Van't Hoff's equation for a chemical reaction under equilibrium is given by standard reaction enthalpy at temperature T and K is the equilibrium constant . Predict how K will vary with temperature for an exothermic

Options

A. K decreases as the temperature rises
B. K is remain unchanged
C. K increases as the temperature rise
D. Information is not sufficient

Answer

To solve the problem regarding how the equilibrium constant \( K \) varies with temperature for an exothermic reaction, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Van't Hoff's Equation**:
Van't Hoff's equation relates the change in the equilibrium constant \( K \) with temperature \( T \) and the standard reaction enthalpy \( \Delta H \). The equation is given by:
\[ \log\left(\frac{K_1}{K_2}\right) = -\frac{\Delta H}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]
where \( R \) is the universal gas constant.

2. **Identify the Nature of the Reaction**:
For an exothermic reaction, the standard reaction enthalpy \( \Delta H \) is negative (\( \Delta H < 0 \)).

3. **Consider Two Temperatures**:
Let’s consider two temperatures, \( T_1 \) and \( T_2 \), where \( T_2 > T_1 \).

4. **Analyze the Equation**:
Since \( \Delta H \) is negative, the right side of the equation becomes positive when \( T_2 > T_1 \):
\[ \log\left(\frac{K_1}{K_2}\right) = -\frac{\Delta H}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) > 0 \]
This implies that:
\[ K_1 > K_2 \]

5. **Interpret the Results**:
Since \( K_1 > K_2 \) and \( T_2 > T_1 \), it indicates that as the temperature increases, the equilibrium constant \( K \) decreases for an exothermic reaction.

6. **Conclusion**:
Therefore, we conclude that for an exothermic reaction, the equilibrium constant \( K \) decreases as the temperature increases.

### Final Answer:
For an exothermic reaction, the equilibrium constant \( K \) decreases with an increase in temperature.