294
Keq [5.4]; C02 formation 0.0008 0.0021 0.0054 0.0131 0.0203

Chapter 5 Stoichiometry and the Chemical Equation 111

Solution: Inspection of the above Кщ values shows that most of the ZnO is reduced to form CO,
so maybe Equation [5.4] could be neglected. However, if accuracy is important, it should be
included. There are three unknowns (the partial pressures of each gas), so we need three equations.
Two equations are available from the two Кщ expressions. The third equation comes by
remembering that in Section 5.3 we used this process as an example of a reaction that had a unique
stoichiometric relation. A three-equation set can be written for each temperature and solved for the
partial pressures. For example at 910 °C, the equations are:

(pCO)(pZn) = 0.297; (pC02)(pZn)2 = 0.0021; pZn =pCO + 2pC02

Solver obtained the following results: pZn = 0.552; pCO = 0.538; pC02 = 0.007.
SuperSolver was then used to calculate gas pressures for the other temperatures. Figure 5.7 shows
the results; the reaction temperature should be 915 °C to obtain P = 1.2 atm.

For all conditions, the #>C02 remains nearly constant at 0.006, which confirms the initial
conjecture that very little ZnO is reduced to form C02. Over the range of temperatures examined,
the production of one mole of Zn requires 0.99 moles of С

960 Total P from ZnO Reduction by С

950 I ^^^^^^^^^^

940 ! ^^ t, °C = -14.45P2 + 91 .IP + 827

930 f,^r^

о 0.9 1.1 1.3 1.5 1.7 1.9 2.1 2.3
P, atm
' 920

910

900

890 .7
0

Figure 5.7 Total pressure (atm) generated by the carbothermic reduction of ZnO at different
temperatures. Text box equation generated using Excel's Trendline tool.

Assignment. ZnO can also be reduced with H2:
ZnO(c) + H2(g) -» Zn(g) + H20(g); Кщ at 950 °C = 0.0197.

Calculate the amount of ZnO reduced when one mole of H2 is passed through a bed of ZnO at 950
°C at P = 1.2 atm. Is there a unique stoichiometric relation between the partial pressures of the
reaction products? What is the effect of P on the amount of ZnO reduced? Explain by the
application of Le Chatelier's principle.

5.9.2 Calculations in Gas-Phase Processes

Materials production and processing operations often require carefully controlled gas
atmospheres. The reduction of oxides, the bright annealing of alloys, and the carburizing of steels
are just some of the processes that rely on generating special gas mixtures and controlling their
composition. The most common gases used in oxide reduction and metal carburization
atmospheres are CO, C02, H2, H20, CH4, NH3, N2, and 02. The quantity of each gas produced or
consumed is determined by the extent of reaction between them, as was shown in Example 5.6.
One of the more important reactions involving four of the above gases is the so-called water gas
shift reaction (WGR):

CO(g) + H20(g) -> H2(g) + C02(g) [5.62]

278 Chapter 5 Stoichiometry and the Chemical Equation

The WGR is a reversible reaction, and can be written in either direction; there is no "correct"

direction for it. If you do the balance arithmetic correctly, it doesn't make any difference which
way you write it. CO can remove oxygen from H20, or H2 can remove oxygen from C02. The
use of CO to deoxidize H20 is an important part of various processes for the production of
hydrogen. You might want to use FREED's Reaction option to see how temperature affects the

value of ^eq, and thereby the progress of the WGR.

The Кщ expression for the WGR is:

Кщ= O H 22))K(Fj ? C 0 2)
^
{рСО){рЩО) [5.63]

Since the equation coefficient for each species is one, either volume fraction or mole fraction

may be substituted for partial pressure. Similarly, total system pressure has no effect on the

equilibrium position because there is no volume change.

A material balance involving a mixture of gases coming to equilibrium may require the use of
the WGR equilibrium as one of the system constraints. Suppose a gas mixture had the following
composition:

<pH2 = 0.47; <pH2O = 0.25; <pCO = 0.18; <pCO2 = 0.10

and was heated to a temperature where Кщ for the WGR was 1.45. (The presence of a trace of
CH4 is neglected here). The gas composition will readjust itself to the equilibrium position, with
values consistent with Кщ for the reaction. There is no change in volume, and the sum of species
amounts remains the same. In addition, the atom ratio of C/O and H/O remain the same during
readjustment. Therefore, the readjusted gas composition can be calculated by solving four
equations: the two atom ratio equations, the sum-of-species equation, and the Кщ expression:

H. 2φΗ2+2φΗ20 _1·44=2286

О 2срС02 + срСО + φ Η 2 0 0.63

С . срС02+срСО 0.28 - 0.4444

О 2ФС02 + ФС0 + Ф Н 2 0 0.63

φΗ2 + φΗ20 + <рСО + фС02 = 1.00
(gH2)(g)C02)=115
foCO)foH20)

The equations were solved using ExcePs Solver tool, with the following result:

<pH2 = 0.485; φΗ20 = 0.235; <pCO = 0.165; <pC02 = 0.115

Alternatively, three material balance equations could have been used instead of the two atom
ratio and sum-of-species equation.

Sometimes the residence time for the gas reactor is too short for equilibrium to be reached. In
that case, we must know from experience the extent of reaction for the limiting reactant, and use
that as a constraint on the material balance instead of Кщ.

EXAMPLE 5.9 — Steam Reforming of Methane.

The production of a reducing gas consisting mainly of CO and H2 is an essential step in the
production of direct-reduced iron. Such a reducing gas can be prepared by heating a mixture of
steam and methane at elevated pressure in contact with a catalyst. Equilibrium is approached, but
not reached. There are five reactive species and three elements, so an independent reaction set
consists of two reactions. Various independent reaction sets can be written, but when possible
inclusion of the WGR is always a good idea:

Chapter 5 Stoichiometry and the Chemical Equation 279

CH4 + H20 -> CO + 3H2 [5.64]

H20 + CO -+ C02 + H2 [5.65]

At a certain reforming temperature and pressure, Xi?CH4 for Equation [5.64] is estimated as
0.96, and XRR20 for Equation [5.65] is estimated as 0.65. Calculate the product gas composition
for a steam-to-methane ratio of 1.5.

Solution. The reaction set indicates a mixed-reaction process. Use of the XR concept here is

unambiguious because Equation [5.64] must take place (at least to some extent) before [5.65] can

initiate. For [5.64], one mole of CH4 enters and 0.96 fraction of that reacts, leaving 0.04 mole
CH4 in the product gas. 1.5 mole of H20 enters and 0.96 mole is consumed, which leaves 0.54
mole of H20 as a reactant for the second reaction. The products of Equation [5.64] are 0.96 mole
of CO and 2.88 mole of H2.

The reactant amounts for [5.65] are 0.54 mole of H20 and 0.96 mole of CO. The amount of
H20 reacting in [5.65] is obtained by multiplying X7?H20 times the amount available after [5.64]
is complete, or (0.65)(0.54) = 0.351 moles. This leaves 0.189 moles of H20 and 0.609 moles of
CO unreacted, while producing 0.351 moles of C02 and H2. The net result is a product gas
containing 0.609 moles of CO, 0.189 moles of H20, 3.231 moles of H2, 0.351 moles of C02, and
0.04 moles of CH4 for a total of 4.42 moles of product gas. The composition of the product gas is:

φΗ2 = 73.1%; φΗ20 = 4.3%; <pCO=13.8%; <pC02 = 7.9%; <pCH4 = 0.90%

Note that R-R for Equation [5.64] - 0.384 and for [5.65] R-R is 0.1404.

Assignment. A reducing gas is prepared under different circumstances using a ratio of
steam/methane =1.4. A gas sample had the following composition:

<pH2 = 68.3%; φΗ20 = 8.4%; <pCO=19.8%; <pC02 = 2.2%; <pCH4=1.3%
Calculate J3?CH4 andXRH20 according to the above reaction sequence, and also R-R.

With special catalysts, it's possible to carry out reforming such that a desired reaction
approaches equilibrium, while less-desired ones don't. This is done for specific purposes, such as
maximizing the amount of H2. Reforming can be carried out in two stages, with different catalysts
at different temperatures, and maybe even different pressures. Calculating the final gas
composition may involve the use of both Keq and extent of reaction equations. If reforming is
carried out above 1000 °C with an effective catalyst, XRCH4 « 1.

5.10 Use of Chemical Reactions in FlowBal

Chapter 4 introduced FlowBal as a material balance calculation tool, and showed how to use it
on non-reactive systems. FlowBal in Chapter 4 did deal with physically-reactive systems (i.e.,
those involving phase transformations). FlowBal can treat phase changes as "reactions" in order to
keep track of the quantity of changed-phase material moving between different streams. The main
phase change application example was the vaporization of water, where the difference between
phases was indicated by species descriptors (/) and (g). Now we show briefly how FlowBal uses
chemical reactions to make material balances in chemically-reactive systems. Chapter 6 covers the
subject in greater depth, and the User's Guide for FlowBal has additional examples. It's a good
idea to review the User's Guide before trying to use FlowBal on this Chapter's examples.

FlowBal requires you to write and balance chemical reactions that describe how species react
and new species form. You will need to create a set of independent reactions (please review
Section 5.8) for each reactor device. Usually you will know the reaction direction from the
problem statement or by experience. But sometimes you may not know for sure which way a
reversible reaction will go in the process. Or, it goes in one direction at one temperature and the
opposite direction at a different temperature. When the reaction goes in the opposite direction to
that inserted in FlowBal, it won't converge because by default, the Solver Options for Assume

280 Chapter 5 Stoichiometry and the Chemical Equation

Non-Negative box will be checked. You must uncheck that box to allow convergence to a
negative value for one of the R-R equation values. However, doing so can sometimes tempt Solver
to converge at negative stream properties. Alternatively, you can use the built in FlowBal solver,
which sets all stream values as non-negative, but not the R-R values.

FlowBal uses two main techniques, one explicit and one implicit, to add information about
reactions. First, for reactions that don't go to equilibrium, you can explicitly define how much
reactant is consumed with FlowBal's X-R dialog box. Second, for reactions that do go to
equilibrium, you can use FlowBal's Insert Equation tool to write а Кщ expression. Each
technique creates a subsidiary relationship, designated as a SR in the DOF calculation routine. In
the first case, FlowBal requires the user to identify the device as a reactor, list the reactions, the
species, and the reactant's XR. The user is prompted for this information in the X-R dialog box. In
the second case, the user must write the Кщ expression for the reaction, find the numerical value
of Кщ from FREED, and enter this information into FlowBal's InsEqn dialog box. We'll show
how both of these methods are used in a process for the oxidation of carbon by air. This system
was explored in detail in Example 5.6; please review this example before reading on.

In some cases, information is available that indicates the extent to which a reaction
approaches equilibrium. Suppose, for example, that you were told, or had data that indicated that a
certain reaction proceeded "80 % of the way to equilibrium". You could incorporate that
information by inserting in FlowBal the Кщ expression with a multiplicand of 0.8. This requires
that you know for certainty the direction of the equation.

5.10.1 FlowBal's Extent of Reaction Tool

Consider the passage of air through a bed of continuously-replenished carbon. Above about
1000 K, this produces a СО-rich gas useful for oxide reduction. Кщ values for all С - N

compounds are very small, so all С - N reaction products are negligible at elevated temperatures.

The O2 - С reaction is spontaneous and complete, and with excess C, 0 2 is the limiting reactant.
The reaction products are known to be a mixture of CO and C02. In Example 5.6 we noted that
the process has two independent reactions. Different reaction sets can be used to specify the

system. Here we choose the sequential reaction set [5.52] and [5.54] as the independent set for
FlowBal. Since 02(g) is the limiting reactant, it is completely consumed. Therefore, XR02(g) = 1
(Equation [5.52]). The CO(g) in the product gas is produced via Equation [5.54].

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

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

It's not necessary to use the X-R tool for substances with an XR of one. Instead, simply omit
the substance from the reactor outstream, which tells FlowBal that the substance is completely
consumed by the reaction*. In this process, X7?C02(g) is less than one because Equation [5.54] is
stoichiometrically reversible. You must have some expert knowledge or experimental data to tell
you the extent to which C02(g) reacts with excess C(c) to use FlowBal's X-R tool. Suppose
laboratory tests showed that at 1050 К and P= 1.15 atm, the extent of C02(g) reaction with carbon
is 0.8, and you want to know how much carbon and air to put into a reactor to produce 100
mol/min of product gas. Figure 5.8 shows the flowsheet for one minute of operation.

FlowBal is set up as usual, with ? entered for unknown stream flows and compositions. There

is no entry for 0 2 in the list of species for S3 because, as mentioned earlier, the reaction extent of
0 2 via Equation [5.52] is one. If we had put a ? for 02 for S3, we would have needed to use the X-
R tool to enter 1 for the reaction extent of 02. Figure 5.9 shows the empty input array for the
process.

* Actually, 02(g) is not completely consumed in Equation [5.52]. There is still a trace present, but in an
amount much less than the convergence limit for Solver, so we omit 02 as a product specie. If you want to
know the amount of 02 or its partial pressure in the outstream, you must calculate it from the Кщ expression.

Chapter 5 Stoichiometry and the Chemical Equation 281

Air _ J _ Carbon 3 Burner gas
F1 = ?n burner F3 = 100n

Carbon 2_
F2 = ?n

Figure 5.8 Flowsheet for the production of high-CO gas. Carbon is supplied as fast as it is
consumed; therefore carbon is always present even though no carbon exits the burner. Dashed
lines indicate gas phase flow, and solid line indicates solid phase flow.

P (atm) 1.15 1.15 1.15

T(K) 310 300 1050

Str-unit Amt (kg-mol) Amt (kg-mol) Amt (kg-mol)

Spec-unit Mol pet Mol pet Mol pet

Str-name Air Carbon BrnrGas

Streams 1 2 3 R#1 R#2
? ? 100 -1 -1
Flow -1
С? 1 2
? -1
N2 79
02 ?
CO ?
C02 ?

I3urner (RX Heat
llnstreams Outstreams Reactions

13 1

22

Figure 5.9 Empty FlowBal input array for the oxidation of carbon by air. The coefficients for R
#1 refer to the molar balancing numbers for Equation [5.52] and R #2 for Equation [5.54].

There are five stream unknowns (we disregard as unknowns the x02 in SI and xC in S2
because FlowBal calculates them by difference from 100). FlowBal included two more unknowns,
the R-R for each reaction (designated R-Rl and R-R2 by FlowBal). FlowBal wrote six equations
for the seven unknowns, indicating the need for an additional equation. In this case, we have
information on the XRCO2 which is entered using FlowBal's X-R tool. Figure 5.10 shows the
results array for the process. The burner gas has a pCO/pC02 ratio of 8.0 and the burner requires
2.64 moles of air per mole of carbon burned.

P(atm) 1.15 1.15 1.15

T(K) 310 300 1050

Str-unit Amt (kg-mol) Amt (kg-mol) Amt (kg-mol) Stream Amounts (kg-mol)

Spec-unit Mol pet Mol pet Mol pet

Str-name Air Carbon BrnrGas Str-name Air Carbon BrnrGas
Streams 1 2 3
Streams 1 2 3 85.616
Flow 0 32.363 100
Flow 85.616 32.363 100 С 67.637 32.363 0
17.979
С 0 100 0 N2 0 0 67.637
02 0 0 0
N2 79 0 67.637 CO 0
C02 0 28.767
0 2 21 0 0 3.596

CO 0 0 28.767

C02 0 0 3.596

R-R1 17.979
R-R2 14.384
Figure 5.10 FlowBal results of carbon reacting with air with XRC02(g) in Equation [5.54] = 0.8.

282 Chapter 5 Stoichiometry and the Chemical Equation

The R-R values tell what happens in the process. An R-Rl of 17.979 means that the rate of
reaction of carbon and oxygen via the first reaction (R#l in FlowBal) is 17.979 mol/min, and
produces 17.979 mol/min of C02. An R-R2 of 14.384 means that the rate of reaction of carbon and
C02 via the second reaction (R#2) produces 2 χ 14.384 = 28.768 mol/min of CO. You can use this
information to (for example) calculate thatZRC for the first reaction is 0.5555. Since carbon is fed
at the rate it is consumed, XRC for the second reaction is one.

In using the X-R tool, remember that the extent-of-reaction concept assumes that the reactions
occur in the order written. This is particularly important when XR of a species in two reactions is
less than one. Use of the X-R tool requires care in the selection and sequencing of reactions for
FlowBaPs analysis of a process. Using the X-R tool intelligently means you must have some
rationale for selecting the set of independent reactions, and the order you list them. When the X-R
tool is not used, you may write the reactions in any order. Please review the FlowBal User's Guide
for examples on selecting reactions for FlowBal and the use of the X-R tool.

5.10.2 FlowBal's Insert Equation Tool
Under certain conditions, Equation [5.54] may proceed very close to equilibrium. XR C02 is

then governed by the numerical value of Кщ as given by Equation [5.56]*. FREED's Reaction
tool gives 4.63 for Кщ at 1050 K.

^eq = (pcoy = 4.63 [5.56a]

pC02

FlowBal's input array is set up the same as before, but this time, we need to use the InsEqn
tool to express the partial pressures of CO and C02 in the form required by FlowBal. First,
Equation [5.56a] must be formatted to eliminate pC02 in the denominator to prevent the division
by zero error that would occur if Solver tried zero for pC02. Next, the expression must be written
to sum to zero. The correct expression is shown below. Figure 5.11 shows the FlowBal result.

|4.63*{pS3-CQ2} - ({pS3-CO})/42 |

P (atm) 1.15 1.15 1.15

T(K) 310 300 1050

Str-unit Amt (kg-mol) Amt (kg-mol) Amt (kg-mol) Stream Amounts (kg-mol)

Spec-unit Mol pet Mol pet Mol pet

Str-name Air Carbon BrnrGas Str-name Air Carbon BrnrGas
Streams 1 2 3
Streams 1 2 3 84.594
Flow 0 33.171 100
Flow 84.594 33.171 100 С 66.829 33.171 0
17.765
С 0 100 0 N2 0 0 66.829
02 0 0 0
N2 79 0 66.829 CO 0
C02 0 30.813
0 2 21 0 0 2.358

CO 0 0 30.813

C02 0 0 2.358

R-R1 17.765
R-R2 15.406

Figure 5.11 Material balance results from FlowBal for the equilibrium combustion of excess
carbon by air. The S3 composition conforms to Equation [5.56a]. The burner gas has a/?CO/pC02
ratio of 13.06, and 2.55 moles of air are required per mole of С burned.

