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 1000K.

[Given : R = 0.0831 L atm mol^–1 K^–1]

K_P for the reaction at 1000 K is

The correct option is 0.033.

Let us analyze the chemical equilibrium and determine the equilibrium constant $K_P$ step-by-step.

First, we write down the given reaction:
$A\left(g\right)\rightleftharpoons 2B\left(g\right)$

We are given that the backward reaction rate constant ($k_b$) is higher than the forward reaction rate constant ($k_f$) by a factor of 2500 at $T=1000\text{K}$.
This relationship can be written as:
$k_b=2500·k_f$

The equilibrium constant in terms of concentration, $K_C$, is defined as the ratio of the forward rate constant to the backward rate constant:
$K_C=\frac{k_f}{k_b}$

Substituting the given relation into this equation, we get:
$K_C=\frac{k_f}{2500·k_f}=\frac{1}{2500}=4\times {10}^{-4}{\text{mol L}}^{-1}$

Next, we relate the equilibrium constant in terms of partial pressures, $K_P$, to $K_C$ using the relation:
$K_P=K_C{\left(RT\right)}^{\Delta n_g}$

Here, $\Delta n_g$ is the change in the number of moles of gaseous products and reactants:
$\Delta n_g=n_{\text{products}}-n_{\text{reactants}}=2-1=1$

Now, we substitute the given values:
$R=0.0831{\text{L atm mol}}^{-1}{\text{K}}^{-1}$
$T=1000\text{K}$
$\Delta n_g=1$

Let us calculate $K_P$:
$K_P=(4\times {10}^{-4})·{(0.0831\times 1000)}^1$

Simplifying the terms:
$K_P=4\times {10}^{-4}\times 83.1$
$K_P=332.4\times {10}^{-4}$
$K_P=0.03324\approx 0.033$

Thus, the value of $K_P$ for the reaction at 1000 K is approximately 0.033.