The relationship between the equilibrium constants $K_p$ and $K_c$ is given by the equation $K_p = K_c(RT)^{\Delta n}$ . Here, $\Delta n$ is the difference between the number of moles of gaseous products and the number of moles of gaseous reactants ($\Delta n = n_{\text{products}} - n_{\text{reactants}}$) . For $K_p$ and $K_c$ to be equal, the value of $\Delta n$ must be zero, since $(RT)^0 = 1$ . If $\Delta n \neq 0$, then $K_p$ and $K_c$ will not be equal.

Let's calculate $\Delta n$ for each of the given reactions:

• Option A: $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$ $\Delta n = 2 - (1 + 1) = 0$. Therefore, $K_p = K_c$ .

• Option B: $\text{CO}(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g) + \text{H}_2(g)$ $\Delta n = (1 + 1) - (1 + 1) = 0$. Therefore, $K_p = K_c$.

• Option C: $2\text{BrCl}(g) \rightleftharpoons \text{Br}_2(g) + \text{Cl}_2(g)$ $\Delta n = (1 + 1) - 2 = 0$. Therefore, $K_p = K_c$.

• Option D: $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$ $\Delta n = (1 + 1) - 1 = 1$. Since $\Delta n \neq 0$, $K_p \neq K_c$.

Thus, $K_p$ and $K_c$ are NOT equal for the reaction $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$.