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For the reaction $A(g) \rightleftharpoons 2B(g)$, the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at $1000\text{ K}$. [Given: $R = 0.0831\text{ L atm mol}^{-1}\text{K}^{-1}$]. $K_P$ for the reaction at $1000\text{ K}$ is:
Explanation
$K_C = k_f / k_b = 1/2500 = 4.0 \times 10^{-4}$. Since $\Delta n_g = 2 - 1 = 1$, $K_P = K_C(RT)^{\Delta n_g} = (4.0 \times 10^{-4})(0.0831 \times 1000) = 4.0 \times 10^{-4} \times 83.1 \approx 0.033$.
Detailed Solution
For the reversible reaction $A(g) \rightleftharpoons 2B(g)$: Equilibrium constant $K_C = \frac{k_f}{k_b} = \frac{1}{2500} = 4.0 \times 10^{-4}$. Change in gaseous moles: $\Delta n_g = 2 - 1 = 1$. The relationship between $K_P$ and $K_C$ is $K_P = K_C (RT)^{\Delta n_g}$. At $T = 1000\text{ K}$, $RT = 0.0831 \times 1000 = 83.1\text{ L atm mol}^{-1}$. Therefore, $K_P = (4.0 \times 10^{-4}) \times 83.1 = 0.03324 \approx 0.033$.