Question

Determine the rate law for a reaction, \(2\text{NO}+\text{Br}_{2}\rightarrow 2\text{NOBr}\) that occurs in following two steps.
i) \(\text{NO}+\text{Br}_{2}\rightarrow \text{NOBr}_{2}\) (fast)
ii) \(\text{NOBr}_{2}+\text{NO}\rightarrow 2\text{NOBr}\) (slow)

• A. \(r=k[\text{NOBr}_{2}][\text{NO}]\)
• B. \(r=k'[\text{NO}]^{2}[\text{Br}_{2}]\)
• C. \(r=k'[\text{NO}][\text{Br}_{2}]\)
• D. \(r=k[\text{NO}][\text{Br}_{2}]^{2}\)

Hint:

The slow step decides the rate, and any intermediate in it is replaced using the fast equilibrium step.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The overall rate is controlled by the slowest step, called the rate-determining step.
If this step contains an intermediate, the intermediate must be written in terms of the original reactants.

Step 2: Rate of the Slow Step
The slow step is \(\text{NOBr}_{2}+\text{NO}\rightarrow 2\text{NOBr}\).
So \(\text{rate}=k[\text{NOBr}_{2}][\text{NO}]\).

Step 3: Remove the Intermediate
The first step is fast and reversible, so it reaches equilibrium.
\(K=\dfrac{[\text{NOBr}_{2}]}{[\text{NO}][\text{Br}_{2}]}\), which gives \([\text{NOBr}_{2}]=K[\text{NO}][\text{Br}_{2}]\).

Step 4: Final Rate Law
Substituting, \(\text{rate}=kK[\text{NO}]^{2}[\text{Br}_{2}]=k'[\text{NO}]^{2}[\text{Br}_{2}]\).
This matches option (B).

Final Answer:
The rate law is \(r=k'[\text{NO}]^{2}[\text{Br}_{2}]\), so the answer is option (B). \[ \boxed{\text{(B) }r=k'[\text{NO}]^{2}[\text{Br}_{2}]} \]