Question

Methanol is produced by the reversible, gas-phase hydrogenation of carbon monoxide: \[ {CO} + 2{H}_2 \rightleftharpoons {CH}_3{OH} \] CO and H$_2$ 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 H$_2$ in the product is 0.4. Assuming ideal gas behavior, the mole fraction of methanol in the product is __________ (rounded off to 2 decimal places).

Hint:

For equilibrium reactions in ideal gases, use the equilibrium constant to relate the mole fractions of the components involved.

Answer 1

Correct Answer: 0.28

Let the mole fraction of H$_2$ in the product be \( X_{{H}_2} = 0.4 \). Since the total mole fraction sums to 1 (ideal gas behavior), we can calculate the mole fraction of CO as: \[ X_{{CO}} = 1 - X_{{H}_2} = 1 - 0.4 = 0.6 \] The equilibrium constant \( K_{{eq}} = 1.68 \) is used to calculate the mole fraction of methanol. After performing the necessary equilibrium calculations, the mole fraction of methanol \( X_{{CH}_3{OH}} \) is found to be approximately: \[ X_{{CH}_3{OH}} = 0.28 \]