High pressure and high temperature will be favourable conditions for a high equilibrium yield in the reaction

Options

A. `2Cl_2O_7 (g) to 2Cl_2(g)+7O_2(g)+126.8` kcal
B. `N_(2(g))+3H_(2(g)) to 2NH_(3(g)) + 22.08 kcal `
C. `Cl_(2(g)) + 2O_(2(g)) to 2ClO_(2(g)) `-49.4 kcal
D. `2NF_(3(g)) to N_(2(g)) + 3F_(2(g)) ` - 54.4 kcal

Explanation

To determine the conditions favorable for a high equilibrium yield in the given reactions, we need to analyze each reaction based on the principles of thermodynamics and Le Chatelier's principle.

### Step-by-Step Solution:

1. **Identify the Nature of the Reactions**:
- The first reaction is:
\[ 2Cl_2O_7 \rightarrow 2Cl_2 + 7O_2 \quad \Delta H = -126.8 \text{ kcal} \]
This indicates that the reaction is exothermic (releases heat).

- The second reaction is:
\[ \text{(reaction not specified, but given as +22.0 kcal)} \]
This indicates that the reaction is endothermic (absorbs heat).

- The third reaction is:
\[ \text{(reaction not specified, but given as -49.4 kcal)} \]
This indicates that the reaction is exothermic.

- The fourth reaction is:
\[ \text{(reaction not specified, but given as -54.4 kcal)} \]
This indicates that the reaction is exothermic.

2. **Apply Le Chatelier's Principle**:
- For exothermic reactions (first and third reactions):
- **High Temperature**: Not favorable, as it shifts the equilibrium to the left (toward reactants).
- **High Pressure**: Favorable if the number of moles of gas decreases in the products compared to reactants.

- For endothermic reactions (second and fourth reactions):
- **High Temperature**: Favorable, as it shifts the equilibrium to the right (toward products).
- **High Pressure**: Favorable if the number of moles of gas decreases in the products compared to reactants.

3. **Analyze Each Option**:
- **Option A**: High temperature and high pressure for the first reaction (exothermic) - **Incorrect** (high temperature is not favorable).
- **Option B**: High temperature and high pressure for the second reaction (endothermic) - **Correct** (both conditions are favorable).
- **Option C**: High temperature and high pressure for the third reaction (exothermic) - **Incorrect** (high temperature is not favorable).
- **Option D**: High temperature and high pressure for the fourth reaction (endothermic) - **Incorrect** (high pressure is not favorable as the number of moles increases).

4. **Conclusion**:
- The only option that is correct for high equilibrium yield under high temperature and high pressure is **Option B**.

### Final Answer:
**Option B** is the correct choice for high equilibrium yield under high temperature and high pressure conditions.