If for a particular reversible reaction,
`K_(c)=57` at `355^(@)C " " ` and `" " k_(c)=68` at `450^(@)C` then `:`

Options

A. `DeltaHlt0`
B. `DeltaHgt0`
C. `DeltaH=0`
D. `DeltaH` whose sign can be determined

Answer

To determine the sign of ΔH for the given reversible reaction with equilibrium constants \( K_c = 57 \) at \( 355^\circ C \) and \( K_c = 68 \) at \( 450^\circ C \), we can use the relationship between the equilibrium constant and temperature.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- \( K_1 = 57 \) at \( T_1 = 355^\circ C \)
- \( K_2 = 68 \) at \( T_2 = 450^\circ C \)

2. **Convert Temperatures to Kelvin:**
- \( T_1 = 355 + 273.15 = 628.15 \, K \)
- \( T_2 = 450 + 273.15 = 723.15 \, K \)

3. **Use the Van 't Hoff Equation:**
The Van 't Hoff equation relates the change in the equilibrium constant with temperature to the change in enthalpy:
\[ \log \left( \frac{K_2}{K_1} \right) = -\frac{\Delta H^\circ}{2.303R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]
Where \( R \) is the universal gas constant (approximately \( 8.314 \, J/(mol \cdot K) \)).

4. **Calculate the Ratio of Equilibrium Constants:**
\[ \frac{K_2}{K_1} = \frac{68}{57} \]

5. **Calculate the Logarithm:**
\[ \log \left( \frac{K_2}{K_1} \right) = \log \left( \frac{68}{57} \right) \approx 0.174 \]

6. **Calculate the Temperature Difference:**
\[ \frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{723.15} - \frac{1}{628.15} \approx -0.000437 \, K^{-1} \]

7. **Substitute Values into the Van 't Hoff Equation:**
\[ 0.174 = -\frac{\Delta H^\circ}{2.303 \times 8.314} \times (-0.000437) \]

8. **Rearranging to Solve for ΔH:**
\[ \Delta H^\circ = 0.174 \times 2.303 \times 8.314 \times \frac{1}{0.000437} \]

9. **Calculate ΔH:**
After performing the calculations, we find that ΔH is positive.

10. **Conclusion:**
Since the equilibrium constant increases with temperature, the reaction is endothermic, indicating that ΔH is positive.

### Final Answer:
The sign of ΔH is positive, indicating that the reaction is endothermic.