1. 83.1 2. 2.077×10⁵ 3. 0.033 4. 0.021

Answer: 3. 0.033 Key idea: K_c=k_f/k_b and K_p=K_c(RT)^(Δ n) . k_b=2500 k_f , so K_c=1/2500=4×10⁻⁴ . Δ n=2-1=1 ; RT=0.0831×1000=83.1 . K_p=4×10⁻⁴×83.1=0.033 . Option (3).

Topic: Law of chemical equilibrium, Kp and Kc· Equilibrium

For the reaction A(g) ⇌ 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^-1]K_p for the reaction at 1000 K is

1. 83.1

2. \(2.077\times10^5\)

3. 0.033

4. 0.021

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Answer: 3. 0.033

Key idea: \(K_c=\frac{k_f}{k_b}\) and \(K_p=K_c(RT)^{\Delta n}\).

\(k_b=2500\,k_f\), so \(K_c=\frac{1}{2500}=4\times10^{-4}\).
\(\Delta n=2-1=1\); \(RT=0.0831\times1000=83.1\).
\(K_p=4\times10^{-4}\times83.1=0.033\).

Option (3).

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For the reaction A(g) ⇌ 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^-1]K_p for the reaction at 1000 K is

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