Question

The partial pressure of \( CH_3OH(g) \), \( CO(g) \), and \( H_2(g) \) in an equilibrium mixture for the reaction:

\[ CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(g) \]

are 2.0, 1.0, and 0.1 atm respectively at \( 427^\circ C \). The value of \( K_p \) for the decomposition of \( CH_3OH \) to \( CO \) and \( H_2 \) is:

• A. \( 10^2 \, {atm} \)
• B. \( 2 \times 10^2 \, {atm}^{-1} \)
• C. \( 50 \, {atm}^2 \)
• D. \( 5 \times 10^{-3} \, {atm}^2 \)

Hint:

The equilibrium constant \( K_p \) is calculated by the ratio of the partial pressures of products to reactants, each raised to the power of their respective coefficients in the balanced equation.

Answer 1

The Correct Option is D

Step 1: For the decomposition of CH\(_3\)OH, the equilibrium constant \( K_p \) for the reverse reaction is given by: \[ K_p = \frac{{[CO][H}_2]^2}{{[CH}_3{OH]}}. \] Step 2: Using the given equilibrium partial pressures: \[ K_p = \frac{(1.0)(0.1)^2}{2.0} = \frac{1.0 \times 0.01}{2.0} = 5 \times 10^{-3} \, {atm}^2. \]