Question 14.14: Calculate the equilibrium compositions at 1000 K and 1 bar o...
Calculate the equilibrium compositions at 1000 K and 1 bar of a gas-phase system containing the species \mathrm{CH}_4, \mathrm{H}_2 \mathrm{O}, \mathrm{CO}, \mathrm{CO}_2 \text {, and } \mathrm{H}_2 \text {. } . In the initial unreacted state there are present 2 mol of CH_4 and 3 mol of H_2O . Values of \Delta G_{f_l}^{\circ} at 1000 K are:
\Delta G_{f_{\mathrm{CH}_4}}^{\circ}=19,720 \mathrm{~J} \cdot \mathrm{mol}^{-1} \quad \Delta G_{f_{\mathrm{H}_2 \mathrm{O}}}^{\circ}=-192,420 \mathrm{~J} \cdot \mathrm{mol}^{-1}
\Delta G_{f_{\mathrm{CO}}}^{\circ}=-200,240 \mathrm{~J} \cdot \mathrm{mol}^{-1} \quad \Delta G_{f_{\mathrm{CO}_2}}^{\circ}=-395,790 \mathrm{~J} \cdot \mathrm{mol}^{-1}
The required values of A_k are determined from the initial numbers of moles, and the values of a_{ik} come directly from the chemical formulas of the species. These are shown in the following table.
| Element k | |||
| Carbon | Oxygen | Hydrogen | |
| A_k = no. of atomic masses of k in the system | |||
| A_C = 2 | A_O = 3 | A_H = 14 | |
| Species i | a_{ik} = no. of atoms of k per molecule of i | ||
| \mathrm{CH}_4 | a_{\mathrm{CH}_4, \mathrm{C}}=1 | a_{\mathrm{CH}_4, \mathrm{O}}=0 | a_{\mathrm{CH}_4, \mathrm{H}}=4 |
| H_2O | a_{\mathrm{H}_2 \mathrm{O}, \mathrm{C}}=0 | a_{\mathrm{H}_2 \mathrm{O}, \mathrm{O}}=1 | a_{\mathrm{H}_2 \mathrm{O}, \mathrm{H}}=2 |
| CO | a_{\mathrm{CO}, \mathrm{C}}=1 | a_{\mathrm{CO}, \mathrm{O}}=1 | a_{\mathrm{CO}, \mathrm{H}}=0 |
| CO_2 | a_{\mathrm{CO}_2, \mathrm{C}}=1 | a_{\mathrm{CO}_2, \mathrm{O}}=2 | a_{\mathrm{CO}_2, \mathrm{H}} = 0 |
| H_2 | a_{\mathrm{H}_2, \mathrm{C}}=0 | a_{\mathrm{H}_2, \mathrm{O}}=0 | a_{\mathrm{H}_2, \mathrm{H}}=2 |
At 1 bar and 1000 K the assumption of ideal gases is justified, and each \hat{\phi}_i is unity. Because P = 1 bar, P∕P° = 1, and Eq. (14.43) is written:
\Delta G_{f_i}^{\circ}+R T \ln \left(\frac{y_i \hat{\phi}_i P}{P^{\circ}}\right)+\sum_k \lambda_k a_{i k}=0 \quad(i=1,2, \ldots, N) (14.43)
\frac{\Delta G_{f_i}^{\circ}}{R T}+\ln \frac{n_i}{\sum_i n_i}+\sum_k \frac{\lambda_k}{R T} a_{i k}=0
The five equations for the five species then become:
\mathrm{CH}_4: \frac{19,720}{R T}+\ln \frac{n_{\mathrm{CH}_4}}{\sum_i n_i}+\frac{\lambda_{\mathrm{C}}}{R T}+\frac{4 \lambda_{\mathrm{H}}}{R T}=0
\mathrm{H}_2 \mathrm{O}: \frac{-192,420}{R T}+\ln \frac{n_{\mathrm{H}_2 \mathrm{O}}}{\sum_i n_i}+\frac{2 \lambda_{\mathrm{H}}}{R T}+\frac{\lambda_{\mathrm{O}}}{R T}=0
\mathrm{CO}: \frac{-200,240}{R T}+\ln \frac{n_{\mathrm{CO}}}{\sum_i n_i}+\frac{\lambda_{\mathrm{C}}}{R T}+\frac{\lambda_{\mathrm{O}}}{R T}=0
\mathrm{CO}_2: \frac{-395,790}{R T}+\ln \frac{n_{\mathrm{CO}_2}}{\sum_i n_i}+\frac{\lambda_{\mathrm{C}}}{R T}+\frac{2 \lambda_{\mathrm{O}}}{R T}=0
\mathrm{H}_2: \quad \ln \frac{n_{\mathrm{H}_2}}{\sum_i n_i}+\frac{2 \lambda_{\mathrm{H}}}{R T}=0
The three atom-balance equations [Eq. (14.41)] and the equation for \sum_i n_i are:
\sum_i n_i a_{i k}=A_k \quad(k=1,2, \ldots, w) (14.41)
\text { C : } n_{\mathrm{CH}_4}+n_{\mathrm{CO}}+n_{\mathrm{CO}_2}=2
\mathrm{H}: \quad 4 n_{\mathrm{CH}_4}+2 n_{\mathrm{H}_2 \mathrm{O}}+2 n_{\mathrm{H}_2}=14
\text { O : } n_{\mathrm{H}_2 \mathrm{O}}+n_{\mathrm{CO}}+2 n_{\mathrm{CO}_2}=3
\sum_i n_i=n_{\mathrm{CH}_4}+n_{\mathrm{H}_2 \mathrm{O}}+n_{\mathrm{CO}}+n_{\mathrm{CO}_2}+n_{\mathrm{H}_2}
With RT = 8314 J mol^{-1}, simultaneous computer solution of these nine equations ^{10} produces the following results \left(y_i=n_i / \sum_i n_i\right):
y_{\mathrm{CH}_4}=0.0196
y_{\mathrm{H}_2 \mathrm{O}}=0.0980 \quad \frac{\lambda_{\mathrm{C}}}{R T}=0.7635
y_{\mathrm{CO}}=0.1743 \quad \frac{\lambda_{\mathrm{O}}}{R T}=25.068
y_{\mathrm{CO}_2}=0.0371 \quad \frac{\lambda_{\mathrm{H}}}{R T}=0.1994
y_{\mathrm{H}_2}=0.6710
The values of \lambda_k / R T are not physically meaningful, but they are included for completeness.
^{10} Example computer code for this problem is available in the Connect online learning center.