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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) \(\left(R=0.083\, L\, \operatorname{bar} \,\mathrm{K}^{-1} \,\mathrm{~mol}^{-1}\right)\)

  1. A \(2\)
  2. B \(4\)
  3. C \(6\)
  4. D \(8\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(2\)

Step-by-step Solution

Detailed explanation

\(\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{1}\) \(\mathrm{~K}_{\mathrm{c}}=\frac{\mathrm{K}_{\mathrm{p}}}{\mathrm{RT}}=\frac{47.9}{0.083 \times 288}=2\)