High temperature and high pressure (as per Lechatelier principle) favour

Options

A. `N_(2) + 3H_(2) hArr 2NH_(3).Delta H = - Q_(1)`
B. `CaCO_(3)(s) hArr CaO(s) + CO_(2) , Delta H = + Q_(2) `
C. `3 O_(2) hArr 2O_(3), Delta H = + Q_(3)`
D. `N_(2) + O_(2) hArr 2NO, Delta = + Q_(4)`

Answer

To determine which reaction is favored by both high temperature and high pressure according to Le Chatelier's principle, we need to analyze each reaction provided in the question.

### Step-by-Step Solution:

1. **Understand the Principle**:
- High temperature favors the endothermic direction of a reaction.
- High pressure favors the side of the reaction with fewer moles of gas.

2. **Analyze Each Reaction**:
- **Reaction 1**: \( N_2 + 3H_2 \rightleftharpoons 2NH_3 \) (Exothermic)
- High temperature: Not favored (exothermic reaction).
- High pressure: Favored (3 moles of gas on the left, 2 moles on the right).
- Conclusion: Not favored by high temperature.

- **Reaction 2**: \( CaCO_3 \rightleftharpoons CaO + CO_2 \) (Endothermic)
- High temperature: Favored (endothermic reaction).
- High pressure: Not favored (1 mole of gas on the right, 1 mole on the left; no change in moles).
- Conclusion: Not favored by high pressure.

- **Reaction 3**: \( 3O_2 \rightleftharpoons 2O_3 \) (Endothermic)
- High temperature: Favored (endothermic reaction).
- High pressure: Favored (3 moles of gas on the left, 2 moles on the right).
- Conclusion: Favored by both high temperature and high pressure.

- **Reaction 4**: \( N_2 + O_2 \rightleftharpoons 2NO \) (Endothermic)
- High temperature: Favored (endothermic reaction).
- High pressure: Not favored (2 moles of gas on both sides; no change in moles).
- Conclusion: Not favored by high pressure.

3. **Final Conclusion**:
- The only reaction that is favored by both high temperature and high pressure is **Reaction 3**: \( 3O_2 \rightleftharpoons 2O_3 \).

### Answer:
**The reaction favored by both high temperature and high pressure is the formation of ozone: \( 3O_2 \rightleftharpoons 2O_3 \).**

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