Handbook_on_Material_and_Energy_Balance_Calculat1

Chapter 6 Reactive Material Balances 375

The ledger for output species is shown next. Clearly, none of the stream compositions sum to
100 %, which is not unexpected for the reasons mentioned earlier for input streams. The dust tends
to be more concentrated in the minor elements that were omitted from analysis, which may explain
why it has the greatest shortfall.

Output Streams

Hot metal Fe С Si Mn S Total % of total

kg 44,520 2120 300 260 30 47,230 99.9%

Slag Si02 CaO MgO MnO A1203 1 S 1 Total % of total
kg 3890 4100 1090 110 950 160 10,300 99.2%

Dust Fe203 CaO MgO С Total % of total Total species mass out:
kg 1190 50 30 220 1490 90.3% 59,020 kg + top gas

The next step is to examine balances for individual species to determine (and insofar as
possible, correct) any discrepancies. We start with Fe. It enters with the ore, BOF slag and coke,
and leaves in the dust and hot metal. Fe203 has wFe = 0.6994 and FeO has wFe = 0.7773, so the
mass balance for Fe is:

In: 0.6994(69,800)(0.918) + 0.7773(2240)(0.2915) + 0.0085(22,970) = 45,520 kg

Out: 0.6994(1650)(0.72) + 0.942(47,260) = 45,350

This gives an apparent "loss" of 170 kg of Fe, which is understandable because hot metal is
unavoidably lost by splashing and sticking to the runners during tapping. If we trust the
composition values more than the mass values, the above loss is corrected by increasing the hot
metal mass by 170 kg, to 47,430 kg. The mass of Si remains at 300 kg, but the mass of Mn is
upped to 270 kg by the change in hot metal mass.

Now we turn our attention to the slag, which must contain all of the CaO, MgO and A1203
from the raw materials (minus the small amount of CaO and MgO in the dust), the unreduced Si02
and MnO, and some S. The new mass of Si in the hot metal is thus 0.0064(47,430) = 300 kg,
which means that 640 kg of Si02 was reduced in the hearth. The slag weight calculation is based
on the main constituents:

CaO in slag = 0.0125(69,800) + 0.388(2240) + 0.546(3760) + 0.3355(970) + 0.0024(22,970)
-0.03(1650) = 4130 kg

MgO in slag = 0.0095(69,800) + 0.105(2240) + 0.004(3760) + 0.1905(970) + 0.0015(22,970)
-0.02(1650) =1100 kg.

A1203 in slag = 0.0045(69,800) + 0.0412(2240) + 0.0242(22,970) = 960 kg.

Si02 in slag = 0.0484(69,800) + 0.133(2240) + 0.0388(22,970) - 640 = 3930 kg.

The mass of these species is 10,120 kg, and they make up 96.73 % of the mass of the slag.
This gives 10,460 as the calculated mass of slag, as compared to the listed mass of 10,380 kg.
Apparently, 80 kg of slag was "lost" by splashing or sticking to the slag runner. The amount of
MnO and S in the slag remains the same at 110 kg and 160 kg respectively.

With this corrected information at hand, we can calculate the solution loss, which requires a
carbon balance. The solid С available for use in the BF is the amount brought in with the coke
minus the amount carried out in the dust. This is 0.911(22,970) - 0.13(1650) = 20,710 kg. Ofthat,
the hot metal carries out 0.0448(47,430) = 2120 kg, the reduction of 300 kg of Si consumes
300(24/28) = 260 kg, and the reduction of 270 kg of Mn consumes 270(12/55) = 60 kg. This
leaves 20,710 - 2120 - 260 - 60 = 18,270 kg of С to reduce C02 and H20 in the shaft and react
with the blast in the combustion zone. The mass of С reacting in the combustion zone is calculated

376 Chapter 6 Reactive Material Balances

by noting that 12 kg of С are consumed for every 16 kg of О entering the combustion zone. The
air, steam and oxygen bring in 15,350, 440, and 3300 kg of О respectively, for a total of 19,090 kg
of O. The 1640 kg of С in the fuel oil will react with 2190 kg of O, leaving 16,900 kg of О to react
with the coke carbon. This consumes 12,680 kg of coke C, which means that 18,270 - 12,680 =
5590 kg of coke С (30 % of the coke С available) is consumed by the solution loss mechanism.

A calculation of the top gas mass requires a nitrogen balance. The N2 from the air has a mass
of 48,550 kg, and by difference, the top gas has wN2 = 0.4266, which gives a top gas mass of
113,810 kg. This gives a total mass out of 172,830 kg, which you can pencil in on the text box for
the output stream ledger. (For dusty and hard-to-sample gas streams, the composition of N2 is
usually obtained by difference). The mass of (CO + C02 + H2 + H20) is then 65,260 kg. This can
be checked by calculating the mass of C, H, and О rising from the BF to the top gas. A
simplifying assumption is that the C02 from carbonate decomposition, and the H20 from coke and
ore moisture leave the BF without undergoing any chemical reaction. This is valid because С is a
limiting reactant in the shaft (only 5590 kg is consumed there), and it therefore makes no
difference in the final result where the molecules of H20 or C02 come from that are consumed by
the solution loss reaction. Bookkeeping is simpler if we pass all the carbonate C02 and burden
H20 directly to the top gas. .

The mass of О removed to the top gas is calculated from the mass of О removed from each
reactive species, and shown in the table below.

Ore - dust BOF slag Si02 redn MnO redn Blast Total О to top gas

18,900 150 350 80 19,090 38,570

The mass of С removed to the top gas is obtained from the previous С balance as 20,230 kg.
The mass of H removed to the top gas is the mass of H in the air (60 kg), plus the mass of H in the
fuel oil (230 kg), plus the mass of H in the steam (60 kg), for a total of 350 kg of H to the top gas.
The ledger summarizing the calculation of top gas mass is shown below. The mass of carbonate-
decomposition C02 is obtained from the gravimetric factor for C02 from CaO (0.7843) and from
MgO (1.092). The mass of the ore and coke water were obtained directly from Figure 6.54

0 С H co2 н2о N2 Top gas mass

38,570 20,230 350 2080 4030 48,550 113,810

The result is in surprisingly good agreement with the top gas mass obtained by the N2 balance
alone. The revised BF mass balance is shown below in ledger format. There are still some minor
discrepancies in the overall mass balance for reasons mentioned earlier. Additional uncertainties
are the moisture content of ore and coke (which are often stored outside and can change between
the time of sampling and use), and the typical errors in chemical analysis. The reconciled mass
balance brings each element or species within ±20 kg.

Note that the mass of the top gas is about 2.4 times that of the hot metal. After cleaning and
cooling to condense much of the water, the top gas is used as a fuel for heating the air blast. This
"thermal recycling" makes the BF an efficient producer of iron.

Total mass in, kg
Ore BOF slag Flux Coke Water Blast Streams Species D
69,800 2240 4740 22970 4030 69,640 173,420 173,080 -340

Total mass out, kg

Hot metal Slag 1 Dust Top gas Streams Species □

47,430 10,460 1650 113,810 173,350 172,830 -520

Chapter 6 Reactive Material Balances Ъ11

6.9 The Use of Distribution Coefficients in Material Balance Calculations

Many industrial processes involving complex solution phases operate without a clear
knowledge of the chemical reactions taking place. For example, solute species are not defined
when elements such as S, Cu, or P are dissolved in slags. So rather than writing chemical
equations involving hypothetical species, it is common to designate a distribution coefficient for an
element between phases. For example in copper smelting, the material balance for As can use
distribution coefficients between the slag, matte and gas. Using distribution coefficients is not the
only way to handle solution phases, but it is the simplest. Distribution coefficients are obtained
from plant data, laboratory experiments, or calculated from thermodynamic data listed in one of
the software packages cited in General References.

6.9.1 Use of Tabulated Distribution Coefficients

Up to this point, we have used SR data such as XR values, Äeq, and other correlations derived
from plant measurements to reach a DOF = 0. We now turn our attention to certain correlations
between the slag, metal and gas phases involved in steelmaking, called distribution coefficients.
These coefficients are based on an understanding of the chemical thermodynamics that underlie the
process. A number of distribution coefficients have been obtained by plant and laboratory
measurements for the oxygen steelmaking process (Fruehan 1998). In basic oxygen furnace
steelmaking, impurities in BF hot metal are oxidized to produce steel. Scrap is added as a coolant,
and flux is added before and during the oxygen blow to make a slag. Figure 6.55 depicts the BOF
steelmaking vessel and oxygen lance.

The injection of oxygen into the BF hot metal oxidizes Si and C, both of which are removed
nearly completely in the production of low-C steels. In addition, a considerable fraction of the Mn
is also oxidized. As the carbon content goes below about 0.2 %C, increasing amounts of Fe are
oxidized to the slag. The impurities distribute themselves between the slag and metal according to
the final carbon content and the composition of the slag. There is considerable variation in the
distribution coefficient values because of the number of different ways oxygen is blown into the
melt and how flux is added, so different practices give different values. Equations for some typical
distribution coefficients are shown below. The symbols in parentheses mean composition of
(slag), and in brackets, composition of [metal]. The expressions are valid for wSi02 in the slag
between about 11 and 30 %Si02.

(%FeO)V%C~ = ^^ [6.108]

0.0001(%SiO2) +0.001

[%Mn] =0.15 [6.109]
[6.110]
(%MnO)V%C

(%s)

[%s]V%c = -2.3(%SiO2) + 70

Equation [6.108] is used in the yield calculations for Fe, and Equation [6.109] is used in a
similar way for Mn. Equation [6.110] is used to determine the distribution of sulfur between slag
and steel. Adding more burnt lime affects the [%S] in the steel in two ways. First, it lowers the
%Si02 in the slag, and thus distributes the sulfur preferentially to the slag. Second, it increases the
mass of slag, further lowering the [%S].

378 Chapter 6 Reactive Material Balances

Gases
Fume (CO, C02)

Air
Burnt lime

Dolomitic lime

Figure 6.55 Sketch of a BOF steelmaking vessel. In some plants, additional oxygen and flux are
injected through a tuyere in the bottom. A hood collects the gas and fume from the process. Air is
drawn into the hood to oxidize the CO.

EXAMPLE 6.15 — Material Balancefor BOF Steelmaking.

BF hot metal is desulfurized to 0.011 %S, and charged to a BOF along with scrap and flux.
Oxygen is blown to produce steel containing 0.055 %C at the end of the blow. During the blow,
about 12 % of the S leaves the vessel as S02, and the CO/C02 ratio in the offgas is about 10.
Calculate the distribution of sulfur between steel and slag as a function of the composition of Si02
in the slag. Calculate the flux and oxygen required for the process, the volume of the offgas at
1600 °C, and the % yield of Fe to steel. Calculate the amount of ferromanganese (wMn = 0.75) to
be added to the steel after tapping to bring the wMn to the same value as the incoming scrap.

Solution. Figure 6.56 depicts the flowsheet for the process and lists the specified stream mass and
composition. The reported accuracy of the gas composition allows the фСО and <pC02 to be set
because <pS02 is negligible. The unspecified constituents in the fume consist of ZnO from the
galvanized steel scrap, other volatile tramp elements, and particles of flux and slag. The amount of
non-analyzed substances in the slag is estimated as 0.02(W). The very small wSi and wO in the
steel can be neglected. The wFe in the steel is set at 0.998 for the Fe balance. The wFe in the
scrap and hot metal is calculated by difference.

Chapter 6 Reactive Material Balances 379

Scrap (B) Fume (F)
mF = 5.5t
mB = 40.0t wFeO = 0.50

ivC = 0.0011, wMn = 0.0036 Gas (G)
wSi = 0.00021, wS= 0.00022
mG = ?
Hot metal (C)
m c = 150.0 t <pC0 = 0.90, <pCO2 = 0.10
wC = 0.0448, w Mn = 0.0056 ¥>SO2~0
wSi = 0.0064, wS= 0.00011
Steel (H)
Dolo lime (D) mH = ?
m -r • wC = 0.00055, wMn = ?
wMgO = 0.405, wCaO = 0.580 w S = ?, w S i « 0
wSi02 = 0.0085, ivS = 0.0025
Slag (I)
Burnt lime (E)
mE = ? m'-?
wCaO = 0.935, ivSi02 = 0.015
ivS = 0.0027 w MgO = 0.083, wCaO = ?
wSi02 = ?, wFeO = ?
wMnO = ?, ivS = ?

Figure 6.56 Flowsheet for the production of steel in the BOF.

Elemental mass balance equations were written as In = Out on Fe, S, Si, Mn, CaO, and MgO
in the scrap, hot metal, flux, slag, steel and fume, m refers to the mass of each stream in tonnes.
Gravimetric factors are listed in parentheses where needed. The three distribution coefficient
equations were used to establish the slag/metal relationships at the final carbon content of 0.055
%C. The sum of slag mass fractions was the 10th equation. The oxygen requirement is calculated
after the ten equations are solved.

1. Fe balance: 0.9947(mB) + 0.9426(mc) = 0.5(mF)(0.777) + wFeO(W)(0.777) + 0.998(wH)

2. Si balance: 0.00021(mB) + 0.0064(mc) + 0.0085(mD)(0.467) + 0.015(wE)(0.467) =
wSiO2(mI)(0.467)

3. S balance: 0.00022(mB) + 0.0001 l(mc) + 0.0025(mD) + 0.0027(mE) = wS(mH) +
wS(ml)

4. Mn balance: 0.0036(mB) + 0.0056(mc) - wMn(mH) + wMnO(W)(0.775)

5. CaO balance: 0.58(wD) + 0.935(wE) = wCaO(W)

6. MgO balance: 0.405(wD) = 0.083(W)

7. Equation [6.108]: wFeO^lOOXVÖÖSS KO.OlOvSiOz1) + 0.001} = 0.012

8. Equation [6.109]: wMnH = wMnO^VO.OSS )(0.15)

9. Equation [6.110]: wSl = wSH(VÖÖ55){-230(wSiO2I) + 70}

10. Slag composition: 0.98 = 0.083 + wCaO + wSi02 + wFeO + wMnO

The ten equations were written in Solver format for different values of WS1O21 and
SuperSolver was used to obtain several solutions. Figure 6.57 shows the flux requirements, slag
mass, and %FeO in the slag. The sulfur distribution between slag and steel is shown in Figure
6.58. As expected from the distribution coefficient expression, decreasing the %Si02 decreases the
%S in the steel, which is further enhanced by the increasing mass of slag. Figure 6.59 shows that
the ferromanganese requirement is decreased at higher %Si02 because of lower slag mass.

380 Chapter 6 Reactive Material Balances

г25 Flux and Slag

20 — -o- — -o-

ΔV _ -0
а% 15
4>— DoLime —D— Lime • - - Δ - 3

?"8SσP io ■Д - Slag —О - %FeO D-

--?
0 c^

11 12 13 14 15 16 17
mass fraction Si02 in slag, %

Figure 6.57 Flux requirements, slag mass, and wFeO in slag as a function of wSi02.

Sulfur in Slag and Steel
0.0148
0.0146
0.0144 ^o
0.0142 1-

<f> 0.0140 %
с 0.0138 ^
со 0.0136 й

0.0134 £
P 0.0132

0.0130
11 12 13 14 15 16 17

mass fraction Si02 in slag, %

Figure 6.58 Sulfur distribution between steel and slag as a function of wSi02 in slag. Increasing
Si02 in slag increases wS in steel because of the change in distribution coefficient, and a decrease
in the mass of slag.

98.2% % Yield and FeMn Required c/> Figure 6.59 Yield of Fe

98.0% 12 13 14 15 16 <D from scrap and hot
mass fraction Si02, %
97.8% Ссо metal to steel as a
U. 97.6% function of wSi02 in

4- Έсω slag. More Fe and Mn
are lost to slag as the
υ 97.4% wSi02 decreases (which

■σ 97.2% causes the mass of slag

.>2. 97.0% 17 to increase).

96.8%
11

Chapter 6 Reactive Material Balances 381

The oxygen requirement is calculated from an О and С balance:
О balance: mA = wFeO(WX0.223) + wMnO(W)(0.225) +0.5(/HF)(0.223) + mCO(0.571) +
mCO2(0.727)

С balance: 0.001 l(wB) + 0.0448(mc) = mCO(0.429) + mCO2(0.273) + 0.00055(mH)

Gas composition: 9(28)/44 - mCO/mC02

The three oxygen balance equations were solved by hand to yield the tonnes of 0 2 (mA) as a
function of the wSi02 in the slag. The results, as STP m3 02/tonne steel, are shown in Figure 6.60.
The offgas volume (stream G) at 1600 °C is, within the accuracy of the data, about 85 300 m3 at all

values of wSi02. In practice, air is added to the offgas well in excess ofthat required to oxidize the
CO, which can triple this volume.

47.0 Oxygen Requirement η Figure 6.60 Oxygen
requirement for the
о 46.5 12 13 14 15 16 production of one tonne
of steel in the BOF. The
(Л 46.0 mass fraction of Si02, % effect of increased wSi02
сαсо> is to lower the mass of
45.5 slag, thus decreasing the
45.0
1 | m02 consumed in
44.5 oxidizing Fe.

см 44.0 17

о 43.5

11

Assignment. Iron ore is sometimes used as a coolant in the BOF with a corresponding decrease in
the amount of scrap and oxygen. Make a mass balance for the addition of 2.9 tonnes of iron ore as
a replacement for 13.0 tonnes of scrap, for wSi02 in the slag = 0.15. Set the ore composition the
same as that in Section 6.4.

6.9.2 Thermodynamic Databases as a Source of Distribution Coefficient Data

Unfortunately there is no freely-available single source for distribution coefficients for
substances between solution phases like alloys, slags, and mattes. However, distribution
coefficients can be calculated from various thermodynamic programs that can be purchased from
commercial vendors. One of the most useful is the FactSage program developed at Ecole
Polytechnique in Montreal, Canada, (FactSage and other useful data sources are cited in General
References). FactSage has thermodynamic models for slag, matte, halide, and alloy phases, which
are used in their Equilib calculation program. These models are rather complex, and cannot be
transported to Excel. However, it is feasible to make multiple FactSage calculations on a particular
system, and use the results to estimate distribution coefficients that are valid over a practical range
of composition. Here we illustrate the use of FactSage to develop the necessary relationships
between slag, matte, and gas composition for the flash smelting of copper concentrate. These
relationships can then be used by FlowBal to make practical material balances for the process.

Copper smelting was mentioned earlier in Section 6.5.5 as preceded by a roasting step.
Modern copper smelting techniques dispense with roasting, and instead smelt the concentrate
directly by mixing the concentrate with enriched air and flux, and burning the mixture in a
specially-designed flash smelting furnace (Davenport 2001). Figure 6.61 is a sketch of an
Outokumpu flash smelting furnace.

382 Chapter 6 Reactive Material Balances

Offgas
(S02, N2, etc.]

