The partial pressure of $\mathrm{CH}_{3} \mathrm{OH}(\mathrm{g}), \mathrm{CO}(\mathrm{g})$ and…

The partial pressure of $\mathrm{CH}_{3} \mathrm{OH}(\mathrm{g}), \mathrm{CO}(\mathrm{g})$ and $\mathrm{H}_{2}(\mathrm{~g})$ in equilibrium mixture for the reaction, $\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) ightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})$ are $2.0,1.0$ and $0.1$ atm respectively at $427^{\circ} \mathrm{C}$. The value of $\mathrm{K}_{\mathrm{p}}$ for the decomposition of $\mathrm{CH}_{3} \mathrm{OH}$ to $\mathrm{CO}$ and $\mathrm{H}_{2}$ is
  1. $10^{2} \mathrm{~atm}$
  2. $2 \times 10^{2} \mathrm{~atm}^{-1}$
  3. $50 \mathrm{~atm}^{2}$
  4. $5 \times 10^{-3} \mathrm{~atm}^{2}$

Solution

$\mathrm{K}_{\mathrm{p}}=\frac{\mathrm{p}_{\mathrm{CH}_{3} \mathrm{OH}}}{\mathrm{p}_{\mathrm{CO}} \times \mathrm{p}_{\mathrm{H}_{2}}}=\frac{2}{1 \times(0.1)^{2}}=200$;
For reverse reaction
$\frac{1}{\mathrm{~K}_{\mathrm{p}}}=\frac{1}{200}=5 \times 10^{-3} \mathrm{~atm}^{2}$ *

Asked in: JEE-TOPICTESTS-CHEMISTRY

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