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At a certain temperature in a 5L vessel, 2 moles of carbon monoxide and 3 moles of chlorine were allowed to reach equilibrium according to the reaction, $\mathrm{CO}+\mathrm{Cl}_2 \rightleftharpoons \mathrm{COCl}_2$. At equilibrium, if one mole of $\mathrm{CO}$ is present then equilibrium constant $\left(K_{\mathrm{c}}\right)$ for the reaction is:
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The correct answer is:
$2.5$
$2.5$
Initially 2 moles of $\mathrm{CO}$ are present.
At equilibrium, 1 mole of $\mathrm{CO}$ is present Hence, $2-1=1$ moles of $\mathrm{CO}$ has reacted. 1 mole of CO will react with 1 mole of $\mathrm{Cl}_2$ to form 1 mole of $\mathrm{COCl}_2$. $3-1=2$ moles of $\mathrm{Cl}_2$ remains at equilibrium The equilibrium constant
$$
K_{\mathrm{c}}=\frac{\left[\mathrm{COCl}_2\right]}{[\mathrm{CO}]\left[\mathrm{Cl}_2\right]}=\frac{\frac{1 \mathrm{~mol}}{5 \mathrm{~L}}}{\frac{1 \mathrm{~mol}}{5 \mathrm{~L}} \times \frac{2 \mathrm{~mol}}{5 \mathrm{~L}}}=2.5
$$
At equilibrium, 1 mole of $\mathrm{CO}$ is present Hence, $2-1=1$ moles of $\mathrm{CO}$ has reacted. 1 mole of CO will react with 1 mole of $\mathrm{Cl}_2$ to form 1 mole of $\mathrm{COCl}_2$. $3-1=2$ moles of $\mathrm{Cl}_2$ remains at equilibrium The equilibrium constant
$$
K_{\mathrm{c}}=\frac{\left[\mathrm{COCl}_2\right]}{[\mathrm{CO}]\left[\mathrm{Cl}_2\right]}=\frac{\frac{1 \mathrm{~mol}}{5 \mathrm{~L}}}{\frac{1 \mathrm{~mol}}{5 \mathrm{~L}} \times \frac{2 \mathrm{~mol}}{5 \mathrm{~L}}}=2.5
$$
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