a) The stoichiometric coefficients of the reaction,
CO_2(g) + C(s) \rightleftarrows 2 CO(g)
are ν_{CO_2(g)} = −1, ν_{C(s)} = −1 and ν_{CO(g)} = 2 . According to relation (8.6), the time evolution of the number of moles of CO_2(g) and CO(g) are given by,
N_A (t) = N_A (0) + ν_{αA} ξα (t) (8.6)
N_{CO_2(g)} (t) = N_{CO_2(g)} (0) − ξ (t)
N_{CO(g)} (t) = 2 ξ (t)
since N_{CO(g)} (0) = 0 . The total number of moles of air is equal to the total number of moles of gas,
N_{air} (t) = N_{gas }(t) = N_{CO_2(g)} (t) + N_{CO(g)} (t) = N_{CO_2(g)} (0) + ξ (t)
The density δ of the gases relative to air at chemical equilibrium occurring at t = t_f is,
δ = \frac{N_{CO_2(g)} (t_f) M_{CO_2} + N_{CO(g)} (t_f) M_{CO}}{N_{gas} (t_f) M_{air}}
= \frac{\Bigl(N_{CO_2(g)} (0) − ξ (t_f)\Bigr) M_{CO_2} + 2 ξ (t_f) M_{CO}}{\Bigl(N_{CO_2(g)} (0) − ξ (t_f)\Bigr) M_{air}}
Thus, the extent of the reaction is,
ξ (t_f) =\frac{N_{CO_2(g)} (0) \Bigl(M_{CO_2} − M_{air} δ\Bigr)}{M_{air} δ + M_{CO_2} − 2 M_{CO}} = 0.043 mol.
since M_{CO} = 28 g and M_{CO_2} = 48 g. The gas pressure p (t_f) inside the reactor is given by the ideal gas law,
p (t_f) = \frac{N_{gas} (t_f) R T_0}{V_0} = \frac{\Bigl(N_{CO_2(g)} (0) + ξ (t_f)\Bigr) R T_0}{V_0} = 1.51 · 10^6 Pa
b) The concentrations of gases at equilibrium are,
c_{CO} = \frac{N_{CO(g)} (t_f)}{N_{gas} (t_f)} = \frac{2 ξ (t_f)}{N_{CO_2(g)} (0) + ξ (t_f)}
c_{CO_2} = \frac{N_{CO_2(g)} (t_f)}{N_{gas} (t_f)} = \frac{N_{CO_2(g)} (0) − ξ (t_f)}{N_{CO_2(g)} (0) + ξ (t_f)}
By applying the mass action law (8.80), we determine that the equilibrium constant is,
K_\alpha = \prod\limits_{A=1}^{r}{c^{\nu \alpha A}_A } (chemical equilibrium) (8.80)
K = \frac{(c_{CO})^2}{c_{CO_2}} = \frac{4 ξ^2 (t_f)}{\Bigl(N_{CO_2(g)} (0)\Bigr)^2 – ξ^2 (t_f)} = 0.91
c) There are three substances, CO(g), CO_2(g) , C(S), i.e. r = 3, two phases (g), (s), i.e. m = 2, and one chemical reaction, i.e. n = 1. Since the temperature T_0 is fixed, there is an additional constraint. The expression for the variance is,
v = r − m − n + 1 = 1