At a certain temperature , only 50% HI is dissociated at equilibrium in the following reaction:
`2HI(g)hArrH_(2)(g)+I_(2)(g)`
the equilibrium constant for this reaction is:

Options

A. `0.25`
B. `1.0`
C. `3.0`
D. `0.5`

Answer

To find the equilibrium constant for the reaction \(2 \text{HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g)\) when 50% of HI is dissociated at equilibrium, we can follow these steps:

### Step-by-Step Solution:

1. **Define Initial Conditions**:
- Let the initial concentration of HI be \(a\) moles. At time \(t = 0\), the concentration of HI is \(a\), and the concentrations of H2 and I2 are both 0.

2. **Determine Change at Equilibrium**:
- Given that 50% of HI is dissociated at equilibrium, we can denote the degree of dissociation as \(\alpha = 0.5\).
- Therefore, the amount of HI that dissociates is \(a \cdot \alpha = 0.5a\).
- The remaining concentration of HI at equilibrium will be \(a - 0.5a = 0.5a\).

3. **Calculate Concentrations at Equilibrium**:
- For every 2 moles of HI that dissociate, 1 mole of H2 and 1 mole of I2 are produced. Thus:
- Concentration of H2 at equilibrium = \(\frac{0.5a}{2} = 0.25a\)
- Concentration of I2 at equilibrium = \(\frac{0.5a}{2} = 0.25a\)

4. **Write the Expression for the Equilibrium Constant**:
- The equilibrium constant \(K_c\) for the reaction is given by:
\[ K_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2} \]
- Substituting the equilibrium concentrations:
\[ K_c = \frac{(0.25a)(0.25a)}{(0.5a)^2} \]

5. **Simplify the Expression**:
- This simplifies to:
\[ K_c = \frac{0.0625a^2}{0.25a^2} = \frac{0.0625}{0.25} = 0.25 \]

6. **Conclusion**:
- Therefore, the equilibrium constant \(K_c\) for the reaction is \(0.25\).

### Final Answer:
The equilibrium constant for the reaction is \(K_c = 0.25\).

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