For a reaction, the value of `K_(p)` increases with increase n temperature. The `DeltaH` for the reaction would be :

Options

A. positive
B. negative
C. zero
D. cannot be predicted

Answer

To determine the sign of ΔH for the reaction given that Kp increases with an increase in temperature, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship Between Kp and Temperature**:
The equilibrium constant Kp is related to the change in enthalpy (ΔH) of the reaction. According to the van 't Hoff equation:
\[ \ln\left(\frac{K_{p2}}{K_{p1}}\right) = \frac{\Delta H}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]
where R is the gas constant, T1 is the initial temperature, and T2 is the final temperature.

2. **Identify the Conditions**:
It is given that Kp increases with an increase in temperature. This means:
\[ K_{p2} > K_{p1} \]
Therefore, the ratio \(\frac{K_{p2}}{K_{p1}} > 1\), which implies:
\[ \ln\left(\frac{K_{p2}}{K_{p1}}\right) > 0 \]

3. **Analyze the Temperature Change**:
Since the temperature is increasing, we have:
\[ T_1 < T_2 \]
This means that \(\frac{1}{T_1} > \frac{1}{T_2}\), leading to:
\[ \frac{1}{T_1} - \frac{1}{T_2} > 0 \]

4. **Evaluate the Right-Hand Side**:
From the van 't Hoff equation, we have:
\[ \ln\left(\frac{K_{p2}}{K_{p1}}\right) > 0 \quad \text{and} \quad \left(\frac{1}{T_1} - \frac{1}{T_2}\right) > 0 \]
Since R (the gas constant) is always positive, for the right-hand side to be positive, ΔH must also be positive:
\[ \Delta H > 0 \]

5. **Conclusion**:
Therefore, since Kp increases with an increase in temperature, we conclude that the ΔH for the reaction is positive.

### Final Answer:
The ΔH for the reaction is **positive**.