!£?2«&!£*2*&*&!«2^^ О Slag
ΐ > Matte

Figure 6.61 Sketch of an Outokumpu flash smelting furnace. Raw materials usually include
recycled slag, plant scrap, other residues, and fuel. Offgas contains combustion products, dust, and
various impurities. The offgas is cleaned and the SO2 used to make sulfuric acid. The matte is
further refined in a converter to produce blister copper, and the slag is discarded. The slag
typically has about 1% of its mass in the form of suspended matte particles.

To minimize the complexity in setting up FlowBal, we make certain simplifying assumptions:
no recycle of material to the flash furnace, no production of dust, and no fuel. The raw material
composition is:

Cu concentrate: wCu = 29.7 %, wFe = 29.3 %, wS = 33.1 %, wSi02 = 7.3 %, balance CaO.

Flux: wSi02 = 88 %, wCaO = 11 %, balance Fe203.

Oxygen-enriched air: φ02 = 47 %, balance N2.

The selected basis is 175 kg of concentrate, which is about 0.01% of the daily capacity of a
typical furnace. We seek to find the amount of enriched air and flux required to produce a matte
having wCu between 50 % and 65 %, while keeping wSi02 in the slag at 33 %. Figure 6.62 shows
a flowsheet for a simplified flash smelting process. Copper is lost from the system as Cu20 in the
slag and as matte particles suspended in the slag.

Gas [4 Furnace

1

v-Mcn air ■ 2 Flash 5 Slag J
Furnace
s* -±J Spltr И
uonc " w
Flux- Furnace L ' wJ\
Matte
8 „ Final

w Matte

Figure 6.62 Flowsheet for the production of copper matte in a flash furnace. The presence of
suspended particles of matte in the final slag is simulated by using a splitter to transfer sufficient
matte to the slag such that the final slag stream (S9) contains 1 % matte.

The FactSage program Equilib produces the needed data. Equilib minimizes the free energy
of all substances consistent with the thermodynamic model of the non-ideal solution phases. If the
user enters the amount and composition of raw materials, specifies the temperature and pressure of
the system, and selects an appropriate slag model, Equilib calculates the equilibrium composition.

Chapter 6 Reactive Material Balances 383

Several calculations were made to determine the range of matte, slag and gas compositions to
determine the major species present in these phases. The main slag species were FeO, Si02, CaO,
and Fe203, with minor amounts of other species, particularly Cu20, which needs to be tracked to
determine copper losses to slag. The matte consisted of Fé, Cu, and S, while the gas phase
contained N2, S02, and a small amount of S2. A sample output from a FactSage calculation is
shown on the Handbook CD in the Copper Smelting folder.

Four FactSage calculations were made over a range of input amounts. The flux amount was
adjusted by trial-and-error to give wSi02 in the slag at 33 %. The results were used to obtain four
different relationships that were needed as part of the Fe, S, O, and Cu mass balance equation set.
Figure 6.63 shows the results. Each of the parameters was expressed as a function of the wCu in
the matte. Three of the four smelting parameters follow a nearly-linear trend with wCu in matte.
Considering the small variation in wFe203 in the slag and its minor effect on the overall material
balance, a linear equation was also used for this parameter.

Smelting Parameters - Gas Smelting Parameters - Slag
7.5
1.7
О
1.5 ^^^^^ О %S2 in gas 7.3 О %Fe203inslag
^(Я 1.3 — Linear (%S2 in gas) \_ Linear (%Fe203 in slag)

rσo> 1 1 7.1

с 0.9 О 6.9
&см CM Л^* О
0.7 %S2 = -0.0683(%CuM) + 5.03 Ъ^^* %Fe203 = 0.050(%Cu) + 4.00
со 54 58 62 0) 6.7
ivCu in matte, % 6.5
0.5
50 66 50 54 58 62 66
wCu in matte, %

0.40 Smelting Parameters - Slag Smelting Parameters - Matte
* 0.35
О wCu20 in slag О %Fe in matte
I 0.30 ^ — Linear (wCu20 in slag) — Linear (%Fe in matte)

- 0.25 %Cu2Os = 0.0117(%CuM) - 0.400
Ü
* 0.20 54 58 62 66 54 58 62 66
wCu in matte, % tv Cu in matte, %
0.15
50

Figure 6.63 Relationship between wCu in matte and four smelting parameters based on data

points obtained from the FactSage Equilib program at 1300 °C and wSi02 in slag = 33 %. Textbox
equation obtained from Excel's Trendline tool.

As usual, FlowBal was initiated by specifying the known stream flow and composition, and
writing five chemical reactions.

1. 2Cu + У2О2 ~> Cu20 (copper oxidized to cuprous state in slag).

2. Fe + У2О2 -* FeO (iron oxidized to ferrous state in slag).

3. 2Fe + IÌ/2O2 —» Fe203 (iron oxidized to ferric state in slag).

4. S + O2 —» S02 (sulfur oxidized to dioxide in gas).

5. 2S —» S2 (sulfur polymerized to diatomic state in gas).

The four smelting parameter equations, plus a fifth equation relating the mass of matte
suspended in slag (here, wmatte in S9 was set at 1.0%) were entered in FlowBal via the Insert
Equation feature. FlowBal converged to unique results for the specified value of wCu in matte
which meant the chemical reactions were truly independent and the equation set was correctly

384 Chapter 6 Reactive Material Balances

written. FlowBal's Repetitive Solve feature was then used to make five calculations for wCu in
matte between 50 % and 65 %

Figure 6.64 shows the FlowBal result for wCu in matte of 60.0 %. The loss of copper to the
final slag (S9) varies between about 1 - 1.6% (Figure 6.65). The mass of flux and volume of
enriched air increase in a nearly linear manner with the intended matte grade (Figure 6.66).

P (atm) 1 111 1 1 1 11

T(K) 300 300 300 1570 1570 1570 1570 1570 1570

Str-unit Voi (шЗ) Mass (kg) Mass (kg) Voi (m3) Mass (kg) Mass (kg) Mass (kg) Mass (kg) Mass (kg)

Spec-unit Voi pet Mass pet Mass pet Voi pet Mass pet Mass pet Mass pet Mass pet Mass pet

Str-name Enrchd Air Cone Flux Offgas Fee Slag Fee Matte Susp Matte Fnl Matte Fnl Slag

Streams 1 2 34 5 6 7 89

Flow 76.07 175 14.80 355.41 78.18 86.28 0.79 85.49 78.97

N2 53 0 0 59.36 0 0 0 00

0 2 47 0 0 0 0 0 0 00

Cu 0 29.7 0 0 0 | 60.0 60.00 60.00 0.60

Fe 0 29.3 0 0 0 15.48 15.48 15.48 0.15

S0 33.1 0 0 0 24.52 24.52 24.52 0.25

Si02 0 7.3 88.0 0 33.0 0 0 0 32.67

CaO 0 0.6 11.0 0 3.43 0 0 0 3.39

S02 0 0 0 39.70 0 0 0 00

S2 0 0 0 0.93 0 0 0 00

FeO 0 0 0 0 56.27 0 0 0 55.71

Fe203 0 0 1.0 0 7.00 0 0 0 6.93

Cu20 0 0 0 0 0.30 0 0 0 0.30

Figure 6.64 FlowBal result for smelting copper concentrate to attain wCu in matte of 60.0 %
(boxed cell). Final slag (S9) consists of 99% fornace slag (S5) and 1% suspended matte (S7).

Loss of Copper to Final Slag Figure 6.65 Loss of copper
in final slag. About two-
1.7% thirds of the copper is lost as
suspended matte particles,
1.5% % Cu loss = 0.0631 mg - 2.47 and the rest as Cu20
dissolved in the slag.
О R2= 1.00
66
о 1.3%
О</>)
-О 1.1%

0.9%

54 56 58 60 62 64
matte grade, %Cu

6.10 Time-Varying Processes

Most of the flowsheets discussed in this Handbook deal with continuous processes operating
at steady-state. However, as pointed out in Section 4.14, many important processes are either
batch or semi-continuous, and for those, we need to make a material balance as a function of time.
If the chemical reaction kinetics are well-understood, we might be able to apply the principles
discussed in Section 5.5.2. However, we need alternate techniques when a process is too complex
to be handled in this way. One technique involves iterative stagewise approximation, which was
illustrated for the batchwise distillation of a Cd-Zn alloy (Example 4.14). This technique involves
dividing a time-varying process into small time steps. The smaller the time period, the better the
approximation is to the real process.

Chapter 6 Reactive Material Balances 385

Air and Flux RequiredI

85 г II I Ж\ 20
83 D mass flux
О voi air 18

81 IJ Linear (voi air) Linear (mass flux) ^ 16 σ)

I I, ^ ^ X
14 33=
79 Nm3 air =1.49mg-13.5
77 j 12
R2 = 0.9987 —=^—X ^^^
75 r 10
о TK \ Ж, *» ^ ^ > kg flux = 0.725mg - 28.9
> 71 R2 = 0.9976 66
у ^ ^^^^
fiQ J

fi7 I 1
65 \

54 56 58 60 62 64
matte grade, %Cu

Figure 6.66 Quantity of enriched air and flux required to smelt 175 kg of copper concentrate to
specified matte grade. Relationship between the required feed is adequately represented by a linear
equation as shown by R2 value in excess of 0.99.

The desulfurization of steel is an example of a batch process. Sulfur is removed from steel in
a ladle process by using one or more reagents that preferentially combine with the sulfur and
transfer it to a separate phase, such as a slag. Sulfur transfer is enhanced by stirring. For example,
argon injected in the bottom of the ladle rises to the slag/steel interface, and bubbles break through
the interface to create a large area for mass transfer. One of the ways to approximate the rate of
change in sulfur content of the slag and steel is to assume that the slag and steel reach equilibrium
at their interface. Fluid flow then moves the interface layers back into the bulk phase. Over a
certain time interval, say one minute, an estimate is made of the fraction of each phase that reacts
to equilibrium at the interface. This fraction is then mixed back into the bulk phases, and the
process is repeated. Figure 6.67 shows a sketch of this concept. A typical steel temperature before
desulfurizing is 1600 °C.

Steel — 1 12--»>Offgas
Slag - -6 3 - Bulk DeS
►( 1st Steel — 4 - " * Steel

Slag

Mixer Bulk S-enriched
" 9 " ~ * Slag

Figure 6.67 Flowsheet depicting short time interval during the batch desulfurizing of steel. Solid
lines represent the flow of steel, dashed lines the flow of slag, and dotted lines the flow of gas.
Double lines represent the fraction of initial steel and slag that bypasses the desulfurization
interface. A time-varying flowsheet sketch would be depicted as a sequence of splitters, reactors,
and mixers attached end-to-end.

A common desulfurizing reagent is a high-lime slag. The initial time period is depicted by
splitters that divide the initial phases into an interface portion (streams 2 and 7) and a bulk portion
(streams 5 and 10). The interface portions react to equilibrium in the first desulfurizing vessel and
exit as streams 3 and 8. The interface and bulk portions are then mixed and leave the first time
period as streams 4 and 9. These two streams become the input to a second time interval period,
and so forth. Thus a 20 minute desulfurization process might be simulated by 20 one-minutes
steps, or 10 two minute steps, and so forth. Trying for a better simulation by using a large number

386 Chapter 6 Reactive Material Balances

of short time steps is probably not justified in view of the number of approximations and
assumptions built into the model.

The mechanism of steel desulfurization by a high-lime slag is described in terms of three
solute species in the steel, three solute species in the slag, and CO gas. This requires three
chemical reactions, where the superscript designates the phase.

SStl + C a O s l g - C a S s l g + 0Stl [6.111]

_sti xStl FeOslg [6.112]

csti + 0sti CO(Gas [6.113

Typically, 16 tonnes of slag briquettes are added to 100 tonnes of steel, which drops the steel
temperature to about 1560 °C. As sulfur transfers to the slag, oxygen transfers to the steel, and
depending on the initial steel and slag composition, oxygen distributes itself to or from the slag (as
FeO). Oxygen also reacts with the dissolved carbon to form CO. As the argon rises through the
steel it captures the CO and the gas bubbles pass through the interface. Over a small composition
range the relationships between the slag and steel solutes can be expressed as distribution
coefficients in mass pet terms. This technique was discussed in Section 6.8.2. Equations [6.114] -
[6.116] show the three distribution coefficient equations at 1560 °C as obtained from the FactSage
program.

360(%CaSslg)(%OStl) - (%CaOslg)(%Sstl) = 0 [6.114]

1.42(%Ostl)(%Festl) - (%FeOslg) = 0 [6.115]

^ c o u a s _ (0/оСxSМtl-1)(о/о0,S^tk)360 = 0 [6.116]

Before setting up the system for multiple steps, it's good practice to explore selected process
variables by making a single desulfurization calculation for a condition where all of the steel and
slag undergo desulfurization for a sufficient time to attain steel/slag/gas equilibrium. In effect, this
means that the split fractions to S5 and S10 are zero. The following table shows the composition
of a typical steel and desulfurizing slag at 1560 °C, with the basis changed to one tonne of steel.
The system pressure is set at 1.2 atm to reflect the above-atmospheric pressure just before the gas
bubbles exit the slag. The Si02 and A1203 do not participate in the desulfurization process, hence
for material balance purposes, can be combined into one value, whose mass does not change.

Steel Slae
%c %S % 0 %CaO %CaS %FeO % Si02 A1203
0.23 0.05 0.012 60 0.2 1.8 30 8

The system has four steel composition variables, three slag composition variables, and one
gas composition variables. There are also three quantity variables, the mass of each phase. For
uniformity, all compositions and quantities are in mass units. We therefore need eleven equations
to bring DOF = 0. First, we can write a mass balance around the reactor for each reacting element
(Fe, C, S, O, and Ca). Next, we have three equations stating that the sum of percent compositions
for each phase equals 100. Finally, we have the three distribution coefficient equations listed
previously. Once we set the mass of Ar, we have all of the necessary information needed for the
mass balance. The first five equations are the element balances, the next three are the distribution
coefficient equations, and the last three are the sum of percent terms. Various factors in the
equations are gravimetric factors for elements from species. The values of 30 and 8 in Equation 10
refer to the % A1203 and %Si02 in the original slag. All equations were written in mass units, and
set to sum to zero.

1. Fe: 1000(99.708) + 160(1.8)(0.7793) - [(mstl)(%FeStl) + (wslg)(%FeOslg)(0.7793)] = 0

2. C: 1000(0.23) -[((ms\%Cst) +(тОа")(%СО^)(0Л2Щ] = 0

3. S: 1000(0.05) + 160(0.2)(0.4444) - [(mStl)(%Sstl) + (mslg)(%CaSslg)(0.4444)] = 0

Chapter 6 Reactive Material Balances 387

4. O: 1000(0.012) + 160[(60)(0.2853) + (1.8)(0.2207)] - [(mStl)(%Ostl) +
wslg[(%FeOslg)(0.2207) + (%CaOslg)(0.2853)] + (wGas)(%COGas)(0.5712)] = 0 +

5. Ca: 160[60(0.7147) + 0.2(0.5556)] - mslg[(%CaOslg)(0.7147) + (%CaSslg)(0.5556)] = 0
6. Slag equilibria: 360(%CaSslg)(%OStl) - (%CaOslg)(%Sstl) = 0
7. Slag/Steel equilibria: 1.42(%0Stl)(%Festl) - (%FeOslg) = 0
8. Gas/steel equilibria: (mGa^(%COGas)(0.8002)(1.2)/[(mGas)(%COGas)(0.8002)(1.2)
100(mAr)(0.5611)] - (%CStl)(%Ostl)360 = 0
9. Sum of steel pet: 100 - (%FeStl + %Cstl + %Sstl + %Ostl) = 0
10. Sum of slag pet: 100 - [%CaOslg + %FeOslg + %CaSslg + 160(30 + 8)/(mslg)] = 0
11. Sum of gas pet: 1 0 0 - [%COGas + mAr(100)/(mGas)] = 0

The eleven equations were written into an Excel spreadsheet, and estimates made for each
unknown value. A useful objective is to find out the effect of mass of injected Ar on steel
desulfurization. SuperSolver is a good choice for this task. It was set up for five different Ar
injections quantities, ranging from 0.1 to 0.5 kg/tonne of steel. Note that 0.1 kg of Ar has an STP
volume of about 55 L. Solver found a solution, which confirmed that the material balance
equations were written correctly. Figure 6.68 shows the results.

Effect of Ar on Steel DeSulf 80
75
- O - %S
- О - %0

•Vol%CO

0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

Ar injected, kg/tonne steel

Figure 6.68 Effect of Ar quantity on equilibrium desulfurization results at 1560 °C and 1.2 atm.
160 kg of desulfurization slag is added to one tonne of steel containing initially 0.05 %S and 0.012
%0. Over the same Ar range, the slag composition showed a change of about 1.8 %FeO to 1.4
%FeO.

The Ar injection mass has a definite, but small, effect on the equilibrium %S. At 0.1 kg Ar
injected, the %0 is slightly larger than in the initial steel. This is because at low Ar, the CO
evolved is not adequate to lower the initial % 0 . At higher Ar, the CO evolved increases, thus
lowering the % 0 , and via Equation [6.114], lowers the %S. In fact, anything that lowers the % 0
causes more sulfur transfer to the slag, thus lowering the %S. In practice, it's common to add a
certain amount of a strong deoxidizing substance (like Mg) to enhance sulfur transfer to the slag.

We now return to the main objective of the process simulation, which is to calculate the rate
of sulfur transfer from steel to slag. The mass balance equations must be reconfigured as time-
defined stages to simulate the batch process described earlier. For this, we use SuperSolver again,
but this time to obtain the equilibrium composition of the slag and steel in each desulfurization
time-defined stage. The steel and slag split fractions to the desulfurizing reactor were arbitrarily
set at 0.30 and 0.25 respectively, with 0.05 kg of Ar added at each stage. Seven time stages were
used, each one simulating one-seventh of the total desulfurizing time. The material balance

388 Chapter 6 Reactive Material Balances

equations for each time-defined stage occupied one column on the spreadsheet. The desulfurizer
vessel (equilibrium) outstreams were directed to the mixer, the streams mixed, and the bulk mixer
outstreams were again split and used as instreams to the next desulfurizer vessel, as depicted in
Figure 6.67 for the first stage.

Figure 6.69 shows the % 0 and %S in the steel, and the %FeO in the slag, for each time-
defined stage. Over the same time, the offgas went from about 76 %CO to 58 %CO, while the CaS
in the slag remained near 0.75%. Figure 6.70 shows the most important result, which is the bulk
steel and slag composition entering the system, and leaving each mixer.

Composition of Steel and Slag out of Desulfurizer

0.017 ^4w J - < b %S 1.7
0.016 ■ -D. - % 0 1.65
4£*. ] —Δ— %FeC»
В 0.015 - о--. * о^ \>w«% 1.55 I
с 0.014 ""* ^<^V^ T
1.45 £
0.013 * * - ia . ' * a . ■ » na 1.4
сш 0.012 1.35
со 0.011 -a .

