The equilibrium constant of a reaction at 298 K is `5 xx 10^(-3)` and at 1000 K is `2 xx 10^(-5)` What is the sign of `triangleH` for the reaction.

Options

A. `triangleH=0`
B. `triangleH` is negative
C. `triangleH` is positive
D. None of these

Answer

To determine the sign of the enthalpy change (ΔH) for the reaction based on the given equilibrium constants at two different temperatures, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Equilibrium constant at 298 K (K1) = \(5 \times 10^{-3}\)
- Equilibrium constant at 1000 K (K2) = \(2 \times 10^{-5}\)

2. **Understand the Relationship Between Temperature and Equilibrium Constant:**
- According to Le Chatelier's principle, if an increase in temperature causes the equilibrium constant to decrease, it indicates that the reaction is exothermic. This is because, for exothermic reactions, increasing temperature shifts the equilibrium to the left (favoring reactants).

3. **Compare the Values of K1 and K2:**
- Since \(K1 > K2\), we observe that the equilibrium constant decreases as the temperature increases. This suggests that the reaction shifts to the left (towards reactants).

4. **Apply the Van 't Hoff Equation:**
- The Van 't Hoff equation relates the change in the equilibrium constant with temperature to the enthalpy change:
\[ \ln\left(\frac{K_2}{K_1}\right) = -\frac{\Delta H}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]
- Here, \(R\) is the universal gas constant.

5. **Calculate the Sign of ΔH:**
- Since \(K_2 < K_1\), \(\frac{K_2}{K_1} < 1\), which means \(\ln\left(\frac{K_2}{K_1}\right) < 0\).
- The term \(\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\) is positive because \(T_2 > T_1\).
- Therefore, the product \(-\frac{\Delta H}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\) must also be negative, implying that \(\Delta H\) must be positive.

6. **Conclusion:**
- Since the reaction shows a decrease in equilibrium constant with an increase in temperature, we conclude that the reaction is exothermic. Thus, the sign of ΔH is negative.

### Final Answer:
The sign of ΔH for the reaction is **negative**.