The material balance R-Rl shows that 17.77 moles of 0 2 produced 17.77 moles of C02 via
Equation [5.52] while consuming 17.77 moles of С R-R2 shows that 15.41 moles of C02 were
consumed by 15.41 moles of С during [5.54], leaving 2.36 moles of C02. The equilibrium XRC02

* The Кщ expression should include the activity of carbon in the denominator, but when carbon is present,
its activity is one, and therefore needn't be included in the expression.

Chapter 5 Stoichiometry and the Chemical Equation 283

via Equation [5.56a] is thus 0.867. The oxidation of С by CO2 as shown in the Figure 5.10 results
therefore approached, but did not reach, equilibrium.

Another use of the InsEqn tool is to use stream sample data to derive an equation in FlowBal.
Suppose the burner gas stream was sampled at 1050 К over a range of system pressures, and an
equation fitted to express the pCO/pC02 as a function of pressure. The system may or may not
have reached equilibrium. You could use the InsEqn tool to express this relationship in FlowBal
for making a material balance on the combustion of carbon with air at 1050 К in the presence of
carbon as a function of system pressure.

FlowBal's Repetitive Solve tool can calculate the material balance as a function of any
stream variable, such as the temperature or pressure of S3, or the degree of oxygen enrichment of
SI. If equilibrium is closely approached, we need an equation relating the change in Ä^eq for
Equation [5.56a] vs. T. Data from FREED's Reaction tool was used to obtain an equation for
log(ATeq) vs. reciprocal Г, and is reformatted as shown below for use by FlowBal's InsEqn tool.

(10^(9.134-8895/<T-S 3>))*{pS3-C02} - ({pS3-CO})A2

Figure 5.12 shows the effect of 7-S3 between 950 and 1100 К on the % C02 in S3 and the
mole ratio of air/C into the burner. Excel's Trendline tool was used to obtain a quadratic equation

relating the variables.

Burning of Carbon with Air at P = 1.15 atm

10 г * nair/nC = 2.347 χ 10_5(T2)- 0.05214T + 31.41 3.1
9 %C02 = 2.395 x 1(Γ*(Ί*) - 0.5395T + 304.9
13
8 "^^^ * к 2.9 ö
7
X 2-8 1
6CO 2.7-52
c/> 5
4^ 2 6 о
ООCM 4 — 0 — % C02 ^ ^ ^^ J>v* * ^ * * * 4 .
3 - □ - nAir/nC 2.5E
-Λ 2.4
2
' 2.3
11 1100

950 975 1000 1025 1050 1075

burner temperature, deg К

Figure 5.12 Process parameters for the equilibrium combustion of carbon with air as a function of
burner temperature. Textbox equations were created by the Trendline tool.

So far, we've described the use of the Кщ term in calculating the equilibrium position of a
process in fairly general terms. The correct calculation of chemical equilibrium requires a more
detailed study of chemical thermodynamics than presented in this Chapter. Unfortunately, it's all
to easy to make erroneous calculations about the equilibrium position of a system, whether doing it
by a hand calculation, using the Reaction option in FREED, or using FlowBal. Novices may wish
to seek expert advice about the validity of their Кщ calculations before using the results in a
practical situation.

5.11 Balancing Aqueous (Ionic) Reactions

Aqueous reactions are important in materials processing where leaching and electrolytic
processes take place. In the simplest case, an electrolyte dissolves and completely ionizes to
clearly defined species, in which case there is no problem writing and balancing the dissolution
and ionization reaction. Other processes involve reactions where there is a change in the valence

284 Chapter 5 Stoichiometry and the Chemical Equation

of atoms or groups of atoms. These chemical reactions are called oxidation-reduction or redox
reactions. They can be balanced in the same manner as other reactions, but the charge must also be
conserved. There are special definitions and rules for balancing redox reactions.

Oxidation is defined as a chemical change in which electrons are lost by an atom group of
atoms, or ions, resulting in more positive (or less negative) valence of at least one reacting
constituent. Reduction is a chemical change in which electrons are added to an atom or group of
atoms, resulting in a more negative (or less positive) valence of the atom, or at least one atom in
the group. Oxidation and reduction always occur simultaneously and the number of electrons
gained by the atom or ion being reduced must be the same number of electrons given up by the
atom or ion being oxidized, i.e., electrons are conserved. The most common valences of the
elements are shown in standard electrode potential (or standard reduction potential) tables in
chemistry texts or on-line (Wikipedia 2010).

We illustrate these definitions for a process of recovering copper from solution by reduction
with iron. This process is called cementation, and copper is described as being cemented out of
solution. Copper is usually present in the divalent form, and is reduced according to Equation
[5.66]. Simultaneously, iron is oxidized to produce the electrons required to reduce the copper, as
shown in Equation [5.67]. Each reaction must be balanced for both the element and the charge.

Cu2\aq) + 2e~ -> Cu(c) [5.66]

Fe(c) -» Fe2+(aq) + 2e" [5.67]

Adding both reactions together and canceling the two electrons that appear on both sides of
the reactions yields the chemical equation for the overall process. The partial reactions show that
copper gains electrons (its charge is less negative) hence copper is being reduced. Iron loses
electrons, and hence is oxidized.

Cu2+(aq) + Fe(c) -> Cu(c) + Fe2~(aq) [5.68]

Confusing terminology is used in describing redox reactions. The substance that contains the
element being oxidized is called the reducing agent. The substance that contains the element being
reduced is called the oxidizing agent. Here, metallic iron is the reducing agent, and divalent copper
ions are the oxidizing agent.

College chemistry textbooks give various methods for balancing redox reactions. The
following summarizes one set of rules.

1. Write an equation that includes those reactants and products that contain the elements
undergoing a change in valence.

2. Determine the change in valence that some element in the oxidizing agent undergoes. The
number of electrons gained is equal to this change times the number of atoms undergoing
the change.

3. Determine the change for some element in the reducing agent.

4. Multiply each principal formula by such numbers as to make the total number of electrons
lost by the reducing agent equal to the number of electrons gained by the oxidizing agent.

5. By inspection, supply the proper coefficients for the rest of the equation.

6. Check the final equation by counting the number of atoms on both sides of the equation.

EXAMPLE 5.10 — Controlled Oxidation of Pyrite.

Much of the sulfur found in coal is present as pyrite, FeS2. A possible way to remove some of
the sulfur is to oxidize the pyrite with water and oxygen to form water-soluble FeS04 and H2S04.
Balance the process equation using the method described above.

Solution. The six steps are:

Chapter 5 Stoichiometry and the Chemical Equation 285

1. Write the unbalanced process equation:*

FeS2(c) + H20(/) + 02(g) -> Fe2+(tf#) + U30+(aq) + S042~(aq) [5.69]

2. The oxidizing agent is 02, which goes from an initial valence of 0 to -2 in S042~ ions.
Thus, four electrons must be gained per mole of 02.

3. Sulfur is oxidized from a charge of-1 to +6. Consequently, 14 electrons must be given up
by sulfur per mole of FeS2.

4. In order to equate the number of electrons given up by sulfur in FeS2 with the number
gained by oxygen, there must be 14/4 or 3.5 moles of 02(g) reacting per mole of FeS2. Thus the
partially balancedreaction is:

FeS2(c) + H20(/) + 3V202(g) -► ¥e2+(aq) + U30+(aq) + S042~(aq) [5.70]

5. Balancing the H and S atoms yields the final balanced reaction for the process:

FeS2(c) + 3H20(/) + 3V202(g) -> F e 2 + H ) + 2H30+(aq) + 2S042~H) [5.71]

6. Check if each side of the reaction has the same number of Fe, H, S, and О atoms. It does!
(If you use is H+(aq) instead of the hydronium ion, only one mole of H20 is needed.)

Assignment. Once the aqueous solution is separated from the coal, it would be nice to separate the
iron in some way. Write and balance an ionic equation for the oxidation of the Fe2+(aq) using
additional 0 2 to produce Fe3+(ag) and H20. Write a different ionic equation for the 0 2 oxidation of
Fe2+(ag) to produce Fe(OH)3(c) and H+ ions.

EXAMPLE 5.11 — Dissolution of Gold in Cyanide Solution.

Gold ores contain finely disseminated gold particles that can be leached by a basic cyanide
solution in the presence of oxygen. Write a series of half-reactions to obtain a final balanced ionic
equation for the process.

Solution. The unbalanced equation is:

Au(c) + CW(aq) + 02(g) -* Au(CN)2 [5.72]

A series of half-reactions will be written and balanced as if H+ ions were present. Then, a
number of OH- ions will be added to both sides of the equation to eliminate the H+ ions (this will
eliminate the H+ ions by forming H20). Species present on both sides of the equation will be
cancelled, and a final check will make sure that the elements and charges balance.

1. The oxidation half-reaction is:

CW(aq) + Au(c) -> Au(CN)2 + e~ [5.73]

This is balanced for carbon, nitrogen and the charge:

2CW(aq) + Au(c) -► Au(CN)2~ + e~ [5.74]

2. The reduction half-cell reaction is:

02(g)->H20(/) [5.75]

This is balanced for oxygen, hydrogen, and charge as follows:

4e" + 02(g) + m\aq) -> 2H20(/) [5.76]

The proton formed when water ionizes is hydrated, which is winrdititceanteHd +b(yaqt)h,eorusseimopflyHH30++, (bauqt)w(thheen
hydronium ion) in this Example. More commonly, the proton is
written that way, it's good to remember that it is hydrated in aqueous solutions.

286 Chapter 5 Stoichiometry and the Chemical Equation

3. The balanced oxidation half-reaction (Equation [5.74]) must be multiplied by four to
balance the electrons, and then added to the reduction half-reaction. After this, and canceling
identical species:

SCW(aq) + 4Au(c) + 02(g) + 4R+(aq) -+ 4Ag(CN)2"(^) + 2H20(/) [5.77]

4. Add four OH~ ions to each side of the reaction, and then eliminate as many H20 molecules
as possible:

SCW(aq) + 4Au(c) + 02(g) + 2H20(/) -> 4Ag(CN)2>#) + 4 0 H " H ) [5.78]

5. See if the elements and charges balance. They do!

Assignment. In practice, sodium cyanide (NaCN) is used in the process. Write and balance a non-
ionic equation for the cyanide dissolution of gold.

5.12 Summary

The relationship between the amount of a substance and its mass is based on a reference of
exactly 12 grams assigned to one mole of 12C. The molar mass of all other atoms is fixed relative
to this reference. The formula mass of a molecule is the sum of the masses of the individual atoms
that make up the molecule. The formula for a molecule of known structure represents the actual
atomic aggregation. An empirical formula is used for a molecule of unknown structure to
represent the relative atomic ratios of the atoms, with simple integers where possible.

A gravimetric factor can be used to convert between amount and mass. The program MMV-C
(on the Handbook CD) can generate gravimetric factors, or can make conversions between amount,
mass, and volume units for stream species. MMV-C also generates a workbook where cells are
renamed with the atomic mass of all elements. This makes it easy to calculate the molecular mass
of a species, or convert between the mass of a species and its molar amount.

The chemical equation represents the relationship between the number of moles of the
participating reactants and products. For any element, the amount or mass present in the reactants
must equal the amount or mass present in the products. The chemical equation provides a variety
of qualitative and quantitative information essential to the calculation of the material balance for
systems undergoing chemical reactions. The chemical equation tells us about stoichiometric
relationships by means of the coefficients of the species in the reaction. The absolute values of
these coefficients are not as important as their ratio in the calculation of the material balance.
Some processes involve reactions that together have a unique stoichiometric relation.

Chemical reactions are useful only if they are correctly balanced. Two guidelines are
important here. First, chemical species should appear on only one side of a reaction. This
precludes the use of inert substances as part of a reaction. Second, each valid reaction must only
be balanceable in one unique way.

The chemical equation does not indicate the true mechanism of the reaction or how fast or
how far it will proceed. The occurrence of a chemical reaction does not exclude other
unanticipated reactions from occurring, nor does a reaction itself tell us if it is more or less favored
than any other reaction. The only thing indicated by the chemical equation is the stoichiometric
amounts required and products obtained // it proceeds in the manner in which it is written.
Notwithstanding these limitations, the first step in evaluating a process is to write and balance all
feasible chemical reactions.

A material balance requires the construction of a set of independent reactions. As a rule of
thumb, the number of independent reactions needed to characterize a system is the number of
reacting species minus the number of reacting elements. The independent reaction set must
include every reactive species in at least one reaction, and to be independent, a reaction cannot be
obtainable by manipulating the other reactions in the set.

Chapter 5 Stoichiometry and the Chemical Equation 287

Information about the progress of a reaction is indicated in different ways. The fractional
conversion, yield, and reaction rate are some of the indicators used to describe the amount of
products formed. These terms are functions of the conditions present in the reacting device.
Spontaneous reactions tend to proceed until one or more limiting reactants are consumed, and if so,
are deemed irreversible. Reversible reactions have a theoretical limit on reaction progress based
on the value of the equilibrium constant, so long as all of the species in the equilibrium constant
expression are present. The value of the equilibrium constant is a function of temperature, so some
reactions go from practically irreversible to reversible as the temperature changes. The total
pressure P (where P = sum of the partial pressures of the chemically involved species) has an
effect on the equilibrium position when there is a different number of gas molecules on the two
sides of a reaction. Le Chatelier's principle describes the effect of system pressure on the
equilibrium position of a reaction involving a change in volume.

A reaction (or a set of reactions) tends to continue until the products attain a certain
relationship as defined by the equilibrium constant (Кщ) expression, which is a function of
temperature only. The equilibrium position is a function of temperature and amount of reactants,
and in some cases, pressure. Two factors can prevent a process from reaching equilibrium. First,
reaction kinetics govern how fast a reaction proceeds. Second, mass transfer governs how rapidly
reactants reach the place where the actual reaction takes place. Even if insufficient time is allotted
for reactions to go to equilibrium, it's still useful to calculate the equilibrium position of a process
to compare with its observed position.

Reactions can be grouped according to their expected equilibrium position. A
stoichiometrically-irreversible (spontaneous) reaction is one that proceeds until the limiting
reactant is completely, or at least stoichiometrically, consumed. Such reactions are identified by
having а Кщ value above about 400, such as the reaction between H2(g) and Cl2(g) at 1000 К. А
reversible reaction is one in which (at equilibrium) a stoichiometrically significant change in
product amount occurs as a result of a moderate change in temperature (or pressure, if there is a
volume change). The water-gas shift reaction (WGR) at 1000 К is such a reaction. A negligible
reaction has such a small Кщ value that there is no limiting reactant. The extent of reaction is so
slight that the amount of products formed is negligible, so the amount of reactants doesn't
appreciably change. The decomposition of MgC03(c) to form MgO(c) and C02(g) at 800 K is
such a reaction, as is the reaction of 02(g) and N2(g) to form NO(g).

The database program FREED was introduced as a source of equilibrium constant data.
FREED's Reaction tool can be used to calculate a table of values of Кщ, or given the value of
Кщ, the calculator option will find the temperature. Log Кщ shows a near-linear relationship with
respect to reciprocal absolute temperature. The Reaction tool displays this relationship in a chart
which allows the use of the Trendline tool to obtain the parameters of a linear equation

The extent-of-reaction term as used in this Handbook has a special definition: it refers to a
reactant, not the overall reaction. The substance involved must be stated, such as "the extent of
CH4 reaction is 0.85", which means that 85 % of the incoming CH4 is consumed by the stated
reaction. The XR concept requires that the amount of reactant must also be stated. When the
stated substance is involved in two or more reactions, some assumption must be made about which
reaction takes place first, even if the reactions actually occur in some other sequence. The first
listed reaction proceeds to the stated reactant extent, after which the second reaction proceeds, etc.
The choice of reaction order must be carefully chosen, based on experience or by inference from
data on similar processes.

In certain well-defined conditions, it may be possible to quantify the rate of a reaction with
time by employing the principles of chemical reaction kinetics. The rate law states that the rate of
reaction of a substance is a function of its composition. The rate of the forward reaction decreases
as the reactant(s) are consumed and the rate of the reverse reaction increases and the composition
of the products increase. According to the law of mass action, equilibrium is reached when the
forward and reverse rates are the same.

288 Chapter 5 Stoichiometry and the Chemical Equation

FlowBal uses reaction equation coefficients to make material balance calculations on reactive
systems. The reaction set for each reactor device must be independent. Chemical reactions in
FlowBal occur only in the reactor device. FlowBal can use both XR and Кщ equations as part of
the equation set for Solver. Alternatively, any empirical relationship may also be used by inserting
it as an equation.

Ionic chemical reactions that involve a change in valence in atoms of some of the reacting
species are called oxidation-reduction (redox) reactions. Both amount and charge are conserved,
and both must be balanced.

References and Further Reading

About.com: Chemistry. Stoichiometry & Reactions, 2010.
http://chemistry.about.com/od/stoichiometry/Stoichiometry.htm.

Bhatt, B. I., and Vora, S. M., Stoichiometry, 2nd edition, McGraw-Hill (1984).

CAcT, University of Waterloo. Chemical Stoichiometry. On-line tutorial, 2010.
http://www.science.uwaterloo.ca/~cchieh/cact/c 120/stoichio.html.

Chemistry Fundamentals Program, Department of Chemistry, UNC-Chapel Hill, Stoichiometry,
2002. http://www.shodor.org/UNChem/basic/stoic/.

ChemTutor, Mols, Percents, and Stoichiometry. http://www.chemtutor.com/mols.htm. 2009.

Encyclopedia Britannica, Oxidation-reduction reaction, 2005.

FactWeb Programs, FactSage, www.crct.polymtl.ca/equiwebmenu.php, 2010.

Rao, Y. K., Stoichiometry and Thermodynamics of Metallurgical Processes, Cambridge Univ
Press, 1985.

Wikipedia contributors, "Stoichiometry", "Chemical reaction", "Redox", "Steam reforming",
"Water gas", Water gas shift reaction", "Equilibrium constant", "Le Chatelier's principle",
"Standard electrode potential" Wikipedia, the Free Encyclopedia, December 2010.
http://en.wikipedia.org/wiki/Main_Page.

Exercises

5.1. A residue from the extraction of zinc concentrate contains several percent franklinite
(ZnFe204). Calculate the mole and mass fractions of the elements in franklinite. Calculate a
gravimetric factor that can be used to determine the mass of franklinite in 1 kg of the residue if the
chemical analysis is reported as mass fraction zinc. Check your answer with MMV-C.

5.2. An iron ore contains hematite, magnetite, and fayalite (Fe203, Fe304, and Fe2Si03). A
chemical analysis of the ore gives wFe = 68.40 %, wSi = 2.36 %. Calculate the mass fraction of
each ore mineral.

