Question

At a certain temperature, only $50\%$ $HI$ is dissociated into $H_2$ and $I_2$ at equilibrium. The equilibrium constant is :

• A. 1
• B. 3
• C. 0.5
• D. 0.25

Answer 1

The Correct Option is D

To solve this problem, we need to determine the equilibrium constant (K_c) for the dissociation of HI into H_2 and I_2. The dissociation reaction is as follows:

2HI \rightleftharpoons H_2 + I_2

It is given that at equilibrium, 50\% of HI is dissociated. Let's assume the initial concentration of HI is C.

1. Initial Concentrations:
   • [HI] = C
   • [H_2] = 0
   • [I_2] = 0
2. Changes in Concentrations:
   • 50\% dissociation implies \frac{C}{2} moles of HI dissociate into \frac{C}{4} moles of H_2 and \frac{C}{4} moles of I_2.
3. Equilibrium Concentrations:
   • [HI] = C - \frac{C}{2} = \frac{C}{2}
   • [H_2] = \frac{C}{4}
   • [I_2] = \frac{C}{4}
4. Expression for Equilibrium Constant K_c:

   K_c = \frac{[H_2] \cdot [I_2]}{[HI]^2}

5. Substitute the Equilibrium Concentrations:

   K_c = \frac{\left( \frac{C}{4} \right) \left( \frac{C}{4} \right)}{\left( \frac{C}{2} \right)^2} = \frac{\frac{C^2}{16}}{\frac{C^2}{4}} = \frac{1}{4}

Thus, the equilibrium constant K_c is 0.25.