0.01

0.009 4 111 i 1.3

345
time defined stage

Figure 6.69 Equilibrium composition of steel and slag streams from the desulfurizer at intervals in
the desulfurization process at 1560 °C and 1.2 atm. 0.05 kg of Ar were added at each stage The
instreams to the first interval were the starting slag and steel, while the other six desulfurizer
instreams came from a prior mixer. The desulfurizer instreams came from a splitter with a steel
and slag split fraction of 0.30 and 0.25 respectively.

Composition of Steel and Slag out of Mixer

0.055 A -a- - %0 1.8
^o—%CaS 1.6
0.05 \ -Δ— %FeO
«s1 4 - 5
<D 0.045 ^r
1 u.
s 0.04 X
° · 160.8 T3
С D - 4- D - - D -
0.4 Ü
О 0.035 3456
^5 0.03 0.2
time-defined stage
С

03 0.025
СО
0 х 0.02

0.015
0.01

Figure 6.70 Composition of bulk steel and slag out of post-desulfurizer mixers. Stage 1
represents the starting steel and slag composition, while the other seven stages represent the bulk
phase composition from the mixers. After seven stages of desulfurization the steel contains 0.017
%S, so about 70% of the incoming sulfur has been removed.

Chapter 6 Reactive Material Balances 389

The results show a steady (but diminishing) decrease in %S in the steel during the
desulfiirizing process. The shape of the %S and %0 trendlines is typical for a mass-transfer
controlled process: the mass transfer rate decreases as the reactive solutes approach the equilibrium
position. However, there is no limit to how low the %S can be so long as Ar continues to be
injected. Each bubble of Ar will carry out some finite amount of CO. The oxygen thus removed
lowers the %FeO in the slag and the %0 in the steel, which causes Equation [6.111] to proceed.
Increasing the Ar flow rate removes more oxygen per stage, and hence more sulfur is transferred to
the slag in each stage. In addition, the increased Ar flow causes more turbulence and mixing at the
slag/metal interface, which can be reflected in the model by increasing the split fraction for each
stage.

If the material balance equation sets are properly structured, it's easy to change the split
fractions, the mass of Ar injected, or any other process variable. One of the main benefits of using
SuperSolver is that it allows a primary and a secondary process variable to be changed at each time
stage. For example, it world be interesting to study the effect of %C in the incoming steel on the
progress of desulfurization.

FlowBal can also simulate the desulfurization process, but with some difficulty. Each stage
requires ten streams, and FlowBal writes about 50 equations per stage, so Solver begins to have
trouble at more than three stages. One way around this is to set up the system in multiple
worksheets. The output of the third stage in the first worksheet can be used as the input to the
fourth stage in the second workbook, and so on. Of course, this works only for systems without
recycle from one worksheet to a previous one.

6.11 Systems Containing Aqueous Electrolytes

Many of the processes used to extract, treat, or dispose of materials involves water, and
solutions of these materials in water. There are two major technologies that use aqueous solutions
in materials processing: hydrometallurgy and electrometallurgy. The first technology covers unit
operations of leaching, solution purification, and metal recovery. The second technology involves
the use of electromotive force to recover metals from solution, or to refine metals already
recovered by other means. Both technologies have been applied to the treatment of waste
materials, and the recovery of valuable products from recycled materials.

All of the technology for hydro- and electrometallurgy is based on the unique properties of
water, which has the ability to dissolve many different substances. When substances with ionic
bonding dissolve in water, the ions become hydrated, and appear in solution as cations and anions.
The differences in solubility of "salts" depend on the relative attraction of the ions for each other,
versus their tendency to become hydrated into cations and anions. In Section 2.7.5, we looked at
procedures for calculating the solubility of sparingly-soluble salts in water.

One of the most useful characteristics of ionic solutions is their ability to conduct electricity.
The passage of an electric current through an aqueous solution causes ions to migrate to one
electrode or the other, and when a positive ion (a cation) receives an electron at the cathode, it is
reduced to a cation with a lower charge, or an element. Similarly, an anion can be oxidized at the
anode. Solutions formed from substances which ionize completely when dissolved in water are
called strong electrolytes, and the solutions are relatively good conductors of electricity. Acids
such as WC\(aq) and H2S04(ag) and bases such as KOH(aq) are strong electrolytes, and ionize
completely.

Mole fraction and mass percent are in common use as designators of solute composition.
However, there are two additional terms in common use: molarity, designated M, and molality,
designated 6.*

The official SI symbol for molality is b to avoid confusion with the symbol m for mass. However, virtually
all chemistry texts use molarity (M). They designate m for molality.

390 Chapter 6 Reactive Material Balances

7 . moles of solute
M = molanty =
[6.117]

liters of solution

7 .. moles of solute r^,,o-,
о = molanty = [6.118]

kg of solvent

The molarity is used often by chemists and in some tables of solubility and equilibrium

constants, but unfortunately, is less useful for making a material balance because the mass of the

solvent is undefined (but can be calculated if the density of the solution is known). The two

concentration terms approach each other in value as the solution becomes more dilute (assuming

that one kg of water has a volume of one liter). Since one kg of water contains 55.5 moles of

water, Xsoiute = 0.0177 in a 1 molai solution.

The activity of a solute was given in Equation [2.23] as the product of an activity coefficient
and the mole fraction, where the activity = 1 at unit mole fraction. The activity of a dissolved ion
is more complex than for a metallic solution. The first complexity is that the standard state for an
ionic solute cannot be unit mole fraction, so an alternate standard state must be chosen. As a
general rule, the activities of ions are given in terms of molalities. Since ionic solutions are non-
ideal, the activity is not one at unit molality. Instead, we refer to a reference state of infinite
dilution, where ionic activity coefficient approaches one as the molality approaches zero. This is a
form of Henry's law as discussed in Section 2.7.2. The second complexity is that there is no way
to measure the activity of one ion in solution. Instead, one measures an entity called the mean
ionic activity coefficient, γ±, which value indicates the deviation from ideal behavior of the
electrolyte. The third complexity is that the activity coefficient of the electrolyte can deviate
significantly from one as the molality exceeds about 0.004, and there is no simple model that that
accurately correlates ionic activity coefficients in solutions. Figure 6.71 shows some typical values
of γ± in aqueous solution at 25 °C. The fourth complexity is that most aqueous process solutions
contain multiple ionic species, and the activity of one electrolyte is affected by the presence of
others. Finally, the activity of the solvent water can also deviate significantly from ideal behavior.

Typical engineering chemistry texts devote several chapters to the properties of aqueous
solution, the electrochemical reactions between them, and the thermodynamic equations that relate
electrode voltage and values of the equilibrium constant. Therefore, there is no need to repeat that
coverage here. The discussion to follow reviews the main principles, with a few examples of their
application to materials processes. There are several excellent references on aqueous electrolytes
and electrochemical engineering that cover the subject in depth, and contain many examples of
ionic calculations (Wright 2007, Prentice 1991, Harned et. al 1958, Lobo 1989, Hamann et. al.
2007, Bagotsky 2005). References are also available on the applications of these techniques to
hydrometallurgy and electrometallurgy (Burkin 2001, Havlik 2007, Osseo-Asare 2007. Han 2002,
Venkatachalam 1998).

6.11.1 The Stability of Ions

We have previously used the equilibrium constant as an indicator of the activities of reactants
and products in a chemical reaction For gases, we used partial pressure as an indicator of the
activity, assuming ideal gas behavior. The equilibrium constant is also a valuable function for
ionic solutions, but the activity term is replaced by the molality, which is designated in a Keq
expression in brackets, such as [H+]. For reasons discussed in the previous Section, tabulated ^eq
data are accurate only for dilute solutions. More concentrated solutions require specialized non-
ideal solution models to calculate deviations from ideality. However, the solution models have
limitations, so for the most critical applications, laboratory data must be used. But even though
Кщ data may not be very accurate for concentrated solutions, using it is better than nothing, and is
often adequate under circumstances where the inaccuracy can be tolerated.

Chapter 6 Reactive Material Balances 391

Mean Ionic Activity Coefficient at 25 °C

1.4

—O—HCI О H2S04
_+_CuS04 — X - NaCI
- -Ж- -ZnCI2 —О -ZnS04

*~->k--*"

-ж. Ж . -ж

■o- оa
О--o-o

0.001 0.01 °·1 molality 1 10

Figure 6.71 Mean ionic activity coefficients of some common strong electrolytes as a function of
molality.

FREED does not contain data on ionic species, but several other sources are available, such as
Fact-Web's Reaction-Web program (see citation in General References). Most convenient are the
typical data-source handbooks listed in the that section of the Handbook. However, the most
convenient source of ionic data are commercial thermodynamic data programs (see citations in
General References) which have tabulations for virtually all species of interest, and can calculate
the value of .Keq for ionic reactions at different temperatures. There is a convention for assigning a
reference for all ionic reactions: the value for Keq of H+(aq) from H2(g) at one atm. is arbitrarily
set at zero. Any requirement by the software that the Н7Н2(§) couple is required in the ionic
reaction to balance the charges has no effect on the value of Keq. The two programs require
slightly different ways of writing the formation reaction:

FactSage: H2 + S + 202 = H[+] + HS04[-]; iogKeq at 25 °C = 132.421 [6.119]

HSC: H2(g) + 2S + 402(g) + 2e- = 2HS04(-a); IogKeq at 25 °C = 264.650 [6.120]

Here, the two programs give very similar (but not identical) results, but sometimes the
differences are greater. Once we know Keq for the standard formation of the ionic species from
the elements to an ideal solution, we can combine the formation reactions to find Keq for any ionic
reaction. Another source of useful data on the degree of ionization of sparingly-soluble ionic
solids is the solubility product, discussed earlier in Section 2.8.5.

Another source of data are tables of standard reduction potential (designated E°) for half
reactions. Such tables are found in many chemistry texts and handbooks, usually only for 25 °C.
By definition, E° for the H2/H+ couple is zero. Suppose we were interested in the ionic equilibrium
between ferrous and ferric iron, and had values for the reduction potentials for ferrous and ferric
ions forming Fe.

¥e1+(aq) + 2e- -»■ Fe; E° = -0.440 [6.121]

Fe3>tf) + 3e- -» Fe; E° = -0.036 [6.122]
Fe3+(o?) + e- -»■ ¥e2+(aq); E° = -2(-0.440) +3(-0.036) = 0.772 [6.123]

The conversion between the standard reduction potential and logÄTeq is: [6.124]
2.303ÄTlog£eq = nE°F

392 Chapter 6 Reactive Material Balances

where n = number of electrons transferred, and F = the faraday constant. F equals 96 485
coulombs of charge per mole of electrons when R is in J/(mol · deg), and 23 060 when 7? is in
cal/(mol · deg). For the reduction of Fe3+(aq) to Fe2+(ag), one electron is transferred, so logA^eq =
13.05 for Equation [6.123].

We have two choices for finding (and using) data for making material balances involving
ionic reactions. First, use one of the commercial thermodynamic programs to find a value for Кщ
for any ionic species that's in the database, and then combine the Кщ values for any reaction of
interest*. Second, refer to typical chemistry texts or handbooks to find suitable half reaction
voltages or Кщ values, and combine them appropriately for the overall reaction. (Remember that
in combining reactions, we add/subtract values of the logarithm of Кщ.) Unfortunately, most
handbook tables are at 25 °C, so aren't useful for processes at other temperatures. For short range
temperature extrapolations of Кщ, you can use the van't Hoff equation:

LogKeqj -LogKeqTo + ^ 2 ^ {Υτο'Υτ) ^6Λ25^

where R = 8.315, and To refers to the reference temperature of 298.15 K.. This equitation neglects
the ACp of the reaction, which is seldom known to any degree of accuracy. In the absence of ACp
data, the commercial databases use the Criss-Cobble method to estimate ACp in calculating ionic
properties at higher temperatures.

6.11.2 Aqueous Processes

Virtually all hydrometallurgical processes can be put into one of four types of chemical
reactions involving solutions.

• Dissolution reactions

• Precipitation reactions

• Acid - base reactions

• Oxidation - reduction reactions.

Dissolution reactions are important in leaching, while precipitation reactions are used to
remove certain (desirable or undesirable) components. Acid-base and oxidation-reduction
reactions are used to control the composition of certain species to facilitate selective leaching,
solution purification, and component recovery from solution by means of dissolution and
precipitation. Writing and balancing ionic reactions is a little more difficult because electrons must
be conserved. Please review the subject by looking back to Section 5.10.

In the simplest case, the dissolution of a strong electrolyte results in complete ionization. For
value of Кщ
example, the ionization is for the dissolution of A gaN10m3 oislaliarsgoelu, tsioonwohfenAgo+n(eaqm)oalendisNa0d3d~e(dagto).onIne
kg of water, complete, and we form

terms of an equilibrium constant:

[Ag+][N03-] [6Л26]
a(AgN03)

where the denominator is the activity of AgN03(c). At saturation, the activity of AgN03(c) is one.

Similarly, if we add one mole of HC1 to one kg of water, we would expect the solution to ionize
completely to H+(aq) and Cl~(aq) with a unit molai concentration of each ion. We mentioned
earlier in Section 5.10 that the proton Yf{aq) is actually hydrated, and is more precisely described
Н30+(ад), but it is usually designated simply as H+(aq). The negative log
as a hydronium ion, is given a special designation: pH. If the mean ionic activity coefficient of
the U+(aq) molality for
HC1 was 1, we would expect the pH of a 1 b solution of HC1 to be 0.00. However, as shown in

Figure 6.71, the mean ionic activity coefficient for HC1 is about 0.8 at unit molality, so the pH is

Both FactSage and HSC can calculate E° and Ä^eq for ionic reactions and ionic equilibria.

Chapter 6 Reactive Material Balances 393

actually about 0.09. We designate HCl(ag) as a strong acid because HCl(g) ionizes completely on
dissolution.

Water itself dissociates only very slightly to ¥t(aq) and 0¥T(aq), with Кщ = 1.01 x 1(Г14 at
25°C. The dissociation increases with temperature, which means that the pH of water decreases
with temperature.

The dissolution of some acids is not as simple as that of HC1 if ionization occurs in steps.
Sulfuric acid for example is diprotic, which means that it ionizes in two steps. Suppose various

amounts of H2S04 are added to one kg of water. The first step is the complete dissociation of the
strong electrolyte acid to H+(aq) and RS04~(aq), so we are sure that the molality of U+(aq) is at
least equal to the molality of the acid. But what about the dissociation of HSOfiaq) to H+(aq) and
S042~(aq); could it occur to the extent that it contributes significantly to the molality of YC{aq)l

To find out, we need to calculate the equilibrium molalities of all three ionic species:

+ S042~(aq); [H + ] [ S Q 2 - ]
4
HSCVH) -> n\aq) Keq = [6.127]
[HSO4-]

If è is the sulfuric acid molality, then 26 = [H+] + [HS04~] and b = [HS04~] + [S042~].

Keq (26-[HS04"])(6-[HS04~]) [6.128]
[HSO4-]

Now we need a value of Кщ for the dissociation equation. Section 6.10.1 showed three
methods for finding Кщ. From HSC, Кщ = 0.012 at 25 °C. Alternatively, we can use handbook
reference data to manipulate half reactions to give E° for Equation [6.127] = -0.113, so from
Equation [6.124] logfeq = -1.911, and Кщ = 0.012, as above. Goal Seek can find a solution for

Equation [6.128] for any input value of 6, from which we can calculate the molality of each ion.

Figure 6.72 shows the molality of each ionic species and the pH of the solution of sulfuric acid

containing up to 0.02 molai. These results show that as acid molality increases, HSS00424~~((aagq))
becomes the predominant sulfur-containing ionic species, with a nearly constant

molality. You may want to check these calculations using the Fact-Web program Aqualib-Web

(see citation in General References).

0.030 Ionization of H 2 S 0 4 at 25 °C 2.5
ж 2.4
0.025 2.3
жч 2.2
$ 0.020 2.1

1 0.015
с
2 0.010

0.005

0.000

0.004 0.008 0.012 0.016 0.02

molality of sulfuric acid

Figure 6.72 Species molality resulting from the ionization of H2S04(ag) at 25 °C. Results based
on the assumption of ideal behavior of the ionic solutes.

394 Chapter 6 Reactive Material Balances

The legend of Figure 6.72 contains a very important caveat: the results are based on assuming

ideal ionic behavior. But in fact, as shown by Figure 6.71, H2S04(ag) departs significantly from
ideal above about 0.015 molai. Thus Equation [6.128] increasingly misrepresents the relative
proportion of HS04~(ag) and S042~(aq) above this molality. For accurate work at higher molality,
we must resort to a non-ideal solution model such as employed by FactSage, or make laboratory

measurements.

There are wide variations in the properties of solutions formed when salts or oxides dissolve
in water, and the nature of these variations are well-covered in typical chemistry texts. Some
substances have no effect on pH, while others raise or lower it from the neutral value of 7.0. We
now look at some of these variations in a situation common in hydrometallurgy — the leaching of
a solid. A good example is the leaching of pyrite, a common mineral present in transition metal
ore bodies. Water is percolated through a heap of ore, collected as it exits, and returned to the top
of the heap. After several months, the ore is depleted in soluble value, and the pregnant solution is
treated to recover the valuable solutes. We discussed the balanced chemical reaction for the
leaching of pyrite in Example 5.10. You may wish to review this example before reading further,
especially Equation [5.71].

If the ore body is amenable to concentrating, the fine concentrate particles dissolve much
more rapidly, especially if leaching is carried out at elevated temperatures. Here we look more
closely at the quantity of ions produced by oxidative leaching of pyrite in water at 75 °C, and see if
balanced reaction [5.71] gives an adequate picture. Our calculations will be based on an initial
mass of one kg of water to make it easy to calculate the molality. Pyrite is added in excess and
oxygen is the limiting reactant, so the extent of pyrite reaction is controlled by the amount of 0 2
added:

FeS2(c) + H20(/) + 3!/202(g) -+ Fe2+H) + m\aq) + 2S042"H) [6.129]

An alternate reaction is:

FeS2(c) + H20(/) + 31/202(g) -► Fe2+(aq) + 2HSOi(aq) [6.130]

In either case, the addition of one mole of oxygen causes 0.2857 moles of pyrite to dissolve to
form 0.2857 moles of Fe2+(aq) and consume 0.2857 moles of water. Equation [6.129] will
decrease the pH of the solution, but Equation [6.130] should have no effect on pH. From an
experimental standpoint, the pH could be used to calculate the relative extents of the two reactions,
provided that only [6.129] and [6.130] occur.

