Equilibrium constant for the reaction $H_{2}(g)+I_{2}(g)\rightleftharpoons2HI(g)$ is $K_{C}$ then what will be the equilibrium constant for the reaction $1/2 H_{2}(g)+\frac{1}{2}I_{2}(g)\rightleftharpoons HI(g)$ ?

Q: Equilibrium constant for the reaction $H_{2}(g)+I_{2}(g)\rightleftharpoons2HI(g)$ is $K_{C}$ then what will be the equilibrium constant for the reaction $1/2 H_{2}(g)+\frac{1}{2}I_{2}(g)\rightleftharpoons HI(g)$ ?

To determine the equilibrium constant for the reaction $ \frac{1}{2} H_{2}(g) + \frac{1}{2} I_{2}(g) \rightleftharpoons HI(g) $, we need to understand how equilibrium constants change when a chemical equation is manipulated. 1. **Explanation for Option 4 ($\sqrt{K_{C}}$)**: - The original reaction is $ H_{2}(g) + I_{2}(g) \rightleftharpoons 2HI(g) $ with an equilibrium constant $ K_{C} $. - When the stoichiometry of a reaction is changed, the equilibrium constant is affected accordingly. Specifically, if the coefficients of a balanced chemical equation are divided by a factor, the equilibrium constant is raised to the power of the reciprocal of that factor. - In this case, the original reaction is divided by 2 to obtain $ \frac{1}{2} H_{2}(g) + \frac{1}{2} I_{2}(g) \rightleftharpoons HI(g) $. - Therefore, the new equilibrium constant is $ K_{C}^{1/2} $, which is equivalent to $ \sqrt{K_{C}} $. - This is why Option 4 is correct. 2. **Why the other options are incorrect**: - **Option 1 ($K_{C}$)**: This option suggests that the equilibrium constant remains the same, which is incorrect because the stoichiometry of the reaction has changed. The equilibrium constant must be adjusted according to the change in the coefficients. - **Option 2 ($\frac{1}{\sqrt{K_{C}}}$)**: This option implies an inverse relationship, which is incorrect. The equilibrium constant for a reaction is not inverted when the reaction is halved; it is instead the square root of the original constant. - **Option 3 ($K_{C}^{2}$)**: This option suggests that the equilibrium constant is squared, which is incorrect. Squaring the equilibrium constant would be appropriate if the reaction were doubled, not halved. In conclusion, the correct adjustment for the equilibrium constant when the reaction is halved is to take the square root of the original equilibrium constant, which supports Option 4 as the correct answer.

Question

Accepted Answer

To determine the equilibrium constant for the reaction $ \frac{1}{2} H_{2}(g) + \frac{1}{2} I_{2}(g) \rightleftharpoons HI(g) $, we need to understand how equilibrium constants change when a chemical equation is manipulated.

1. **Explanation for Option 4 ($\sqrt{K_{C}}$)**:
   - The original reaction is $ H_{2}(g) + I_{2}(g) \rightleftharpoons 2HI(g) $ with an equilibrium constant $ K_{C} $.
   - When the stoichiometry of a reaction is changed, the equilibrium constant is affected accordingly. Specifically, if the coefficients of a balanced chemical equation are divided by a factor, the equilibrium constant is raised to the power of the reciprocal of that factor.
   - In this case, the original reaction is divided by 2 to obtain $ \frac{1}{2} H_{2}(g) + \frac{1}{2} I_{2}(g) \rightleftharpoons HI(g) $.
   - Therefore, the new equilibrium constant is $ K_{C}^{1/2} $, which is equivalent to $ \sqrt{K_{C}} $.
   - This is why Option 4 is correct.

2. **Why the other options are incorrect**:
   - **Option 1 ($K_{C}$)**: This option suggests that the equilibrium constant remains the same, which is incorrect because the stoichiometry of the reaction has changed. The equilibrium constant must be adjusted according to the change in the coefficients.
   - **Option 2 ($\frac{1}{\sqrt{K_{C}}}$)**: This option implies an inverse relationship, which is incorrect. The equilibrium constant for a reaction is not inverted when the reaction is halved; it is instead the square root of the original constant.
   - **Option 3 ($K_{C}^{2}$)**: This option suggests that the equilibrium constant is squared, which is incorrect. Squaring the equilibrium constant would be appropriate if the reaction were doubled, not halved.

In conclusion, the correct adjustment for the equilibrium constant when the reaction is halved is to take the square root of the original equilibrium constant, which supports Option 4 as the correct answer.