5.3. Certain ceramic parts are made of sintered alumina, and are to contain between 2 and 4 %
B203. The alumina powder is mixed with boric acid (H3B03), which acts as a binder and sintering
aid. During sintering, the boric acid dehydrates, and forms A15(B03)06. Use a spreadsheet to
develop a graphical relationship between the amount of boric acid to be added to 100 lb of pure
alumina to obtain parts in the specified range of B203 composition. Add to the graph the mass
fraction of unreacted A1203 present.

5.4. A furnace refractory is prepared by mixing dolomite (CaC03MgC03) and tar. During firing,
the tar decomposes to a volatile hydrocarbon and a solid carbon residue having 64 % of the mass
of the tar. The dolomite decomposes to the oxides. How much tar should be added to one ton of

Chapter 5 Stoichiometry and the Chemical Equation 289

dolomite to obtain a refractory brick containing wC = 4.3 % after firing? What is the weight loss
per ton of brick produced?

5.5. A refractory is made by mixing magnesite (MgC03) and gibbsite (A103H3) in proportions
such that the refractory will have wspinel (MgAl204) 88 %, balance A1203. How many kg of each
mineral should be used to produce one tonne of refractory?

5.6. What is the empirical formula of a fuel oil with wC = 84.5 %, balance H? Check your answer
with MMV-C.

5.7. X-ray analysis showed that a titanium ore consists of three minerals: Fayalite (Fe2Si04),
ilmenite (FeTi03) and pseudobrookite (Fe2Ti05). Chemical analysis of the ore is reported in terms
of the mass fraction of Fe203 and Ti02. A sample of titanium ore analyzed 62.2 %Fe203 and 35.56
%Ti02. Calculate the mass of each mineral in 1 kg of ore.

5.8. Write a balanced chemical equation for the following reactions. If you cannot express the
reaction with one equation, use two.

a) The oxidation of chalcopyrite (CuFeS2) to form CuS04, Fe203, and S02.

b) The oxidation of cementite (Fe3C) to form hematite (Fe203) and C02.

c) The reduction of a mixture of rutile (Ti02) and silica by carbon to form titanium disilicide
and CO.

d) The oxidation of an aqueous solution of FeCl2 and MnCl2 to form MnFe204 and HCl(ag).

e) The removal of sulfur from pyrite (FeS2) by H2 to form H2S and pyrrhotite (Fe0.875S).

f) The roasting of cinnabar (HgS) to form a gas containing Hg, S02, S03, and 02.

g) The roasting of chalcopyrite in the presence of CaC03 to form CaS04, Cu, Fe304, and

co2.

h) The reduction of Si02 with С to produce Si and a gas containing CO and SiO.

/) The formation of sulfuric acid by the oxidation of S02 in aqueous solution.

j) The oxidation of ethyl alcohol (C2H5OH) to form C02 and H20.

k) The oxidation of ammonia to form NO and H20.
/) The action of sulfuric acid on fluorapatite [CaF2'3Ca3(P04)2] to produce phosphoric acid,
R¥(aq% and gypsum (CaS04*2H20).

J) The reaction of CaO with S02 to produce CaS04 and CaS.

5.9. The items below list input and output substances, and an amount or mass of one of them.
Write a balanced chemical reaction for reactants and products, and calculate the mass of the bold-
faced substance.

a) Input: Silica, graphite, nitrogen. Output: silicon nitride (Si3N4, 1 kg), CO.

b) Input: Cobaltous chloride (CoCl2, 800 g), water, oxygen. Output: Co304, HC1.

c) Input: Cryolite (Na3AlF6, 180 kg), sodium hydroxide. Output: Sodium fluoride, sodium
aluminate (NaA102), water.

d) Input: Titanium tetrachloride, magnesium, nitrogen. Output: Titanium mononitride
(1500 g), magnesium chloride.

è) Input: Magnetite (1 tonne), hydrogen, methane. Output: Iron carbide (Fe3C), steam.

f) Input: Sodium jarosite [NaFe3(S04)2(OH)6, 1200 kg], sulfuric acid. Output: Ferric
sulfate, sodium sulfate, water.

290 Chapter 5 Stoichiometry and the Chemical Equation

g) Two reactions required. Input: Zinc ferrite (ZnFe204, 300 kg), methane. Output from
first reaction: Zinc oxide, wustite (Feo.9470), CO, H2. Output from second reaction: Zinc oxide,
wustite (Feo.9470), C02, H20. Extent of each reaction = 0.5.

h) Input: Wolframite (FeW04, 5600 kg), carbon. Output: Iron, tungsten carbide (WC), CO.

/) Input: Molybdenite (MoS2, 8000 kg), sodium carbonate, oxygen. Output: Sodium
molybdate (Na2Mo04), sodium sulfate, carbon dioxide.

j) Input: Arsenopyrite (FeAsS, two tonnes), oxygen, calcium chloride, water. Output: Iron
arsenate (FeAs04), calcium sulfate, hydrogen chloride.

5.10. Natural gas contains a volume fraction of components as follows: 94.1 %CH4, 3.0 %C2H6,
balance N2. How much air is required to completely oxidize the hydrocarbons in one ft3 (STP) of
natural gas? Use MMV-C to calculate the elemental composition of the natural gas.

5.11. A residue from steelmaking contains a mixture of franklinite (ZnFe204), magnetite (Fe304),
and silica. Its composition is: wZn = 20.5 % and wFe = 33.7 %. How much carbon should be
added to 1 tonne of residue to reduce all of the zinc to metal, and all of the iron to Feo.9470, while
producing CO?

5.12. Methanol (CH3OH) can be used to make a cleaner-burning gasoline. It is prepared by
catalytic reaction between H2 and CO. The plant mixes CO and H2 in a mass ratio of 11:2.
Calculate the limiting reactant, and the theoretical yield of methanol. In practice, the plant feeds
80 kg/s of reactants to produce 37 kg/s of methanol. What is the % yield?

5.13. Impure Ni can be purified by reacting it with CO(g) under pressure to form Ni(CO)4(g). In a
laboratory experiment, 150 ft3/min (STP) of CO was fed into a fluidized bed furnace along with 5
lb/min of impure Ni containing wNi = 88 %, balance inert solid. What is the limiting reactant?
What is the volume and composition of the product gas and solid if XRCO was 0.87?

5.14. A reducing gas is prepared by mixing steam and methane in equal proportions and heating in
contact with a catalyst with the aim of forming a gas having a large fraction of CO and H2. The
product gas also contains C02? H20, and CH4. The process can be represented by two parallel
reactions:

CH4(g) + H20(g) -> CO(g) + 3H2(g)

CH4(g) + 2H20(g) - C02(g) + 4H2(g)

For the first reaction, ZZ?CH4 was found to be 0.838, and for the second reaction JLKCH4 was
0.34. Calculate the composition of the reducing gas.

5.15. Ozone (03) is used in hydrometallurgy to oxidize certain ions to precipitate them as oxides.
Ozone must be produced on-site because it spontaneously and irreversibly decomposes to oxygen
at normal temperatures. An ozone producer generates a mixture of O3 and 0 2 having w03 = 0.035.
38 minutes later, the w03 = 0.023. How long will it take for the gas to reach an ozone composition
half of the initial (produced) value? Assume the decomposition reaction is first order in ozone
composition.

5.16. The reaction between iron, steam, wustite (Feo.9470) and hydrogen is reversible. When both
solids are present, the equilibrium constant expression can be expressed in logarithmic terms.
Over the temperature range 800 - 1200 K, the equation is:

Л PH2 J T

Calculate the equilibrium gas composition for four temperatures over the range 850 - 1150 K.
If 3.6 moles of H2 and one mole of wustite per second are fed into an equilibrium reactor, which
reactant is limiting?

Chapter 5 Stoichiometry and the Chemical Equation 291

5.17. Write and balance a reaction for the formation of CaS04 from lime, S02 and 02. Calculate
JL7?S02 based on the fact that when 160 m3 (25° C, 1 atm) of a gas mixture containing <jpN2 = 0.56,
<pS02 = 0.12, balance 0 2 was passed across 240 kg of hot CaO (t = 710 °C), the solid gained 18.7
kg.

5.18. Silicon carbide (SiC) is produced by a reaction between carbon and silica to produce a gas
containing 90 %CO(g) and 10 %SiO(g). The main reaction is thought to be the formation of SiC,
while a second (undesirable) reaction occurs between SiC and Si02 to produce SiO and CO. Write
and balance the two reactions, and calculate XR for the limiting reactant in each. Calculate the
selectivity of the process.

5.19. Molybdenum trioxide (M0O3) is reduced at 900 °C with hydrogen in a fluidized bed. The
overall reduction process occurs by two reactions in series:

Mo03(c) + H2(g) -+ Mo02(c) + H20(g)

Mo02(c) + 2H2(g) -> Mo(c) + 2H20(g)

The M0O3 in the first reaction always has XR = 1. The progress of the second reaction is a
function of feed rate of M0O3 into the furnace. At M0O3 feed rates below about 0.5 tonnes/h,
XRR2 is constant at 0.223. However, at higher feed rates, the reaction extent has the following
empirical relationship, where F = feed rate of M0O3 in tonnes/h:

XRR2 with Mo02 = -0.0064F2 + 0.0045F + 0.222.

Make calculations (using a spreadsheet) for flowrates of M0O3 between 0.5 and 1.5 tonnes/hr
to determine the minimum volumetric flowrate of H2 that will produce oxide-free Mo. Plot the
results.

5.20. FREED gives the equilibrium constant at 1150 К for the following three reactions:

C(c) + Vi02{g) -► CO(g); ^eq = 4.93 x 109

C(c) + CbCg) -> C02(g); ^eq = 9.68 x 1017

Sn(/) + 02(g) -► Sn02(c); Кщ = 3.46 x 1015

Use the above ÄTeq values to calculate Кщ for the following reaction, and check your answer
using FREED's Reaction program. :

Sn02(c) + 2CO(g) -► Sn(/) + 2C02(g)

5.21. One mole of CO is passed through a bed of ZnO at 930 °C. Write a balanced chemical
reaction for what happens by algebraically combining Equations [5.3] and [5.4], and calculate Ä^eq.
How much ZnO is reduced by one mole of CO?

5.22. One mole of CO is passed through a bed of wustite, Feo.9470, at various temperatures
between 900 and 1500 K. Use FREED's Reaction tool to calculate the value of Кщ in this range
as a function of temperature. Use the results to calculate the amount of CO required to reduce one
mole of wustite under equilibrium conditions, and plot the results. Use Excel's Trendline tool to
seek an equation to fit the data.

5.23. The most efficient heat utilization for the combustion of natural gas is with the
stoichiometric amount of oxygen to form C02 and H20. However, unless excess air (XSA) is
used, a small amount of CO can remain unburned. With any excess air amount, XR02 for the
formation of H20 is 1. However, the 0 2 reaction for the oxidation of CO to C02 has an extent
given by the following equation:

XR02 for oxidation of CO to C02 = 1 - 0.03/%XSA

What %XSA is required for the combustion of CH4 to produce a combustion gas having 0.01
%CO?

292 Chapter 5 Stoichiometry and the Chemical Equation

5.24. A process gas known to be at equilibrium is analyzed and found to contain <jpCO of 27 %,
<pC02 of 18 %, φΗ2 of 37 %, balance H20. Ten moles of H20 are added to 90 moles of the gas,
and the gas readjusts its composition to conform to Кщ for the water-gas shift reaction. Calculate
the volume fraction of each gas species.

5.25. Process engineers are attempting to model the steam reforming of methane using the
following reaction sequence:

CH4(g)+ 2H20(g) - C02 (g)+ 4H2(g)

CH4(g) + C02(g) - 2CO(g) + 2H2(g)

The flow of methane is 2400 L/s, measured at 30 °C and 1.08 atm. The flow of steam is 6100
L/s, measured at 280 °C and 1.12 atm. The reformer operates at 1 atm and a temperature where
Кщ for the WGR =1.10. For the first reaction, XRCR4 with H20 is estimated to be 0.544, and for
the second reaction, XRCU4 with C02 is 0.947. Calculate the gas composition produced by the
process, and determine if it conforms to the WGR.

5.26. Air is passed through a large amount of Cu20(c) at various temperatures to produce CuO(c).
Use FREED's Reaction tool to calculate log(^eq) for the reaction, and plot this function vs. 1/Г.
Use Excel's Trendline tool to find an equation that adequately represents the data, and calculate the
lowest temperature where the air will not oxidize Cu20(c).

5.27. Calculate the effect of system (total) pressure on the partial pressures of H2(g), 02(g), and
H20(g) resulting from the equilibrium reaction between the elements at 1000 К and at 3000 K, and
see if the results conform to expectations based on Le Chatelier's principle. :

a) One mole of H2 is mixed with VA mole of 02.

b) One mole of H2 is mixed with one mole of 02.

5.28. Write a reaction for the thermal decomposition of H2S(g) to form H2(g) and S2(g), and use
FREED's Reaction tool to obtain a value of Кщ for the reaction at 1300 K. The instream reactor
flow is 10 mol/min, with <jpH2S(g) = 90 % balance H2(g). Use FlowBal to calculate the flowrate
and equilibrium composition of the reactor outstream at four reactor pressures between 1 and 4
atm. Use the results to calculate JÖ?H2S at each pressure, and satisfy yourself that the results
conform to Le Chatelier's principle.

5.29 CuS is roasted to produce a calcine containing Cu2S and CuO and some unroasted CuS. The
flowsheet is shown below.

Oxidizing gas 2_ -► Roaster Gas
x 0 2 = ?; x S 0 2 = ?; x N 2 = ?
F2 = 22 л/min
-► Calcine
x 0 2 = 18%, x S 0 2 = 4%, x N 2 = ? Roaster xCuS = ?; xCuO = 68.3%; xCu2S = 21.9%

CuS 1

F1 = 2.5 n/min

Two chemical reactions occur in the roaster. The first one forms Cu2S, which is oxidized in
the second reaction to form CuO. Write and balance the two reactions, and make a material
balance on the process in FlowBal. Interpret the meaning of R-Rl and R-R2. Calculate X7?CuS in
the first reaction, and XR02 for both reactions.

5.30. A reducing gas prepared from sulfur-contaminated natural gas has the following composition
(voi pet):

CO co2 H2 H20 H2S

27.0 2.4 65.5 3.8 1.3

Chapter 5 Stoichiometry and the Chemical Equation 293

The H2S is removed by passing the gas through a fluidized bed furnace at 1000 К along with
some limestone at the rate of 0.2 mol limestone/mol of reducing gas. The object is to remove 98 %

of the H2S by reaction with lime. A flowsheet sketch is shown below.

Limestone — 1 De-sulf • - 3 ■ - ► Clean outgas
Contaminated ingas 2 -► reactor — 4 —►Solid residue
(1000 K)

Three chemical reactions define the system. The first reaction is the thermal decomposition
of limestone to form lime and carbon dioxide, which generates the equilibrium partial pressure of
C02 in the clean outgas. The second reaction is between H2S and CaO to form CaS and H20, and
the XRH2S = 0.98. The third reaction is the water gas shift reaction, Equation [5.58], which
proceeds to equilibrium. Use FlowBal to make a material balance on the process.

Start by writing the three reactions, and then use FREED's Reaction tool to obtain the log Кщ
for the first and third reaction at 1000 K. This will give you the/?C02 in the clean outgas and an
equation relating the partial pressures of the four gases involved in the water gas shift reaction.
FlowBal will write ten equations for twelve unknowns. You'll need to use the X-R and the Insert
Equation tools to generate two more relationships. Remember that we don't know which direction
the water gas shift reaction will actually go, so to be sure, uncheck the Assume Non-Negative box
on the Solver Options display screen. Alternatively, use the built-in FlowBal solver, which allows
negative R-R values, while constraining stream property values to non-negative.

5.31. Balance each of the following redox reactions:

a) Cu(c) + HN03(aq) -+ Cu2+(aq) + NO(g)

b) Мп2+Ц) + NaBi03(c) -+Bi3+(aq) + Mn04~H)

c) Al(c) + MnCVH) -> Mn02(c) + А1(ОН)4~И)

d) Al(c) + N C V H ) -+ NH3(g) + A102>4)

5.32. Gold will not dissolve in nitric or hydrochloric acid. However, it will dissolve in a mixture
of the two (called aqua regia) to form the AuCLf ion and NO(g). Write a balanced equation for
this process.

5.33. Ferrous iron is oxidized by the permanganate ion in aqueous solutions. Write and balance an
ionic reaction for this process, which produces Fe3+ and Mn2+ ions (among other things).

CHAPTER 6

Reactive Material Balances

Chemical reactions are common occurrences in the production and processing of materials.
Reactions may take place during extraction and refining of a raw material, during final processing,
and even in service by being exposed to a reactive environment. Along the way, many less-
valuable and waste materials will be subjected to chemical reactions to extract whatever value they
may have, or to render them inert for permanent disposal. This Chapter will show how material
balances are constructed for chemically-reacting systems by combining the techniques developed
in Chapters 4 and 5. We start by expanding on the concept of independent reactions introduced in
Chapter 5. Next, we outline two different techniques for making material balance calculations: one
using molecular species, and one using atomic species. The rest of the Chapter greatly expands
our use of Excel's tools on reactive systems, and examines in some detail a number of important
processes for material treatment and metal production.

The presence of a chemical reaction changes the degree of freedom of a system. As discussed
in Section 2.2.1, the Gibbs phase rule for a one-mole non-reactive gas-phase system consisting of
several component species, states that the degrees of freedom D = С + 1, where С = number of
independent components. For example when three gas species are present, D = 4. This allows us
to independently vary the temperature, pressure, and two of the three gas phase compositions (the
mole fraction of the third is obtainable by subtracting from one). However, as discussed in Section
2.2.3, D is decreased by one for every chemical reaction that occurs. For three species with one
independent chemical reaction that occurs as a result of the system moving towards the equilibrium
position, D = 3. We can independently vary the temperature, pressure, and one of the gas phase
compositions. The stoichiometry and occurrence of the chemical reaction controls the relationship
between the three reactive species. In effect, specifying the extent of the chemical reaction
between the three species "uses up" one of the degrees of freedom.

Application of these concepts to material balance systems means that the existence of a
chemical reaction requires an additional supplementary relationship as compared with the same
system without a chemical reaction. If a mixture of unreactive gases enters and leaves a device, we
only need to specify the instream OR the outstream composition to get DOF = 0. If a chemical
reaction takes place we need one additional supplementary relationship. The relationship might be
explicit (by specifying a species XT?) or implicit (by specifying an extra stream composition).

If the reaction closely approaches equilibrium, we can use an SR involving the equilibrium
constant expression to indicate the relationship between gas partial pressures in a stream. The
development and use of reaction-related SRs is an important part of this Chapter.