Equation [6.127] can be used to calculate the molality of S042~(aq), US04~(aq), and U+(aq)
from S and H balance equations. It's appropriate to make the quite reasonable approximation that
the moles of each ion are equal to their molality by neglecting the small change in mass of water as
the oxidation takes place. On a basis of one mole of 0 2 added:

6HS04" + 6S042_ - 2/3.5 = 0.5714 [6.131]
bU+ = bS042~ [6.132]

Next we need data at 75°C. We choose (arbitrarily) values from the HSC program at 75 °C to
obtain the following equation:

484=0.5714-fV-] [6.133]
[so42-]2

This gives [S042_1 = 0.0333, [H+] = 0.0333, and [HS04~] = 0.538*. This shows that 5.8 % of
the leached pyrite went to form S042", so when one mole of 02 is added, the extent of pyrite
reaction according to Equation [6.129] was 0.0167. This shows that Equation [5.71] accounts for

only a small amount of the leaching process but all of the change in pH.

Remember that a term enclosed in brackets [ ] refers to its molality.

Chapter 6 Reactive Material Balances 395

The pH of the solution is calculated from Equation [6.132] as 1.48. This calculational
procedure was repeated for different amounts of added oxygen up to the point where one mole of
pyrite was leached, with the results shown in Figure 6.73.

Aqueous Leaching of Pyrite at 75 °C 9%

1.7 | - 1 1
1.6
Δ ^^pH 8%
1.5
\4 — о - % to form [S04--] ω
a 1·4 .оCсD
чE > 6%о7% <NM-
1.3 CD

1.2 CO CO
CD +о5
1.1 +■ b% LL
0
^ '—-w 4% ^
3%
^ ^ — c ■^ D

0 2 added, moles

Figure 6.73 Calculational analysis of the oxidative leaching of pyrite at 75 °C as a function of the

amount of oxygen added to a system containing initially 1 kg of water. One mole of FeS2
dissolves when ЪУг mole of 0 2 is added. The right-hand axis represents the percentage of pyrite
leached according to Equation [6.129].

There are other ions that might form during the dissolution of pyrite that might affect the
assumption we made that all of the iron dissolves as ¥c2+(aq). Hydrated ferrous iron, FeOH+(ag)
may form, and possibly even Fe3+(aq). We can check if FeOH+(ag) is stoichiometrically
significant with Equation [6.134].

Fe2+H) + H20(Z) -> FeOH+H) + U\aq) [6.134]

HSC data gives logÄ^eq for [6.134] as -8.016. For the condition of one mole of 02 added, we
know the molality of Fe2+(aq) and H+(aq). Since the solvent water has a mol fraction >0.98, we

can reasonably set its activity equal to one.

9.64x10- = [FeOH-][H-] = [FeOH+][0.033]
[Fe2+](>H20)
[0.286](l)

This gives [FeOH+] = 8.4 χ 10"8. Therefore, we can neglect FeOH+(ag) as a factor in the
molality of iron-containing ions formed when pyrite dissolves. A similar calculation shows that
the molality of ferric iron or hydrated ferric iron is even less than that of FeOH+(ag).

EXAMPLE 6.16 — Leaching of Scrubber Dust.

A pollution control residue consisting of CuO, Fe203, and other components is collected in
scrubber water. The solids are filtered and sent to a copper smelter for treatment. However, owing
to increased shipping cost and low payment for the residue, the plant engineer would like to
recover the copper values on site. A metallurgy text indicates that if the residue can be leached in
sulfuric acid, the copper sulfate can be recovered by passing it over scrap iron, a process called
cementation. A lab test indicated that the CuO dissolves rapidly and completely if enough sulfuric
acid is added to give the leach solution a pH of 2.0 at 30°C, but some of the Fe203 might also go
into solution. The other components are inert in the leaching process. If the pH gets much below
2.0, too much scrap iron starts dissolving during cementation, and scrap might consumed in
reducing the ferric iron back to ferrous. Make calculations at 30 °C to estimate how much Fe203

396 Chapter 6 Reactive Material Balances

goes into solution and what the solution pH is as a function of the amount of H2S04 added to the
scrubber water. Base your calculations on a sample which contained 1.90 kg of water and 200 g of
solids. The solid contains wCu = 21.8 % and wFe 18.1 %. A preliminary check on ferric ion
species in sulfate solutions indicated that FeOH and Fe are negligible compared to FeS04

Data. Log(A^eq) was obtained from FactSage at 30 °C for all relevant species.

Species HS04["] so4[2] FeS04[+] Cu[2+] Fe203(c) CuO(c) H20(/)

Log(A:eq) 129.86 127.73 132.66 -11.29 127.95 22.03 40.73

Solution. The solid contains nCuO = 0.686 and /?Fe203 = 0.324 (54.6 grams of CuO and 51.7
grams of Fe203). The other solids are inert. For orientation purposes, the following two reactions
describe the dissolution of CuO and Fe203 in the acidified water:

CuO(c) + H2S04(/) -> C u S 0 4 H ) + H20(/) [6.136]

Fe203(c) + 3H2S04(/) -> Fe2(S04)3H) + 3H20(/) [6.137]

We glean from the problem statement that CuO dissolves preferentially compared to Fe203,

and that the extent of H2S04 reaction in [6.136] is effectively one. It must actually be slightly less
than one because a little acid simply dissolves to form HSOzf(ag), H+(aq), and S042~(aq).

However, considering the error introduced by assuming ideal ionic activity, the assumption of the

extent of dissolution of CuO being one is quite reasonable. The stoichiometry of CuO dissolution

indicates that 18 g of water are produced per mole of CuO dissolved. This information allows us

to calculate the pH of the leach solution for any addition of H2S04 according to the following

equilibria:

CuO(c) + m\aq) -> Си2+И) + H20(/) [6.138]

Kveq- 2if.,6x1ι0π7 = [Cuz+](aH2Q—) [6.139]
[H+]2(aCuO)

The [Си,2+]Ί is obtained by dividing the moles of CuO dissolved by the mass of water, which
can be set as 1.91 kg throughout dissolution to account for the small amount of water that is
produced via Equation [6.136]. For example, 50 g of H2S04 will dissolve 0.51 moles of CuO, so
the [Cu2+] = 0.267 b. Referring to Equation [6.137], since we haven't dissolved all of the CuO by
the addition of 50 g of H2S04, the activity of CuO = 1, and the activity of water is only slightly
diminished from one because the ionic solutes are so dilute. Therefore, [H+] = 1.0 χ КГ4, for a pH
of 4.00. The pH continues to drop as more acid is added (and more CuO dissolves), until all of the
CuO dissolves by the addition of about 67.3 grams of H2S04. At the last trace of CuO, [Cuz2+η] =
0.359, and Equation [6.139] gives pH = 3.93.

We now check to see how much Fe203 has dissolved at the point where CuO has just
disappeared or when about 67.3 g (0.686 moles) of H2SO4 was added. We can do that by
calculating the [FeS04+] in solution. First, we need to know [HS04~] and [S042~].

CuO(c) + H S 0 4 " H ) + H\aq) -> Cu2\aq) + SO4 (a?) + H20(Q [6.140]

Keq = 1.93x10s [Cu2+][SQ42-](aH2Q) [6.141]
(aCuO)[HS04~][H+]

aCuO = 1 (because there is a trace present), the aH20 « 1 (because the solution is so dilute),
[Cu2+] = 0.359 (because «99 % of the CuO has dissolved), and [H+] = 1.18 χ 1QT* (because pH =
3.93). We also know that [S042~] + [HS04~] = 0.686/1.91 = 0.359 (because we added 0.686 moles
of H2S04). Solving, [S042~]= 0.353 and [HS04-] = 0.0054. Next, we write an ionic equation
relating these species, Fe203(c) and [FeS04+]

Fe203(c) + 3HS04"(a^) + 3H\aq) -> 2¥eS04\aq) + S042-(aq) + 3H20(Z) [6.142]

Chapter 6 Reactive Material Balances 397

**-5.1xl(T'- Р^°^]'Р°«'Ч [6.143]
(aFe203)[H*]3[HS04-)3

After making appropriate substitutions, [FeS04+] = 6.3 χ IO"11. This shows that a negligible
amount of Fe203 has dissolved at the point where the last trace of CuO has just dissolved.

Adding a slight excess amount of H2SO4 over the stoichiometric amount would eventually
take all the CuO into solution, while hematite will remain essentially undissolved, as just shown.
However, as a reaction approaches equilibrium, the rate slows, and the problem statement indicates
that in order to attain complete CuO dissolution in an acceptably short time, excess H2S04 is
required, possibly enough to bring the pH to 2 to get all the CuO dissolved. The last question to be
answered is: how much hematite will dissolve as the pH is lowered? The ionization of added
sulfuric acid, will not be affected by the presence of Cu2+ nor by FeS04+ ions because [FeS04+] is
so low. However, it will be affected by the presence of the existing H+, HS04" and S042~.

Sulfuric acid ionizes in two steps, as described in Section 6.10.2. First, [HS04~] will increase

by the molality of the H2S04 that's added in excess ofthat needed to completely dissolve CuO. In
the second step, HS04" will either increase or decrease to form the equilibrium amounts of S042"
H+ according to Equation The new H+, HS04" and 2_
and [6.127]. S0 4 ion molalities will then react
with Fe203 to form FeS04+ according to Equations [6.142] and [6.143].

Suppose we add H2S04 to the solution at the point where CuO has just dissolved, at which
point [S042~] = 0.353, [HS04~] = 0.00548, and [H+] - 1.18 χ Ю-4. We define V as the added
molality and Ab as the change in molality caused by Equation [6.127]. Note that Ab can be
positive or negative; we don't know beforehand. The values of H+, HS04" and S042" can then be
calculated by a procedure similar to the one we used in calculating their amounts when H2S04 by
itself underwent secondary ionization. We write an equation somewhat similar to [6.128]:*

Кщ = 0.000747 = [*042--Δ6][Η++*'-Δ*] ^щ
[HS04"+Z>' + Aò]

If 20 g of H2S04 is added beyond that needed to dissolve all the CuO, then V = 0.107.
Solving, Ab = 0.101, [S042-] = 0.252, [HS04~] = 0.214, and {H+] = 6.12 χ Ю-3, so pH - 2.21.
These values can be inserted into Equation [6.143] to give [FeS04+] = 7 χ IO-6, still a very small
number. This calculation was repeated for a series of H2S04 additions, with the results shown in
Figure 6.74. Even up to a total of 100 g of added H2S04 (32.7 g more than needed for dissolution
of all of the CuO), [FeS04+] was only 4.6 χ 10-5. This is a very small amount, equivalent to the
dissolution of 0.007 g of Fe203. Thus, ferric iron will not be a significant factor in the
consumption of iron during cementation of copper from a leach solution with a starting pH of

about 2.

Assignment. Suppose hydrochloric acid (added as an aqueous solution of wHCl = 35 %) was used

as a leaching agent at 30 °C instead of H2S04. See if CuO completely dissolves before the
dissolution of any appreciable amount of hematite, and calculate the amount of acid required.
What is the pH at the point of complete CuO dissolution? The relevant Keq data is: Cl~(aq) =
22.51; CuCh~(aq) = 41.66; FeCl2'(aq) = 49.34.

The [FeS04 ] is so small that it can be neglected in the ion balance.

398 Chapter 6 Reactive Material Balances

Figure 6.74 Changes in solution properties as a result of leaching CuO from a scrubber dust. The
initial mass of water was 1.90 kg. The CuO is completely dissolved by the addition of 67.3 g of
H2S04.

6.11.3 The Solubility of Ionizable Gases in Water
In Section 2.8.3, we showed that non-ionizable gases like H2 and N2 are sparingly soluble in

water. They dissolve sparingly as the gas molecule, and as such, are slow to react with other
solutes. The low solubility of H2 for example, has frustrated commercial attempts to use it to
reduce metal cations (such as Ni2+) to metal, even at high /?H2. Weakly ionized gases dissolve
primarily as the gas molecule (which may be hydrated in solution), but also form ionic species in
sufficient concentration to participate in ionic reactions. For example, at 30 °C and/?C02 = 2 atm
in equilibrium with water, C02(aq) = 0.060 b. At the same time, the molality of [H+] and [HC03~]
from dissociated C02(aq) = 1.6 χ 10"4. Owing to their predominately molecular dissolution, the
solubility of weakly-ionized gases follows Henry's Law, which states that solubility is a linear
function of the partial pressure of the gas in equilibrium with the water-gas solution*.

Although the solubility of weakly-ionized gases is small, their solubility and ionization plays
an important role in hydrometallurgy. For example, the kinetics of a reaction between an aqueous
ion and a gas is much faster if the gas dissolves (and at least partly ionizes) in water. If necessary,
the system pressure can be raised to enhance the dissolution. The precipitation of heavy metals as
sulfides from process streams can be accomplished by the addition of a small amount of H2S(g).
The precipitation of very small amounts of solids is enhanced by causing the co-precipitation of a
larger amount of some innocuous substance that is easily filtered from the aqueous phase. Heavy
metals either co-precipitate or adsorb on the surface of the finely-divided precipitate. The
precipitate can "flocculate", thus making it easy to filter from the cleaned water.

EXAMPLE 6.17— The Optimum Precipitation ofCaC03 by C02.
Wastewater containing very small amounts of heavy metal solutes of unknown speciation is

being treated by carbonate precipitation. Figure 6.75 shows a simplified flowsheet of the process.
Make a material balance on the process to determine the effect of added C02(g) to the molality of
solutes and the amount of carbonate precipitated in the carbonate precipitation reactor at 25 °C.
The basis is 1 kg of wastewater, so the concentrations are conveniently expressed in molality.

Remember that Henry's Law is a limiting law, so deviations may become significant at higher pressures.

Chapter 6 Reactive Material Balances 399

Lime-satd. Clean discharge
" " * water
CaO Lime water ->| Carbonate
Wastewater dissolution 3 pptn.
+\ reactor
- i j CO,-?- 1reactor
CaC03 + impurities
Ca(OH)2

Figure 6.75 Flowsheet for the removal of trace amounts of heavy metals from a wastewater

stream at 25 °C. Excess lime is added to the lime dissolution reactor to saturate the water with
Ca2+(ag) and OH (aq). Undissolved Ca(OH)2(c) is filtered out in stream #4. Stream #3 carries the
lime-saturated aqueous phase to a precipitation reactor where C02(g) is injected into the water to
precipitate CaC03(c) and co-precipitate heavy metal carbonates. The clean discharge water is
limited to a maximum [Ca2+] of 0.0002.

Data: The table below lists AG0form of relevant species at 25°C in kJ/mol. Кщ for various
relevant reactions is obtained by division of AG°form of the reaction by 5.709.

Ca(OH)2(c) СаСОз(с) C02(g) COi(aq), Са2+И) ОН" И ) R20(liq) HCOf(aq)

-897.5 -1129.1 -394.4 -386.0 -553.5 -157.3 -237.10 -586.85

Solution: CaO(c) forms Ca(OH)2(c) when added to water, and a small amount dissolves according
to Equation [6.145]. A little excess CaO is added via stream #1 to be sure that stream #3 is

saturated with calcium hydroxide.

Ca(OH)2(c) - Ca2+(aq) + 20U (aq); Кщ = 6.97 χ IO"6 [6.145]

Stoichiometry shows that [OH-] = 2[Ca2+], so in stream #3, [Ca2+] - 0.012 and [OH-] = 0.024.
Also, the pH in stream #3 = 12.31, based on an ionization constant for water of 1.01 χ IO"14. The

precipitation process in the second reactor involves adding C02(g) (stream #5) to the aqueous
phase to form CaC03(c). C02(g) dissolves to form C02(aq), which in turn reacts with Ca2+(aq) to

precipitate CaC03(c).

C02(g) - C02(aq); ^eq - 0.034 [6.146]

Czt\aq) + C02(aq) + 20U (aq) - CaC03(c) + H20(/); ^eq = 4.60 x 101 [6.147]

In the early stages of C02 addition, the C02(aq) stoichiometrically "titrates the C2L2+(aq).
The reaction is spontaneous and virtually complete because of the large value of Кщ. Thus (for
example) the dissolution of 0.01 mole of C02(g) lowers [Ca2+] by 0.01, precipitates 0.01 mole of
СаСОз(с), and lowers [OH~] by 0.02. The [Ca2+] and [OH] decrease as more C02(g) is added,

while at the same time, the pH increases in order to keep the ionization constant for water at 1.01 χ
IO14. As the amount of added C02(g) approaches 0.012 mol, the [Ca2+] and [OH-] become very
small, and HC03_(ag) begins
some point, adding "excess" to form in significant amounts, according to Equation [6.148]. At
C02(g) causes an increase in U+(aq) according to Equation [6.149],

and CaC03(c) begins to re-dissolve as shown by Equation [6.150].

CaC03(c) + C02(aq) + H20(/) Са2+И) + 2HCOf(aq); Кщ = 4.12 x 10-5 [6.148]

C02(aq) + H20(/) - llCOf(aq) + n+(aq); Кщ = 4.38 χ 10" [6.149]

CaC03(c) + IT (aq) - Ca2z\+a/ q) + HC03~(aq); Кщ = 94.3 [6.150]

Thus there is a certain minimum value of [Ca2+] when the optimum amount of C02 has been
added to the carbonate precipitation reactor. At this point, the amount of precipitated CaC03(c) is
a maximum, and approaches 0.012 mol. Table 6.15 shows the molality of each solute calculated as
a function of the amount of C02(g) added to the reactor. The assumption is that every increment of
added C02(g) removes the same increment of Са2+(<яд) and double that amount of OH (aq). The
other solute molalities are calculated as dependent variables with an appropriate reaction Кщ.
Values are slightly in error above 0.0115 mol C02(g) added because of the increase in НС03"(яд).

400 Chapter 6 Reactive Material Balances

However, it is clear that the minimum solubility of CaC03(c) occurs when about 0.0119 moles of
C02(g) have been added.

Table 6.15 Solute concentration (molality) in aqueous solution as a function of amount of C02(g)
added to a solution initially containing [Ca2+] = 0.012 and [OH-] = 0.024. In the absence of a gas
phase, all of the added C02(g) enters the solution.

