If one-third of HI decomposes at a particular temperature, `K_C` for `2HI Leftrightarrow H_2+I_2` is

Options

A. `1//16`
B. `1//4`
C. `1//6`
D. `1//2`

Answer

To solve the problem, we need to determine the equilibrium constant \( K_C \) for the decomposition of hydrogen iodide (HI) into hydrogen (H₂) and iodine (I₂) given that one-third of HI decomposes. The reaction can be represented as:

\[ 2 \text{HI} \rightleftharpoons \text{H}_2 + \text{I}_2 \]

### Step-by-Step Solution:

1. **Initial Concentration Setup**:
Let's assume the initial concentration of HI is \( [\text{HI}]_0 \). Since one-third of HI decomposes, the amount of HI that decomposes is:
\[ \Delta [\text{HI}] = \frac{1}{3} [\text{HI}]_0 \]

2. **Change in Concentrations**:
From the stoichiometry of the reaction, for every 2 moles of HI that decompose, 1 mole of H₂ and 1 mole of I₂ are produced. Therefore, if \( \frac{1}{3} [\text{HI}]_0 \) decomposes:
- The change in concentration of HI is:
\[ [\text{HI}] = [\text{HI}]_0 - \frac{1}{3} [\text{HI}]_0 = \frac{2}{3} [\text{HI}]_0 \]
- The change in concentration of H₂ and I₂ is:
\[ [\text{H}_2] = \frac{1}{2} \times \frac{1}{3} [\text{HI}]_0 = \frac{1}{6} [\text{HI}]_0 \]
\[ [\text{I}_2] = \frac{1}{2} \times \frac{1}{3} [\text{HI}]_0 = \frac{1}{6} [\text{HI}]_0 \]

3. **Equilibrium Concentrations**:
At equilibrium, the concentrations will be:
- \( [\text{HI}] = \frac{2}{3} [\text{HI}]_0 \)
- \( [\text{H}_2] = \frac{1}{6} [\text{HI}]_0 \)
- \( [\text{I}_2] = \frac{1}{6} [\text{HI}]_0 \)

4. **Expression for \( K_C \)**:
The equilibrium constant \( K_C \) is given by the expression:
\[ K_C = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2} \]

5. **Substituting the Equilibrium Concentrations**:
Plugging in the equilibrium concentrations into the \( K_C \) expression:
\[ K_C = \frac{\left(\frac{1}{6} [\text{HI}]_0\right) \left(\frac{1}{6} [\text{HI}]_0\right)}{\left(\frac{2}{3} [\text{HI}]_0\right)^2} \]
Simplifying this:
\[ K_C = \frac{\frac{1}{36} [\text{HI}]_0^2}{\frac{4}{9} [\text{HI}]_0^2} = \frac{1}{36} \times \frac{9}{4} = \frac{9}{144} = \frac{1}{16} \]

6. **Final Result**:
Therefore, the equilibrium constant \( K_C \) for the reaction is:
\[ K_C = \frac{1}{16} \]