At certain temperature `50%` of `HI` is dissociated into `H_(2)` and `I_(2)` the equilibrium constant is

Options

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

Explanation

To find the equilibrium constant (K) for the dissociation of hydrogen iodide (HI) at a certain temperature where 50% of HI is dissociated, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Balanced Chemical Equation:**
The dissociation of hydrogen iodide can be represented as:
\[ 2 \text{HI} \rightleftharpoons \text{H}_2 + \text{I}_2 \]

2. **Define Initial Concentrations:**
Let's assume we start with 2 moles of HI in a 1-liter container. Thus, the initial concentration of HI is:
\[ [\text{HI}]_0 = 2 \, \text{mol/L} \]
At the start, the concentrations of H2 and I2 are both 0:
\[ [\text{H}_2]_0 = 0 \, \text{mol/L}, \quad [\text{I}_2]_0 = 0 \, \text{mol/L} \]

3. **Determine the Change in Concentration:**
Since 50% of HI is dissociated, this means that 1 mole of HI dissociates (50% of 2 moles). Therefore, the change in concentration is:
\[ \Delta [\text{HI}] = -1 \, \text{mol/L}, \quad \Delta [\text{H}_2] = +0.5 \, \text{mol/L}, \quad \Delta [\text{I}_2] = +0.5 \, \text{mol/L} \]

4. **Calculate Equilibrium Concentrations:**
At equilibrium, the concentrations will be:
\[ [\text{HI}]_{eq} = 2 - 1 = 1 \, \text{mol/L} \]
\[ [\text{H}_2]_{eq} = 0 + 0.5 = 0.5 \, \text{mol/L} \]
\[ [\text{I}_2]_{eq} = 0 + 0.5 = 0.5 \, \text{mol/L} \]

5. **Write the Expression for the Equilibrium Constant (K):**
The equilibrium constant expression for the reaction is given by:
\[ K = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2} \]

6. **Substitute the Equilibrium Concentrations into the K Expression:**
Plugging in the equilibrium concentrations we calculated:
\[ K = \frac{(0.5)(0.5)}{(1)^2} \]

7. **Calculate K:**
\[ K = \frac{0.25}{1} = 0.25 \]

### Final Answer:
The equilibrium constant \( K \) is \( 0.25 \).

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