The correct answer is: $\boxed{K_1 = K_2^{-1}}$.
The equilibrium constant for a reaction is a measure of the extent to which the reaction proceeds to completion. It is defined as the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of its stoichiometric coefficient.
For the reaction $H_2 + I_2 \rightleftharpoons 2HI$, the equilibrium constant is $K_1$. For the reaction $2HI \rightleftharpoons H_2 + I_2$, the equilibrium constant is $K_2$.
The relationship between $K_1$ and $K_2$ can be determined by considering the following equation:
$$K_1 = \frac{[HI]^2}{[H_2][I_2]}$$
$$K_2 = \frac{[H_2][I_2]}{[HI]^2}$$
Taking the reciprocal of both sides of the second equation, we get:
$$\frac{1}{K_2} = \frac{[HI]^2}{[H_2][I_2]}$$
Substituting this into the first equation, we get:
$$K_1 = \frac{1}{K_2}$$
Therefore, the relationship between $K_1$ and $K_2$ is $K_1 = K_2^{-1}$.
Here is a brief explanation of each option:
- Option A: $K_1 = K_2$. This is not the correct answer because the equilibrium constants for the two reactions are not equal.
- Option B: $K_1 = 2K_2$. This is not the correct answer because the equilibrium constants for the two reactions are not in a simple ratio.
- Option C: $K_1 = K_2/2$. This is not the correct answer because the equilibrium constants for the two reactions are not in a simple ratio.
- Option D: $K_1 = 1/K_2$. This is the correct answer because the equilibrium constants for the two reactions are in an inverse relationship.