Methanol is made from the gas-phase equilibrium $CO (g) + 2 H X_{2} (g) CH X_{3} OH (g)$ at constant temperature. If the total pressure of the equilibrium mixture is doubled (by compression), the new equilibrium mole fractions satisfy: the value of $x (CH X_{3} OH) / [x (CO) x (H X_{2})^{2}]$ becomes

Answer: D

With

p_{i} = x_{i} P

K_{p} = x_{M} P / (x_{CO} P \cdot x_{H X_{2}}^{2} P^{2}) = \frac{x _{M}}{x _{CO} x _{H X_{2}}^{2}} \cdot P^{- 2}

. So the mole-fraction ratio equals

K_{p} P^{2}

. Doubling

P

at fixed

K_{p}

multiplies it by

2^{2} = 4