For a given exothermic reaction ,`K_P and K_p'` are the equilibrium constant at temperature `T_1 and T_2 ` respectively .Assuming that heat of reaction is constant in temperature range between ` T_1 and T_2` , it is readly observed that

Options

A. ` K_p gt K_p `
B. ` K_p lt K_p `
C. ` K_p = K_p `
D. ` K_p = (1)/(K_p')`

Answer

To solve the problem, we need to analyze the relationship between the equilibrium constants \( K_p \) and \( K_p' \) for an exothermic reaction at two different temperatures \( T_1 \) and \( T_2 \).

### Step-by-Step Solution:

1. **Understand the Reaction Type**:
The reaction is exothermic, which means it releases heat. Therefore, the enthalpy change (\( \Delta H \)) for the reaction is negative.

2. **Identify the Relationship Between \( K_p \) and Temperature**:
The relationship between the equilibrium constants at two different temperatures is given by the van 't Hoff equation:
\[ \log \left( \frac{K_p'}{K_p} \right) = \frac{\Delta H}{2.303 R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]
Here, \( R \) is the universal gas constant.

3. **Analyze the Temperature Change**:
Since \( T_2 \) is higher than \( T_1 \) (as stated in the problem), we have:
\[ \frac{1}{T_2} - \frac{1}{T_1} < 0 \]
This means that the term \( \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \) is negative.

4. **Substituting the Values**:
Given that \( \Delta H \) is negative (because the reaction is exothermic), the right side of the equation becomes positive:
\[ \log \left( \frac{K_p'}{K_p} \right) < 0 \]
This implies that:
\[ K_p' < K_p \]

5. **Conclusion**:
From the above analysis, we conclude that \( K_p \) is greater than \( K_p' \):
\[ K_p > K_p' \]

### Final Answer:
The correct option is **A**: \( K_p \) is greater than \( K_p' \).