for a reversible reaction `K_(c )gtK_(p) at 298 K` and
`Delta H = + 200 KJ` the forward reaction is not favoured by :

Options

A. Increase in pressure
B. Increase in temperature
C. Decrease in pressure
D. Increase in concentration of reactants

Explanation

To solve the question, we need to analyze the given conditions for the reversible reaction and determine under which circumstances the forward reaction is not favored.

### Step-by-step Solution:

1. **Understand the Relationship Between Kc and Kp**:
The relationship between the equilibrium constants Kc and Kp is given by the equation:
\[ K_p = K_c (RT)^{\Delta n} \]
where \( \Delta n \) is the change in the number of moles of gas (moles of products - moles of reactants).

2. **Analyze the Given Information**:
We know from the question:
- \( K_c > K_p \) at 298 K
- \( \Delta H = +200 \, \text{kJ} \) (indicating that the reaction is endothermic)

3. **Determine the Sign of \( \Delta n \)**:
Since \( K_c > K_p \), it implies that \( \Delta n \) must be negative. This means that the number of moles of products is less than the number of moles of reactants:
\[ \Delta n < 0 \quad \Rightarrow \quad n_P < n_R \]

4. **Apply Le Chatelier's Principle**:
According to Le Chatelier's Principle, if we increase the pressure of a system, the equilibrium will shift towards the side with fewer moles of gas. Since we have established that \( n_R > n_P \), an increase in pressure will favor the forward reaction.

5. **Consider Temperature Effects**:
For endothermic reactions (which absorb heat), increasing the temperature will favor the forward reaction. Conversely, lowering the temperature will favor the reverse reaction.

6. **Conclusion**:
Given that the forward reaction is favored by an increase in pressure and an increase in temperature, the condition under which the forward reaction is **not favored** is a **decrease in temperature**.

### Final Answer:
The forward reaction is not favored by a decrease in temperature.