Methanol, can be prepared from and as: (g) + 2(g) ⇌ (g); = 6.23 × 10^-3 at 500 K. What total pressur — Chemical Equilibrium Chemistry Question
Question
Methanol, $CH_3OH$ can be prepared from $CO$ and $H_2$ as: $CO$(g) + 2$H_2$(g) ⇌ $CH_3OH$(g); $K_p$ = 6.23 × 10^-3 at 500 K. What total pressure is required to convert 25% of $CO$ to $CH_3OH$ at 500 K, if $CO$ and $H_2$ comes from: $CH_4$(g) + $H_2O$(g) -> $CO$(g) + 3$H_2$(g)?
💡 Solution & Explanation
Reaction: $\text{CO}(g) + 2\text{H}_2(g) \rightleftharpoons \text{CH}_3\text{OH}(g)$; $K_p = 6.23\times10^{-3}$ at 500 K. Syngas from steam reforming: $\text{CH}_4 + \text{H}_2\text{O} \to \text{CO} + 3\text{H}_2$ gives CO : H$_2$ = 1 : 3. \textbf{For 25\% conversion of CO} (basis: 1 mol CO + 3 mol H$_2$): \begin{center} \begin{tabular}{lcccc} & CO & H$_2$ & CH$_3$OH & Total \\ Initial & 1 & 3 & 0 & 4 \\ Change & $-0.25$ & $-0.50$ & $+0.25$ & $-0.50$ \\ Equil. & 0.75 & 2.50 & 0.25 & 3.50 \\ \end{tabular} \end{center} Mole fractions: $x_{\text{CO}} = \dfrac{3}{14},\ x_{\text{H}_2} = \dfrac{10}{14} = \dfrac{5}{7},\ x_{\text{CH}_3\text{OH}} = \dfrac{1}{14}$ \[ K_p = \frac{x_{\text{CH}_3\text{OH}}}{x_{\text{CO}} \cdot x_{\text{H}_2}^2 \cdot P^2} = \frac{1/14}{(3/14)(10/14)^2 \cdot P^2} \] \[ = \frac{1/14}{\dfrac{3\times100}{14^3} \cdot P^2} = \frac{14^2}{300\,P^2} = \frac{196}{300\,P^2} \approx \frac{0.653}{P^2} \] Setting equal to $K_p$: \[ \frac{0.653}{P^2} = 6.23\times10^{-3} \implies P^2 = 104.9 \implies P \approx 10.24\ \text{bar} \] \textbf{Note on answer key:} The textbook answer is listed as B (21 bar). This likely arises from a different assumption about initial gas composition or a calculation discrepancy in the key. The correct derivation gives \textbf{$P \approx 10.24$ bar} (option C). \textbf{Answer key: B (21 bar); correct calculation: C (10.24 bar)}