JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
For the reaction \(\mathrm{N}_2 \mathrm{O}_4(\mathrm{~g}) \rightleftharpoons 2 \ \mathrm{NO}_2(\mathrm{~g})\), \(\mathrm{K}_{\mathrm{p}}=0.492 \mathrm{~atm}\) at \(300 \mathrm{~K}\) . \(\mathrm{K}_{\mathrm{c}}\) for the reaction at same temperature is _______ \( \times 10^{-2}\). (Given : \(\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\) )
- A \(1\)
- B \(4\)
- C \(3\)
- D \(2\)
Answer & Solution
Correct Answer
(D) \(2\)
Step-by-step Solution
Detailed explanation
\( \mathrm{K}_{\mathrm{P}}=\mathrm{K}_{\mathrm{C}} \cdot(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}} \) \( \Delta \mathrm{n}_{\mathrm{g}}=1 \) \( \Rightarrow \mathrm{K}_{\mathrm{c}}=\frac{\mathrm{K}_{\mathrm{P}}}{\mathrm{RT}}=\frac{0.492}{0.082 \times 300}=2 \times 10^{-2}\)
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