6.1 The General Material Balance Procedure for a Reactive System

Systems that involve chemical reactions require the writing and balancing of chemical
reactions using the techniques outlined in Chapter 5. When several possible reactions can be
written, we need some way to select a subset of all possible reactions from which a correct degree-
of-freedom analysis can be made. In particular, as outlined in Section 5.8, we must create a set of
independent chemical reactions for each reactor. The stoichiometric relations in the chosen
reaction set tell us what we need to know about the relative amounts of elements or species
consumed or produced in the reactor. We write material balance equations based on the
conservation of atoms as species, or as elements. The former approach is called a species material
balance, while the second is called an elemental material balance. Each approach leads to the same

Chapter 6 Reactive Material Balances 295

result, and both are used in practice, so it is a good idea to learn both. Non-reactive systems
(covered in Chapter 4) used the molecular species approach; the atomic species method is
generally inadvisable for non-reactive systems. For reactive systems, the molecular species
method requires the use of balanced chemical reactions. The atomic species method places less
reliance on using balanced chemical reactions, but they are still useful in keeping track of the
reaction extent and writing Кщ expressions.

6.1.1 Independent Chemical Reactions, Independent Species, and Independent Elements

The first step in making a reactive material balance is to write possible chemical reactions for
each reactor, and to select from them a subset of independent reactions. These reactions have two
characteristics:

• A reaction set is independent if none of its constituent reactions can be obtained by
adding or subtracting other independent reactions.

• Each independent reaction must contain a species that is not present in any of the other
independent reactions.

Equation [6.1] gives a rule for calculating the number of independent reactions (NIRx):

NIRx = number of independent reacting species - number of independent reacting components [6.1]

When making a material balance, we may choose, for one reason or another, not to include all
possible species, and therefore not write all possible independent chemical reactions. For example,
when making a material balance for the combustion of a hydrocarbon fuel with excess air, we may
choose to ignore the fact that trace amounts of CO, H2 and CH4 are present in the combustion gas
products. And even if we wanted to calculate the amounts of these trace species, we would have to
know (or assume) that the reactions involving them went to equilibrium. Another reason for
omitting trace species from the material balance equation set when using Solver is that it has
trouble converging when one or more of the species is present in compositions <1(T9.

For most chemically reactive systems the number of components = number of reacting
elements. The determination of the NIRx is sometimes difficult if the system has intrinsic
stoichiometric ratios (as discussed in Section 5.3.1) between molecular species or atoms. For
example, if two molecular species are in the same ratio to each other whenever they appear in a
reactor, balances on these species are not independent because these species are not independent.
Similarly, if two elements are in the same ratio whenever they appear in a reactor, balances on
these elements are not independent because these elements are not independent.. The calculation
for the NIRx in Equation [6.1] should be revised to define the reacting components and species as
the number of independent reacting components and species. Therefore, the concept involving an
independent reaction set and independent entities requires some additional discussion.

Figure 6.1 shows a process for the absorption of water vapor from a gas by reaction with lime
to give hydrated lime. Equation [6.2] shows the reaction.

CaO(c) + H20(g) -> Ca(OH)2(c) [6.2]

H20 F1 = ? 1 —■ ► Absorber —3 -►НгО F3 = 5
2" —4
CaO & Ca(OH)2 CaO & Ca(OH)2
F2 = ? xCaO = 0.8 ~ " * V = 100; xCaO = 0.5

Figure 6.1 Flowsheet for the absorption of water vapor by lime.

The system consists of three elements and we consider only three species. According to this
selection of species, if the number of components = number of reacting elements, Equation [6.1]
says that the system does not have an independent reaction. However, there are intrinsic
stoichiometric ratios in this system because one О atom is always associated with one Ca atom,

296 Chapter 6 Reactive Material Balances

and one О is always associated with two H atoms. Therefore, if any two of the three element
balances is carried out, the third balance is automatically redundant. Hence, the system has only
two independent elements. Owing to the stoichiometric ratio, a species balance can be carried out
by selecting any two of the three species, or a total (flow) balance and one species balance.

These concepts are important in making a DOF calculation for a reactive system. Table 6.1
shows a DOF calculation using the molecular balance method for the lime hydration system. The
subsidiary relationship is the existence of an intrinsic stoichiometric ratio in the system, as
discussed above. The SR originating from an intrinsic stoichiometric ratio is sometimes hard to
detect. A clue in the lime hydration system is that Ca(OH)2(c) is simply a "mixture" of the other
two species in a fixed stoichiometric ratio.

Table 6.1 DOF calculation strategy for a reactive one-device system as applied to the special case
of the hydration of lime.

+ number of independent stream variables 6

+ number of independent chemical reactions +1

- number of independent molecular species balance equations -2

- number of independent compositions -2

- number of independent stream flows -2

- number of subsidiary relationships -1

= degrees of freedom 0

DOF = 0 so the system is defined and the two unknown stream flows can be calculated. The
simplest method is to notice that F2 = F4 = 100. The water balance is then trivial, with Fx = 35.

Most other systems give the correct NIRx when the reacting elements are chosen as

components. For example, consider the carbon-oxygen system, which was discussed in detail in

various parts of Chapter 5, especially Example 5.6. We determined that the species C, CO, C02
and 0 2 could be present and of interest in the system. We see that NIRx = 4 - 2 = 2. Although
four reactions can be written involving these species, only two are independent.

C(c) + V202(g)^CO(g) [6.3]

C(c) + 0 2 ( g ) ^ C 0 2 ( g ) [6.4]
CO(g) + V202(g)->C02(g) [6.5]

C(c) + C02(g)^2CO(g) [6.6]

The selection of two independent reactions from the above set is easy because the С - О
system is simple; any two of the four is an independent set. It's not as easy to select independent
reactions in systems with three or four elements and several species.

If the system is constrained by temperature, pressure, or amount restrictions, not all species
may be present. In a multi-unit process, one device may have only CO(g), 02(g), and C02(g), in
which case NIRx = 1. An adjacent device (connected to the other by streams) may have only C(c),
C02(g), and CO(g) present. Here again, NIRx = 1, but the independent reaction will be different.
We should apply Equation [6.1] device-by-device to calculate the reactor DOF.

6.1.2 Molecular Species Material Balance Method

The general molecular balance equation for a reactive steady-state system has four terms:

input + generation = output + consumption [6.7]

The generation and consumption terms are often expressed as a ratio, such as "one mole of
species A consumed per Vi mole of species В generated". These ratios are obtained by noting the
stoichiometric balancing numbers for the chemical reaction involving the selected species.

Chapter 6 Reactive Material Balances 297

The S - О - N system is a good example to use for illustrating the application of Equation
[6.7] because we've already used it as an example in Chapter 5. Please review this system by
looking back at Section 5.5.5. When a measurable amount of oxygen is present, only three
reacting species are present in significant amounts: 02, S02, and S03.* Nitrogen reacts
insignificantly with the other three species (i.e., it is inert), so from Equation [6.1], NIRx = one.
Therefore only one independent reaction exists between the three reacting species.

S 0 3 ( g ) ^ S 0 2 ( g ) + 1/202(g) [6.8]

The progress of Equation [6.8] is favored by increasing temperature. The (equilibrium)
reaction is stoichiometrically reversible in the temperature range where Кщ is between 0.005 and

500 (700 - 2400 K). If S03 is heated above about 700 K, a noticeable amount of it will
decompose. If a mixture of 0 2 and S02 is heated to a temperature below about 2400 К, а
noticeable amount of S03 will form. If a mixture of S03, 0 2 and S02 is heated to any temperature
in the range, a reaction occurs, but its direction is undetermined without additional information.

The simplest situation is the thermal decomposition of S03 in the presence of N2. Since four
independent species are present, we can make four species balances around the reactor. The
flowrate of the instream is 100 mol/sec of a gas containing 114 times as much S03 as N2. The gas
mixture is heated to 920 К in a reactor at P = 1.1 atm. We seek to calculate the flow rate of the
outstream and its composition by making molecular species balances around the reactor. The first
step is to sketch and label a flowsheet, as shown in Figure 6.2.

Fin = 100n/s Reactor F°ut = ?
x S 0 3 = 0.60 (920 K, 1.1 atm) (composition
xN2 = 0.40
unknown)

Figure 6.2 Flowsheet for the thermal decomposition of S03 in the presence of N2.

The next step is to calculate the DOF. The strategy for a single-device system is very similar
to that for a non-reactive system. The main difference is the addition of a term for the number of
independent chemical reactions. The DOF equation for a reactive one-device molecular species
balance system is

DOF = SV + NIRx - IB - 1С - F - SR

Table 6.2 DOF analysis for molecular species method for thermal decomposition of S03 in
the presence of N2.

+ number of independent stream variables 6
+ number of independent chemical reactions +1
- number of independent molecular species balance equations -4
- number of independent compositions -1
- number of independent stream flows -1
- number of subsidiary relationships
= degrees of freedom 0
+1

The system is underspecified, so we need one additional piece of information, such as the
extent of reaction of a reactant (XR), knowledge that the reaction goes to equilibrium (Кщ
expression), a stream flow rate, or an outstream compositions or amount.

Suppose a sample of the outstream showed the flow of 0 2 to be 10.8 mol/s. We make four
independent species balances, three for the reactive species, and one for the non-reactive species.

* The reaction between sulfur and oxygen is spontaneous and complete, so if oxygen is present, sulfur isn't.

298 Chapter 6 Reactive Material Balances

The oxygen (02) balance is made first since it is not present in the input stream, and we know the
amount out:

O2 balance: generation = output. 0 2 generation = 10.8 mol/sec.

Next, the SO3 balance. No SO3 is generated, so input = output + consumption.

S03 input = 60.0 mol/s.

SO3 consumption: 1 mole SO 3 consumed (10.8 moles 02 generated) = 21.6 wS03/sec
V2 mole О 2 generated

S03 output = 60.0 21.6 = 38.4 mol/sec S03 out

Next the S02 balance: generation = output
f
1 mole SO 2 generated ^
S02 generation: Vi mole О generated (10.8 moles 02 generated) = 21.6 mol/sec S02
2

S02 output 21.6 mol/sec S02 out.

Last the N2 balance: input = output.

Output = 40.0 mol/sec N2.

These four species balance outputs sum to 110.8 mol/sec. The results, in percent units, are:

<pS03 = 34.7%, <pS02=19.5%, φ02 = 9.7%, balance N2

Consider a different situation in which data is available on how much of a reactant is
consumed by one of the independent reactions. This data can be used to generate an extent of
species reaction (or consumption) value, designated XR. This is expressed as the fractional
amount of a reactant consumed by the specified reaction. This concept was first introduced in
Section 5.5.1 and extended in Section 5.5.3 and 5.6. XR of a certain species is often a function of
temperature, pressure, or reaction time. If a reaction has two or more reactants,, each species may
have a different XR. In the present case, SO3 is the only reactant, so Xi?S03 is the only possible
extent of reaction species.

In one experiment, measurements showed that 36 % of the SO3 decomposed in passing
through the furnace. X7?S03 is then 0.36, so this gives the SR we need to get DOF = 0. An XÄSO3
of 0.36 means 21.6 moles of SO3 decompose per second. From the stoichiometry of Equation
[6.8], this forms 21.6 mol/sec of S02 and 10.8 mol/sec of 02, leaving 38.4 mol/sec of S03
unreacted. Here we show how to balance the system based on composition units instead of amount
units. There are four unknown species compositions and one outstream flow, so we need to write
five equations: four species balances and a defining relationship for mole fraction. Table 6.3
shows the four species balance equations.

Table 6.3 Species balance relationships expressed as stream compositions for the thermal
decomposition of SO3 in the presence of nitrogen.

Species Input Generation = Output Consumption

S03 60 0 ^ut(xS03) 21.6
S02 0 21.6 F°a\xS02) 0
10.8 ru,(x02) 0
o2 0 rut(xN2) 0
0
N2 40

These four equations plus the defining relationship for mole fractions (i.e., Σχ = 1) were
solved by Solver to obtain values for all five unknowns. The reactor outstream flow is 110.8
mol/sec with the same composition obtained from the previous case where we knew the outflow of
0 2 to be 10.8 mol/sec.

The results from the two methods are consistent but the calculational technique is different
because we had different types of system information. The main point is that the consumption and

Chapter 6 Reactive Material Balances 299

generation terms require use of the independent reaction(s), based on independent species. In both
cases, R-R = 21.6.

Molecular species balance equations can be written in a slightly different way than in Table
6.3. Recall that we introduced the rate of reaction concept (R-R) in Section 5.5.1 as defined in
Equation [5.36]. Please review this section in order to follow the discussion below. The use of an
R-R term requires the addition of one more material balance equation, and produces one more
material balance description. Recall that R-R is defined as the net molar rate of change of a species
divided by the stoichiometric reaction coefficient. In the case of Equation [6.8], R-R is the rate of
change of S03 divided by 1, the rate of change of S02 divided by 1, and the rate of change of 0 2
divided by 0.5. The table below shows the material balance equations that include the R-R term.

Species Input Generation = Output Consumption

S03 60 0 ^ut(xS03) R-R
^ut(xS02) 0
S02 0 R-R F°u\x02) 0
F°u\xN2) 0
o2 0 V2(R-R)
N2 40 0

We need two more equations to solve for the six unknowns. As before, the defining
relationship for mole fractions (i.e., Σχ = 1) is one equation, and we need to use the fact that R-R =
X7?SO3(60) = 0.36(60). Solving these six equations gives the same answer as before, plus the
solution includes a value for R-R, which is 21.6. The difference in the two molecular species
methods is trivial in this case, but is not when two or more reactions occur in a device.

Finally, what if there wasn't any measured data on the process — is it still possible to
calculate the expected product gas composition? The answer is yes if we assume (or are told) that
Equation [6.8] goes to equilibrium. Кщ values for SO3 and S02 at 920 К are needed in order to
calculate Кщ for the reaction. The database program FREED (on the Handbook CD) has Кщ data
for these species. Section 5.5 outlined the procedure for calculating Кщ for a reaction.

Species so3 so2 Reaction

Keq at 920 К 2.67 x 1017 5.31 x 1016 0.199

A variation of the molecular balance method is used here. The table below shows the moles

of species in and out, using D as the moles of S02, T as the moles of S03, X as the moles of 02,
and N as the moles of N2.

so3 so2 o2 N2 total
100
moles in 60 0 0 40

moles out T D X 40 T + D + X + 40

From the stoichiometry of Equation [6.8], T + D = 60, and X = ιΛΌ. All of the amount
variables can be expressed in terms of D, in which case the total moles in the product gas mixture
= 100 + ViD. The Кщ expression for Equation [6.8] involves the partial pressure of each gas. Let
t, d and x = partial pressures of S03, S02, and 02:

Keq = ^ * = 0 . 1 9 9
t

Mixed units (amount and partial pressure) require expressing one as a function of the other.
Recall from Equation [2.7] that the partial pressure of a gas is equal to its mole fraction times the
total pressure. This relationship for S02 at a system pressure of 1.1 atm is:

Dxl.l
"100 +540

300 Chapter 6 Reactive Material Balances
Equivalent expressions for t and x are also written and substituted into the Кщ expression:

LID/ Ί I.IC/2D)/
/ ( 1 0 0 + У1РЩ / ( 1 0 0 + УтР)
0.199 =
1.1(60-D)/

/ ( 1 0 0 + !/2D)

Goal Seek can solve for D, with T and X obtained algebraically:
^ u t S 0 2 = 22.4; 7^utS03 = 37.6; 7^*02=11.2

The values of 7^utS02 and F°utÖ2 from the equilibrium calculation are higher than those
calculated from XRSO3 or the gas measurement values. This indicates that SO3 decomposition
didn't quite reach equilibrium in the time allowed. However, if equilibrium is reached, the
maximum possible XRSO3 is 0.373.

6.1.3 Atomic Species Method

An atomic species balance takes the form input = output because atomic species can neither
be generated nor consumed in chemical reactions. The atomic species method does not require the
writing of an independent reaction(s), but it's still a good practice. The DOF for an atomic species
balance does not include a term for the NIRx. The DOF calculation is relatively straightforward
and uncomplicated. Generally, inert species are treated in whatever form they occur, even if (as in
the case of nitrogen) it's a molecule. For a reactive system single-device atomic species balance::

DOF-SV -EB-IMB-IC-F-SR

EB represents the number of independent element balance and 1MB represents the number of
independent inert molecular species balances. Table 6.4 shows a DOF analysis for an atomic
species balance for the thermal decomposition of SO3 in the presence of N2 as depicted in Figure
6.2.

Table 6.4 DOF analysis for atomic species balance for the thermal decomposition of SO3.

+ number of stream variables +6
- number of independent atomic species balances -2
- number of molecular balances on independent inert species -1
- number of independent components -1
- number of stream flows -1
- number of other subsidiary relationships 0
= degrees of freedom +1

The system is underspecified by one factor, just as was shown by the molecular species DOF
in Table 6.2. Suppose in this case the 0 2 outflow was found to be 10.8 mol/sec.

S balance: input = output.
S input: (1 mole S/l mole SO3)(60.0 mol/sec SO3) = 60.0 mol/sec S
S output: (1 mole S/l mole S03)F(S03) + 1 mole S/1 mole S02)FS02

S balance: F(S03) + FS02 = 60.0 mol/sec
О balance: input = output.

О input: (3 moles O/l mole SO3)(60 mol/sec S03) = 180.0 mol/sec О
О output: (2 moles O/l mole S02)FS02 + (3 moles O/l mole S03)FS03 +

(2 moles O/l mole O2)(9.0 mol/sec 02).

Chapter 6 Reactive Material Balances 301

О balance: 180.0 = 2FS02 + 3FS03 + 21.6

The two atomic species balance equations have two unknowns, and it is readily apparent that
i^utS02 - 21.6 n/s and 7^utS03 = 38.4 mol/sec. The remaining element is N, which is handled the
same way as the S and О balance (except we leave nitrogen as N2). This gives the same output
stream composition as with the molecular species balance. The F°ut = 110.8 mol/sec. You should
satisfy yourself that you could make a molecular or atomic species balance if the F°ut or the φ02
had been given instead of XRSO3 reaction or the F02. You should satisfy yourself that you could
use the atomic species method with Кщ data to make the same calculation.

An alternative technique for the atomic species method is based on writing an expression for
the atom ratio of two elements, or between one element and the sum of all elements. Section
5.5.5, introduced this technique, which is particularly useful when data for gases is given in partial
pressure units. Since there are no condensed-phase species in the system, all of the S and О remain
in the gas. In this example, an expression for the O/S ratio is:

Q _ 2nQ2 + 2nSQ2 + 3nSQ3
S ~ nS02 + nS03

Also, 2Ю2 = 21.6, as was specified in order to bring the DOF = 0. A second equation is the
total element balance for S and O:

240 = 3nS02 + 4nS03 + 21.6

which allows the calculation of the amount of S02 and S03 with the same results as before.

