a. At 1223 K,
\Delta G^{\circ }_{(i)}=177,800-(111.2\times 1223)=41,800 J
and thus,
K_{(iii) 1223 K}=exp\left\lgroup\frac{-41,800}{8.3144\times 1223} \right\rgroup =0.0164=\frac{p_{Zn}p_{CO_2}}{p_{CO}}
From the stoichiometry, p_{Zn} =p_{CO_2} and the total pressure P is
P=p_{CO}+p_{Zn}+p_{CO_2} (iv)
Thus, with P = 1 atm, p_{CO} =1-2p_{Zn} and
0.0164=\frac{p^{2}_{Zn} }{1-2_{Zn}} (v)
which has the solution p_{Zn}=0.113 atm. Therefore, p_{CO_2}=0.113 atm and p_{CO}=1-(2\times 0.113) =0.775 atm. At P = 1 atm, the mole fractions of the species in the gas phase are equal to their partial pressures.
b. Condensation occurs when the partial pressure of Zn reaches the value of the saturated vapor pressure of liquid zinc at 1223 K. The saturated vapor pressure of liquid zinc is given by
\ln p^{\circ }_{Zn,(l)}(atm)=\frac{-15,250}{T}-1.255\ln T+21.79
which gives p^{\circ }_{Zn,(l)}=1.49 atm at 1223 K. Therefore,
K_{(iii) 1223 K}=0.0164=\frac{1.49^2}{P-(2\times 1.49)}
which has the solution P = 138 atm. Thus, p_{Zn}=p_{CO}=1.49 atm and p_{CO_2}=138-(2\times 1.49)=135 atm. At the point of condensation of zinc, the mole fractions of the species in the gas phase are X_{Zn}=X_{CO_2}=1.49/138=0.011 and X_{CO}=135/138=0.978.
c. The system now contains three phases and the equilibrium thus has two degrees of freedom. Condensation of the zinc eliminates the stoichiometric requirement that p_{Zn}=p_{CO_2}, but phase equilibrium between liquid zinc and zinc vapor at 1223 K requires that the partial pressure of zinc be the saturated value of 1.49 atm. Thus,
K_{1223 K} =0.0164=\frac{1.49\times p_{CO_2}}{p_{CO}} (vi)
and
p=150=1.49+p_{CO}+P_{CO_2} (vii)
Simultaneous solution of Equation (vi) and (vii) gives p_{CO}=146.9 atm and p_{CO_2}=1.61 atm. The mole fractions of the species in the gas phase are thus X_{Zn}=1.49/150=0.01 , X_{CO_2}=1.61/150=0.011 , and X_{CO}=146.9/150=0.98 .
Further considerations :
i . Consider, now, the reduction of ZnO by graphite to form zinc vapor, CO, and CO_2 according to
ZnO_{(s)}+C_{(gr)}=Zn_{(v)}+CO_{(g)} (viii)
and
2ZnO_{(s)}+C_{(gr)}=2Zn_{(v)}+CO_{2(g)} (ix)
For 2C+O_2=2CO:
\Delta G^{\circ }_{(x)}=-223,400-175.3T J (x)
For C+O_2=CO_2:
\Delta G^{\circ }_{(xi)}=-394,100-0.84T J (xi)
Combination of \Delta G^{\circ }_{(ii)} and \Delta G^{\circ }_{(x)}/2 gives
\Delta G^{\circ }_{(viii)}=348,500-285.7T J
for ZnO + C = Zn + CO, and combination of 2\Delta G^{\circ }_{(ii)} and \Delta G^{\circ }_{(xi)} gives
\Delta G^{\circ }_{(ix)}=526,300-396.8T J
for 2ZnO+C = 2Zn + CO_2.
ii . The equilibrium involves three components and three phases (ZnO, graphite, and a gas phase) and thus, according to the phase rule, has two degrees of freedom. However, as stoichiometric ZnO is the only source of oxygen and zinc in the gas phase, one of the degrees of freedom is used by the requirement that equal numbers of moles of Zn and O occur in the gas phase. Alternatively, as ZnO has a fixed composition, the system can be considered to be the quasi-binary ZnO– C, in which case the phase rule gives one degree of freedom to the equilibrium. Thus, fixing either (1) the temperature, (2) the total pressure, (3)_{p_{Zn}} , (4)_{p_{CO}} , or (5)_{p_{CO_2}} fixes the equilibrium. Determine the equilibrium state at 1223 K.
At 1223 K, \Delta G^{\circ }_{(viii)1223 K}=-850 J and thus,
K_{(viii)1223 K}=exp\left\lgroup\frac{850}{8.3144\times 1223} \right\rgroup =1.087=p_{Zn}p_{CO} (xii)
and \Delta G^{\circ }_{(ix)1223 K}=40,960 J , in which case
K_{(ix)1223 K}=exp\left\lgroup\frac{-40,960}{8.3144\times 1223} \right\rgroup =0.018=p^{2}_{Zn}p_{CO_2} (xiii)
The requirement that n_{Zn}/n_O=1 in the gas phase leads to
\frac{n_{Zn}}{n_O} =1=\frac{n_{Zn}}{n_{CO}+2n_{CO_2}}= \frac{p_{Zn}}{p_{CO}+2p_{CO_2}}
Thus,
p_{Zn}=p_{CO}+2p_{CO_2} (xiv)
Substitution of Equations (xiv) into (xii) gives
(p_{CO}+2p_{CO_2})p_{CO}=1.08 (xv)
and substitution of Equations (xiv) into (xiii) gives
(p_{CO}+2p_{CO_2})^2p_{CO_2}=0.018 (xvi)
Simultaneous solution of Equations (xv) and (xvi) gives p_{CO}=1.023 atm and p_{CO_2}=0.016 atm, and Equation xiv gives p_{Zn}=1.023+(2\times 0.016)=1.055 atm. The total pressure at which the equilibrium exists at 1223 K is thus 1.055 + 1.023 + 0.016 = 2.094 atm.
iii . Consider now the temperature at which the total pressure is 1 atm. Rewriting Equations (xii) and (xiii) to include temperature as a variable gives
K_{(viii),T}=exp\left\lgroup\frac{-\Delta G^{\circ }_{(viii)} }{RT} \right\rgroup =p_{Zn}p_{CO}
or, substituting for \Delta G^{\circ }_{(viii)} and p_{Zn},
exp\left\lgroup\frac{-348,500}{8.3144T} \right\rgroup exp\left\lgroup\frac{285.7}{8.3144} \right\rgroup =(p_{CO}+2p_{CO_2})p_{CO} (xvii)
and
K_{(ix),T}=exp\left\lgroup\frac{-\Delta G^{\circ }_{(ix)} }{RT} \right\rgroup =p^{2}_{Zn}p_{CO_2}
or, substituting for \Delta G^{\circ }_{(ix)} and p_{Zn},
K_{(ix),T}=exp\left\lgroup\frac{-526,300}{8.3144T} \right\rgroup exp\left\lgroup\frac{396.8}{8.3144} \right\rgroup =(p_{CO}+2p_{CO_2})^2p_{CO_2} (xviii)
The third equation is
P = 1 =p_{Zn}+p_{CO}+p_{CO_2}
or
1 =(p_{CO}+2p_{CO_2})+p_{CO}+p_{CO_2} (xix)
Simultaneous solution of Equations (xvii), (xviii), and (xix) gives
T = 1172 K
p_{CO}=0.489 atm
and
p_{CO_2}=0.007 atm
Thus, p_{Zn}=0.489+(2\times 0.007)=0.503 atm, and the total pressure is 0.489 + 0.503 + 0.007 = 1 atm.