In which of the following reaction, the formation of product is favoured by increase in pressure?

Options

A. `CO_2(g) rArr 2CO(g)+O_2(g)`
B. `3O_2(g)implies2O_3(g)`
C. `CO_2(g)+C(s) implies 2CO(g)`
D. `CH_4(g)+H_2O(g)implies CO(g)+3H_2(g)`

Answer

To determine which reaction favors product formation with an increase in pressure, we can utilize Le Chatelier's Principle. This principle states that if an external change is applied to a system at equilibrium, the system will adjust to counteract that change.

### Step-by-Step Solution:

1. **Understand Le Chatelier's Principle**:
- When pressure is increased, the equilibrium will shift towards the side with fewer moles of gas.

2. **Analyze Each Reaction**:
- We will examine the number of gaseous moles on both the reactant and product sides for each reaction provided.

3. **Option 1**:
- Reaction: \( \text{1 CO} + \text{O}_2 \rightarrow 2 \text{CO} + \text{O}_2 \)
- Reactants: 1 mole of CO (gas) + 1 mole of O2 (gas) = 2 moles of gas.
- Products: 2 moles of CO (gas) + 1 mole of O2 (gas) = 3 moles of gas.
- **Conclusion**: Increase in pressure shifts equilibrium to the left (backward direction), favoring reactants. **Not favored**.

4. **Option 2**:
- Reaction: \( 3 \text{O}_2 \rightarrow 2 \text{O}_3 \)
- Reactants: 3 moles of O2 (gas).
- Products: 2 moles of O3 (gas).
- **Conclusion**: Increase in pressure shifts equilibrium to the right (forward direction), favoring products. **Favored**.

5. **Option 3**:
- Reaction: \( \text{CO}_2 (g) + \text{solid} \rightarrow 2 \text{CO} (g) \)
- Reactants: 1 mole of CO2 (gas).
- Products: 2 moles of CO (gas).
- **Conclusion**: Increase in pressure shifts equilibrium to the left (backward direction), favoring reactants. **Not favored**.

6. **Option 4**:
- Reaction: \( \text{CH}_4 (g) + \text{H}_2\text{O} (g) \rightarrow \text{CO} (g) + 3 \text{H}_2 (g) \)
- Reactants: 1 mole of CH4 (gas) + 1 mole of H2O (gas) = 2 moles of gas.
- Products: 1 mole of CO (gas) + 3 moles of H2 (gas) = 4 moles of gas.
- **Conclusion**: Increase in pressure shifts equilibrium to the left (backward direction), favoring reactants. **Not favored**.

7. **Final Conclusion**:
- The only reaction that favors product formation with an increase in pressure is **Option 2**: \( 3 \text{O}_2 \rightarrow 2 \text{O}_3 \).

### Summary:
- The reaction that favors product formation with an increase in pressure is **Option 2**.