When an inert component (N2 in this case) is present, it's better to keep it as a molecular
species rather than go to the trouble of converting it to an atom and back again to a molecule. The
practice of leaving inerts as molecules in atomic species balances is followed in this text.

The SO3 decomposition process is rather simple because S03 was the only reactant.
However, what if the input stream contains both reactants and products? Consider the flowsheet
below, also operating at 920 К and 1.1 atm.

Fin = 100.00 n/s Reactor _^Fout = 98.00 n/s
xS03 = 0.200 xS02 = 0.200
x 0 2 = 0.100 xN2 = 0.500 xS03 = ? xS02 = ?
xN2 = ? x02 = ?

The DOF for the system is zero, based on::

SV = 8, EB = 2,1MB = 1,1С = 3, and F = 2

The next step is to discern the direction of Equation [6.8]. Note that fewer moles exit than
enter, so Equation [6.8] must be going in the reverse of the direction written, so (for example) less
oxygen will exit than enter. (Knowing this helps us check our answer.) Next, calculate the amount
of each species by writing two atomic species balance equations, one molecular species balance,
and one sum-of-molecular species equation. As before, let D = moles of S02, T = moles of SO3
and X = moles of 02:

S balance: D + T = 40.00

О balance: 2D + 3T + 2X = 120.00

N2 balance: N2 = 50.00

Sum of species: D + T + X + 5 0 - 98.00
Solving, n02 = 8.00; nS02 = 16.00; nS03 = 24.00.

The composition of the product gas is:

φ02 = 8.16 %; <pS02 - 16.33 %; <pS03 - 24.49 %; bai. N2

302 Chapter 6 Reactive Material Balances

The atomic and molecular balance methods require a different approach and different
arithmetic techniques. Atomic species balances do not require writing independent reactions,
although it's good practice to write them anyway, especially if you're going to use the XR factor or
the Кщ expression. Processes involving complex substances with empirical formulae require
special care when using a molecular balance method. When using the atomic species method, the
arithmetic may be easier if you pick as a basis a stream with the most completely defined
composition. Either technique can use stream flow, stream composition, extent of species reaction,
or Кщ data. This text generally adopts the atomic balance method in making material balances in
reactive systems, but FlowBal uses the molecular species method. It's sometimes advantageous to
combine the approaches. In cases where certain reactions are known to come close to equilibrium,
the j^eq expression can be used instead of XR as a subsidiary relationship.

6.1.4 Atomic and Molecular Species Balance Examples

The atomic and molecular species material balance methods are illustrated on a more complex
system in two examples. A shortened version of the DOF calculation is used here to combine the
first terms involving SV, 1С and F.

EXAMPLE 6.1 — Production of Sulfur by Reduction of Sulfur Dioxide.

Ever-more stringent limits on the emission of sulfur dioxide into the atmosphere have led to
the exploration of techniques to convert it to elemental sulfur, which can be stored safely and
indefinitely in the open. Hydrogen is an effective reducing agent for sulfur dioxide but in addition
to producing sulfur, also produces hydrogen sulfide, which requires separation and a different
treatment to obtain sulfur. Calculate the product stream composition for the S02 reduction system
shown below, using a molecular species and an atomic species balance. The process is carried out
at 1300 К and 1 atm.

г — о шъ Reduction B F B = ? nls
F*H2 = 2 n/s" reactor
F*S02 = 1 nls " - " * > B H 2 S = 0.2741 nls
FBS02 = 0.1728 nls FBH2 = ?
FBS2 = ? FBH20 = ?

Solution. The three-element S - О - H system has five major species (ignoring trace-amount
species), therefore, NIRx = two. Various two-equation independent reaction sets can be devised,
but the two independent chemical reactions selected for this example are:

3H2(g) + S02(g) - H2S(g) + 2H20(g) [6.9]

H2S(g)-H2(g) + l/2S2(g) [6.10]

They qualify as independent reactions because they meet the two criteria mentioned earlier
(one not derived from the other, and each contains a unique species). The material balance will be
solved first using the molecular species balance method. A DOF analysis for the molecular
balance is shown below. Since DOF = 0, the system is properly specified.

+ number of unknown labeled variables +3
+ number of independent chemical reactions +2
- number of independent molecular species balances -5
- number of other subsidiary relationships 0
= degrees of freedom 0

The following terminology is adopted to help keep track of the input, consumption and
generation terms, all as moles.

G69X refers to the rate of generation of species X via Equation [6.9].

Chapter 6 Reactive Material Balances 303

C610X refers to the rate of consumption of species X via Equation [6.10].

F®X refers to the rate of output of species X (stream B).
FAX refers to the rate of input of species X (stream A).

The first step is to write five species balances for the system, using the molecular species
material balance input-output relationship expressed by Equation [6.7].

H2S balance: G6 9H2S = 7^H2S + C610H2S = 0.2741 + C610H2S.
S02 balance: FAS02 = 1 - C*9S02 + F B S 0 2 - C*9S02 + 0.1728.
H2 balance: FAH2 + G610H2 = 3 + G610H2 = C6 9H2 + FBH2.

S2 balance: G610S2-FBS2.
H20 balance: G6 9H20 = i^H20.

The next step is to write equations relating the various generation and consumption terms for
each species, using the stoichiometric factors from Equations [6.9 - 6.10].

C*9H2 - 3G69H2S
C6 9S02 = G6 9H2S
C610H2S = G610H2
C* 10H2S = 2G6-10S2
G6 9H20 = 2G6 9H2S

These equations are solved by sequentially substituting relationships from the second set into
the first set. The results for the product stream in moles are:

7^H20 = 1.6544. i^S2 = 0.2766. FBU2 = 0.0715.

The product stream flow is 2.449 mol/sec, and has the following composition:

67.5 % H 2 0 , 11.3 % S 2 , 11.2 %H2S, 7.1 % S02, 2.9 % H2*

A DOF for the atomic species balance isn't needed because the system has already been
shown to be correctly specified. We need three atomic element balances as follows:

S balance: 1 = 27^S2 + 7^H2S + F B S 0 2 = 2FBS2 + 0.2741 + 0.1728

О balance: 2 - ^ H 2 0 + 2/^S02 = / ^ H 2 0 + 2(0.1728)
H balance: 4 - 27^H20 + 2FBH2 + 2i^H2S = 2 ^ H 2 0 + 2FBR2 +2(0.2741)

The above three equations are readily solved for the specified flows, which are the same as
obtained from the molecular species balance.

Assignment. The process temperature was decreased to 1250 K, the FR2 increased to 2.5 mol/sec,

while FS02 remained the same. A partial analysis of the product stream indicated that <pH2S =
0.17962 and <jpS02 = 0.01966. Calculate the composition and total flowrate of the product stream
using the molecular and element species balance method.

The above molecular species method procedure is a rather detailed technique. But there are
less-complex ways of writing molecular species equations when the system is fairly simple. For
the system just discussed, it may already have occurred to you that there are really only three
unknown stream variables: FBS2,7^H20, and 7^H2. If we can calculate these, we can sum them to
get F8. You should be able to write a molecular balance equation for each of the three species

A plant using H2 for S02 reduction would optimize the composition of stream A to maximize the
production of S2 and minimize that of H2S. Also, the product gas would be cleaned and the unreacted S02
and H2 recycled.

304 Chapter 6 Reactive Material Balances

flows. Set up a table for amount in, amount produced, amount consumed, and amount out. The
amount out = in + produced - consumed. These amounts are deduced from the stoichiometric
reaction coefficients used to balance Equations [6.9] and [6.10]. The table below shows the
balance for each species. Four simple species balance equations can be written but there are only
three unknowns, so one equation is redundant. (We cannot write a simple S2 balance equation
because we've already used the two possible S2 production terms elsewhere.) We note that the
H20 and S02 balance equations have only one unknown, so from either one of them, ^ H 2 0 =
1.6544. From the H2S balance, we get 7^S2 - 0.2766, and from the H2 balance, 7^H2 - 0.0715.
The results, of course, are the same.

Species In Produced Consumed Out

H2 2 2(^82) 1.5(^Η20) FBU2

S02 1 0 0.5(^Н2О) 0.1728

H2S 0 0.5(^H2O) 2(^S2) 0.2741
s2 0 —
0 ^S2

H20 0 2(1-0.1728) 0 f*H20

The "simplified" method used in the above table contains a condensed version of the
equations written in the example, with the produced and consumed equations obtained by
inspection of the two NIRx. It's your choice which technique to use. Alternately, you can include
the R-R terms for both of the independent reactions as described earlier. The next example
illustrates the atomic species method when the data is given in mass units.

EXAMPLE 6.2 — Chlorination of Silicon.

Silicon produced in the submerged electric arc furnace must be refined before being used by
the semiconductor industry. The Siemens process involves chlorinating the impure Si with HCl
and processing the mixture of chlorides to remove the impurities. A sketch of the chlorinator part
of the process is shown below. The subsequent steps to separate the various species are not shown.
Use the atomic species balance method to make a material balance on the process to determine the
mass of Si chlorinated and the mass of each gas leaving the reactor.

mingas = 1 k g H C | Chlorination : ш01*да8 = ?
reactor
mSiCI4= 598.7 g; mSiCI3 =148.2 g
■ffiaaaaaaaaaaai mSiCI2 = 144.8 g; mSiHCI3 = ?
VH2 = 262.8 L; /nHCI = ?

Data. In the first step, HCl is passed into a reactor filled with Si, which produces a number of
silicon chlorides. A laboratory investigation was carried out in which 1 kg of HCl was passed
through a bed of crude Si and the product gas sampled. The sample was cooled to condense the
SiCl4, S1CI3, and SiCl2, and the mass of each of these constituents was measured. The remaining
gaseous sample was passed through an absorption column where HCl and S1HCI3 were absorbed,
allowing the volume of H2 to be measured.

Solution. An atomic species DOF analysis on the process is shown below. The system is correctly
specified.

+ number of unknown labeled variables +3
- number of independent atomic species balances -3
- number of molecular balances on independent inert species 0
- number of other subsidiary relationships 0
= degrees of freedom 0

Chapter 6 Reactive Material Balances 305

To minimize the complexity of formula writing, always assign a symbol to the mass of each
species for which product data is not specified. The mass units are grams. The mass of H was
determined from the volume of H2, assuming it was expressed at STP. The element balances are
written, as usual, in the form In = Out. The number in parenthesis after the mass of each species is
the gravimetric factor for the conversion of the mass of the species to the mass of the contained
element.

mSiCl4 = 598.7, mSiCl3 = 148.2, mSiCl2 = 144.8, тЯ2 = 23.65

mHCl = L, /uSiHCb = C, mSi - M

Si balance: M = 598.7(0.1653) + 148.2(0.2089) + 144.8(0.2838) + C(0.2074).

Cl balance: 1000(0.9724) = 598.7(0.8347) + 148.2(0.7911) + 144.8(0.7162) + C(0.7852) +
Ц0.9724).

H balance: 1000(0.02765) - 23.65 + C(0.00743) + Ц0.02765)

This is a simple equation set to solve, with С = 212.0 g, L = 87.7 g, and M = 215.0 g. Thus, 1
kg of HC1 will chlorinate 215 g of Si, and produce 1215 g of gas.

The molecular species balance method needs a set of independent reactions for the system.
(Selecting a set of independent reactions is important in making atomic element balances and
essential in making molecular species balances.) There are several chemical reactions that can be
written between the various molecular species, but of course, not all are independent. For a system
of three elements, six gas-phase species, and one solid, NIRx = four. Suppose you arbitrarily
chose the following four equations:

4HCl(g) + Si(c) - SiCl4(g) + 2H2(g) [6.11]

SiCl4(g) + VH2(g) - SiCl3(g) + HCl(g) [6.12]

sici3(g) + im2{g) - sici2(g) + Hcife) [6.13]

sici3(g) + vabfe) - siHci3(g) [6.14]

Truly independent reactions must contain at least one species not found in any other reaction.
This is not true for Equation [6.11], hence the set is not independent. You must write other
reactions in place of one or more of the above to get an independent set. Trial-and-error reaction
writing shows that you have to delete equations [6.12] and [6.13] and write two new ones to get a
set of four independent reactions.

4HCl(g) + Si(c) - SiCl4(g) + 2H2(g) [6.11]
3HCl(g) + Si(c) -> SiCl3(g) + Vm2{g) [6.15]

2HCl(g) + Si(c) - SiCl2(g) + H2(g) [6.16]
SiCl3(g) + V2U2(g) - SiHCl3(g) [6.14]

Try using these four reactions to make a molecular-species balance on the system.

Assignment. A laboratory investigation of the chemical vapor deposition of Si was carried out by

mixing H2 and SiHCl3 and heating to 1300 K. A mixture of 10 moles of H2 and one mole of
SiHCl3 deposited 0.270 moles of Si. The product gas was analyzed for H2, HC1 and SiHCl3 with
the following results:

φΗ2, = 88.04%, <pHCl = 5.57%, <pSiHCl3 - 2.34 %

Calculate the DOF analysis to determine if the problem is correctly specified. If so, calculate
the volume fraction of the rest of the constituents. Work your answer using both the atomic and
molecular species balance methods.

306 Chapter 6 Reactive Material Balances

If you are unfamiliar with a system, it's hard to choose the independent chemical reactions
that dominate the system from a set of all possible reactions. Sometimes there is no clear way to
determine the direction of a potentially reversible* reaction, or to determine if a reaction occurs to
any significant extent at all. A good way to start sorting all this out is to use FREED's Reaction
option, which was described in Section 5.5.5. Create a table of Кщ for all feasible reactions to
identify the reaction characteristics (reversible, stoichiometrically complete, and stoichiometrically
negligible). This information can help eliminate certain species as reaction products, or identify
the most likely direction of a reaction. Try to find out if there is a limiting reactant, and if so, what
it is. Finally, be sure you haven't missed an important species in writing your reactions. Valuable
guidance on selecting the important species can be obtained by the use of thermodynamic software,
such as THERBAL, FactSage, or HSC. The Fact-Web program Equilib-Web is particularly useful
(see citation in General References). Results from equilibrium calculations are particularly
important when attempting to design a new process.

6.2 The Use of Excel-based Computational Tools in Reactive System Balances

Chemical reactions in a system can add considerable complexity to material balance
calculations, especially where multiple units are involved. Often the balance contains non-linear
relationships, and multiple answers are required to show trends. Chapter 4 introduced two
enhanced Excel programs that can help make such calculations: SuperGoalSeek (SuperGS) and
SuperSolver. Both programs are used in this and subsequent Chapters, and are described in more
detail in the Appendix. Please keep the User's Guides for these programs handy while reading
through the examples that use them.

Another computational tool is FlowBal, a specialized Excel-based program for materials
balance calculations and flowsheet simulation. Chapter 4 introduced FlowBal for non-reactive
systems, and is extended in this Chapter to reactive system material balances. FlowBal does the
arithmetic, but you still must write and balance the proper number of independent equations for
each reactor.

6.2.1 Application of SuperSolver

SuperSolver is well suited for repetitive calculations involving systems with several reactors

and reactions. SuperSolver is applied here to a single reactor system where S02 is being oxidized
by the 0 2 in air. Section 6.1 described this system in detail. There is only one independent
chemical reaction, Equation [6.8], which in this case goes in the reverse of the direction written

previously. The sketch below shows a single device flowsheet for the oxidation of a gas (the

"Ingas") with air to form a product gas (the "Outgas") containing a mixture of S02, S03, 02, and
N2. The reactor contains an effective catalyst to coax the reaction to near-equilibrium before exit.
The basis is an outgas flow of 100 mol/min at 900 K. The objective is to calculate the amount of

S03 formed in the reactor as a function of the amount of air and the system pressure.

Flngas = ? x S O 2 = 0 7 7

xSO3 = 0.12 xO2 = 0.11 Ingas Outgas F0ut9as = 10o mol/min
xS02 = ? xS03 = ?
F = ? xN2 = 0.79 Air Reactor x02 = ? xN2 = ?
x02 = 0.21
T = 900 К
P = ? atm

The term "reversible reaction " means one that is devoid of spontaneity. If spontaneity is removed, the
system is at all times in equilibrium. Any change in temperature or amount of reactant causes a reversible
reaction to move from its initial to its final condition along a path of continuous equilibrium states. This
Handbook includes an additional criterion: A reversible reaction is one that "closely approaches"
equilibrium with stoichiometrically-significant amounts of reactants and products present.

Chapter 6 Reactive Material Balances 307

The equilibrium constant expression can be used to write a relationship between the partial

pressures of S02, S03, and 02. Equation [6.8] goes in the direction that forms S03(g), here written
with twice the amount of each species to avoid square root functions :

02+2S02^2S03; Keq = (Р*°ъ) [6.17]
P02(pS02)2
LJ

FREED's Reaction tool indicates that Кщ for Equation [6.17] - 45.0 at 900 K. A DOF
analysis for an element balance indicates that SV = 9. There are three independent balance
equations (one for each element), 1С = 3 and F = 1, and one SR (the Кщ expression), so DOF =
+1. This leaves one variable unspecified. The problem statement states that one variable should

be the flow of air. In principle, this would make DOF = 0.

However, the independent reaction has a volume change, so mole fraction can't be used in the
Кщ expression except for the singular condition that P = 1 atm. Equation [6.17] can be written in
terms of moles by recognizing the conversion between moles, mole fraction, partial pressure, and
total pressure. SuperSolver requires designation of one variable as the primary variable, and the
other as the secondary variable. Here, the flow of air is arbitrarily selected as the primary variable,
with pressure as the secondary variable. As the amount of air increases, the amount of ingas
decreases so as to keep the outgas flow constant at 100 mol/min. Seven air flows were selected,
from 0 to 50 mol/min, and five pressures, from 1.0 to 2.2 atm. This gives 35 different calculations
of the material balance for the system. The unknowns are the amounts of the four outgas
components, and the flow of ingas.

There are four material balance equations and one SR equation for an atomic species balance:
S balance: rcS02 in outgas + ttS03 in outgas = (0.12 + 0.77)(i^ngas)

О balance: 2Ю2 in outgas = 2T?S02 in outgas + 3«S03 in outgas
- (3 x 0.12 + 2 x 0.11 +2 x 0.77)(Fngas) + 2 x 0.21(FAir).

N2 balance: nN2 in outgas = 0.79(FAir).
Outgas flow: T?S02 in outgas + «S03 in outgas + n02 in outgas + nN2 in outgas = 100.
Кzrщ (,wri·tten i.n terms ofr.molies)ч: (/ «oS^03)x2 = 4 5 x ( w S 0—2 ) 2 x w 0 2 x P

SuperSolver varied the amount of each outgas component and the ingas flow to get solutions
for all 35 conditions. Figure 6.3 shows the amount of S03 that formed via Equation [6.17] at 900
К for different amounts of added air over a range of system pressure.

