For the reaction `H_(2)(g) + I_(2)(g)hArr2HI(g)K_(c) = 66.9` at `350^(@)C` and `K_(c) = 50.0` at `448^(@)C` . The reaction has

Options

A. `DeltaH=+ve`
B. `DeltaH=-ve`
C. `DeltaH=" zero"`
D. `DeltaH` whose sign cannot be predicted

Answer

To solve the problem regarding the reaction \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \) with given equilibrium constants \( K_c \) at two different temperatures, we will use the van 't Hoff equation, which relates the change in the equilibrium constant with temperature to the enthalpy change (\( \Delta H \)) of the reaction.

### Step-by-Step Solution:

1. **Identify the given values:**
- \( K_1 = 66.9 \) at \( T_1 = 350^\circ C = 623 \, K \)
- \( K_2 = 50.0 \) at \( T_2 = 448^\circ C = 721 \, K \)

2. **Convert temperatures to Kelvin:**
- \( T_1 = 350 + 273 = 623 \, K \)
- \( T_2 = 448 + 273 = 721 \, K \)

3. **Use the van 't Hoff equation:**
\[ \log \left( \frac{K_1}{K_2} \right) = -\frac{\Delta H}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
Where \( R \) is the universal gas constant, \( R = 8.314 \, J/(mol \cdot K) \).

4. **Calculate \( \frac{K_1}{K_2} \):**
\[ \frac{K_1}{K_2} = \frac{66.9}{50.0} = 1.338 \]

5. **Calculate the logarithm:**
\[ \log(1.338) \approx 0.128 \]

6. **Calculate \( \frac{1}{T_1} - \frac{1}{T_2} \):**
\[ \frac{1}{T_1} = \frac{1}{623} \approx 0.001606 \, K^{-1} \]
\[ \frac{1}{T_2} = \frac{1}{721} \approx 0.001386 \, K^{-1} \]
\[ \frac{1}{T_1} - \frac{1}{T_2} \approx 0.001606 - 0.001386 = 0.000220 \, K^{-1} \]

7. **Substitute values into the van 't Hoff equation:**
\[ 0.128 = -\frac{\Delta H}{8.314} \times 0.000220 \]

8. **Rearranging to find \( \Delta H \):**
\[ \Delta H = -0.128 \times 8.314 \times \frac{1}{0.000220} \]

9. **Calculate \( \Delta H \):**
\[ \Delta H \approx -0.128 \times 8.314 \times 4545.45 \approx -4,800 \, J/mol \approx -4.8 \, kJ/mol \]

10. **Conclusion:**
Since \( \Delta H \) is negative, the reaction is exothermic.

### Final Answer:
The reaction has \( \Delta H < 0 \) (exothermic).