AP EAMCET · Chemistry · Chemical Equilibrium
At \(1000 \mathrm{~K}\), the equilibrium constant. \(K_C\) for the reaction \(2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_2(g)\) is \(4.0 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}\). The \(K_P\) (in bar) at the same temperature is \(\left(R=0.083 \mathrm{~L} \mathrm{bar} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right)\)
- A \(3.32 \times 10^{-6}\)
- B \(3.32 \times 10^4\)
- C \(3.32 \times 10^{-4}\)
- D \(3.32 \times 10^{-3}\)
Answer & Solution
Correct Answer
(C) \(3.32 \times 10^{-4}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Given, } K_C=4 \times 10^{-6} \mathrm{~mol} / \mathrm{L} \\ & \quad 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_2(g) \\ & \Delta n=\text { product mole }- \text { reactant mole } \\ & =3-2=1\end{aligned}\)
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