Added C02 [Ca2+] [OH'] CaC03 [H+] pH [HCO3-] [COaq\ pC02
0
0.0000 0.01200 0.02400 0 4.21E-13 12.4 0 0

0.0020 0.01000 0.02000 0.0020 5.05E-13 12.3 4.76E-09 5.43E-15 1.60E-13

0.0040 0.00800 0.01600 0.0040 6.31E-13 12.2 7.44E-09 1.06E-14 3.12E-13

0.0060 0.00600 0.01200 0.0060 8.42E-13 12.1 1.32E-08 2.52E-14 7.40E-13

0.0080 0.00400 0.00800 0.0080 1.26E-12 11.9 2.98E-08 8.49E-14 2.50E-12

0.0100 0.00200 0.00400 0.0100 2.53E-12 11.6 1.19E-07 6.79E-13 2.00E-11

0.0110 0.00100 0.00200 0.0110 5.05E-12 11.3 4.76E-07 5.43E-12 1.60E-10

0.0115 0.00050 0.00100 0.0115 1.01E-11 11.0 1.90E-06 4.35E-11 1.28E-09

0.0117 0.00030 0.00060 0.0117 1.68E-11 10.8 5.29E-06 2.01E-10 5.92E-09

0.0118 0.00020 0.00040 0.0118 2.52E-11 10.6 1.19E-05 6.79E-10 2.00E-08

0.0119 0.00010 0.00020 0.0119 5.05E-11 10.3 4.76E-05 5.43E-09 1.60E-07

In order to more accurately calculate the minimum value of [Ca2+], we must properly calculate
the solute concentrations of other species that are present in significant amounts. We do this at a
certain condition, which is equivalent to the solute concentrations when CaC03(c) is immersed in
pure water. Four solutes are expected to be present in amounts sufficient to be included in a
material balance: Ca2+(ag), OH (aq), HCOf(aq), and C02(aq). We therefore need four equations,
two of which can be obtained from stoichiometry considerations and two Keq equations. The
stoichiometry relations are written as mole ratio equations, recognizing that one mole of H20(/) is
consumed for each mole of CaC03(c) that dissolves:

Ca [Ca2+] 1/ [6.151]
H [OH-] + [HC03-]

Ca [Ca 2 + 1 =1 [6.152]

<- [HC03-] + [C02«7] [6.153]
[6.154]
CaC03(c) + H20(/) Ca2\aq) + НС03~И) + OH'(flg); Кщ = 9.83 χ 10~13

СО2И) + OH'(aq) - nCOf(aq); Кщ = A25 x 107

Solver was able to find a solution to these four equations, as shown below. The pH was 9.99,
which verifies that it is safe to neglect [H+] as a factor in this particular situation. The equivalent
/?C02is6.92 x 10~7atm.

[Ca2+] [C02aq] [HCO3-] [OH"]

9.944E-05 2.352E-08 9.942E-05 9.946E-05

At the point of maximum precipitation of CaC03(c), the precipitation tank contains 0.0119
mol of CaC03(c) and [Ca2+] is only about 0.0001. With the addition of the next trace of C02(g),
the pH increases slightly such that OH ~(aq) is no longer a significant contributor to the material
balance. From that point on, each increment of added C02(g) causes more CaC03(c) to dissolve,
and increases the amount of HC03~(ag), C02(aq), and Ca2+(ag). We can use Equation [6.148] to
express the CaC03(c) dissolution reaction.

First, from a Ca balance we note that the increment in [Ca2+] equals the increment in
CaC03(c) dissolving, and since at the starting point, the CaC03(c) remaining un-precipitated is
equal to [Ca ], we can set the [Ca,2+-] equal to the amount of CaC03(c) "in solution". Second, the

Chapter 6 Reactive Material Balances 401

stoichiometry of [6.148] indicates that two moles of HC03~(ag) form for every mole of Ca2+(aq).
Third, from a carbon balance, the amount of CC>2(g) added plus the amount of CaC03(c) dissolved
equals the amount of HC03~(ag) + the amount of C02(aq) in solution. These relationships allow
various algebraic substitutions, which results in one final expression for the Кщ of [6.148].

Keq = 4A2xlO~:> =- V2[HCOfY [6.155]
C02added-1/2[HC03"]

Goal Seek was used to calculate [HC03~] for a range of amounts of added C02(g), from which
all other solute compositions were calculated. Table 6.16 shows the results up to a/?C02 of nearly
1 atm. Figure 6.76 shows the solute changes that take place during limestone precipitation and

dissolution.

Table 6.16 Effect of added C02(g) to a solution containing initially [HC03~] and [Ca,2z"+]Ί = 0.0001
at 25 °C. Initial solution composition generated by dissolution of CaC03(c) in pure water; amount
of CaC03(c) initially present = 0.012 mol. No gas phase is present; the pC02 represents the
escaping pressure of the gas.

C02(g) added [HCO3-] [Ca2+] СаСОз(с) [C02(aq)] pC02 [H+] pH
0 0.0001 0.0001 0.0119 2.40E-08 6.90E-07 1.00E-10 10
0.00185 0.00092 0.0111 0.00008 1.78E-08 7.75
0.001 0.00321 0.00161 0.0104 0.00039 0.002 5.36E-08 7.27
0.002 0.00501 0.00250 0.0095 0.00150 0.012 1.31E-07 6.88
0.004 0.00622 0.00311 0.0089 0.00289 0.044 2.03E-07 6.69
0.006 0.00794 0.00397 0.0080 0.00603 0.085 3.33E-07 6.48
0.01 0.00948 0.00474 0.0073 0.01026 0.177 4.74E-07 6.32
0.015 0.01068 0.00534 0.0067 0.01466 0.302 6.01E-07 6.22
0.02 0.01254 0.00627 0.0057 0.02373 0.431 8.29E-07 6.08
0.03 0.01399 0.00700 0.0050 0.03300 0.698 1.03E-06 5.99
0.04 0.971

Precipitation and Dissolution of Limestone

0.010

0.000 a- 0.01 0.02 0.03 0.04
0.00 C02 added, mol/kg of solution

Figure 6.76 The effect of added C02(g) on the solubility of limestone in water. The initial
addition of 0.012 moles of C02(g) (Table 6.15) causes most of the Са2+(я#) to precipitate as
limestone. Further C02(g) additions (Table 6.16) cause a dissolution of the limestone.

402 Chapter 6 Reactive Material Balances

The addition of C02(g) to a solution containing initially [Ca ] = 0.012 causes a linear
decrease in [Ca2+] with a corresponding increase in the amount of precipitated CaC03(c). From a
process control standpoint, the desired condition of maximum CaC03(c) precipitation should be to
inject C02(g) into the solution until the pH drops to about 10. Any further C02(g) addition causes
СаСОз(с) dissolution.

Assignment: Calculate the solubility (i.e., molality) of PbS04(c) in pure water at 25 °C. How is its
solubility affected by the addition of 0.001, 0.002, 0.003 etc. moles of H2S04(/j to one kg of water?

The effect of pH and C02 on the solubility of limestone in water is well understood by
geologists. The phenomenon is responsible for the formation of caverns and stalactites in
limestone. Surface water trickling into the ground, with different levels of dissolved C02, cause
the dissolution and precipitation of limestone. Other carbonates are similarly affected, but to a
lesser extent.

The solubility of C02, like all other gases, decreases with increasing temperature. This is
because the hydrate (water-gas) bond becomes weaker at higher temperature. Owing to the non-
ideality of the solute C02(ag) and a small amount of C02(g)/H20(g) interaction as the /?H20
increases, the solubility/temperature relationship is complex.

The solubility of completely ionized gases is much higher than for weakly-ionized gases. For
example, when one mol of HCl(g) is injected into one kg of water at 30 °C, it dissolves
completely. The molalities of both ¥t(aq) and CV(aq) = 1.00, while the/?HCl in equilibrium with
the one molai hydrochloric acid solution is 5.4 χ Ю-7 atm. The solubility of HC1 does not follow
Henry's law because HC1 is completely ionized in solution. S02(g) shows a more complex
dissolution process than HC1. It is also completely ionized in solution, and forms the same three
ionic species that form when H2S04(/) dissolves: U+(aq), RS04~(aq), and S042~(aq).

Admittedly, the ionic process examples in this section are somewhat contrived in that all of
the calculations were based on using thermodynamic data rather than measured data. The ionic
reactions were assumed to proceed to equilibrium, certain stable solids were ignored, and
deviations from ideal ionic solution behavior were ignored. In practice, hydrometallurgical
processes often produce metastable phases that persist indefinitely in place of stable phases whose
kinetics of formation are extremely slow. For example, elemental sulfur may form as an
intermediate phase in the acid leaching of some sulfide concentrates. However, the mass balance
principles are valid and the results of the calculations correct if the relevant stability data is correct.

6.12 Summary

This Chapter outlined a systematic approach to material balance calculations by illustrating
the use of the flowsheet and reaction-writing procedures developed in Chapters 4 and 5. The
number of independent chemical reactions (NIRx) that can be written for a system is equal to the
number of independent reacting species minus the number of independent reacting elements. For
spontaneous reactions that do not go to completion because of time or mixing constraints, the
extent of reaction of the limiting reactant must either be specified or determined from process
measurements. For spontaneous reactions that do go to completion, the amount of the limiting
reactant determines the amount of product(s) produced. For reversible reactions, the extent of
reaction must be specified or determined by plant measurements. If equilibrium is approached
closely, the extent may be calculated using the equilibrium constant. FREED is a good source of
A^eq values for substances and reactions. It's also possible to define the fractional approach to
equilibrium by using a multiplying factor on the value of ATeq. These principles were applied to
calculations involving different systems. In particular, A^eq for the water-gas shift reaction (WGR)
was used as an example to determine the relationship between amounts of CO, H20, C02 and H2 in
process gases containing these species.

Chapter 6 Reactive Material Balances 403

Material balance techniques were illustrated using two approaches. Where chemical reactions
occur, the molecular species balance approach is more cumbersome and is usually used for simple
systems. The atomic species balance is the preferred approach. Atom balances should be used for
species undergoing chemical changes, while molecular balances may be used for inert gases or
substances that do not change composition. Extensive use was made of ExcePs Solver tool and the
SuperSolver program in finding solutions to complex equation sets. Writing material balance
equations for systems with multiple units and streams is aided by a degree of freedom analysis.

The FlowBal program offers an alternate way of using chemical reactions in a material
balance. FlowBal uses the molecular species balance method because that method is more
amenable to being programmed. Reaction coefficients are entered for all of the independent
chemical reactions, and FlowBal checks that the reactions are correctly balanced. The degree of
progress of a reaction can be defined by an "extent of reaction" factor for a substance, or an
equation for Кщ can be inserted. Additional constraints on the system can also be included based
on plant data or anecdotal experience. FlowBal requires care in the selection and sequencing of the
independent reactions.

A material balance on the combustion of a fuel illustrated different techniques for reaction-
writing and equation-solving. The composition of a stack gas was calculated from data on the
amount of fuel burned with excess air, oxygen-enriched air, and moist air. Techniques for using a
"reverse" material balance showed how a stack gas analysis could indicate the amount of excess air
used. The amount of process gas used in the reduction of iron oxide is influenced by the extent of
reaction and the composition of the process gas. The amount of new gas required for iron ore
reduction is greatly decreased if the spent gas is treated and recycled. The FlowBal program was
used to make a materials balance on a two-stage fluidized-bed reduction process. Reference was
made to certain Excel-based specialized combustion-calculation templates on the Handbook CD.

Processes often require a reducing gas for the reduction of a metal from an oxide ore. The
principle raw material source is natural gas, mainly CH4, which must be reformed to a gas rich in
CO and H2. Reforming occurs in a catalytic reactor by reaction with steam or spent gas from an
ore reduction furnace. The reformed gas composition is constrained by the thermodynamics of the
C-O-H system. Calculation of the aC and p02 of reducing and carburizing gases requires the
application of chemical thermodynamics as illustrated in folder Atmosphere on the Handbook CD.
The Rist diagram was used as a tool for illustrating the change in gas and solid composition during
the reduction of iron oxide in a counter-current flow reactor. Material balance techniques were
applied to the partial and complete roasting of a sulfide concentrate.

Discrepancies in a material balance can be corrected if redundant data is present. This
requires identification of the most likely source of error in the data and an assessment of the
relative accuracies of the material flow vs. composition data. Published distribution coefficients
can be used to develop equations relating the amount of a species that is distributed between two or
more phases. This technique was illustrated for the BOF steelmaking process where chemical
equations cannot be used because of the unknown nature of the reacting species. Thermodynamic
programs such as those cited in General References may be used for more complex systems where
distribution coefficients are not available in the open literature.

For reactions that don't proceed to equilibrium, data may be available on reaction kinetics.
We can sometimes use that information in connection with the reaction rate law to determine what
fraction of a reactant is consumed as a function of the residence time in the reactor. Unfortunately,
most processes are either too complex to be described by a simple rate equation, or the presence of
particulate matter in the gas stream acts in an undefined catalytic manner, and the rate-controlling
step is undefined.

Aqueous solutions pose a different challenge to material balance calculations because of the
lack of readily-available thermodynamic data (except at 25 °C) and the need to account for the
possible presence of multiple ionic species. The activities of ionic species can deviate significantly
from ideal, and there is no simple activity coefficient model that correlates the deviations with ion

404 Chapter 6 Reactive Material Balances

concentration. The possible presence of metastable phases and the absence of stable ones in
aqueous processing shows the importance of not relying entirely on thermodynamic data for
aqueous-system material balance calculations.

References and Further Reading

Bagotsky, Vladimir S. Editor, Fundamentals of Electrochemistry, 2nd Edition, John Wiley, 2005.
Barthel, Josef M. G et. al., Physical Chemistry of Electrolyte Solutions: Modern Aspects, Springer,
1998.
Baukal, Charles E., Editor, Oxygen-Enhanced Combustion, CRC Press, 1998.
Baukal, Charles E., Editor, The John Zink Combustion Handbook, CRC Press, 2001.
Besik, Ferdinand K. et. al, Low NOx Burner, United States Patent 5,772,421, June 30, 1998.
Biswas, A. K., Principles of Blast Furnace Ironmaking, Cootha Publishing House, 1981.
Burkin, Alfred R., Chemical Hydrometallurgy: Theory and Principles, Imperial College Press,
distributed by World Scientific Publishing Co., 2001.
Davenport, William G, Jones, D. M., King, M. J., and Partelpoeg, E.H., Flash Smelting: Analysis,
Control, and Optimization, TMS, Warrendale, PA, 2001.
Fruehan, Richard, Editor, Steelmaking and Refining Volume, The Making, Shaping, and Treating
of Steel. 11th Edition, AISE Steel Foundation, Pittsburgh, PA 1998.
Hamann, Carl H. et. al., Electrochemistry, John Wiley, 2007.
Han, Kenneth N, Fundamentals of Aqueous Metallurgy, SME, 2002.
Harned, Herbert S. and Owen, Benton B, The Physical Chemistry of Electrolytic Solutions, 3rd
Edition, Reinhold Publishing, 1958.
Havlik, Tomas, Hydrometallurgy: Principles and Applications, Cambridge International Science,
2007.
Kamal, Azfar and Christenson, Danny L., Low NOx Burner Assembly, United States Patent
5,957.682, September 28, 1999.

Learning Resource on Steel Technology. World Steel Association, 2010, [Online].
http://www.steeluniversity.org.
Lobo, Victor M., Handbook of Electrolyte Solutions, Oxford : Elsevier, 1989.
Osseo-Assare, Kwadwo, Aqueous Processing of Materials: an Introduction to Unit Processes with
Applications to Hydrometallurgy, Materials Processing, and Environmental Systems, Academic
Press, 2007.
Prentice, Geoffrey, Electrochemical Engineering Principles, Prentice Hall, 1991.
Venkatachalam, S., Hydrometallurgy, Alpha Science International, 1998.

Wikipedia contributors: "Mass balance", "Combustion", "Calcination", "Stoichiometry", "Direct
reduced iron", "Hydrometallurgy", "Leaching", "Electrolyte", Wikipedia, the free encyclopedia,
December 2010. http://en.wikipedia.org/wiki/Main_Page.
Wright, Margaret R., An Introduction to Aqueous Electrolyte Solutions, John Wiley, 2007.

Chapter 6 Reactive Material Balances 405

Exercises

6.1. One mole of 0 2 is passed through a bed of (excess) carbon at a temperature such that all of the
0 2 is consumed, and the product gas has pCO/pC02 = 10. Write independent reactions for the
process, and make a material balance to determine the amount of С consumed by the 02. Calculate
the extent of each independent reaction.

6.2. A gas mixture with <jpN2 = 0.200 and <jpH2 = 0.800 flow into a reaction chamber at 10 mol/min.
Write an independent reaction for the formation of ammonia (NH3). The extent of N2 reaction is
0.264. Calculate the flowrate of ammonia and the composition of the gas leaving the reactor.

6.3. One mole of H2S is burned with 10 moles air (assume 21 %02) at 810 °C and 1.3 atm. The
products are H20, N2, 02, S02 and S03. Calculate the composition of the product gas assuming
equilibrium is reached. Values of Кщ for formation from the elements are:

^eq S02 = 3.55 x 1013 Кщ S03 = 2.49 x 1013

6.4. S02 is passed through a reactor filled with carbon at 1300 K. The product gas consists of CO,
C02, COS, and S2. A partial analysis gives:

<pCO = 0.6569; <pC02 = 0.0082

Write two independent reactions for the process. Use either the molecular species or atomic
element method to determine the volume fractions of COS and S2 in the gas, and the amount of С
consumed by the passage of one mole of S02 through the reactor.

6.5. Gold often occurs as fine particles embedded in pyrite (FeS2). In order for the gold to be
leached, the pyrite must be roasted to magnetite (Fe304). A pyrite concentrate contains various
inert gangue minerals, and analyses 48 %S on a dry basis. The roaster is fed with 22 t/h of a pyrite

concentrate that contains 8.3 % water, and roasted with 200 t/h dry air. The exit gas has a <pS03 =
0.38 %. Calculate the complete exit gas composition.

6.6. Silane (SiH4) and methyl imide (CH3NH2) gases are mixed in a 1:3.5 molar ratio and heated.
The products consist of a powder containing wSi3N4 = 66% and the rest SiC, and a gas containing
H2, NH3 and CH4. Calculate the mass of powder expected from 1 STP m3 of gas mixture, and the
composition of the product gas.

6.7. Scrap automobile batteries are recycled by breaking the battery and separating the various
components. The metallic lead grids contain 2 % Sb, balance Pb. The grid metal is melted and
refined by stirring in two tons of PbO per 100 tons of grid metal. The products are a slag that
contains PbO and Sb203 that analyzes 20 % Sb, and metal of lowered Sb (with no dissolved O).
Calculate the Sb content of the metal, and the % loss of Pb to the slag.

6.8. Gas from a BOF steelmaking furnace is collected by a sealed hood such that a small amount
of air enters the hood during collection. During the 17-minute blow, 11.2 tonnes of CO and 2.8
tonnes of C02 are evolved from the BOF, and 1.7 tonnes of air leak in. Calculate the volumetric
gas composition assuming air is 21 %02, 79 %N2.

6.9. Liquefied petroleum gas (LPG) contains wC2H6 = 0.042, wC4Hi0 = 0.057, balance C3H8. The
LPG is vaporized and metered into a burner. Calculate the volume of dry air and vaporized LPG

for combustion at 15 % XSA. What is the volume fraction air in the air/LPG gas mixture?

6.10. An operator has two sources of fuel oil. The higher quality #2 oil has a specific gravity of
0.865, and the heavier #6 oil has a specific gravity of 1.013. The analysis of the two oils is listed
below.

#2 oil: wC-0.873, wH-0.125, wS = 0.002.
#6 oil: wC = 0.884, wH-0.094, wO - 0.007, wN-0.003, wH20 = 0.002, wS = 0.010.

The #2 oil burns satisfactorily with 8 %XSA, while the #6 oil requires 11 %XSA. Calculate
the air requirement in STP ft3 per gallon of each oil.

