Sure, let's go through this step by step.
### (a) Writing the Overall Reaction
The overall reaction can be obtained by adding the steps of the mechanism together.
Step 1:
\(\[ \text{NO(g)} + \text{NO(g)} \rightarrow \text{N}_2\text{O}_2 \text{(g)} \]\)
Step 2:
\(\[ \text{N}_2\text{O}_2 \text{(g)} + \text{H}_2 \text{(g)} \rightarrow \text{N}_2\text{O(g)} + \text{H}_2\text{O(g)} \]\)
To find the overall reaction, combine the reactants and products from both steps:
\(\[
2 \text{NO(g)} + \text{H}_2 \text{(g)} \rightarrow \text{N}_2\text{O(g)} + \text{H}_2\text{O(g)}
\]\)
### (b) Writing the Rate Law for Each Elementary Reaction
The rate law for an elementary reaction is determined by the molecularity of the reaction, which refers to the number of molecules colliding in that step.
Step 1:
\(\[ \text{NO(g)} + \text{NO(g)} \rightarrow \text{N}_2\text{O}_2 \text{(g)} \]\)
This step is given as fast. The rate law for this elementary reaction will depend on the concentration of NO. Since two molecules of NO are involved:
\(\[ \text{rate} = k_1[\text{NO}]^2 \]\)
where \(\( k_1 \)\) is the rate constant for Step 1.
Step 2:
\(\[ \text{N}_2\text{O}_2 \text{(g)} + \text{H}_2 \text{(g)} \rightarrow \text{N}_2\text{O(g)} + \text{H}_2\text{O(g)} \]\)
This step is slow and is the rate-determining step. The rate law for this elementary reaction will depend on the concentration of \(\( \text{N}_2\text{O}_2 \)\) and \(\( \text{H}_2 \)\). Since one molecule of each is involved:
\(\[ \text{rate} = k_2[\text{N}_2\text{O}_2][\text{H}_2] \]\)
where \(\( k_2 \)\) is the rate constant for Step 2.
### Summary:
- Overall Reaction:
\(\[ 2 \text{NO(g)} + \text{H}_2 \text{(g)} \rightarrow \text{N}_2\text{O(g)} + \text{H}_2\text{O(g)} \]\)
- Rate Laws for Each Step:
- Step 1 (Fast):
\(\[ \text{rate} = k_1[\text{NO}]^2 \]\)
- Step 2 (Slow, Rate-Determining):
\(\[ \text{rate} = k_2[\text{N}_2\text{O}_2][\text{H}_2] \]\)
By understanding these, we get a clear picture of how the reaction proceeds and the kinetic dependencies of each step.
