If `K_1` is the equilibrium constant at temperature `T_1 and K_2` is the equilibrium constant at temperature `T_2 and` If `T_2 gt T_1` and reaction is endothermic then

Options

A. `K_2 lt K_1`
B. `K_2 = K_1`
C. `K_2 gt K_1`
D. All of these

Answer

To solve the problem, we need to analyze the relationship between the equilibrium constants \( K_1 \) and \( K_2 \) at two different temperatures \( T_1 \) and \( T_2 \) for an endothermic reaction.

### Step-by-Step Solution:

1. **Understanding Endothermic Reactions**:
- An endothermic reaction absorbs heat, which means that the enthalpy change (\( \Delta H \)) for the reaction is positive.

2. **Using the Van 't Hoff Equation**:
- The Van 't Hoff equation relates the change in the equilibrium constant with temperature:
\[ \log \frac{K_2}{K_1} = \frac{\Delta H}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
- Here, \( R \) is the universal gas constant, which is always positive.

3. **Analyzing the Temperature Condition**:
- We are given that \( T_2 > T_1 \). Therefore, \( \frac{1}{T_1} > \frac{1}{T_2} \). This means that the term \( \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \) is positive.

4. **Substituting into the Van 't Hoff Equation**:
- Since \( \Delta H \) is positive (for endothermic reactions) and \( R \) is positive, the entire right-hand side of the equation is positive:
\[ \log \frac{K_2}{K_1} > 0 \]

5. **Interpreting the Result**:
- If \( \log \frac{K_2}{K_1} > 0 \), it implies that:
\[ K_2 > K_1 \]

6. **Conclusion**:
- Therefore, for an endothermic reaction where \( T_2 > T_1 \), the equilibrium constant at the higher temperature \( K_2 \) is greater than the equilibrium constant at the lower temperature \( K_1 \).

### Final Answer:
- The correct option is that \( K_2 \) is greater than \( K_1 \).