406 Chapter 6 Reactive Material Balances

6.11. A heavy fuel oil is vaporized into a burner by steam at a ratio of 1 lb of steam per 4 lb of oil.
The fuel oil has wC = 88.5% and wH = 11.5%. The combustion air is enriched with oxygen to φ02
= 27.5 %, balance N2. Calculate the STP ft3 O-enriched air per lb of fuel oil to give a stack gas of
<jp02 = 3.2 %. What % of stoichiometric air is required to attain this condition?

6.12. A plant uses natural gas for use as a process fuel for the incineration of sewage sludge. The
NG composition is:

<pC2H6 = 2.5%, <pC02 = 3.5%, <pN2 = 2.7%, balance CH4.

In order to assure complete sludge oxidation, combustion takes place with a large amount of
excess air. The combustion gases must contain oxygen in excess of that necessary to completely
oxidize the sludge to C02, H20, HC1, and S02, which by experience means the exit stack gas must
have at least φ02 = 2.8%. The composition of the sludge is:

wC = 13.8%, wR = 1.95%, wN = 3.25%, wO = 18.3%, wCl = 7.6%, wS = 1.1%, wH20 -
17.5%, balance inert ash.

The plant has a contract to incinerate 3 tonnes of wet sludge per hour. Calculate the volume
flowrate of NG and air required, and the composition of the stack gas.

6.13. A plant has two supplies of natural gas for use as a process fuel. NG #2 is the cheapest, so
the operator wants to use as much ofthat fuel as possible. However, the environmental regulations
limit the emission to 0.015 %S02 in the stack gas. The NG is burned with sufficient excess air
such that the stack gas has 2.2 %02. What fraction of the total NG used can be NG #2?

NG#1: <pC2H6 = 2.5%, <pC02 = 3.5%, φΝ2 = 2.7%, <pH2S = 0.11%, balance CH4.

NG#2: <pC2H6 = 2.9%, <pC02 = 3.7%, φΝ2 = 3.2%, <pH2S - 0.28%, balance CH4.

6.14. Natural gas of the composition listed in Exercise 6.12 is burned with 114 % of
stoichiometric dry air. The material being heated generates various particulate impurities which
are carried out with the combustion gases, and which act as catalysts for the formation of NO.
Calculate the volume fraction of the equilibrium amount of NO at 900 °C. How much CH4 should
be added to 100 m3 (STP) of stack gas to lower the NO to 10 vppm?

6.15. Natural gas of the composition listed in Exercise 6.12 is burned with dry air, and alternately
with oxygen-enriched air having <p02 = 35%. In both cases, the amount of oxidant was
proportioned so that the stack gas had <jp02 = 2.5%. For ordinary air, the stack gas temperature was
1000 °C, and for oxygen-enriched air, it was 1100 °C. Calculate the mg of NO produced per kg of
NG burned if the reaction to form NO went 50 % of the way to equilibrium.

6.16. Air having a dew point temperature of 21 °C is used to burn coal in a boiler. The air and
coal are proportioned so that a sample of the completely dry stack gas has a φ02 of 3.35%.
Calculate the mass of air per lb of as-received coal, which contains 3.74 % water. The mass
fraction of constituents in dry coal is:

86.4 %C, 4.93 %H, 1.6 %N, 3.6 % 0 , 0.62 %S, balance inert ash.

6.17 A hydrated iron oxide ore contains 15 % gangue minerals, 80 %Fe02H, and 5 %H20. A kiln
dryer, fired with natural gas (assume all CH4) and twice the stoichiometric air, produces an offgas
having <jpH20 = 40 %. Calculate the volume of natural gas at the burner temperature of 33 °C and
1.4 atm required to dry 1 tonne of ore.

6.18. A producer gas furnace burns coke with an air/steam mixture in order to obtain a producer
gas having <pH2 = 11 %. The extent of 0 2 reaction with the coke carbon to produce CO is 1, while
the extent of H20 reaction with coke carbon to produce CO and H2 is 0.96. What mass of steam is
required per m3 (STP) of air?

6.19. The spent gas from an iron ore reduction process is to be converted into reducing gas by
cooling to 18 °C to remove water, adding CH4, and reforming at 925 °C. The objective is to
prepare a reducing gas having a volume fraction of (CO + H2) = 0.91. Calculate the volume of

Chapter 6 Reactive Material Balances 407

CH4 to add to 1 m3 of cooled spent gas to attain the objective. The composition of the spent gas in
volume fraction of constituents before cooling is:

CO = 0.224, C02 = 0.183, R2 = 0.413, H20 = 0.151, balance N2.

6.20. A reducing gas is prepared by reforming a mixture of steam and NG at 940 °C. The
steam/NG mole ratio is 1.4, and the catalyst is so effective that a negligible amount of CH4 remains
in the reformed gas. The composition of the NG is that given in Exercise 6.12. Calculate the
composition of the reducing gas, and the value of X as defined by Equation [6.83]. What would
the composition of the gas be if it were cooled to a dpt of 15 °C and reheated to attain the WGR
equilibrium at 850 °C?

6.21. The LPG described in Example 6.9 is reformed with steam at 900 °C. The objective is to
prepare a reducing gas having a value of X as defined by Equation [6.83] of 0.07. Past experience
with the reforming process indicates that <pCH4 will be 2.3 %. Calculate the lb of steam per lb of
LPG to be fed to the reformer.

6.22. A reducing gas is prepared with the volume fraction of constituents listed below.

CO = 0.284, C02 = 0.043, H2 = 0.552, H20 = 0.071, balance N2.

The reducing gas is fed to a fluidized bed iron ore reduction furnace where hematite is being
reduced at 855 °C to DRI having an O/Fe atom ratio of 0.06. The spent gas from the reduction
furnace has a/?H20/pH2 ratio that is 0.892 times the value defined by the iron/wustite equilibrium.
Calculate the extent of wustite reaction in the reactor, and the amount of hematite that can be
reduced by 1000 ft3 of reducing gas. How much change in hematite reduction would occur by
adding 50 ft3 of CH4 to 1000 ft3 of reducing gas?

6.23. Scrap engine blocks are charged to the top of a cupola for remelting, along with coke and
flux. The flowrate and composition of the charged materials is:

10 t/hr scrap: 3.6 %C, 2.0 %Si, 0.7 %Mn, 0.7 %P, 2.0 %"dirt", balance Fe. (The "dirt"
is75%Si02, 25%A1203)

1.25 t/h coke: 5.0%SiO2, 3.0%Al2O3, 2.0 %FeO, balance С

250 kg/h flux: 1.0%SiO2, 1.0%Al2O3, balance CaC03.

Dry air is blown into the bottom of the cupola to burn the coke and provide heat for the
process. The metal from the cupola contains 4.8 %C and some Si and Mn, but negligible amounts
of anything else. The slag contains 10 %FeO, 46 %Si02, and 3 %MnO, plus CaO, A1203 and P205.
The cupola stack gas has a pCO/pC02 ratio of 7.5, plus all of the nitrogen from the air. Calculate:
a) the mass of metal and slag produced per hour; b) the composition of slag and metal; and c) the
mass of air required per hour.

6.24. Under special circumstances, magnesium can be produced economically by using

ferrasilicon (18 %Fe, 4 %A1, balance Si) as a reducing agent for magnesium oxide (the

Magnetherm process). The MgO is added in the form of burnt dolomitic lime (57 %CaO, 40

%MgO, 3 %Si02) with a flux of lime-alumina (30 %CaO, balance A1203). The Mg vapor is
extracted by vacuum until the ferrosilicon drops to 20 %Si, balance Fe. The slag contains CaO,

A1203, MgO and Si02. The charge to the furnace is 10,000 kg of dolomitic lime, 1870 kg of lime-
alumina, and 1540 kg of ferrosilicon. Calculate: a) the mass and % recovery of Mg to metal, and;
b) the mass and composition of the slag.

6.25. Suppose the process described in Section 6.5.3 and 6.5.4 (Figure 6.35) operates with stream
4 as follows: 11.0 moles of reducing gas containing <jpH20 = 0.04. Evaluate the process and
material balance equation set to determine if it is possible, without specifying XRM or XRw, to
calculate the composition of the solid streams. If not, what other process parameter(s) need to be
defined? Try to solve the material balance using FlowBal.

6.26. Tungsten concentrate contains 85.2% scheelite (CaW04), balance gypsum (CaS04-2H20).
The concentrate is roasted at 660° С with various amounts of natrite (Na2C03) to convert the
scheelite to water-soluble sodium tungstate (Na2W04). The roasted concentrate (calcine) is

408 Chapter 6 Reactive Material Balances

leached with water to dissolve all of the sodium tungstate. The flowsheet for the process is shown
below. Solid lines refer to solid-phase streams, dashed lines refer to liquid-phase streams, and
dotted lines refer to gas streams.

Scheelite 1' ^ Offgas 4■ 3 Water " Leach
concentrat*1eP6 2^ solution
Roast Г4ППП n*\''+ 5
(3000 g) ^^P Insoluble
4 . .._*. Leach solids
Na2C03 Calcine
■■^

Four chemical reactions take place. The first two occur in the roaster, and the second two in
the leach tank. The first reaction does not go to completion, but the other three do.

CaW04 + Na2C03 -> Na2W04 + CaO + C02

CaS04-2H20 -> CaS04 + 2H20

CaO + H20 -+ Ca(OH)2

CaS04 + 2H20 -> CaS04-2H20

Five laboratory tests were carried out using different amounts of Na2C03 to determine the
dissolution of Na2W04 in the leach solution. The results are shown in the following table.

Mass of Na2C03, g 660 940 1320 1645 1980

wNa2WC>4 in leach solution, % 22.38 30.21 29.75 29.60 29.11

Use FlowBal to make a complete mass balance on the system for each test. Determine the

percent of stoichiometric Na2C03 added, the fractional yield of CaW04, the extent of CaW04
reaction in the roaster, and the actual volume (at 660 °C and 0.95 atm) of the offgas.

6.27. A process reduces hematite to DRI in a shaft furnace. The reducing gas is prepared by

mixing steam and oxygen and passing the mixture through a bed of carbon at 900 °C and 2 atm

pressure. The resulting gas is passed into the shaft furnace, still at 2 atm. A sketch of the process

is shown below. The operator determines the quality of the reducing gas by continuously

removing 100 1/min (STP) of reducing gas via a sampling probe, and chills the gas to 0 °C to

condense out virtually all the water. The flowrate of water is used as a basis for calculating how

effectively the steam and oxygen react with the carbon. The complete composition of the reducing

gas is calculated by assuming it attains internal equilibrium according to the water-gas shift

reaction, even though it likely does not come to equilibrium with the carbon. During one period of

operation, steam and oxygen are mixed in the volumetric ratio of four, and the flow of water from

the sampling tube is 1.808 g/min. Calculate the composition of the reducing gas. Plot a Rist

diagram for the shaft furnace, assuming the wustite/iron temperature is 880 °C. Draw an operating

line for DRI having an O/Fe of 0.05, if the amount of gas is 110% of the theoretical amount for ore

reduction in a shaft furnace. Try simulating the process using FlowBal.
aOre
Top
gas
1
h►t♦t♦t♦i♦1 Shaft
Oxygen
И+++1 furnace
1DRI
Steam Mixer

Water

6.28. Make a materials balance on a conceptual DRI production process for the reduction of
magnetite in a single fluidized bed connected to a C02-absorption reactor, where lime absorbs C02

Chapter 6 Reactive Material Balances 409

to form CaC03. The process uses carbon as a reactant, with added oxygen to provide the needed
thermal energy. The entire system operates at 2.0 atm. Solid lines indicate solid phase flow, while

dashed lines indicate gas phase flow.

Bleed gas <*-- 9 - H

Return gas ι - 1 0 - - - [ Recy
« gas
Fe304 1 CaO C02
С absorption -► Solid
Reduction Ì Redn
Oo reactor reactor
DRI 5 gas
M

300 kmol/min of magnetite enter the furnace. Reduction takes place at a temperature such
that the /?CO//?C02 in the reduction gas = 2.0, and the DRI has an O/Fe mole ratio of 0.07. The
absorption reactor temperature is such that/?C02 in S7 is 0.03 atm, and the extent of CaO reaction
is 0.90. The solid (S8) is calcined in a separate process (not shown), and the lime is recycled via
S6. The concept is that recycling a СО-rich gas back to the reduction reactor will decrease the
amount of С required. However, the maximum split fraction of S7 to S10 is 0.9 because some
bleed is necessary to prevent buildup of various impurities. Five reactions are necessary to
characterize the system. One feasible set is shown below, but others are possible.

1. Fe304 + 0.82CO - 3FeO106 + 0.82CO2

2. FeOi.o6 + 1.06CO — Fe + 1.06CO2

3. 2CO + 0 2 - 2 C 0 2

4. С + C02 - 2CO

5. CaO + C02 - СаСОз

Calculate the amount of carbon and lime needed as a function of split fraction to S10 for 20
and 30 moles of 02 added.

6.29. Ground water at 40 °C trickling from an abandoned battery recycling plant contains wPb =
0.268%, apparently as dissolved PbCl2 (wCl = 0.092%). The environmental requirement is that the
wPb < 0.004% in the discharge water to allow it to be discharged into the nearby creek. The
environmental engineer in charge of the cleanup reads that the solubility product of PbCl2 is a very
small number, and thinks that if he can add common salt to the ground water in a holding tank,
possibly the solubility of the PbCl2 will decrease to the desired level. The precipitated PbCl2 will
collect in the bottom, and be removed at monthly intervals. There is no restriction on discharging
salty water to the creek. Make calculations to determine how much salt should be added to each
m3 of water to attain the desired wPb. Use the following Кщ data, all at 40 °C.

PbCl2(c) = 1.05 x 1052; CY{aq) = 3.90 χ 1021; РЬ2+И) = 1.82 χ 104

CHAPTER 7

Energy and the First Law of Thermodynamics

Every process route for a material involves at some point a physical or chemical treatment, with a
corresponding exchange of energy from one type to another, or between the system and its
surroundings. An engineer working on process improvement or design needs to account for the
energy that flows into or out of each step of the process and to determine an overall energy balance
for the process. This is done by writing energy balance equations in much the same way as
material balance equations. Our understanding and control of energy relies on both experimental
measurements and practical observations. A theoretical framework exists which explains these
measurements and observations, and which allows a quantitative prediction of the energy required
to produce or process a material with specified characteristics. This framework is based on the
engineering science of thermodynamics. This Chapter consists of a brief overview of the first law
of thermodynamics with a description of various formulae that are particularly useful for energy
balance calculations. The type and source of thermodynamic data is described, and examples
given for using the data to make simple energy balances. Subsequent chapters will contain more
extensive application of the first law to energy balances in non-reactive and reactive systems.
There are many excellent texts on thermodynamics for materials engineers (Gaskell 2008, DeHoff
2006), so no attempt is made here to present a complete and rigorous treatment of the first law.

7.1 Principles and Definitions

Chapter 2 introduced a number of concepts relating to a system and its properties. Terms
were introduced to describe the nature of a system, and to define the conditions imposed when it
changes state. The reason we need to be very clear on what defines a system is that
thermodynamics requires that the universe be divided into two parts: the part being studied and
everything else. The part being studied is defined as the system, and is bounded by real or
imaginary boundaries and thereby separated from the rest of the universe (the surroundings). The
laws of thermodynamics apply only to systems that contain so many molecules that the laws of
probability hold to a high degree of certainty; i.e., macroscopic systems. The permeability of the
boundary describes how the system interacts with the surroundings. An open system has a
boundary that is permeable to mass, while a closed system boundary is not. A closed system is
completely isolated when the boundary is impermeable to energy in any form, is thermally isolated
when the boundary is impermeable to the flow of heat, and mechanically isolated when the
boundary is rigid.

Systems are also defined according to the state of aggregation of the contained substances. A
homogeneous system is one where the properties of the contents are the same throughout, with no
apparent surfaces of discontinuity. A homogeneous system contains only one phase. A
heterogeneous system contains two or more phases, which are separated from each other by
boundaries across which occur sharp abrupt changes in properties. If surface tension, molecular
absorption on surfaces etc. are not involved, the phase boundaries may be considered surfaces of
discontinuity of infinitesimal thickness. In a heterogeneous system, each phase is an open system.

The state of a system at any instant is defined by specifying a sufficient number of
independent variables (such as temperature, pressure, volume, viscosity, vapor pressure, surface
tension, composition, etc.) such that the system has zero degrees of freedom. Recall that for an
ideal gas, only three variables (or properties) of the named gas need be specified, and all other
properties are then determined. The state properties (or state functions) are those whose values in
different states are independent of the path in going from one state to another.

Chapter 7 Energy and the First Law of Thermodynamics 411

The properties of substances vary greatly with their states of aggregation. Pure substances
commonly exist as a plasma, gas, liquid, crystal, or glass. The solution state is important for
mixtures, and the dispersed state is important when a substance is disseminated in the form of
small aggregates of molecules (a colloid) throughout a phase.

Of the many interrelations between state properties, the most important is the equation of
state, the relationship between temperature, pressure, and volume. The general functional
relationship is given by: V=f(T, P, n), where К is the volume, P is the pressure, T is the absolute
temperature, and n is the number of moles of the substance. (Note: in this Chapter, equations will
refer to one mole of a substance, unless otherwise indicated.) It is quite common to assume that all
pure substances (or a solution phase of specified composition) obey an equation of state. A
substance that exists in a single aggregation state is completely specified by fixing any two state
properties (and the composition if it is a solution phase). In Chapter 2, we discussed the equation
of state for an ideal gas, and the van der Waals (vdW) equation of state for a real gas.

Although there is no doubt about the validity of the equation of state for fluids, the state of
solids is not completely specified by an equation of state. For example, the specific volume of a
solid is a function of its previous mechanical and thermal treatment. Although this situation may
be explainable in terms of dislocations or other departures from perfect crystallinity, no state
variable has yet been invented to adequately describe this behavior. Furthermore, there is no
known universal method for completely removing the effects of past processing such that every
solid sample of a specified substance will obey the same equation of state. Additional
complications arise if the solid material is anisotropie. Fortunately, processing history and
anisotropy have an extremely weak effect on many thermodynamic properties, so it's possible to
define a standard state for any substance such that, within experimental error, all state functions
can be specified at a given temperature and pressure.

A standard state (sometimes called a reference state) for a substance is defined by specifying
certain conditions. The normal standard state is commonly defined as the most stable physical
form of the substance at a specified temperature and at 1 atm pressure (or in some cases, 1 bar).
However, since any non-normal condition could be chosen as the standard state, it must be defined
in the context of use. A physical standard state is one that exists for a time sufficient to allow
measurements of its properties. The most common physical standard state is one that is stable
thermodynamically — it has no tendency to transform into any other physical state. If a substance
can exist but is not thermodynamically stable (for example, a supercooled liquid), it is called a
metastable state. A non-physical standard state is one whose properties are obtained by
extrapolation from a physical state (for example, a solid superheated above the normal melting
point, or an ideal gas at a condition where the real gas is non-ideal). Metastable liquids and solids
are important in materials engineering because some substances can persist and be used in that
state indefinitely.

