GATE 2025 CH – Question 61
Methanol is produced by the reversible, gas-phase hydrogenation of carbon monoxide as
$CO + 2H_2 \longleftrightarrow CH_3OH$
$CO$ 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 behaviour, the mole fraction of methanol in the product is ____ (rounded off to 2 decimal places).
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Correct answer: 0.30 to 0.32
Explanation
For ideal gases, $K = \frac{y_{M}P}{(y_{CO}P)(y_{H_2}P)^2}\cdot P^{0}$ with all the pressures in atm, which gives $K = \frac{y_M}{y_{CO}\,y_{H_2}^2P^2}$. So $\frac{y_M}{y_{CO}y_{H_2}^2} = KP^2 = 1.68 \times 4 = 6.72$. With $y_{H_2} = 0.4$: $y_M = 6.72 \times 0.16\,y_{CO} = 1.0752\,y_{CO}$. The mole fractions add to 1, so $y_{CO} + y_M = 0.6$, so $y_{CO} = \frac{0.6}{2.0752} = 0.2891$ and $y_M = 0.3109 \approx 0.31$.