The results can be explained qualitatively in terms of two principles. First, as a consequence
of the law of mass action, the addition of air increases the amount of 0 2 available to oxidize S02,
and thus increases the p02. This causes a shift in the equilibrium position (but not the value of
Кщ), so (at least initially) more S02 is oxidized. However, since the outgas flow is constrained to
100 mol/min, adding more air decreases the flowrate of ingas, so less S02 enters. Eventually, a
maximum rate of S03 production is reached, after which it decreases. The second principle is that
of Le Chatelier, which favors the equilibrium position of a decreasing-volume reaction in the
forward direction at increased sum of the reacting species pressure**. So for any given air flow,
the amount of S03 formation increases with pressure. However, the two factors are working in the
opposite direction at constant system pressure when the air flow changes. The law of mass action
prevails during the initial increase in air flow, so the production rate of S03 increases. In a

This prevents Solver from trying to take the square root of a negative number. Even if you specify
"Assume non-negative", Solver may try a negative number to see what happens.

Notice the distinction between the system pressure (the sum of the partial pressures of all species) and the
sum of the reacting species pressure (pS02 +pSOi + p02). Le Chatelier's Principle applies to the sum of the
reacting species pressure.

308 Chapter 6 Reactive Material Balances

countervailing trend, additional air lowers the total pressure of the reacting species by dilution with
nitrogen, so Le Chatelier's principle has the effect of lowering S03 production as air flow
increases. Le Chatelier's principle decreases the effect of the law of mass action up to the point
where the production rate of S03 peaks, and enhances the effect past that point.

Production Rate of SO3 at 900 К

10 20 30 40 50 60
air flow, moI/min

Figure 6.3 Rate of S03 formation at 900 К as a function of the air flow and system pressure.
Outgas flow set at 100 mol/min. The sum of air flow plus ingas flow varied between about 111

and 115 mol/min.

Solver can also optimize an aspect of a system by finding a solution at a maximum (or a
minimum) of one of the system variables. For example, Solver can find the condition for
maximum SO3 production. However, this requires eliminating an independent variable; here, the
air flow. SuperSolver solved six material balance equations at five system pressures to find values
for the maximum production rate of SO3. Figure 6.4 shows the results. The Trendline tool found
an equation relating air flow to system pressure at the maximum SO3 formation rate. For an outgas
flow of 100 mol/min at 900 K, the air flow for maximum SO3 formation = -1.92P2+ 1Ì3P + 25.5.

45 Maximum SO3 Production at 900 К 79
78
43 — O ^ S 0 3 formed
— o— Airflow
с 41 - -Δ- - Ingas flow

F 39

с

S 37

Έ0 35
33

О 31

со
О
СЛ 29

27

25

1.6

pressure, atm

Figure 6.4 Conditions for maximum production rate of SO3 at 900 К as a function of system
pressure. Outgas flow set at 100 mol/min. The sum of air flow plus ingas flow varied between
about 113 and 115 mol/min.

Chapter 6 Reactive Material Balances 309

6.2.2 Reactive System Material Balances Using FlowBal

FlowBaPs application to simple reactive systems was introduced in Section 5.9, which
showed how to add reaction equation coefficients, use the X-R tool, and insert equations developed
from SRs. The FlowBal User's Guide explains each feature in detail, and includes many
examples. This section has two text examples that compare the use of the molecular and atomic
species material balance techniques to the method used by FlowBal, which uses the molecular
species technique. The objective is to point out differences in the way a balance can be set up in
FlowBal as compared to writing and solving equation sets in Excel.

The first example is a process already studied: the chlorination of silicon (Example 6.2, a
semi-batch process). The material balance there was carried out using the atomic species method.
FlowBal is primarily designed for a continuous process, so you must modify the flowsheet slightly
to be continuous instead of semi-batch, as shown in Figure 6.5. Two devices were added, one for
condensation of the silicon chlorides and one for absorption of SiHCl3 and HCl.

2 Sichlor SiHCI3/HCI VH2 =
condenser absorber 262.8 m3
► mSiCI4 = 598.7 kg

1000 kg HCl Chlorination

1 reactor

► /nSiCI3 = 148.2 kg mSiHCU = ?

?kgSi mSiCI2 = 144.8 kg ~ ' *ШНС1 = ?

Figure 6.5 Silicon chlorination flowsheet modified for continuous process.

For a continuous process, the flow of stream 1 should be such that all of the Si entering is
completely consumed by HCl; no silicon exits the chlorinator. A new basis of 1000 kg for the
mass of HCl (stream 2) is used rather than one kg so that FlowBal will more easily display a
reasonable number of significant figures. In order to conserve text space, FlowBaPs application to
this system is indicative rather than descriptive. You can set up FlowBal yourself for the process
to better follow this explanation.

FlowBal requires you to enter a variety of information in an opening dialog box. Before
starting a new system, review the help text on the Title page, review the instructions in Chapter 5
on writing chemical equation coefficients in FlowBal, and look back at the material balance results
for the silicon chlorination process in Example 6.2. Notice the stream units: mass units for SI -
S6, and volume units for S7, Check out the reaction coefficients in the column cells under each
reaction number; reactants are negative numbers and products are positive. Figure 6.6 shows the
FlowBal setup array.

The next step is to have FlowBal read the data and construct the relevant material balance
equations. FlowBal created 20 unknowns related to various process characteristics, and wrote 20
equations. There are as many unknowns as equations, so DOF = 0 (however, FlowBal does not
calculate the DOF in the same way as outlined previously in the text). If the number of unknowns
and number of equations aren't equal, either the system is under-specified (you need to specify
some additional information) or over-specified. However, this system is correctly specified, so it's
solvable by clicking on the Solve Equations toolbar button.

FlowBal first converts all input units to moles, and then invokes Solver to solve the equations.
FlowBal uses the species balance method for its calculations. One or two output arrays are
generated, according to check boxes on the initial dialog screen. The first is always in units
specified by the user during input and the other one is always in mole amounts. Figure 6.7 shows
the original-units output. The results are (within round-off errors) the same as obtained in
Example 6.2; 1000 g of HCl consumes 214.9 g Si.

310 Chapter 6 Reactive Material Balances

P (atm) 1 1 11 1 11

T(K) 273.15 273.15 273.15 273.15 273.15 273.15 273.15
Str-unit Mass (kg) Mass (kg) Mass (kg) Mass (kg) Mass (kg) Mass (kg) volume (m3)

Spec-unit Mass (kg) Mass (kg) Mass (kg) Mass (kg) Mass (kg) Mass (kg) Volume (m3)

Str-name Si Feed HCI ChlorGas SiCILiq SiCIGas AbsdGas HydrGas

Streams 1 2 3 4 5 6 7 R#1 R#2 R#3 R#4
Flow ? 1000 ? ? ? -1
? ? -1 -2 -0.5
-3 1 -1
Si ? -1 1.5 1
HCI ? ? ?? 1
-4 1
? ?
H2 262.8 2
? 598.7
SÌCI4 ? 148.2 1

SÌCI3 ? 144.8

SÌCI2 ? ??

SÌHCI3

Si Chlorinator (RX) SiChlor Condenser (RX) SÌHCI3 & HCI Absorber (RX)
Instreams Outstreams Reactions Instreams Outstreams Reactions Heat
Instreams Outstreams Reactions Heat Heat
34 56
13 1 5 7

22

3

4

Figure 6.6 Initial setup array created by FlowBal for the chlorination of silicon and subsequent
stream processing. Basis changed to 1000 kg of HCI in. All devices were designated reactors,
although the condenser and absorber could just as well have been designated mixers. The stream
temperatures and pressures have no effect on the balance except for S7, which was left at STP
conditions.

P (atm) 1 1 1 1 1 1 1

T(K) 273.15 273.15 273.15 273.15 273.15 273.15 273.15

Str-unit Mass (kg) Mass (kg) Mass (kg) Mass (kg) Mass (kg) Mass (kg) Volume (m3)

Spec-unit Mass (kg) Mass (kg) Mass (kg) Mass (kg) Mass (kg) Mass (kg) Volume (m3)

Str-name Si Feed HCI ChlorGas SiCILiq SiCIGas AbsdGas HydrGas

Streams 1 2 3 4 5 6 7

Flow 214.8 1000 1214.8 891.7 323.1 299.5 262.8

Si 214.8 0 0 0 0 0 0

HCI 0 1000 88.1 0 88.1 88.1 0

H2 0 0 23.6 0 23.6 0 262.8

SÌCI4 0 0 598.7 598.7 0 0 0

SÌCI3 0 0 148.2 148.2 0 0 0

SÌCI2 0 0 144.8 144.8 0 0 0

SÌHCI3 0 0 211.4 0 211.4 211.4 0

Figure 6.7 FlowBaPs results array in the original input units for the chlorination of silicon. One
kg of HCI will chlorinate 0.215 kg of Si.

Sometimes a process has stream analyses reported in mixed units. FlowBal uniquely suited to
this situation because it can handle systems where each stream is expressed in a different set of
units. We illustrate the difference between making an atomic element balance and a FlowBal
balance on a coal gasification system with mixed stream units. The steam, coal, and ash streams
are in mass units, while the offgas stream is expressed in volume units. The MMV-C program can
convert mass to amount units to obtain a consistent set of units for a system. Figure 6.8 shows the
flowsheet. The material balance objective is to determine the mass of steam that will produce an
offgas with <jpCO + <jpH20 = 90 %. Table 6.5 gives the composition of the coal in mass units.

Table 6.5 Coal composition to gasifier. wFeS2 wSi02 wH20
3 6 8
wC wU wN
76 5 2

Chapter 6 Reactive Material Balances 311

Steam 1 Gasifier ungas
Ash
Coal 2^ (1200 °C,

F2 = 20 kg/s 1.5 atm)

Figure 6.8 Steam gasification of coal to produce a feed gas for iron ore reduction. Dashed lines

depict gas phase streams, while solid lines are solid phase streams. The ash consists of Si02,
Fe0.877S, and unburned carbon; wC = 8.0 %. To minimize complexity, the offgas was assumed to
consists of H2, H20, CO, C02, N2, and H2S, but in fact, small amounts of S02 and S2 will also be
present.

The Si02 and N2 are inert in the system, hence the system has six reactive elements. There
are nine reactive species, so NIRx = 3. Three independent reactions are shown as Equations
[6.18], [6.19], and [6.20]. You may recognize Equation [6.19] as the water-gas shift reaction
(WGR), which we introduced in Chapter 5. At 1200 °C, the WGR very likely approaches
equilibrium. FREED's Reaction tool shows that Кщ for the WGR at 1200 °C = -0.41.

C(graph) + 2H20(g) - 2H2(g) + C02(g) [6.18]

COfe) + H20(g) - H2(g) + C02(g) [6.19]

FeS2feyr) + aU2(g) - bFe0.siiS(c) + aH2S(g) [6.20]

A slight difficulty exists in getting stoichiometric balancing numbers for three species in
Equation [6.20]. The value of b is 1/0.877 « 1.140, while the value of a is 2 - 1/0.877 « 0.86.

Fortunately we don't need these values when making an atomic species balance.

Usually the best way to handle mixed units systems is to convert to mole or mole fraction
units, make the material balance, and convert back to the initial units. This is the procedure used
here, where MMV-C converted the coal feed to mole fraction units. Table 6.6 shows the results.
The amount units for hydrogen and nitrogen were converted to diatomic gases because both are
present diatomically, combined or not.

Table 6.6 Amount units for coal composition to gasifier as calculated from MMV-C.

n Coal xC xH2 xN2 xFeS2 xSi02 xH20
1.8897 0.6697. 0.2625 0.00756 0.00265 0.01057 0.0470

The hurdle for an atomic species balance is that the ash contains wC = 8.0 %. However, the
mass of the other two ash species (Si02 and Feo.swS) are easily calculated from the coal
composition, and then converted to amount units. The amount of ash is 0.03771 moles, and its
composition is:

xSi02 = 0.5297; xFeo.svvS = 0.1512; xC = 0.3191

With this information, we don't need a silica or iron balance, but we still need five element
balances and three SR equations to solve for two flow amounts and the six S3 compositions. One
SR is simply that the stream mole fractions must sum to 1. The other two SR are that the gas
composition must conform to the WGR equation, and that the reducing gas content xCO + xH20 =
0.90. The five atomic species balance and three SR equations are listed below in a format that is
conducive for Solver use.

С balance: 1.8896(0.6697) - S3 flow(x3CO + x3C02) - 0.03771(0.3191) = 0

H2 balance: SI flow + 1.8896(0.2625 + 0.047) - S3 flow(x3H2 + x3H2S + x3H20) = 0

N2 balance: 1.8896(0.00756) - S3 flow(x3N2) = 0

S balance: 1.8896(2)(0.00265) - S3 flow(x3H2S) - 0.03771(0.1512) = 0

312 Chapter 6 Reactive Material Balances

О balance: SI flow + 1.8896(0.047) - S3 flow(x3H20 + (JC3CO + 2x3C02) = 0
Σ mole fractions: x3CO + JC3C02 + x3H20 + x3H2S + x3H2 + x3N2 - 1 = 0
WGR: 0.41(x3H2O)( x3CO) - (x3H2 )(JC3C02) = 0
Reducing gas content: x3CO + x3H2 - 0.90 = 0
Solver found a solution as shown in Table 6.7.

Table 6.7 Results of atomic species material balance calculation for steam and offgas flow, and
composition of offgas (voi pet).

Sl-flow(H20) S3 flow (OG) S3H2 S3N2 S3H2S S3H20 S3 CO S3C02
1.479 moles 3.332 moles 54.38% 0.43% 0.13% 7.44% 35.62% 2.00%

The FlowBal version of the process is set up in the usual way, as shown in Figure 6.9.

P(atm) 5 1.6 1.5 1.5

T(C) 900 35 1200 1200
Str-unit Mass (kg) Mass (kg) Volume (m3) Mass (kg)

Spec-unit Mass pet Mass pet Volume pet Mass pet

Str-name Steam Coal Offgas Ash

Streams 1 2 3 4 R#1 R#2 R#3
? ? ? -1
Flow 20 2

С 76 8 -2
H2 5 ? 1
N2 2 ? 1 -0.85975

FeS2 3 ? -1

Si02 6 ? 1.140251
0.859749
Fe.877S -1
H2S ? -1
H20 ? ? ? 1
CO ?
C02 ?

Figure 6.9 Starting array for a FlowBal material balance on gasifying 20 kg/s of coal. FlowBal
checks that all reactions are balanced. To avoid an error message caused by entering an irrational
number value via the keyboard, enter instead a formula in the R #3 columns for the stoichiometric
coefficients for H2, Fe.swS and H2S. For example, enter = -(2 - 1/0.877) for H2. Notice the stream
units, and the Celsius temperature. The FlowBal details for the gasifier device (a reactor) are not
shown.

FlowBal wrote 12 equations for the eleven stream variables and included three reaction rate
factors (R-R), so DOF = +2. We use two other SR equations, one of which relates the stream
compositions to Ä'eq for the WGR, and one which expresses the reducing gas content. Both of
these equations are entered with the Insert Equation tool. FlowBal comments these equations as
shown below.

{S3-H2} + {S3-CO} - 9(T| |{S3-CQ2}*{S3-H2} ■ 0.41*{S3-H2O}*{S3-COy|

If the setup units for S3 had been in mass fraction or mass pet instead of volume pet, you
would have had to convert them to volume or moles beforehand. Even though the Кщ expression
requires partial pressure, the setup volume units are acceptable because the WGR has no volume
change. Otherwise, volume units can be converted to partial pressure on the Insert Equation dialog
box by clicking the PP box. However, FlowBal cannot convert mass units to partial pressure. This
is particularly important when entering a A^eq expression equation for a reaction that has a change
in volume.

Chapter 6 Reactive Material Balances 313

Initially, FlowBal couldn't find a solution to the equation set. The most likely reason is that
the WGR direction is wrongly written. To test this hypothesis, return to the Solver Options screen,
and uncheck the default Assume Non-Negative box. This allows R-R2 to be a negative value.
This resulted in a solved balance, as shown in Figure 6.10. An alternative is to use the built-in
FlowBal solver, which constrains stream property values to non-negative, but not the R-R values.
In any case, the result is the same as obtained by the atomic species balance method.

P (atm) 5 1.6 1.5 1.5

T(C) 900 35 1200 1200 Stream Amounts (kg-mol)
Str-unit Mass (kg) Mass (kg) Volume (m3) Mass (kg)
Spec-unit Mass pet Mass pet Volume pet Mass pet
Coal Offgas Str-name Steam Coal Offgas Ash
Str-name Steam Ash Streams 1 2 34
Streams 2 3
1 4 Flow 1.4793 1.8896 3.3319 0.0377

Flow 26.65 20 268.52 1.81 С0 1.2655 0 0.0120
H2 0 0.4961
С0 76 0 8 N2 0 0.0143 1.8119 0
FeS2 0 0.0050
H2 0 5 54.38 0 SÌ02 0 0.0200 0.0143 0
N2 0 2 0.429 0
FeS2 0 3 0 Fe.877S 0 0 00
0 0
Si02 6 0 66.42 H2S 0 0.0888 0 0.0200
0 H20 1.4793 0 0 0.0057
Fe.877S 0 0 0 25.58 CO 0
0 0.1290 0 C02 0 0 0.0043 0
H2S 0 7.44
H20 100 8 0 0.2480 0
1.1869 0
CO 0 0 35.62 0 0.0666 0
0 1.999 0
C02 0

Figure 6.10 Results of FlowBal material balance on one second of operation of a coal gasification
system. The mass of steam required is about 1.3 times the mass of the coal, which itself has 8 %
moisture. S3 volumes are ambient.

FlowBal also calculates the rates of each reaction (R-RN) as part of the material balance
calculation method. These values can be used to calculate the extent of reaction of a species. R-Rl
was calculated at 1.2535, which means that the rate of carbon consumption was 1.2535 mol/s. The
amount of entering (coal) carbon was 1.2655 mol/s, so XRC = 0.99. The unreacted carbon exits at
0.012 mol/s in the ash. If you didn't know the composition of the ash, you'd enter a ? for the wC
in the ash. This increases the DOF, so you must then enter another relationship. For example,
enter a value for XRC for reaction 1. FlowBal will then calculate the wC in the ash.