A process consists of changes imposed upon or occurring spontaneously in a specified system
or group of systems. A process takes systems from an initial to a final condition, each of which is
specified by a minimum number of selected state properties. The difference in state properties
between condition 1 and condition 2 is independent of the path followed by the system between the
two conditions. A reversible process occurs when the restoring force differs from the deforming
force by an infinitesimal amount, such that after completion of the process, both the system and the
surroundings can be restored to their original conditions. No degradation of energy occurs during
a reversible process, and the availability of energy in the combined system plus surroundings
remains constant. No real process is strictly reversible, but the concept of reversibility is extremely
valuable as a standard against which a real process can be compared. Some processes are
inherently irreversible, such as the flow of heat between two bodies, the mixing of two pure
substances to form a solution, or any process involving friction (or more generally, the conversion
of work into heat).

For systems undergoing chemical reactions, we may speak of chemical irreversibility, which
is characterized by the occurrence of one or more highly-energetic chemical reactions that are

412 Chapter 7 Energy and the First Law of Thermodynamics

spontaneous. Once initiated, they proceed to completion, with the virtual exhaustion of the
limiting reactant. This and other types of chemical reactions was discussed in Section 5.5.5.

Various restrictions may be placed on the operation of a process. A process operating at
constant-temperature is isothermal, at constant-pressure is isobaric, and at constant volume is
isochoric. A thermally isolated process is adiabatic. While no real process operates exactly
according to these restrictions, the restricted condition may be approached closely enough to
justify making calculations as if the condition existed.

A non-isolated system undergoing a process interacts with its surroundings and exchanges
energy. Heat is energy crossing the system boundary under the influence of a temperature
gradient. The amount of heat Q has a positive sign when heat is added to the system. Work is
energy in transit between a system and its surroundings caused (among other things) by the
displacement of an external force acting on the system. The amount of work W has a positive sign
when work is done by the system. The objective of engineering processes is to maximize the
desirable effects of this interchange of energy and to minimize the undesirable effects.

Although the word energy is used frequently, there is no generally acceptable definition of the
term in a thermodynamic sense. Various forms of energy exist, each of which can be defined
rigorously. Different processes may have associated with them quite different amounts of energy,
in quite different forms. The following examples illustrate the magnitude of the change in energy
that accompanies each process for 1 mole of substance.

1. Nuclear fission of uranium: -1.6 x 1010kJ.
2. Ionization of a hydrogen atom: +2 x 103 kJ.

3. Oxidation of a hydrogen molecule to water vapor: -240 kJ.

4. Vaporization of water: +40 kJ.

5. Heating water vapor from room temperature to 1000 K: +20 kJ.

6. Lifting nickel an elevation of 300 meters: +160 J.

7. Raising the velocity of nickel by 100 km/hr: +20 J.

Even though thermodynamics is not concerned with the mechanism of a process, it can be
helpful to get a picture of how a system possesses energy. There are two general ways to classify
the energy of a system. First, because of a change in the position or velocity of the system relative
to the surroundings. This is potential and kinetic energy, and is taken to include changes owing to
electrical and magnetic fields, and also includes changes in surface energy. Second, a system
acquires or loses energy because of a change in internal chemical structure. The atoms and
molecules change structure; included here are the changes in translational, vibrational, and
rotational energy of the atoms and molecules. In most industrial processes, the second class of
energy is overwhelmingly large compared to the first class, so the first class is ignored. However,
when dealing (for example) with very fine particles undergoing weak phase changes, neglect of the
first class of energy may sometimes lead to significant errors. The second class of energy is called
the internal energy (given the symbol U) to distinguish it from the first class.

The total energy of a system therefore has three components:

1. Kinetic energy: Energy associated with the translational motion of a system relative to
some frame of reference, which is usually the surface of the earth.

2. Potential energy: Energy associated with the position of a system in an electrical,
gravitational, or magnetic field.

3. Internal energy: All energy possessed by the system other than kinetic and potential, such
as energy stored within a system due to the relative motion and position of atoms within
the system. This includes the so-called P-V energy of a gas, which is a measure of its
ability to do work by reversible expansion.

Chapter 7 Energy and the First Law of Thermodynamics 413

Many of the terms used in the discussion of the energy in a system and its transport to and
from the surroundings were defined in Chapter 2, especially Section 2.1. In a closed system,
energy may be transferred across the boundary in two ways:

1. As heat, or energy that flows as a result of a temperature difference across the system
boundary. Heat always flows from a higher temperature to a lower one. Heat is defined as
positive when transferred to the system from the surroundings.

2. As work, or energy that flows due to a driving force other than temperature difference.
Examples are mechanical force or electric voltage. Work is defined as positive when done
by the system on the surroundings.

The units of energy are force times distance, such as joules (N · m) and ft-lbf. It is also
common to define energy in terms of the amount of heat required to raise the temperature of a
certain mass of water by one degree over a certain interval. For example, a calorie may be defined
as the heat that will raise the temperature of one g of water from 15 °C to 16 °C. Various types of
calories and Btus have been used over the years, but in this text, we will use the thermochemical
calorie and thermochemical Btu. The calth is defined as 4.184 J, and the value of the Btuth is
similarly defined as 1054.35 J. To convert from calth to Btuth, multiply calth by 252. Conversion
between these and other forms of energy are made using the tables inside the front cover of the
book, or the program U-Converter, described earlier.

7.2 General Statement of the First Law of Thermodynamics

In its most general form, the first law of thermodynamics states that the rate of energy carried
into a system by streams, plus the rate of heat transferred to the system across the boundary, equals
the rate at which energy is carried out by streams plus the rate at which the system does work, plus
the rate of accumulation of energy in the system. This is often stated as the law of conservation of
energy, to wit: the total energy of an isolated system is constant despite internal changes. This law
is analogous to the law of conservation of mass, as introduced in Chapter 4. The first law tells us
nothing about the tendency for a system to undergo chemical change, only that if it does, there is a
certain relationship between the forms of energy involved. The first law in equation form is:

Input of energy to system = output of energy from system + accumulation [7.1]

The mathematical form of the first law takes different forms, depending on the type of system
or process under consideration. Most common applications fall into one of two classes: a closed
vs. an open system (unsteady-state batch process vs. a steady-state continuous process). Consider
a closed system at some initial condition / and some final condition/ As discussed in Example
1.2, the kinetic energy of the system is a function of its mass and velocity:

EK=V2mv2 [7.2]

The system may have potential energy by virtue of its height above a reference plane:

Ep = mgz [7.3]

where g is the gravitational constant discussed in Chapter 1 and z is the height of the system.

The system will have internal energy as a result of energy stored in the form of the motion of
its atoms and molecules (vibrating, rotating, etc), and in the form of atomic bonds between the
different atoms. Internal energy may be considered as a combination of kinetic and potential
energy of the atoms in the system. The internal energy U is the sum of all the atomic interactions
in the system. The first law statement for a closed system is:

AU+AEK + AEP = Q-W [7.4]

where AU = change in total internal energy of the system, AEK and AEP are the change in kinetic
and potential energy of the system, Q = heat; positive for energy entering the system, and W =

work; positive for energy leaving the system. The term Δ refers to the final - initial state. An

414 Chapter 7 Energy and the First Law of Thermodynamics

alternate expression of the first law is that the final system energy - initial system energy = net
energy transferred to/from the system.

Equation [7.4] is a statement of the first law for a closed system. Each of the terms in the
equation deserves additional explanation.

■ Heat and work both represent an energy flux crossing the system boundary. A non-
isolated system undergoing a process interacts with its surroundings to interchange
energy. Heat is energy crossing the system boundary under the influence of a
temperature gradient. Work is energy in transit between a system and its surroundings.
It is convenient to identify two types of work. One type, called P-V work, is caused by
the displacement of an external force acting on the system. Non-P-F work can be a
rotating stirrer that penetrates the system boundary, the passage of an electrical current
through the system, or radiant heat transfer. The objective of engineering processes is to
maximize the desirable effects of this interchange of energy and to minimize the
undesirable effects. The amount of heat and work for a process depends on the path
taken from the initial to the final state.

■ If the system and surroundings are at the same temperature, or if the system is perfectly
insulated from the surroundings, Q = 0. This defines an adiabatic process.

■ If the system is not undergoing acceleration (or deceleration), AEK is zero. If the height
of the system is not changing, AEP is zero.

■ The internal energy U of a system depends almost entirely on the chemical composition,
state of aggregation (solid, liquid or gas), and temperature. U is independent of pressure
for an ideal gas, and nearly so for liquids and solids, t/is a state function, and hence AU
is independent of the path from initial to final state.

In most cases, the system is stationary (or nearly so), so that AEK and AEP are zero. This
simplifies the first law to:

AU=Uf-U{ = Q-W [7.5]

Many processes operate under restrictions of one sort or another. It is of interest to examine
the consequences of fixing one of the independent variables (state properties) while allowing the
rest to vary. In this way, we can examine the nature of isothermal, isobaric, isochoric (sometimes
called isometric) and adiabatic processes.

An isochoric process operates at constant volume; hence the P-V form of work is zero.
Neglecting the non-P-V forms of work, the first law becomes:

ΔΕ/=βν [7.6]

where the subscript V refers to an isochoric process. The change in internal energy becomes a
function of the amount of heat gained or lost by the system. By specifying the path, gv becomes a
state property.

An isobaric process operates at constant P. The work done on the system is PAV, hence the
first law becomes:

AU=QV-PAV [7.7]

where the subscript p refers to an isobaric process. Since many processes are isobaric (or nearly
so), we define a new state property enthalpy as H:

AH=AU + PAV=QV [7.8]

Thus, the enthalpy change during a constant-pressure process is equal to the heat absorbed or
given up in going from the initial to the final state.

Q and W are not state functions, being dependent on the path between the initial and final state.

Chapter 7 Energy and the First Law of Thermodynamics 415

For an adiabatic process Q = 0, and if the process is carried out reversibly, the work is given
by the integral | PdV. For this definedpath, the first law is:

AU=-PAVR [7.9]

where the subscript R specifies a reversible path.

For an isothermal process, AU= 0. Then, Q=W. For a reversible isothermal process, Q and
W are given by the integral |. PdV

For systems containing multiple phases, it is possible to select one, two, or some other number
of phases to treat as a closed system. Consider a system containing a metal, a molten salt, and a
gas. If the metal and salt phase are undergoing a chemical reaction independent of the gas, then
the metal/salt mixture may be considered a closed system. Similarly, if solid carbon is being used
to reduce a metal oxide to a metal while producing a gas, the gas/metal/metal oxide phases can be
considered a closed system if the gas is confined to the reaction vessel. If not, it is an open system.

EXAMPLE 7Л — Work and Heat During the Compression of an Ideal Gas.

Oxygen is placed into a closed system fitted with a piston capable of compressing the gas.

The initial state is 100 L of 0 2 at 1.5 atm pressure and 400 K. The gas is compressed reversibly
and isothermally until the pressure reaches 6.5 atm. Calculate the work done on the system.

Solution. The amount of gas in the system is calculated first. For the initial conditions:

PPi 1.5x100
n = number of moles of 0 2 = — = 0Ш06хШ
= 4.57

For a reversible isothermal process, AC/ = 0, and Q and W are given by the integral |. PdV.

For an ideal gas, this is equal to the integral | RT — per mole of gas. Integrating between the

initial and final states gives:

/ VV. \
R71n
' VQ=W= =
R71n
KVU \Pfj

Thus for an ideal gas, U is constant for an isothermal process and the work done by/on the
system equals the heat absorbed/evolved by the system. Therefore, for 4.57 moles:

Q=W= 4.57x8.3144x400xln —
6.5

Q=W=-22290]

Assignment. Ten liters of an ideal gas at 3.5 atm pressure and 350 К expands reversibly and
isothermally, and in doing so, 580 J of heat is lost. What is the final gas pressure?

7.3 First Law for an Open System

For an open system (steady-state process), where a boundary is drawn around a piece of
equipment such as a furnace, the amount of material enclosed by the boundary and the
thermodynamic properties ofthat material do not change with time. Energy enters and leaves this
system (or control volume) not only as heat and/or work, but also because of the energy contained
by the mass crossing the boundary. Although the volume and pressure of the system are constant,
the volume of the products may well be different from that of the reactants. The expression of the
first law for this case advantageously uses the thermodynamic function enthalpy defined in

416 Chapter 7 Energy and the First Law of Thermodynamics

Equation [7.8]. For an open system, the first law takes the form:

Q - WmQ = Z[(/f ex + gzex + % Κ « > « ] - If/Tin + gzin + У2У^)пцп\ [7.10]
WmQ = mechanical work,
where H = specific enthalpy, U + Px F ,

P = pressure,

i / = specific energy,

F = specific volume,

ex = subscript indicating an exit stream,

in = subscript indicating an entering stream.

and the other terms are the same as those for a closed system. Remember that the term "specific"
is used to define an extensive property per unit of amount or mass, thus H refers to the molar
enthalpy of a substance. The P*V term is required to account for work done on or by the system

owing to a difference in volume between the exit and inlet stream.

The first law expression may be simplified if the terms for kinetic and potential energy are
zero or negligible. Letting delta (Δ) represent the summation for all exit streams minus the
summation for all entering streams, the first law expression for a steady-state system becomes:

Q-Wms = A(mH') = H [7.11]

where H is the total enthalpy.

7.4 Enthalpy, Heat Capacity, and Heat Content

Since a system does not have an intrinsic quantity of H or U, it is the changes in these
functions that are of interest. For example, heat balances use the enthalpy H. The high
temperature heat content (also called the sensible heat) refers to the change in Я of a substance
when heated above a defined reference temperature, usually 298.15 K, or 25 °C (or, usually, 77
°F). This reference temperature is not the standard temperature of 273.15 K. It is common
practice to simply write 298 instead of 298.15 for the reference temperature, and this practice will
be followed henceforth, even though 298.15 is used in more precise calculations. The change in H
above the reference temperature is designated as: Нт-Н2д$. This value has been measured and
tabulated for virtually all known substances, and is commonly expressed in equation form as a
power series with temperature. The heat content of an ideal gas is independent of pressure (or
volume), but the heat content of real gases varies with pressure.

It is now appropriate to define heat capacity C, which is the ratio of heat added to the
temperature increase. For an incremental addition of heat:

C = bQlbT [7.12]

However, the addition of heat Q does not define the final state, since the path has not been
specified. Since processes are often either isobaric or isochoric, we define Cp and Cv as:

Cp = δβρ/δΓ = (HT2-HTì)/dT [7.13]

Cv = bQJbT = (HT2-Hix)ldT [7.14]

For the addition of heat to a closed isobaric system, the integration of Equation [7.13] gives:

Ят2-Ят, = f CpdT [7.15]

the evaluation of which requires knowledge of the variation of Cp with T. The heat capacity is a
continuous function of Г except at phase changes and certain zero-order changes.

Chapter 7 Energy and the First Law of Thermodynamics 417

Cp for a gas is of greater magnitude than Cv because additional heat is required to provide the
work necessary to expand the system isobarically. For one mole of ideal gas, this difference is
equal to the gas constant R. For condensed phases, the difference between Cp and Cv is
vanishingly small. Cp (and Cv) for an ideal gas is unaffected by pressure or volume, and since
gases tend to be more ideal at high temperature, it is then often justified to use the Cp for the
standard state (1 atm) as valid for higher pressures. Cp for an ideal monatomic gas like argon is
2.5(R), or about 5.0 cai deg"1 mol"1.

Cp and Cv are important in calculating work and heat effects in adiabatic processes caused by
reversible expansion or compression of a gas. For an adiabatic process, Q = 0, so dU = -bW = -
PdV = CydT. From various relationships involving the ideal gas law, and for one mole of gas,
integration of the CydT expression (assuming Cv does not vary with temperature) gives:

C v l n \Ά) = RlnΎ^ [7.16]
\y2)

The relationship between pressure and volume during reversible adiabatic expansion is:

PVCp/Cv = constant [7.17]

Values of Cp for substances of invariant composition are usually tabulated as cai deg- mol1-1
or J deg-1 mol-1. For more complex substances like fuel oil, coal, or mineral mixtures, Cp is
tabulated in units of cai deg-1 g"1, kJ deg"1 kg-1, or Btu deg F 1 lb-1 A value of Cp for mixtures is
obtained by adding the values of Cp for the constituent substances.

EXAMPLE 7.2 — Heat Capacity and Enthalpyfor a Flux.

Develop a quadratic equation for Ητ-Η29% in terms of kJ/kg for a steelmaking flux having
wcalcite = 65 %, balance magnesite, valid over the temperature range 500 - 1000 K.

Data. The heat content data for the carbonates, from FREED, in J/mol, is listed below.

Temp, К 500 600 700 800 900 1000
19 376 30 109 41317 52 922 64 881 77 170
Hj-H298, СаСОз 18 106 28 479 39 550 51253 63 552 76 426

tfT-tf298, MgC03

Solution. From MMV-C, one kg of flux contains 6.49 moles of CaC03 and 4.15 moles of MgC03.
HT-H29s was multiplied by these amount values, and merged to create a single Ητ-Η29% table for one
kg of flux. The results are shown below.

Temp, К 500 600 700 800 900 1000
#т-Я298, kJ/kg 200.9 313.6 432.3 556.2 684.8 818.0

A chart was created from this data, and Excel's Trendline tool was used to obtain a quadratic
heat content equation:

#T-#298 = 2.54 x 10^T2 + 0.855T - 290 (kJ/kg) [7.18]

This equation is valid from 500 to 1000 K, and since R2 = 1.00, it may be safely extrapolated

outside this range by 100 degrees. As a check on the extrapolation accuracy, you may want to see
how close the Ητ-Η29% value approaches zero at 298.15 K.

This equation can be differentiated to obtain a linear Cp equation.

Cö = 5.08x 10^T +0.855

Assignment. Develop a Ητ-Η29$ and Cp equation for flux in units of cal/g and °C.

418 Chapter 7 Energy and the First Law of Thermodynamics

7.5 Enthalpy Change of Phase Transformations

The preceding section discussed the change in enthalpy caused by adding heat to a
homogeneous phase of a substance under specified conditions. This was designated the sensible
heat, Я т 2 - # т ь because it is sensed relative to defined zero heat at a reference temperature. When
heat is added to a condensed-phase substance, its temperature increases until a phase change
temperature is reached. Then the temperature remains constant while heat is added and the
transformation takes place. The amount of substance undergoing transformation is a function of
the amount of heat added. After the transformation is complete, adding more heat increases the
temperature. In other words, an enthalpy change occurs isothermally when a substance undergoes
a physical (phase) change. This enthalpy change is designated AH. There are four types of
enthalpy changes resulting from a physical change*. To wit:

• Enthalpy of transformation (also called heat of transformation). This applies to the
transformations from one solid phase to another, such as the transformation from α-Fe (ferrite) to
γ-Fe (austenite). The transformation is designated AHtr.

