The reaction f `NO_(2)(g)` and `O_(3)(g)` is first-order in `NO_(2)` and `O_(3)(g)`
`2NO_(2)(g) +O_(3)(g) rarr N_(2)O_(5)(g) + O_(2)(g)`
The reaction can take place by mechansim:
(P) `NO_(2)+O_(3) overset("slow") rarr NO_(3)+O_(2)`
`NO_(3)+NO_(2) overset("fast")rarr N_(2)O_(5)`
`(Q) O_(3)underset(k_(b)) overset(k_(a) "fast")hArr O_(2)+[O]`
`NO_(2) +O overset("slow")rarr NO_(3)`
`NO_(3) +NO_(2) overset("fast")rarr N_(2)O_(5)`
Select the correct mechanism.

Options

A. P only
B. Q only
C. Both P and Q
D. none of these

Answer

To determine the correct mechanism for the reaction between \( NO_2(g) \) and \( O_3(g) \), we need to analyze both proposed mechanisms (P and Q) and their corresponding rate laws.

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The overall reaction is:
\[ 2NO_2(g) + O_3(g) \rightarrow N_2O_5(g) + O_2(g) \]
We know that the reaction is first-order with respect to both \( NO_2 \) and \( O_3 \).

2. **Analyzing Mechanism P**:
- The first step is:
\[ NO_2 + O_3 \overset{\text{slow}}{\rightarrow} NO_3 + O_2 \]
- The second step is:
\[ NO_3 + NO_2 \overset{\text{fast}}{\rightarrow} N_2O_5 \]
- The rate-determining step (RDS) is the slow step. The rate law for this step can be written as:
\[ \text{Rate} = k_1 [NO_2][O_3] \]
- Since there are no intermediates in the rate law, this mechanism is valid and gives an overall order of 2 (1 for \( NO_2 \) and 1 for \( O_3 \)), which matches the given information.

3. **Analyzing Mechanism Q**:
- The first step is:
\[ O_3 \underset{k_b}{\overset{k_a}{\rightleftharpoons}} O_2 + [O] \]
- The second step is:
\[ NO_2 + [O] \overset{\text{slow}}{\rightarrow} NO_3 \]
- The third step is:
\[ NO_3 + NO_2 \overset{\text{fast}}{\rightarrow} N_2O_5 \]
- The rate law for the slow step is:
\[ \text{Rate} = k_2 [NO_2][O] \]
- Here, \( [O] \) is an intermediate. To express the rate law in terms of reactants, we need to eliminate the intermediate \( [O] \).

4. **Eliminating the Intermediate**:
- From the equilibrium step, we can express \( [O] \) in terms of \( O_3 \) and \( O_2 \):
\[ K = \frac{[O_2][O]}{[O_3]} \implies [O] = \frac{K [O_3]}{[O_2]} \]
- Substituting this back into the rate law gives:
\[ \text{Rate} = k_2 [NO_2] \left(\frac{K [O_3]}{[O_2]}\right) \]
- This results in:
\[ \text{Rate} = \frac{k_2 K}{[O_2]} [NO_2][O_3] \]
- The order with respect to \( O_2 \) becomes negative, indicating that the overall order of the reaction is not consistent with the first-order requirement for both \( NO_2 \) and \( O_3 \).

5. **Conclusion**:
- Mechanism P is valid as it matches the required order of the reaction.
- Mechanism Q is not valid because it introduces an intermediate that leads to a negative order with respect to a product.

### Final Answer:
The correct mechanism is **P**.