FlowBal's Repetitive Solve tool can repeat the material balance for different values of
certain stream properties. For example, suppose you wanted to know the relationship between the
reducing gas composition and the amount of steam added. One useful reducing gas composition
variable is the sum of <pCO + фН20. There is more than one way to set FlowBal to accomplish
this. One procedure is outlined below:

• Return to the setup screen, and enter a steam flow rate higher than the value previously
calculated by FlowBal (see Figure 6.10), such as 30.

• Read the data, and note that only one equation is required. Insert the previous equation for
Keq of the WGR. This equation will appear in cell E41.

• Create a second equation in the Constructed Equation screen that sums the S3-CO and S3-
H2. You need to insert this equation on the worksheet in a column A cell. Do this by
clicking the Insert in Arbitrary Cell screen, and pick for example cell A29. This is
designated by Excel as Rw29Cll. Then click on Insert Equation. You now have an
equation inserted in cell E41 and in A29:

{S3-CQ2}*{S3-H2> - 0.41*{S3-H2O}*{S3-CO> [ [{S3-CO} + {S3-H2} |

• Solve the system. Look at the value in cell A29, which should be 85.24. This says that
with 30 kg/sec of steam, the % CO + % H2 will be 85.24. Do not enter a target value.

314 Chapter 6 Reactive Material Balances

• Click on Repetitive Solve. Enter the Stream 1 flow as the independent variable. Set the
range from 30 to 20, with 11 solves. For output, select Arbitrary, and pick cell A29. Click
the arrow to insert this cell's address into the screen.

• Click OK to solve the system. FlowBal will repeat the material balance calculation at 11
different steam flows, and display the results in a table and chart. Also, FlowBal will fit
the results to a quadratic equation. Table 6.8 shows the results.

Table 6.8 Repetitive solve material balance result as a function of the flow of steam (SI-Flow)
and the <pCO + <pH20 in the offgas (row 29 column 1). The lower limit on steam flow is actually
about 21 kg/sec. FlowBal converged at 20 kg/sec to an impossible result because the Assume
Non-Negative box was unchecked. FlowBal will not do this if you use the built-in FlowBal solver.

Sl-Flow 30 29 28 27 26 25 24 23 22 21 20
Rw29CU 85.2 86.6 88.0 89.5 91.0 92.5 94.2 95.8 97.6 99.4 101.2

The advantages of FlowBal are more apparent as the number of devices and process
restrictions increase. The following examples extend FlowBal's application to such systems.

EXAMPLE 6.3 — Application of FlowBal to Stack Gas Desulfurization.

A coal-fired power plant consumes 108 kg/min of coal and produces 3690 m3/min of
combustion-product gas at 1100 К having the following composition:

<pC02 = 0.152; φΗ20 = 0.061; φ02 = 0.026; <pS02 = 0.0031; bai. N2

Their air pollution permit restricts the total emission of S02 to 1.5 kg/min, which means that
the combustion gas must be desulfurized. The plant uses a fluidized-bed limestone scrubber that
removes 95 % of the S02 from the gas as CaS04. The high scrubber efficiency means that not all
of the stack gas needs to be desulfurized; part can be vented. Figure 6.11 shows a sketch of the
process. Excess CaC03 is used such that the spent solid has wCaC03 = 0.20. Start FlowBal, and
follow the User's Guide instructions to make a material balance on the desulfurization process.

Limestone (CaC03) 6 Clean gas
(lowS02)
Combustion gas Vent gas De
rea 5 » Spent solid
F1 = 3690 m3/min ▲ (CaS04 + CaC03)
(C02, H20, 02
S02, N2) De-S gas I

Figure 6.11 Flowsheet for the desulfurization of stack gas. Gas streams are shown as dashed
lines, and solid streams as solid lines.

Solution. The chemical reaction taking place in the desulfurization unit is:

СаСОз(с) + S02(g) + V202(g) - CaS04(c) + C02(g) [6.21]

There are two special constraints on the material balance. The first is the removal of 95 % of
the incoming S02 in the desulfurizing reactor, which means that the extent of S02 reaction = 0.95.
The second is that the mass of S02 in stream 2 + mass of S02 in stream 6 = 1.5 kg/min. These
constraints suggest that the stream units should be: SI, volume percent. S2 - S4, mass. S5, mass
percent. S6, mass. The only stream where temperature is involved in the material balance is SI,
but the other stream temperatures were entered for informational purposes. Figure 6.12 shows the
input FlowBal array for the system.

Chapter 6 Reactive Material Balances 315

P(atm) 111111

T(K) 1100 1100 1100 300 980 980

Str-unit volume (m3) Mass (kg) Mass (kg) Mass (kg) Mass (kg) Mass (kg)

Spec-unit Volume pet Mass (kg) Mass (kg) Mass (kg) Mass pet Mass (kg)

Str-name CombGas VentGas De-S Gas L-stone SpntSId CleanGas

Streams 1 2 3 4 5 6 R#1
1
Flow 3690 ? ? ? ? ?
-0.5
C 0 2 15.2 ? ? ? -1

H20 6.1 ? ? ? -1
1
0 2 2.6 ? ? ?

S 0 2 0.31 ? ? ?

N2 ? ? ? ?

СаСОЗ ? 20
CaS04 ?

Gas Splitter (SP) Heat I Gas Desulfurizer (RX) Heat
Instreams Outstreams Split Fract Instreams Outstreams Reactions

12? 351
3? 46

Figure 6.12 Input setup array created by FlowBal for the desulfurization of stack gas.

FlowBal defined 24 unknowns and created 22 equations. Two more equations can be
generated from the process specifications: the extent of reaction (0.95) and an equation relating
the 1.5 kg limit for S02 for S2 and S6. This completes the data entry for the system, and FlowBal
found a solution. Figure 6.13 shows the result.

Based on the amounts table, the combustion gas contains 8.13 kg of S02 and 81 % was
captured. Desulfurization requires 13.9 kg/min of limestone, and about 14% of the combustion gas
can be vented. MMV-C can convert the mass or amount units to volume units. Alternatively, it
can calculate the volume composition from the amount table that FlowBal creates. The following
table shows the S6 composition (volume units).

co2 н2о o2 so2 N2
75.90%
15.52% 6.11% 2.46% 0.016%

P (atm) 1 1 1 1 1 1

T(K) 1100 1100 1100 300 980 980

Str-unit Volume (m3) Mass (kg) Mass (kg) Mass (kg) Mass (kg) Mass (kg)

Spec-unit Volume pct Mass (kg) Mass (kg) Mass (kg) Mass pet Mass (kg)

Str-name CombGai5 VentGas De-S Gas L-stone SpntSId CleanGas

Streams 1 2 3 4 5 6

Flow 3690 174.3 1054.2 13.9 17.6 1050.5

C02 15.2 38.8 234.7 0 0 239.2

H20 6.1 6.4 38.6 0 0 38.6

0 2 2.6 4.8 29.2 0 0 27.5

S02 0.31 1.15 6.97 0 0 0.35

N2 75.79 123.1 744.8 0 0 744.8

СаСОЗ 0 0 0 13.9 20 0

CaS04 0 0 0 0 80 0

Figure 6.13 FlowBal results array in original units. FlowBal also creates a second table (not
shown) in amount units.

Assignment. A new emission requirement went into effect that required removal of 90 % of the

incoming S02. Calculate the mass of limestone required, and the fraction of combustion gas that
can be vented.

316 Chapter 6 Reactive Material Balances

The complexity of material balance calculations increases with the number of devices and
chemical reactions. For example, the reforming of natural gas discussed earlier involves several
reactions. Please see Example 5.9 for one way to treat this process. Here we show the difference
between doing the arithmetic with Excel and using FlowBal; the materials balance will be solved
both ways. The NG composition is: <pCH4 = 0.92; <pC2H6 = 0.05; bai. N2. Three devices are
involved, two reformers and a condenser, with a different catalyst and temperature in each
reformer. Extra steam is used in Reformer I to prevent carbon deposits. The extra water requires a
condenser between the two reformers. The process is carried out at P = 2.2 atm. Figure 6.14
shows a flowsheet for the process.

NatGas 1 . 3 Condenser 5* 6/* ►СRг efgoarms ed

= 200 n/min Reformer I ψ (299 K) Dry Reformer II
(1100 K) gas (1170 K)
Steam 2 RfGas

2 мм . 4 1 Water
= 350 n/min

Figure 6.14 Flowsheet for the production of a reduction gas by steam reforming of natural gas.

Several reactions can be written to describe the steam reforming of natural gas hydrocarbons.
There is no a priori way to know which reactions predominate, or which take place first.
Ordinarily, since NIRx = 3, only three reactions would be needed, but sometimes data is available
in a form that requires more than the NIRx number. Here, suppose that the process occurring in
Reformer I can be characterized by estimates of the XR for ethane and methane reactions, and the
process occurring in Reformer II can be characterized by a different XR for methane, plus the
WGR. This requires four chemical reactions. Reaction #4 is the WGR, and proceeds to
equilibrium, but at first, we don't know which way it goes.

C2H6(g) + 2H20(g) - CH4(g) + C02(g) + 3H2(g); X7?C2H6 = 1.00 [6.22]

CH4(g) + 2H20(g) - C02(g) + 4H2(g); XRъKXэZаC, H4 = 0.65 [6.23]
CH4(g) + C02(g) -> 2CO(g) + 2H2(g); Xi?KXJCH4 = 0.75 or 0.30 [6.24]

CO(g) + H20(g) — C02(g) + H2(g); Keq for WGR at 1170 K = 0.800 [6.25]

The first three reactions are assumed to occur sequentially in Reformer I. First, C2H6 is

completely consumed via Equation [6.22] to form CH4, C02 and H2. Then CH4 is partially reacted

according to sequential Equations [6.23] and [6.24], first by steam oxidation, and next by C02
oxidation. JÖ?Rx2CH4 is 0.65 and XRRx3Ctì4 is 0.75. This gas (stream 3) passes to a condenser at

299 K, where most of the H20 is removed (stream 5 is at water vapor saturation). Then it passes to

a hotter reactor (Reformer II) where additional reactions occur in sequence. Some of the
remaining CH4 reacts according to [6.23] at XR^CIU = 0.30, while the CO, H20, H2 and C02

react via the WGR (reaction 4) to equilibrium between these gases (not involving CH4). This

multi-reaction set involves sequential reactions [6.22] - [6.24] in Reformer I, and sequential

reactions [6.23] and [6.25] in Reformer II. If the reformers were operated above about 1250 K, the
XRCR4 would approach one, and we could safely assume the absence of CH4 in S3 and S6.

Remember that when using extent of reaction criteria, reactions are assumed to proceed in the
sequence written, to the defined species extent before the next reaction takes place. Writing the
reactions in the appropriate order makes the amount of each species reacted in each unit
unambiguous. When the XR factor is not involved, it makes no difference in FlowBal which set of
independent reactions is chosen or their order. Even so, it's helpful to use the physical nature of
the system to select the choice of independent reactions. However, the reaction direction might
make a different because the default for the Excel Solver is set to assume non-negative, and if you
write the reaction direction wrongly, the R-R will be negative, and Solver won't find a solution.

Although four reactions and seven species are involved, the system is easy to solve because
there are no recycle or counter-current flow streams. A molecular species material balance was

Chapter 6 Reactive Material Balances 317

used. A ledger was set up for each unit to list the amount of each species in, amount reacted,
amount formed, and amount out. Equation [2.15] was used to calculate that/?H20 in stream 5 =
0.0323 atm. At P = 2.2 atm, stream 5 has xH20 = 0.015.

The DOF for each unit = 0, so we can proceed unit by unit, reaction by reaction, to reach the
final step — the WGR. The rather large amount of C02 (90.3 moles) relative to the amount of
H20 (2.7 moles) in Reformer II indicates that Equation [6.25] proceeds in the reverse direction to
that written. The post-WGR gas composition was calculated with the aid of Goal Seek and the R-R
concept. First, a guessed R-R value (50) was entered in the Reaction #4 WGR table for C02 (bold
font in Figure 6.15), and worksheet formulae written to copy this value to the "reacts" column for
H2 and the "forms" column for H20 and CO. Next, a formula was written for the WGR expression
in a nearby cell. Finally, Goal Seek varied the R-R value for C02 until the correct Кщ value was
reached (0.8). Figure 6.15 shows the completed Excel-calculated ledger for the process.

Reactions and Amounts in Reformer I Condensation of Water Vapor

Reaction #1: ХДС2Н6 = 1 /?H20 = 0.033 atm, xH20 = 0.015

Species In Reacts Forms Out Species In Cndnes Out
с2н6 10
10 0 0 H20 77.8 64.91 12.89

H20 350 20 0 330 CH4 16.98 0 16.98
0 85.18
CH4 184 0 10 194 co2 85.18 0 636.25
0 10 10
co2 0 0 30 30 H2 636.25

H2 0 CO 101.85 0 101.85

CO 0 0 0 0 N2 6 0 6

N2 6 0 06 Total 924.05 859.137
Total 550 570

Reactions and Amounts in Reformer II

Reaction #2: XR CH4 = 0.65 Reaction #2: ХЯСН4 = 0.3

H20 330 252.2 0 77.8 Species In Reacts Forms Out
67.9
CH4 194 126.1 0 136.1 H20 12.89 10.19 0 2.70
co2 10 534.4
0 126.1 CH4 16.98 5.09 0 11.88
0
H2 30 0 504.4 6 co2 85.18 0 5.09 90.27
822.2
CO 0 0 0 H2 636.25 0 20.37 656.62

N2 6 0 0 CO 101.85 0 0 101.85

Total 570 N2 6 0 0 6

Total 859.137 869.322

Reaction #3: ЛЖСН4 = 0.75

H20 77.8 0 0 77.8 Reaction #4: WGR. Net Out
16.98
CH4 67.9 50.93 0 85.18 H20 2.70 0 72.36 75.06
636.25 0 11.88
C02 136.1 50.93 0 101.85 CH4 11.88 0 0 17.91
0 584.258
H2 534.4 0 101.85 6 C02 90.27 72.36
924.05 72.36 174.212
CO 0 0 101.85 H2 656.62 72.36 06
869.322
N2 6 0 0 CO 101.85 0

Total 822.2 N2 6 0

Total 869.322

Figure 6.15 Material balance ledger for the steam reforming of natural gas, based on species
balance with specified reaction extent, and WGR equilibrium being reached in Reformer II. Bold
font in the WGR portion of the ledger represents location of original estimate for the R-R value
used for a Goal Seek solve of the WGR.

318 Chapter 6 Reactive Material Balances

The application of FlowBal to this system starts as usual. First, identification of the number
of species, streams and chemical reactions involved . Next, definition of each device and the
instreams and outstreams, and specification of the chemical reaction taking place in each device.
Since we are using XT? as a criteria, the reactions must be listed in the assumed order of occurrence.
(Please review the FlowBal User's guide for more information on effect of reaction sequencing
when using the XR concept.) Once this initial data is entered, FlowBal presents a starting array
where the user can enter known stream flows and compositions.

The next steps are reading the data and entering the necessary material balance equations. As
expected, there are four fewer equations than unknown variables which requires four system
constraints to obtain DOF = 0. These are the three reaction extents and the value of 0.800 for Кщ
of the WGR. FlowBal has a specific entry button for the XR values, and a general-purpose Insert
Equation feature that is used here for Кщ of the WGR (but can be used for any equation
relationship involving process flows or composition). Since we now know the correct WGR
direction, we can write it properly for FlowBal. To avoid using a ratio for the WGR, enter S6-
C02*S6-H2 - 0.8*S6-CO*S6-H2O.

Notice that we did not need to specify the X/?C2H6 (reaction #1 in Reformer I). This is
because when X7?(C2H6) = 1, we don't place a ? in the C2H6 composition column for stream 3.
FlowBal recognizes the blank C2H6 composition cell to mean no C2H6 in stream 3 (an ХЛ(С2Н6) =
1 condition), so no XR variable was needed. But had we wrongly placed a ? in the C2H6
composition column for stream 3, the system would have had fewer equations than variables, and it
would have been necessary to enter a ZRC2H6 expression, and enter a 1 for the XR value.

Once the equations are correctly in place, Solver can be activated. Figure 6.16 shows the
results. As expected, they are identical to the ones obtained by the Excel arithmetic that was
shown in Figure 6.15. FlowBal would not have converged if we hadn't reversed the WGR
direction. If we don't know beforehand which direction a reaction proceeds, we simply make a
guess, and try to solve the system**. If convergence fails, you can go back and uncheck the
Assume Non-Negative box, or change the signs on the equation coefficients in the FlowBal input
array. Another option is to use the built-in FlowBal solver, which forces stream variables to be
non-negative, but omits that restriction for the R-R values.

In this example, we used the WGR reaction as one of those in the independent set, and also
used the Кщ expression for the WGR as an inserted equation. However, it isn't necessary to use
the WGR as one of the independent reactions to use the Кщ expression for the WGR as an inserted
equation. You can require any stream to conform to а Кщ expression, irrespective of whether or
not the reaction is part of the independent set for the reactor.

FlowBal is particularly useful for making a material balance on a system with a recycle
stream because recycle makes it impossible to solve a system balance device-by-device. Let's look
at the way FlowBal's Repetitive Solve feature can indicate certain limitations to the amount of
recycle a process can have. The application is the reprocessing of spent catalyst material, which
contains valuable material that cannot be discarded.

A material balance is the first step in designing a process for recovering the valuable metals
from a spent catalyst. The catalyst eventually becomes contaminated with sulfur, carbon, and
minor impurities, and must be replaced. The catalyst contains nickel and molybdenum on an
alumina substrate. The first step is oxidation, which burns off the impurities, and converts the
nickel and molybdenum to NiO and Mo03. Next, the two oxides are separated and recovered by
chlorination and re-oxidation/precipitation. The outlet gas needs further treatment in a separate
process. This example concerns the metal separation and recovery part of the process according to

If you prefer, you may clarify the nature of stream #4 by including a fifth reaction for the condensation of
H20(g) to H20(/) in the condenser. FlowBal recognizes the species descriptors for different phases of the
same species. This is optional unless a heat balance is to be made.
** Note that the earlier Excel solution showed that the WGR went in the opposite direction to the direction it
was written as Equation [6.21], but usually you would not know that.

Chapter 6 Reactive Material Balances 319

the flowsheet shown in Figure 6.17. FlowBal can make repetitive calculations on the effect of
changing the amount of recycle gas. This sort of process modeling, or process simulation, is quite
common in industry. Known chemistry with some lab tests is used to design a process "on paper",
after which calculations can be made to determine the effect of process changes.