• Enthalpy for heat) offusion or melting. This applies to the transformation of a solid to a
liquid, and is designated AHm.

• Enthalpy of vaporization. This applies to the formation of a vapor from a liquid and is
designated AHV.

• Enthalpy of sublimation. This applies to the formation of a vapor from a solid, and is
designated AHS.

Values of AH are usually given for the transformation at the standard state temperature for the
two states, and if so, are designated with a superscript °. AH is a weak function of temperature. In
some texts, the standard heats of transition are called latent heats (for example, latent heat of
fusion). A plot of the heat content will show a discontinuity in the line at phase change
temperatures, as shown in Figure 7.1 and 7.2 for Zn and ZnBr2.

Heat Content of Zn(c,/,g)

160,000 f '" —

S 120,000 ΊГ bpZn ]
1300
00 mpZn

£ 80,000 500 700 900 1100
temperature, К
£ 40,000

0
300

Figure 7.1 Heat content above 298 К for one mole of zinc at 1 atm pressure. The discontinuity at
693 К represents the AH°m of Zn (7323 J) at its normal melting point and the discontinuity at 1180
К represents the AH°V of Zn (115 332 J) at its normal boiling point. The slope of the lines is the
heat capacity, which is infinite during phase transformations.

The heat content of a substance can be measured experimentally with a calorimeter, a device
designed to capture the heat given up by a substance when it is cooled from a high temperature to a
lower temperature. The calorimeter consists of a receptacle immersed in an insulated container of
water or other fluid. The fluid and calorimeter heat capacity is measured after being calibrated
against a reference substance. The temperature change of the water and calorimeter is measured
very accurately, which then gives the enthalpy change when the substance is cooled.

At the Curie temperature, Cp shows a sharp discontinuity while the enthalpy has a change in slope.

Chapter 7 Energy and the First Law of Thermodynamics 419

Heat Content of ZnBr2

70,000

^ 60,000

5 50,000

J 40,000

l· 30,000

1 20,000

10,000

0

300 400 500 600 700 800 900
temperature, К

Figure 7.2 Heat content above 298 К for 1 mole of ZnBr2 at 1 atm from FREED's Graphics tool.
The discontinuity at 675 К represents the AH°m of ZnBr2 (15 648 J) at its normal melting point.

EXAMPLE 7.3 — Heat of Fusion of Lead.

Experiments were conducted to determine the A#°f of Pb. One hundred gram samples of Pb
were heated to different temperatures and dropped into a calorimeter calibrated to measure the heat
content of the sample above the reference temperature of 25 °C. One set of experiments gave the
following result:

/,°C 100 140 180 220 260 300 340 380
Hx-H25, J 951 1515 2032 2548 3147 3653 6573 7162

Calculate the standard heat of fusion of Pb (AH°f) in J/mole.

Solution. The normal melting point of Pb is 327.5 °C, so we have six data points below the mp
and two above it. First we plot the data (Figure 7.3) to see if it contains any obvious errors. The
vertical dashed line is at the normal mp of Pb. The object is to extrapolate the solid data up to the
mp and the liquid data down to the mp, and determine the difference at that point.

Heat Content of 100 g of Lead

8000 •Solid Pb H,iq = 14.7t +15701
7000 •Liquid Pb
6000
"I 5000
a? 4000
£ 3000

100 200 300 400
temperature, °C

Figure 7.3 Calorimetrie data for the heat content of 100 g of lead. Vertical dashed line is at the
normal melting point of lead, 327.5 °C.

The data appears to be free of obvious error. Excel's Trendline tool was used to develop a
linear and a polynomial (quadratic) equation for the solid Pb data, and within the expected limits of

420 Chapter 7 Energy and the First Law of Thermodynamics

data error, found no statistical advantage to a quadratic fit. For this reason, and not knowing more
about the accuracy or precision of the calorimeter, the data for solid Pb was represented by a linear
equation*. The two equations for heat content of lead are:

Solid Pb: HrH25 = 13.52(f) - 396 Liquid Pb: # r # 2 5 = 14.7(t) + 1570

Substituting 327.5 °C in both equations, the difference in HrH25 = 2350 J/100 g. Therefore,
A#°f of Pb = 4870 J/mol. The AH°{ of Pb from FREED is 4800 J/mol.

Assignment. Add an additional data point: HrH2s = 0 at 25 °C, then repeat the calculations using a
quadratic fit to the solid Pb data. Consider the general advisability of extrapolating a quadratic
equation above the range of experimental data.

7.6 Enthalpy Change of Chemical Reactions

Chemical energy is converted into heat energy during a chemical reaction. It is convenient to
make a distinction between two types of reactions: formation of a compound from the elements,
and all other reactions. The heat change associated with a chemical reaction is an indefinite
quantity, depending on the path taken. If the reaction is carried out at constant pressure or constant
volume, the heat change has a definite value. Generally, heats of reaction are tabulated at a
constant pressure, usually one atm, in which case Q?** the heat absorbed in an isobaric chemical
reaction is equal to АЯР, the increase in isobaric heat content. The difference between the total
heat content of the products and that of the reactants is the quantity usually referred to as the heat
of reaction. Stated another way, the heat of reaction is that amount of heat that must be exchanged
with the surroundings in order for an isobaric reaction to remain isothermal. If all gases behaved
ideally, the value of AH would be independent of pressure. We generally distinguish between the
reaction heat for any reaction and that to form a compound from the elements.

• Enthalpy (heat) offormation from the elements. This refers to the enthalpy change when
two or more elements react to form a compound. This is designated AHiorm and like most heats of
reactions, is a weak function of temperature. Values of AH{orm are usually given for the case where
the elements and compound are in their standard states, and as such are designated standard heats
offormation. A superscript ° is used to indicate the standard heat: Af/°form. By definition, the heat
of formation of elements in their standard state equals zero. The standard heat of formation of
ZnBr2 from the elements is depicted in Figure 7.4.

Figure 7.4 Standard molar heat of formation of ZnBr2 at 1 atm. The discontinuities represent the
heat effects caused by phase changes in the constituent elements and the compound.

An idea of the accuracy of the equation fit down to room temperature is obtained by calculating the heat
content at 25 °C from the equation for solid Pb. The value should be zero.
** Henceforth, we will dispense using the general heat term g, and use instead the more specific term H

Chapter 7 Energy and the First Law of Thermodynamics 421

• Enthalpy (heat) of reaction. The enthalpy of reaction at a specified temperature (and
pressure if a non-ideal gas is present) is designated A#rx and is a weak function of temperature. If
the value of Δ#ΓΧ is for substances in their standard states, M/°rx is designated as the standard
enthalpy (heat) of reaction. Figure 7.5 depicts the standard heat of reaction for the carbonation of
magnesium hydroxide to form magnesium carbonate:

Mg(OH)2(c) + C02(s) - MgC03(c) + H20(/,g) [7.19]

ΔΗ°ΓΧ of MgC03 from Мд(ОН)2

-30,000 1
-40,000 r - " -
S -50,000 ^ bp H20

o b -60,000 Liquid
< -70,000 - - - steam

-80,000
-90,000 :

300 350 400 450 500 550 600 650

temperature, К

Figure 7.5 Heat of reaction at 1 atm between 1 mole of magnesium hydroxide and carbon dioxide
to form 1 mole of magnesium carbonate and water according to Equation [7.19]. Solid line
represents АЯ°ГХ when water is in its standard state (liquid below 373 K, steam above 373 K).
Dashed line represents ΑΗΐχ where steam is the reaction product at all temperatures. Below 373 К
and at 1 atm, steam is a non-physical state.

Since enthalpy is a state property, the change in enthalpy does not depend on the path of the
process or the reaction sequences; AH is the same if the process takes place by a series of reactions
or one reaction. This principle is known as Hess 's law, and can be demonstrated by examining the
reduction of Mo02 with CO at 900 °C to form Mo2C and C02. The process probably occurs in two
steps. FREED's Reaction tool was used to find the value of AH° for the two postulated reactions:

2CO(g) + Mo02(c) -> Mo(c) + 2C02(g); AH° = 3025 cai

CO(g) + Mo(c) -> l/2Mo2C(c) + ViCOzfe); AH° = -25 672 cai
According to Hess's law, AH° for the overall process of reduction of one mole of Mo02 to lA
mole of Mo2C is the sum of the Δ//° values for the two sequential reactions:

3CO(g) + Mo02(c) ->im.o2C{c) + 2V2C02(g); AH° = 3025 - 2588 = -22 647 cai [7.20]

As a check, FREED's Reaction tool for the overall reaction yielded the same result.

7.7 Thermodynamic Databases for Pure Substances

The starting point for all energy-related calculations is a thermodynamic database, which
starts from a collection of critically evaluated thermodynamic data on the elements and pure
compounds. All thermodynamic data is derived from experimental measurements. Calorimetrie
methods have been used to measure the heat content, heat of formation, and heat of transformation
of a tremendous number of substances. All of the data is correlated by first establishing a database
for the pure elements in their normal and metastable states, and then extending the data to chemical
compounds. Specialized databases are also available for solutions. Heat content data is plotted
and differentiated to derive Cp. The thermodynamic data for non-physical states are calculated or
extrapolated as appropriate.

422 Chapter 7 Energy and the First Law of Thermodynamics

The traditional reference temperature for thermodynamic databases is 298.15 К (listed for
simplicity in this Handbook as 298 K). In order to derive values of Cp from experimental heat
content data above 298 K, measured values of HT-H29s are fitted to polynomials having up to 8
terms, and constrained so that ΗΊ-Η2η is identically equal to 0 at 298 K, and so that the first
derivative of the fitted equation at 298 К was equal to the value of Cp obtained at room-
temperature. This procedure forces the values of Cp derived from high temperature heat content
measurements to join smoothly with the more accurately measured values of Cp obtained by room-
temperature calorimetry. The enthalpy of formation of the elements in their normal standard states
is commonly assigned a value of zero.

For a single phase of a substance, the high-temperature heat content data from 298 to 2500 К
(or even higher) can almost always be adequately fit with six or fewer terms; if not, more than one
equation can be used. Each phase requires a different set of coefficients. A typical 6-term
equation is:

Ят-Я298 = АГ+ B P + С/Г + ΌΤΑ + ΕΓ3 + F [7.21]
from which a five-term equation for Cp is obtained by differentiation: [7.22]

Cp = A + 2 Β Γ - СГ2 + УтОГ* + 3EP

A different equation format is required near a Curie point or other zero-order phase transition
where Cp is discontinuous.

A number of thermodynamic databases are available, the most useful of which are
computerized databases. The database used in the Handbook is FREED, which incorporates all of
the data from the U. S. Bureau of Mines, plus data from other sources. FREED and its use have
already been discussed in earlier chapters, and will be used for heat balances from here on.

A FREED table for MgCl2(c,l,g) is shown in Figure 7.6. The table is too long to be displayed
on a single page, so half of the table extends to the next page. The equations in Figure 7.6 are part
of the whole table, and contain the equation parameters used in creating the table. The
temperatures in the left-hand column are in degrees K, as is the case with all published
thermodynamic data compilations. This is because the degrees К scale is universally used in
thermodynamic equations, and in equations relating pressure and volume of an ideal gas.
However, FREED will create a table in the Celsius scale if desired. (Some of the Handbook
examples use heat content equations in Celsius.) FREED's Calculator tool eliminates the need to
interpolate in the table to obtain data at a temperature not listed in the table. Alternatively, the
table equations can be used for this purpose.

Notice the two types of "gaps" in the table for MgCl2. The single-line gaps (at 922 К and
1363 K) occur at phase transformation temperatures for one of the constituent elements (here, at
the mp and bp of Mg). A three-line gap occurs at phase transformation temperatures for the
species (here, the mp and bp for MgCb). The heat of the species transformation is listed in the
table, but not that of the constituent element. Note that the data and transition temperatures are at 1
atm (i.e., at standard state), even though the table does not explicitly mention this.

The FREED database uses a reference temperature of 298.15 K, and sets the heat content
equal to zero at that temperature. This reference temperature is based on historical precedent, with
the implicit assumption that materials are often encountered first at "room temperature", which
was set for database purposes at 25 °C. The drawback in taking 298.15 К as a reference
temperature is that the database is difficult to use for processes occurring below that temperature,
or accounting for materials entering a system below 298.15 K. It is generally safe to extrapolate
FREED data to 0 °C (273.15 K) for all but the most accurate calculations (providing there are no
phase transformations). Other databases (such as the steam tables) set the heat content value to
zero at 0 °C. It is important to be very careful in taking data from more than one database to use in
process calculations.

Chapter 7 Energy and the First Law of Thermodynamics 423

Compilers of thermochemical databases may use some additional thermodynamic functions.
For example, the absolute enthalpy of a substance H(T) is defined in terms of its formation
enthalpy and its heat content:

ЩТ) = A#°form,298.15 + [Hj - H29s]

For an element, H(T) and [ΗΊ - H29s] are identical at all temperatures because A//°form is zero,
and of course at 298.15 K, H{T) = 0. For a compound:

A//°form = #(7)compound - Z{#(7)elements}

Similarly, the absolute Gibbs energy G(T) is defined by the absolute enthalpy and entropy of a
substance:

G(T) = H(T)-TxS(T)

For a compound:

AG°form = G(7)compound - I{#(7)elements}

MgC12, (c,l,g) GFW: 95.2104 Density 2.316 Tmax (K) 2000

(Magnesium Dichloride) dH298°: -644294 S298°: 89.5376 Reference: 2

Elements: %Mg %C1

Atomic composition 33.3333 66.6667

Weight composition 25.5277 74.4723

Comments: M g : melts at 922 and boils at 1363 K.Calc'd. bp of species at 1643 K.

Unit: (iK, g-mole, J)

~) Cp ,HT-H298 S° (H°298-G°T)/T dHf dGf Log(Kf) Kf
298.15
71.296 0 89.538 89.538 -644293 -594769 104.199 1.582E+104

300.00 71.444 133 89.979 89.534 -644270 -594461 103.503 3.185E+103
500.00 78.303 15304 128.585 97.977 -641365 -562038 58.715 5.184E+58
700.00 80.777 31214 155.329 110.737 -638456 -530829 39.610 4.076E+39
900.00 84.224 47686 176.009 123.024 -635572 -500452 29.045 1.109E+29
922.00 84.716 49544 178.049 124.313 -635241 -497154 28.165 1.462E+28

922.00 84.716 49544 178.049 124.313 -644191 -497154 28.165 1.462E+28
980.00 86.135 54498 183.259 127.649 -643289 -487931 26.007 1.015E+26

980.00 d H 39936 (Melting Pt)
1100.00
1300.00 92.801 94435 224.010 127.648 -603353 -487931 26.007 1.015E+26
1363.00 92.801 105571 234.730 138.756 -600734 -473954 22.506 3.205E+22
92.801 124131 250.233 154.747 -596741 -451252 18.131 1.353E+18
92.801 129977 254.624 159.263 -595574 -444230 17.024 1.057E+17

1363.00 92.801 129977 254.624 159.263 -722140 -444230 17.024 1.057E+17
1500.00 92.801 142691 263.513 168.385 -717465 -416524 14.504 3.194E+14
1643.00 92.801 155962 271.963 177.038 -712599 -388059 12.337 2.173E+12

1643.00 d H 178209 (Boiling Pt)
1700.00
1900.00 62.153 334171 380.429 177.038 -534390 -388059 12.337 2.173E+12
2000.00 62.153 337714 382.548 183.893 -534201 -382985 11.768 5.855E+11
62.153 350144 389.461 205.175 -533556 -365232 10.041 1.098E+10
62.153 356360 392.649 214.470 -533243 -356381 9.308 2.030E+09

Figure 7.6 FREED table showing the standard-state thermodynamic properties of MgCl2. For
first-law calculations, data from only the first three columns is needed. The bottom half of the

table is shown on the next page.

424 Chapter 7 Energy and the First Law of Thermodynamics

Enthalpy and Heat capacity equations:
Cp = A + B*T + С*Тл-2 + D*TA-0.5 + Е*ТЛ2

T(K) A B C DE

298.15-980 95.048 -0.038505 -1339557.8 0 3.1463E-05
980-1643 00
1643-2000 92.801 0 0 00

62.153 0 0

HT-H298 = A*T + В*ТЛ2 C*TA-1 + D*TA0.5 + Е*ТЛ3 + F

T(K) A В CD

298.15-980 95.0483 -0.019253 1339557.81 0 1.0488E-05 -31397
980-1643 00 3489
1643-2000 92.8011 0 0 00
232053
62.1533 0 0

Heat offormation and Free Energy of formation equations:
Del(Hf) = A*T + В*ТЛ2 + С*Тл-1 + D*TA0.5 + Е*ТЛ3 + F

T(K) A В СD E F

298.15-922 30.7926 -0.024215 1221780.3 245.164 1.0347E-05 -659926
922-980 27.1692 -0.022893 126827.5 245.164 1.0347E-05 -665472
24.9220 -0.003640 -1212730.3 245.164 -1.4084E-07 -630585
980-1363 29.2775 0.000845 -109827.91 245.164 -1.4084E-07 -772230
1363-1643 -1.3703 0.000845 -109827.91 245.164 -1.4084E-07 -543666
1643-2000

)el(Gf) = A''Т*!^ T ) + B*T + С*ТЛ2 + D*TA-■ 1 + Ε*ΤΛ0.5 + F*TA3 +G

TJK) A В CD E F G

298.15-922 -30.7926 351.95505 0.02421489 610890.2 490.327981 -5.1734E-06 -659926
922-980: -27.1692 335.0975 0.02289275 63413.75 490.327981 -5.1734E-06 -665472
980-1363: -24.9220 298.55061 0.003640 -606365.2 490.327981 7.0420E-08 -630585
-29.2775 439.72407 -0.000845 -54913.95 490.327981 7.0420E-08 -772230
1363-1643: 490.327981 7.0420E-08 -543666
1643-2000 1.3703 73.685722 -0.000845 -54913.95

Figure 7.6 {continuedfrom previous page)

The majority of enthalpy data needed for material and energy balance calculations relates to
the heat of formation and transformation, and the heat content. However FREED also contains
data on the value of Кщ for species, as shown in the rightmost column of Figure 7.6. Recall that
in Chapter 5 we introduced the use of Кщ as a way to estimate the stability of a substance, and the
tendency for a reaction to proceed. In Chapter 6 we showed how to use the Кщ expression as a SR
in FlowBal, and illustrated the use of Кщ for various processes, including the hydrogen reduction
of wustite and the water-gas shift reaction (WGR). FREED data for Кщ of species was the source
for charts and equations in Chapter 6.

Sometimes we need the heat content of a substance at a temperature not easily generated by

FREED's Table tool, such as 460 °C for MgCl2(c). The easiest method is to use FREED's
Calculator tool, which returns the following information (J):

T (input) T(K)
460 733.15

Solution: T(C) Cp HT-H298

460.00 81.239 33900.278