If the equilibrium constant for the reaction $H_{2}(g)+I_{2}(g)\rightleftharpoons2HI(g)$ is $K_{c}$, then at same temperature, what will be the equilibrium constant for the reaction $2H_{2}(g)+2I_{2}(g)\rightleftharpoons4HI(g)$ ?

Question

Accepted Answer

To determine the equilibrium constant for the reaction $2H_{2}(g) + 2I_{2}(g) \rightleftharpoons 4HI(g)$, given that the equilibrium constant for the reaction $H_{2}(g) + I_{2}(g) \rightleftharpoons 2HI(g)$ is $K_{c}$, we need to understand how equilibrium constants change with the stoichiometry of the reaction.

1. **Explanation for Option 3 ($K_{c}^{2}$):**

The original reaction is:
   $$
   H_{2}(g) + I_{2}(g) \rightleftharpoons 2HI(g)
   $$
   with an equilibrium constant $K_{c}$.

The new reaction is:
   $$
   2H_{2}(g) + 2I_{2}(g) \rightleftharpoons 4HI(g)
   $$

This new reaction is essentially the original reaction multiplied by 2. When a chemical equation is multiplied by a factor, the equilibrium constant for the new reaction is the original equilibrium constant raised to the power of that factor. In this case, the factor is 2. Therefore, the equilibrium constant for the new reaction is:
   $$
   (K_{c})^{2}
   $$

Thus, the correct answer is Option 3, $K_{c}^{2}$.

2. **Explanation for why the other options are incorrect:**

- **Option 1 ($K_{c}$):** This option would be correct if the stoichiometry of the reaction remained unchanged. However, since the reaction is multiplied by 2, the equilibrium constant must be squared, not remain the same.

- **Option 2 ($\sqrt{K_{c}}$):** This option would be applicable if the reaction was divided by 2, meaning the stoichiometry was halved. In such a case, the equilibrium constant would be the square root of the original. However, since the reaction is multiplied by 2, this option is incorrect.

- **Option 4 ($2K_{c}$):** This option suggests a linear relationship between the stoichiometry change and the equilibrium constant, which is incorrect. The equilibrium constant is not simply multiplied by the factor; it is raised to the power of the factor. Therefore, this option is incorrect.

In conclusion, the correct equilibrium constant for the reaction $2H_{2}(g) + 2I_{2}(g) \rightleftharpoons 4HI(g)$ is $K_{c}^{2}$, making Option 3 the correct choice.