Chemegate

GATE 2025 Chemical Engineering, Q61 · Thermodynamics

The correct answer is 0.28 to 0.34.

Question

Methanol is produced by the reversible, gas-phase hydrogenation of carbon monoxide as CO +2H2CH3OH.+ 2H₂ \rightleftharpoons CH₃OH. CO and H2H₂ are charged to a reactor and the reaction proceeds to equilibrium at 453 K and 2 atm. The reaction equilibrium constant, which depends only on the temperature, is 1.68 at the reaction conditions. The mole fraction of H2H₂ in the product is 0.4. Assuming ideal gas behaviour, the mole fraction of methanol in the product is ____ (rounded off to 2 decimal places).

Official answer: 0.28 to 0.34

Why

GATE 2025 official key: 0.28 to 0.34.Kp=yMeOH/(yCOyH22P2)1.68=yMeOH/(yCO×0.16×4),0.34. K_{p} = y_{Me}OH/(y_{C}O\cdot y_{H2}²\cdot P²) \Rightarrow 1.68 = y_{Me}OH/(y_{C}O \times 0.16 \times 4), so yMeOH1.075yCO.y_{Me}OH \approx 1.075\cdot y_{C}O. With yCO+0.4+yMeOH=1,y_{C}O+0.4+y_{Me}OH=1, solving gives yCO0.289y_{C}O\approx 0.289 and yMeOH0.311.y_{Me}OH\approx 0.311.

Explanation cross-checked for consistency. Final answer matches the official IIT answer key.