P (atm) 2.4 2.4 2.3 2.2 2.2 2.1

T (K) 320 550 1100 299 299 1170

Str-unit Amt (kg-mol) Amt (kg-mol) Amt (kg-mol) Amt (kg-mol) Amt (kg-mol) Amt (kg-mol)

Spec-unit Mol frac Mol frac Mol frac Mol frac Mol frac Mol frac

Str-name NatGas Steam RfGas Water DryGas RfmdGas

Streams 1 2 3 456

Flow 200 350 924.1 64.9 859.1 869.3

C2H6, (g) 0.05 0 0 000

CH4, (g) 0.92 0 0.0184 0 0.0198 0.0137

N2, (g) 0.03 0 0.0065 0 0.0070 0.0069

H20, (l,g) 0 1 0.0842 1 0.0150 0.0863
CO, (g) 0 0 0.1102 0 0.1185 0.2004
0 0 0.0922 0 0.0991 0.0206
C02, (g) 0 0 0.6885 0 0.7406 0.6721
H2, (g)

Stream Amounts (kg-mol)

Str-name NatGas Steam RfGas Water DryGas RfmdGas
3 4 5
Streams 1 2 6
924.1 64.9 859.1 869.3
Flow 200 350 0 0 0
0 0
C2H6, (g) 10 0 16.98 0 16.98 11.88
6 6
CH4, (g) 184 0 64.91 6
77.80 0 12.89 75.06
N2, (g) 6 0 101.85 0 101.85 174.21
85.18 0 85.18 17.91
H20, (l,g) 0 350 636.25 636.25 584.26

CO, (g) 0 0

C02, (g) 0 0

H2, (g) 0 0

Figure 6.16 Results of FlowBal material balance calculation for natural gas reforming.

InChlr j. Mixer Chlorine Recycle gas F = ?
F

Precipitator

Catalyst -* (200 °C)

F 2 = 2040 kg/min Chlor solid F3 = ? 5| 6
wNiO = 0.219 wNiO = ? ; Air
wMo03 = 0.282 wNiCI2 = ? F5=?
wAI203 = ? wAI203 = ?

Figure 6.17 Flowsheet diagram for treating spent catalyst after a preliminary oxidation step. Solid
lines represent flow of solid-phase streams, and dashed lines for gas streams. Catalyst is first
roasted (not shown), then chlorinated to convert NiO(c) to NiCl2(c), and Mo03(c) to Mo02Cl2(g).
The Mo02Cl2 is oxidized to Mo03 in the precipitator. The alumina substrate does not chlorinate in
the presence of oxygen, hence it is entirely discharged in stream 3. Air is added to the precipitator
to initiate the re-oxidation reaction and provide a suitable p 0 2 in the precipitator.

Two chemical reactions occur. Equation [6.26] and [6.27] take place in the chlorinator, and
the reverse of Equation [6.27] takes place in the precipitator.

320 Chapter 6 Reactive Material Balances

NiO(c) + Cl2(g) -> NiCl2(c) + l/202(g); log(^eq) = 3390/Γ- 3.05 [6.26]

Mo03(c) + Cl2(g) -+ Mo02Cl2(g) + x/202(g); log(^eq) = -5700/Г + 6.9 [6.27]

Кщ for [6.26] is about 40 at 450 °C, hence in the presence of excess chlorine, the extent of
NiO reaction should be one. However, owing to a film of NiCl2 that forms on the alumina
substrate, lab tests showed that only about 90% of the NiO reacts in the allotted time. Кщ for
[6.27] is about 0.1 at 450 °C, but if the molar ratio of Cl2/02 in the chlorination gas is high enough,
virtually all of the M0O3 should be chlorinated. Кщ for [6.27] is a strong function of temperature,
being about 7 x 10-6 at 200 °C, thus strongly favoring its reverse at that temperature. To enhance
the M0O3 chlorination, the pC\2lp02 in S4 should be two. Furthermore, to enhance Mo02Cl2(g)
oxidation, the pCl2/p02 in S7 should be four. Even so, lab tests showed that owing to kinetic
limitations, XRMo02Cl2(g) is typically 0.96. The table below lists the four system constraints.

Gas stream constraints Extent of reaction constraints
pCl2/p02 in S4 = 2.0 XRNiO(c) in Chlorinator = 0.90
pCl2/p02 in S7 =4.0 XRMo02Cl2(g) in Precipitato^ 0.96

The objective of making a material balance is to determine the amount of chlorine and air
required, and calculate the overall recovery of M0O3 in S6 at different splitter fractions for S8 vs.
S9. As a check on the FlowBal setup, the split fraction to S9 was initially set at 0.05.

As usual, FlowBal requires stream data, species selection and reactor details. The stream
conditions are taken from process data, using as-measured units. Note that the third FlowBal
reaction is the reverse of Equation [6.27]. A 5 % recycle gas flow of S7 to S9 was used in the first
FlowBal test.

FlowBal wrote 31 material balance equations for the system, and listed 35 unknowns. Two
equations were added by specifying XRNiO(c) chlorination and XRMo02C\2(g) oxidation. Two
more equations were written to express the pC\2/p02 constraints for S4 and S7. Since these
constraints are simple ratios, it makes no difference if the species units are partial pressure, volume
fraction, or mole fraction. Figure 6.18 shows the converged FlowBal solution.

P (atm) 1.1 1.1 1.05 1.05 1.1 1.05 1.05 1.05 1.05 1.05

T(K) 305 305 723 723 302 473 473 473 473 320

Str-unit Volume (m3) Mass (kg) Mass (kg) Volume (m3) Volume (m3) MaSS (kg) Volume (m3) Volume (m3) Volume (m3) Volume (m3)

Spec-unit Volume pet Mass pet Mass pet Volume pet Volume pet Mass pet Volume pet Volume pet Volume pet Volume pet

Str-name Chlorine Catalyst Chlr Solid O-ChlrGas Air Pptr Solid Pptr Gas Exit Gas Rcycl Gas InChlr Gas

Streams 1 2 3 4 5 6 7 8 9 10

Flow 419.1 2040 1760.3 1055.1 48.7 553.4 699.0 664.1 35.0 484.3

AI203 0 49.9 57.8 0 0 0 0 0 0 0

NiO 0 21.9 2.5 0 0 0 0 0 0 0

МоОЗ 0 28.2 0 0 0 100 0 0 0 0

NÌCI2 0 0 39.6 0 0 0 0 0 0 0

CI2 100 0 0 52.0 0 0 71.7 71.7 71.7 98.6

02 0 0 0 26.0 21 0 17.9 17.9 17.9 0.88

N2 0 0 0 0.48 79 0 9.5 9.5 9.5 0.46

Mo02CI2 0 0 0 21.4 0 0 0.85 0.85 0.85 0.0

Figure 6.18 Results of FlowBal material balance on spent catalyst processing flowsheet for 5%

recycle gas flow. Volume units are actual, not STP. The catalyst contained 575.3 kg of Mo03,
while the process recovered 553.4 kg in S6. The Mo02Cl2 lost via S8 represents 3.8 % of the input
molybdenum amount.

The effect of varying the splitter fraction was explored by using FlowBal's Repetitive Solve
feature. The user can select any arbitrary variable (here, split fraction to S9) and FlowBal will
solve the equations repeatedly for different values of the variable. Splitter fractions for S7 to S9
from 0 to 0.25 were used, with selected results shown in Figure 6.19.

Chapter 6 Reactive Material Balances 321

The results show some process improvement by recycling up to 27 % of the outgas back to
the chlorinator. The main advantage is to decrease the amount of Cl2 needed (SI). A more
effective process change would be to scrub the outlet gas to recover all of the Mo, then separate the
Cl2 and use it to replace new Cl2.

450 о Effect of Recycle ύ 4.1%
3.9%
ЙК оv 3.7%

10 15 20 25

% pptr gas recycled

Figure 6.19 Effect of increased recycle (S9) on selected process parameters. The amount of
recycle of the precipitator gas (S8 to S9) is limited to about 27% by the fact that some small
amount of air (S5) is required to initiate the oxidation of M0O2CI2 in the precipitator. The use of
maximum percent recycle (minimum air) cuts the Cl2 consumption by about 12 %, and decreases
the percentage of incoming Mo lost to the outlet gas to about 3.0 %.

6.3 Combustion Material Balances

Most materials processing require a source of energy. The most common source of thermal
energy is fuel combustion (Baukal 2001). A broad definition of a fuel means any substance
(organic or inorganic) that is oxidized to produce useful heat. Some examples of fuel are:

• A fossil fuel, which is a hydrocarbon derived from geologic processes. Coal, oil, and
natural gas are examples of fossil fuels.

• A manufactured or by-product fuel, which is any non-naturally-occurring substance
used for providing thermal energy. Hydrogen, coke-oven gas, and acetylene are examples of
manufactured fuels.

• An intrinsic substance, which is anything contained in the raw material that is oxidized
during extraction or refining. Sulfur in a metal sulfide concentrate is one example of an
intrinsic fuel, as is carbon and silicon oxidized in the production of steel.

Chemical fuels can be added to this list, such as iron and aluminum. These are burned in
oxygen lances to pierce refractory substances.

Fossil fuels are the most common source of thermal energy. The three main types are natural
gas, fuel oil, and coal. Natural gas and to a lesser extent fuel oil are mainly used for process
heating. Coal and natural gas are mainly used for the generation of electricity. When used for
electrical generation or process heating, the greatest efficiency is attained when the fuel is
completely combusted to C02 and H20. When fuels have a secondary purpose, such as control of
the oxygen potential in a furnace gas, combustion products will include CO and H2. The three

322 Chapter 6 Reactive Material Balances

main factors involved in assuring complete reaction between the oxidant and fuel are: time,
temperature, and turbulence. Sufficient time must be allowed for the chemical reactions to
proceed to completion. The temperature must be high enough for ignition and rapid reaction
kinetics. In addition, the fuel and oxidant must be well mixed to assure adequate oxidant/fuel
contact.

6.3.1 Material Balance for the Combustion of a Gaseous Fuel

Common gaseous fuels are refined natural gas, coke-oven gas, blast-furnace gas, and various
other manufactured fuels such as acetylene. The only fossil fuel in that list is natural gas, which
occurs as a mixture of hydrocarbon and other gases. The natural gas used by consumers may be
much different from the natural gas brought from underground to the wellhead. Although the
processing of natural gas is less complicated than the processing and refining of crude oil, it is
equally necessary before final use.

At the wellhead, natural gas is primarily methane, but other hydrocarbons (ethane, propane,
butane etc.) may be as high as 20 %. Impurities are C02, N2, H20, and H2S. Refining is required
to remove higher hydrocarbons and impurities before sending to customers. Hydrocarbons such as
propane and butane are partly removed and sold separately, and sometimes used as a process fuel
where natural gas is unavailable. Other impurities are also removed, especially H2S (once
extracted, H2S can produce sulfur, which is then also sold separately). After refining, pipeline-
quality natural gas is mostly methane. Table 6.9 shows the composition of a typical end-user
natural gas from two different sources. An increasing source of natural gas is imported liquefied
natural gas (LNG), which usually has a higher content of non-methane hydrocarbons and is nearly
devoid of non-hydrocarbons.

Table 6.9 Typical composition of major species in end-user natural gas. Top half of table from
North American wellhead source, and bottom half from LNG. Notice the absence of impurities in
the LNG.

CH4 (methane) C2H6 (ethane) СзНв (propane) C4Hio (butane) co2 N2

volume % 94.5 2.9 0.7 0.2 0.8 0.9

mass % 88.9 5.1 1.8 0.7 2.1 1.5

volume % 92.9 4.7 1.9 0.5 0.0 0.0
mass % 85.4 8.1 4.8 1.7 0.0 0.0

Natural gas can be quantified in different ways. As a gas, it can be measured by the volume it
takes up at specified conditions, commonly as ft3 or m3 at STP. A common volume measure is
thousands of cubic feet (Mcf), millions of cubic feet (MMcf), or trillions of cubic feet (Tcf).
Natural gas can also be quantified in terms of the energy it produces when burned. The fuel value
of natural gas in the U.S. is commonly measured and expressed in British thermal units (Btu). One
cubic foot of natural gas can deliver about 1,030 Btu. The gas utility may bill customers in
"therms". A therm is equivalent to 100,000 Btu, or the combustion energy of about 100 cubic feet
of natural gas. The SI unit is kJ/m3; alternatively, kWh/m3.

There is considerable interest in the preparation and use of synthetic gaseous fuels that are
prepared by the destructive volatilization of solids like coal, or (of more recent interest) biomass.
Syngas is rich in hydrocarbons, CO, H2, C02, and H20, and can be upgraded by removal of the
latter two species. The remaining char can be reacted with steam and oxygen to provide a syngas
rich in CO and H2. The most common syngas is produced by the devolatilization of coal in coke
ovens. The condensable substances from the coke oven effluent are a valuable source of organic
chemicals, and the remaining coke-oven gas is a valuable steelplant fuel.

Air is the most common combustion oxidant. A common approximation is that air contains
ф02 = 21.0 % but sometimes a more accurate analysis of air is required. Usually the amount/mass

Chapter 6 Reactive Material Balances 323

of Ne is added to the amount/mass of Ar, and the mass and volume fraction of the four listed gases
are summed to 100. The air composition in Table 6.10 is virtually identical to that presented in
Section 4.8 as calculated by MMV-C. The slight difference in the Ar composition between the two
tables comes from ignoring Ne in Section 4.8. For accurate calculations, the φ02 should be taken
as 20.95 % and the molar mass of air taken as 28.96 g. At STP, the density of dry air is 1.292 g/L
(0.08067 lb/ft3). (See Equation [1.11] for density at other conditions.) Table 6.11 shows how
these values are decreased by the presence of moisture in air. The pH20 calculations were made
using Equation [2.16].

Table 6.10 Composition of dry air; major species and elements.

species N2 o2 Ar C02 element N О Ar С
amount % 78.44 21.07 0.47 0.02
amount % 78.09 20.95 0.93 0.04
mass % 75.52 23.17 1.29 0.02
mass % 75.53 23.15 1.28 0.05

Empirical equations relating the properties of air to the dew point temperature are:

Volume fraction 0 2 in moist air = -6.18 x 10^(dpt)2 - 4.96 x 10"3(dpt) + 20.82 [6.28]
Mass of dry air/mass of moist air = -2.04 x 10~5(dpt)2 - 1.27 x 10^(dpt) + 0.996 [6.29]

Molar mass of moist air = -3.47 x 10^(dpt)2 - 2.36 x 10"3(dpt) + 28.89 [6.30]

Density, kg/m3 at STP = -1.55 x 10"5(dpt)2 - 1.05 x KT*(dpt) + 1.289 [6.31]

where dpt refers to the dew point temperature in °C. The above equations are based on the more
precise volume fraction of 20.95 %02 in air, but are not valid above 35 °C or below 0 °C. To
convert density to lb/ft3, multiply by 0.062427. For handy reference, you may want to copy this
data and paste it near the front cover of your Handbook. The PsyCalc 98 program can calculate
additional moist air properties.

Table 6.11 Effect of humidity on the properties of air.

dew point, °C 0 5 10 15 20 25 30
20.82 20.77 20.70 20.60 20.48 20.31 20.11
<jp02, voi % 0.9962 0.9946 0.9925 0.9895 0.9856 0.9804 0.9736
m dry air/m moist air 28.89 28.87 28.83 28.78 28.71 28.62 28.50
1.289 1.288 1.286 1.284 1.281 1.277 1.272
M moist air 0.08047 0.08039 0.08029 0.08014 0.07995 0.07971 0.07938
p, kg/m3 (STP)
p, lb/ft3 (STP)

The composition of the gaseous fuel/oxidant mixture is commonly expressed in volume
fraction units and flow in volumetric units. The drawback in using volumetric units is the
possibility of errors if there is any ambiguity about the pressure and temperature at which the
volume is stated. It's better to use amount (moles) or mass units to eliminate confusion about the
exact quantity of reactants and products in a material balance. Unless otherwise stated, volumetric
flows in this handbook are expressed at STP of 273.15 К and 1 atm, so that the molar volume is
359.04 ft3 or 22.414 liters.*

The concept of stoichiometric combustion refers to the amount of oxygen required to oxidize
the fuel's combustible constituents to their highest stable oxidation state. For hydrocarbon fuels,
this means H20, C02, and S02. (The thermodynamic stability of S03 and its slow kinetics of
formation mean that the predominant oxidation product of sulfur-containing substances is S02.) It
is convenient to have an expression to calculate a mass ratio of dry air/fuel for stoichiometric
combustion. For a fuel containing C, H, O, and S, the composition should be expressed as

Except for special cases, calculations will use 22.41 and 359 as the STP molar volumes.

324 Chapter 6 Reactive Material Balances

elemental mass fraction. Based on one mole of 0 2 in 4.773 moles of dry air, and taking each
reactive element in turn:

Carbon : mass air 4.773x28.96 = 11.51 [6.32]
mass С 12.01

Hydrogen : mass air = —'/4(4J73)x2S.96 = 3„4,.„28„ [6.33]

mass H 1.008

Sulfur: mass air 4.773x28.96 = 4.31 [6.34]

mass S 32.06

Considering the accuracy of typical analytical procedures, an equation for the mass of air per
mass of fuel at stoichiometric combustion is:

mass of air - 1 1 . 5 0 C ) + 34.30H) + 4.3(wS - wO) [6.35]
mass of fuel

where wC, wH, wS, and wO refer to the mass fraction of each element in the fuel, and sulfur burns

to S02. Since these equations refer to dry air, use of moist air will require a correction based on
Equation [6.29].

At exact stoichiometry, 0 2 should be absent, and the products of hydrocarbon combustion
should be only H20, and CO2. However, a tiny amount of CO, H2, and O2 are present because of
thermodynamic and kinetic factors. A slight change away from stoichiometry causes a very large

change in the volume fraction of these three gases. At the slightest amount of excess air (XSA),

the product gases contain a small amount of 02, and very tiny amounts of CO or H2. At the
slightest amount of excess fuel (XSF), the product gases contain a small amount of CO and H2 and
a very tiny amount 02. Figure 6.20 shows this for combustion of methane (CH4) with air, where
air is assumed to contain 21.00 %02. The stoichiometric volume ratio of air/CH4 is 2/0.21 = 9.524.

Near-Stoichiometric Combustion of CH4 —|
with Air at 800°C

— " " T"

/"*

8 -7 _Q^ ...—·— 1 H20
1
t *O) ^πτ
-11 1
-13 100.4
*
-15 <»
99.4
<»1—-»-XSA

99.6 99.8 100.0 100.2 100.6

% stoichiometric air

Figure 6.20 Change in partial pressure of CO, H2, and O2 near stoichiometric combustion of CH4
with air at temperatures below about 2000 °C. Dashed line indicates log(/?02). An XSF condition
exists at less than 100 % stoichiometric air, and an XSA condition exists at more than 100 %

stoichiometric air. Results based on assumption of equilibrium gas composition.