JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
Value of \(\mathrm{K}_{\mathrm{p}}\) for the equilibrium reaction \(\mathrm{N}_{2} \mathrm{O}_{4(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NO}_{2(9)}\) at \(288 \,\mathrm{~K}\) is \(47.9 .\) The \(\mathrm{K}_{\mathrm{C}}\) for this reaction at same temperature is \(......\) (Nearest integer) \(\left(R=0.083\, L\, \operatorname{bar} \,\mathrm{K}^{-1} \,\mathrm{~mol}^{-1}\right)\)
- A \(2\)
- B \(4\)
- C \(6\)
- D \(8\)
Answer & Solution
Correct Answer
(A) \(2\)
Step-by-step Solution
Detailed explanation
\(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{1}\) \(\mathrm{~K}_{\mathrm{c}}=\frac{\mathrm{K}_{\mathrm{p}}}{\mathrm{RT}}=\frac{47.9}{0.083 \times 288}=2\)
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