Methanol, CH₃OH can be prepared from CO and H₂ as
CO(g) + 2H₂(g) ⇌ CH₃OH(g); K_p = 6.23 × 10⁻³ at 500 K.
What total pressure is required to convert 25% of CO to CH₃OH at 500 K, if CO and H₂ come

Question

Methanol, CH₃OH can be prepared from CO and H₂ as
CO(g) + 2H₂(g) ⇌ CH₃OH(g); K_p = 6.23 × 10⁻³ at 500 K.
What total pressure is required to convert 25% of CO to CH₃OH at 500 K, if CO and H₂ comes from the reaction:
CH₄(g) + H₂O(g) → CO(g) + 3H₂(g)

Prashant Kotian · Accepted Answer

Correct answer: C. Initial mixture has 1 mol CO and 3 mol H₂. 25% of CO converts, so 0.25 mol CO reacts with 0.5 mol H₂. Eq moles: CO = 0.75, H₂ = 2.5, CH₃OH = 0.25. Total moles = 3.5. Partial pressures: P_CO = (0.75/3.5)P, P_H₂ = (2.5/3.5)P, P_CH₃OH = (0.25/3.5)P. K_p = P_CH₃OH / (P_CO × P_H₂²) = (0.25/3.5) / [ (0.75/3.5) × (2.5/3.5)² P² ] = 3.5² / (3 × 6.25 × P²) = 12.25 / (18.75 P²) = 196 / 300 P² = 6.23 × 10⁻³. Solving for P gives P² = 196 / (300 × 6.23 × 10⁻³) ≈ 104.87, so P ≈ 10.24 bar. Therefore, correct answer is C.

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