JEE Chemistry 2023 Question with Solution

Given: The mechanism is

$$NO + Br_2 \leftrightarrow NOBr_2 \quad (\text{fast})$$ $$NOBr_2 + NO \to 2NOBr \quad (\text{slow})$$

Find: The overall order of the reaction.

The slow step is the rate-determining step, so the rate law is first written from that step:

$$\text{Rate} = k[NOBr_2][NO]$$

Since the first step is a fast equilibrium,

$$K = \frac{[NOBr_2]}{[NO][Br_2]}$$

So,

$$[NOBr_2] = K[NO][Br_2]$$

Substituting into the rate law:

$$\text{Rate} = k(K[NO][Br_2])[NO]$$ $$\text{Rate} = kK[NO]^2[Br_2]$$

Therefore, the overall order is $2 + 1 = 3$.

So, the numerical answer is $3$.

Given: A two-step mechanism with a fast equilibrium followed by a slow step.

1. The molecularity of the slow step alone cannot be used directly unless the intermediate is removed using the fast equilibrium.
2. From the slow step,

$$\text{Rate} = k[NOBr_2][NO]$$

3. The intermediate $NOBr_2$ is formed in the fast step, so use the equilibrium relation:

$$NO + Br_2 \leftrightarrow NOBr_2$$ $$K = \frac{[NOBr_2]}{[NO][Br_2]}$$

Hence,

$$[NOBr_2] = K[NO][Br_2]$$

4. Replace the intermediate concentration in the rate law:

$$\text{Rate} = k[NO]\bigl(K[NO][Br_2]\bigr)$$ $$\text{Rate} = kK[NO]^2[Br_2]$$

Now add the powers of reactant concentrations in the final rate law.

Thus, the reaction is of third order, and the answer is $3$.