Consider the reaction :
`2H_(2)(g)+2NO(g)rarrN_(2)(g)+2H_(2)(g)`
The rate law for this reaction is :
Rate = `k[H_(2)][NO]^(2)`
Under what conditions could these steps represent mechanism?
`{:("Step 1:",2NO(g)hArrN_(2)O_(2)(g)),("Step 2:",N_(2)O_(2)+H_(2)rarrN_(2)O+H_(2)O),("Step 3:",N_(2)O+H_(2)rarrH_(2)O+N_(2)):}`

Options

A. Step 1 is the slow step
B. Step 2 is the slow step
C. Step 3 is the slow step
D. Step 1 and 2 are slow steps

Explanation

To determine under what conditions the given steps could represent the mechanism for the reaction \(2H_2 + 2NO \rightarrow N_2 + 2H_2O\) with the rate law \( \text{Rate} = k[H_2][NO]^2 \), we need to analyze the proposed elementary steps of the mechanism.

### Step-by-Step Solution:

1. **Identify the Rate Law**:
The rate law provided is \( \text{Rate} = k[H_2][NO]^2 \). This indicates that the rate of the reaction depends on the concentration of \(H_2\) to the first power and \(NO\) to the second power.

2. **Examine the Proposed Mechanism**:
The proposed steps are:
- **Step 1**: \(2NO(g) \rightleftharpoons N_2O_2(g)\) (equilibrium step)
- **Step 2**: \(N_2O_2 + H_2 \rightarrow N_2O + H_2O\) (forward step)
- **Step 3**: \(N_2O + H_2 \rightarrow H_2O + N_2\) (forward step)

3. **Determine the Rate-Determining Step**:
The rate-determining step (RDS) is the slowest step in the mechanism and dictates the rate law. The rate law derived from the mechanism should match the given rate law.

4. **Analyze Each Step**:
- **Step 1**: This step involves \(NO\) but does not produce any \(H_2\) or \(N_2\). It is in equilibrium, which means it does not contribute directly to the rate law.
- **Step 2**: This step involves \(H_2\) and \(N_2O_2\). However, since \(N_2O_2\) is an intermediate, it should not appear in the final rate law.
- **Step 3**: This step also involves \(H_2\) and produces \(N_2\) and \(H_2O\).

5. **Conclusion on the Rate-Determining Step**:
Since the rate law depends on \(H_2\) and \(NO\), and considering that \(NO\) appears in Step 1, the slow step must be Step 2, as it is the only step that directly involves \(H_2\) and leads to the formation of products. Thus, the condition that Step 2 is the slow step is necessary for the mechanism to be valid.

### Final Answer:
The condition under which these steps could represent the mechanism is that **Step 2 is the slow step**.

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