Equilibrium constant `K_(p)` for
`H_(2)S(g) hArr 2H_(2)(g)+S_(2)(g)`
is `0.0118` atm at `1065^(@)C` and heat of dissociation is `42.4` Kcal. Find equilibrium constant at `1132^(@)C`.

Explanation

To find the equilibrium constant \( K_p \) at a new temperature using the given data, we can use the Van 't Hoff equation, which relates the change in the equilibrium constant with temperature and the heat of reaction.

### Step-by-Step Solution:

1. **Identify Given Values**:
- \( K_{p1} = 0.0118 \) atm at \( T_1 = 1065^\circ C \)
- Heat of dissociation \( \Delta H = 42.4 \) kcal
- New temperature \( T_2 = 1132^\circ C \)

2. **Convert Temperatures to Kelvin**:
- \( T_1 = 1065 + 273 = 1338 \) K
- \( T_2 = 1132 + 273 = 1405 \) K

3. **Convert Heat of Dissociation to Joules**:
- \( \Delta H = 42.4 \) kcal \( = 42.4 \times 4184 \) J = \( 177,000 \) J (approximately)

4. **Use the Van 't Hoff Equation**:
The Van 't Hoff equation is given by:
\[ \log \left( \frac{K_{p2}}{K_{p1}} \right) = \frac{\Delta H}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
where \( R = 8.314 \) J/(mol·K).

5. **Substitute the Values into the Equation**:
\[ \log \left( \frac{K_{p2}}{0.0118} \right) = \frac{177000}{2.303 \times 8.314} \left( \frac{1}{1338} - \frac{1}{1405} \right) \]

6. **Calculate the Right Side**:
- Calculate \( \frac{1}{T_1} - \frac{1}{T_2} \):
\[ \frac{1}{1338} - \frac{1}{1405} = \frac{1405 - 1338}{1338 \times 1405} = \frac{67}{1870490} \approx 0.0000357 \]
- Now calculate:
\[ \frac{177000}{2.303 \times 8.314} \approx \frac{177000}{19.187} \approx 9222.4 \]
- Therefore:
\[ \log \left( \frac{K_{p2}}{0.0118} \right) \approx 9222.4 \times 0.0000357 \approx 0.329 \]

7. **Find \( K_{p2} \)**:
\[ \frac{K_{p2}}{0.0118} = 10^{0.329} \]
- Calculate \( 10^{0.329} \approx 2.14 \)
- Thus:
\[ K_{p2} = 0.0118 \times 2.14 \approx 0.0253 \text{ atm} \]

### Final Answer:
The equilibrium constant \( K_{p2} \) at \( 1132^\circ C \) is approximately \( 0.0253 \) atm.