At $1000 \mathrm{~K}$, the equilibrium constant. $K_C$ for the reaction $2 \mathrm{NOCl}(g) ightleftharpoons…

At $1000 \mathrm{~K}$, the equilibrium constant. $K_C$ for the reaction $2 \mathrm{NOCl}(g) ightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_2(g)$ is $4.0 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}$. The $K_P$ (in bar) at the same temperature is $\left(R=0.083 \mathrm{~L} \mathrm{bar} \mathrm{K}^{-1} \mathrm{~mol}^{-1}ight)$
  1. $3.32 \times 10^{-6}$
  2. $3.32 \times 10^4$
  3. $3.32 \times 10^{-4}$
  4. $3.32 \times 10^{-3}$

Solution

$\begin{aligned} & \text { Given, } K_C=4 \times 10^{-6} \mathrm{~mol} / \mathrm{L} \\ & \quad 2 \mathrm{NOCl}(g) ightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_2(g) \\ & \Delta n=\text { product mole }- \text { reactant mole } \\ & =3-2=1\end{aligned}$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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