Consider the following equilibrium:
`2NO_(2) +O_(2)hArr 2NO_(3), DeltaH=-ve`,
If `O_(2)` is added and volume of the reaction vessel is reduced, the equilibrium

Options

A. Shift in the product side
B. Shifts in the reactant side
C. Cannot be predicted
D. Remains unchanged

Explanation

To solve the problem regarding the equilibrium:

**Given Reaction:**
\[ 2NO_2(g) + O_2(g) \rightleftharpoons 2NO_3(g) \]
**ΔH = -ve (exothermic reaction)**

### Step-by-Step Solution:

1. **Identify the Changes:**
- We have two changes imposed on the equilibrium:
1. Addition of \( O_2 \)
2. Reduction of the volume of the reaction vessel

2. **Effect of Adding \( O_2 \):**
- According to Le Chatelier's principle, if a reactant is added, the equilibrium will shift in the direction that consumes that reactant.
- In this case, adding \( O_2 \) will shift the equilibrium to the right (towards the products) to produce more \( NO_3 \).

3. **Effect of Reducing Volume:**
- Reducing the volume of the reaction vessel increases the pressure.
- Le Chatelier's principle states that the equilibrium will shift in the direction that reduces the number of moles of gas.
- In the given reaction:
- Reactants: \( 2NO_2 + O_2 \) (Total = 3 moles)
- Products: \( 2NO_3 \) (Total = 2 moles)
- Since there are more moles of gas on the reactant side (3 moles) than on the product side (2 moles), reducing the volume will shift the equilibrium to the right (towards the products).

4. **Conclusion:**
- The addition of \( O_2 \) shifts the equilibrium to the right (towards products).
- Reducing the volume also shifts the equilibrium to the right (towards products).
- Both changes favor the production of \( NO_3 \).

### Final Answer:
The equilibrium will shift to the right (towards the products) due to both the addition of \( O_2 \) and the reduction of volume.

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