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:

To find the equilibrium constant $K_p$ for the given reaction at 1000 K, we start by analyzing the relationship between the rate constants and the equilibrium constant of a reaction.

For the reversible gaseous reaction:
$A\left(g\right)\rightleftharpoons 2B\left(g\right)$
Let $k_f$ be the forward reaction rate constant and $k_b$ be the backward reaction rate constant.

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

According to the problem, the backward reaction rate constant $k_b$ is higher than the forward reaction rate constant $k_f$ by a factor of 2500:
$k_b=2500\times k_f$
This gives:
$K_c=\frac{k_f}{2500k_f}=\frac{1}{2500}=4\times {10}^{-4}$

Next, we relate the equilibrium constant in terms of partial pressures, $K_p$, to $K_c$ using the relation:
$K_p=K_c{\left(RT\right)}^{\Delta n_g}$
where:
- $\Delta n_g$ is the change in the number of moles of gaseous products and reactants.
- $R$ is the universal gas constant, given as 0.0831 L atm mol^-1 K^-1.
- $T$ is the absolute temperature, which is 1000 K.

For the reaction $A\left(g\right)\rightleftharpoons 2B\left(g\right)$:
$\Delta n_g=2-1=1$

Substitute the values into the equation for $K_p$:
$K_p=K_c{\left(RT\right)}^1$
$K_p=(4\times {10}^{-4})\times (0.0831\times 1000)$
$K_p=(4\times {10}^{-4})\times 83.1$
$K_p=0.03324\approx 0.033$

Therefore, the value of $K_p$ for the reaction at 1000 K is approximately 0.033.