Nitric oxide (NO) reacts with oxygen to produce nitrogen dioxide
`2NO_((g))+O_2(g)rarr2N_(2(g))`
If the mechanism of reaction is
`NO+O_2overset(K)hArrNO_3` (fast)
`NO_3+NOoverset(K_1)rarrNO_2+NO_2` (slow)
then rate law is

Options

A. Rate `=K'[NO][O_2]`
B. Rate `=K'[NO][O_2]^2`
C. Rate `=K'[NO]^2[O_2]`
D. Rate `=K'[NO]^3[O_2]`

Answer

To derive the rate law for the reaction of nitric oxide (NO) with oxygen (O2) to produce nitrogen dioxide (NO2), we will analyze the provided reaction mechanism step by step.

### Step 1: Write down the reaction mechanism
The mechanism consists of two steps:
1. \( \text{NO} + \text{O}_2 \overset{K}{\rightleftharpoons} \text{NO}_3 \) (fast)
2. \( \text{NO}_3 + \text{NO} \overset{K_1}{\rightarrow} \text{NO}_2 + \text{NO}_2 \) (slow)

### Step 2: Identify the rate-determining step
The slow step of the mechanism is the rate-determining step. This means that the rate of the overall reaction is determined by this step. Therefore, we will focus on the second step to write the rate law.

### Step 3: Write the rate law for the slow step
From the slow step, we can write the rate law as:
\[ \text{Rate} = k_1 [\text{NO}_3][\text{NO}] \]
where \( k_1 \) is the rate constant for the slow step.

### Step 4: Substitute for the intermediate \( \text{NO}_3 \)
Since \( \text{NO}_3 \) is an intermediate and does not appear in the overall reaction, we need to express it in terms of the reactants. We can use the equilibrium expression from the fast step:
\[ K = \frac{[\text{NO}_3]}{[\text{NO}][\text{O}_2]} \]
From this, we can express \( [\text{NO}_3] \):
\[ [\text{NO}_3] = K [\text{NO}][\text{O}_2] \]

### Step 5: Substitute \( [\text{NO}_3] \) back into the rate law
Now, substitute \( [\text{NO}_3] \) into the rate law:
\[ \text{Rate} = k_1 [\text{NO}_3][\text{NO}] = k_1 (K [\text{NO}][\text{O}_2]) [\text{NO}] \]
This simplifies to:
\[ \text{Rate} = k_1 K [\text{NO}]^2 [\text{O}_2] \]

### Step 6: Define a new rate constant
We can define a new rate constant \( k' = k_1 K \), leading to:
\[ \text{Rate} = k' [\text{NO}]^2 [\text{O}_2] \]

### Final Rate Law
Thus, the rate law for the reaction is:
\[ \text{Rate} = k' [\text{NO}]^2 [\text{O}_2] \]