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The partial pressure of ${CH}_{3} OH (g),$ CO(g) and $H_{2} O (g)$ in equilibrium mixture for the reaction $CO (g) + 2 H_{2} (g) ⇌ {CH}_{3} OH (g)$ are 2.0, 1.0, and 0.1 atm respectively at $427^{0} C$ . The value of $K_{P}$ for decomposition of ${CH}_{3} OH$ to CO and $H_{2}$ is

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$50 {atm}^{2}$

$10^{2}$ atm

$2 \times 10^{2} {atm}^{- 1}$

$5 \times 10^{- 3} {atm}^{2}$

answer is D.

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Detailed Solution

$K_{P} = \frac{P_{{CH}_{3} OH}}{P_{H_{2}}^{2} \times P_{CO}} = \frac{2}{1 \times {(0.1)}^{2}} = 200$

$K_{P}$ for reverse reaction will be

$K_{P}^{1} = \frac{1}{K_{P}} = \frac{1}{200} = 5 \times 10^{- 3} {atm}^{2}$

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