Question

Consider the following equilibrium, $$ \text{CO(g)} + \text{H}_2\text{(g)} \rightleftharpoons \text{CH}_3\text{OH(g)} $$ 0.1 mol of CO along with a catalyst is present in a 2 dm$^3$ flask maintained at 500 K. Hydrogen is introduced into the flask until the pressure is 5 bar and 0.04 mol of CH$_3$OH is formed. The $ K_p $ is ...... x $ 10^7 $ (nearest integer). 
Given: $ R = 0.08 \, \text{dm}^3 \, \text{bar} \, \text{K}^{-1} \, \text{mol}^{-1} $ 
Assume only methanol is formed as the product and the system follows ideal gas behavior.

Hint:

When calculating equilibrium constants, use the ideal gas law to determine the total moles and pressure. For equilibrium calculations, ensure that you account for the changes in the concentration of reactants and products.

Answer 1

Correct Answer: 74

The reaction is: \[ \text{CO(g)} + \text{H}_2\text{(g)} \rightleftharpoons \text{CH}_3\text{OH(g)} \] At time \( t = 0 \), moles of CO = 0.1 mol and moles of H\(_2\) = 0.1 mol. At equilibrium, the number of moles is: \[ \text{CO(g)} = 0.1 - x \quad \text{H}_2\text{(g)} = 0.1 - 2x \quad \text{CH}_3\text{OH(g)} = x = 0.04 \, \text{mol} \] Substituting values: \[ x = 0.04 \quad \text{CO(g)} = 0.06 \quad \text{H}_2\text{(g)} = 0.08 \] Given: \[ V = 2 \, \text{L} \quad T = 500 \, \text{K} \quad P_{\text{total}} = 5 \, \text{bar} \] Using the ideal gas law: \[ n_{\text{total}} = 0.25 \, \text{mol} \] \[ P_{\text{total}} = \frac{n_{\text{total}} \times R \times T}{V} \] \[ P_{\text{total}} = \frac{(0.06 + 0.08 + 0.04) \times 0.08 \times 500}{2} = 5 \, \text{bar} \] Thus, \( K_p = 74 \). Now continuing the calculation: \[ K_p = \frac{X_{\text{CH}_3\text{OH}} \times X_{\text{CO}} \times X_{\text{H}_2}}{P_{\text{total}}^2} \] \[ K_p = \frac{0.04}{(0.06)(0.15)^2} = \frac{4}{6 \times 0.15 \times 16} \times \frac{1}{25} \] \[ K_p = \frac{100 \times 100}{24 \times 225 \times 25} = 0.074 \quad \Rightarrow K_p = 74 \times 10^7 \]