For the reaction
`2NO(g) hArr N_2(g) + O_2(g) , DeltaH`=-ve
`K=2.5xx10^20` at 298 K. What will happen to concentration of `N_2` if
(i) concentration of NO is increased
(ii) temperature is decreased to 273 K
(iii) pressure is reduced ?

Answer

To analyze the effects on the concentration of \( N_2 \) for the reaction

\[ 2NO(g) \rightleftharpoons N_2(g) + O_2(g) \]

with \( \Delta H < 0 \) (indicating an exothermic reaction) and \( K = 2.5 \times 10^{20} \) at 298 K, we will apply Le Chatelier's principle to each scenario.

### Step-by-Step Solution:

#### (i) If the concentration of \( NO \) is increased:

1. **Identify the change**: The concentration of \( NO \) is increased.
2. **Apply Le Chatelier's principle**: The system will respond to counteract this change by shifting the equilibrium position.
3. **Direction of shift**: Since \( NO \) is a reactant, increasing its concentration will shift the equilibrium to the right (forward direction) to produce more products.
4. **Effect on \( N_2 \)**: As the reaction shifts to the right, the concentration of \( N_2 \) will increase.

**Conclusion**: The concentration of \( N_2 \) will increase.

#### (ii) If the temperature is decreased to 273 K:

1. **Identify the change**: The temperature is decreased.
2. **Determine the nature of the reaction**: The reaction is exothermic (\( \Delta H < 0 \)).
3. **Apply Le Chatelier's principle**: Lowering the temperature favors the exothermic direction (the forward reaction in this case).
4. **Direction of shift**: The equilibrium will shift to the right to produce more heat (and thus more products).
5. **Effect on \( N_2 \)**: As the reaction shifts to the right, the concentration of \( N_2 \) will increase.

**Conclusion**: The concentration of \( N_2 \) will increase.

#### (iii) If the pressure is reduced:

1. **Identify the change**: The pressure is reduced.
2. **Count the moles of gas**: On the reactant side, there are 2 moles of \( NO \) and on the product side, there are 2 moles (1 mole of \( N_2 \) and 1 mole of \( O_2 \)).
3. **Apply Le Chatelier's principle**: When the pressure is reduced, the equilibrium will shift towards the side with more moles of gas. However, in this case, the number of moles is equal on both sides (2 moles on the reactant side and 2 moles on the product side).
4. **Direction of shift**: Since the moles are equal, there will be no shift in the equilibrium position.
5. **Effect on \( N_2 \)**: The concentration of \( N_2 \) will remain constant.

**Conclusion**: The concentration of \( N_2 \) will remain constant.

### Summary of Effects:
- (i) Increasing \( NO \) → Concentration of \( N_2 \) increases.
- (ii) Decreasing temperature to 273 K → Concentration of \( N_2 \) increases.
- (iii) Reducing pressure → Concentration of \( N_2 \) remains constant.