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ChemistryNEET 2025Law of mass action, Kc and Kp, Le Chatelier

Equilibrium

For the reaction A(g)⇌2B(g)\mathrm{A(g) \rightleftharpoons 2B(g)}, the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500 at 1000 K. [Given R=0.0831 L atm mol−1 K−1R = 0.0831\ \mathrm{L\,atm\,mol^{-1}\,K^{-1}}.] KpK_p for the reaction at 1000 K is:

  1. A.83.1
  2. B.2.077×1052.077 \times 10^{5}
  3. C.0.033✓
  4. D.0.021

Solution

Kc=kfkb=12500=4×10−4K_c = \frac{k_f}{k_b} = \frac{1}{2500} = 4\times10^{-4}. Δng=2−1=1\Delta n_g = 2 - 1 = 1, so Kp=Kc(RT)=4×10−4×0.0831×1000≈0.033K_p = K_c(RT) = 4\times10^{-4} \times 0.0831 \times 1000 \approx 0.033.

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