JEE Chemistry 2025 Question with Solution

Given: The mechanism is

$$\text{A}_2 \xrightarrow{k_1} \text{A} + \text{A}$$

(fast)

$$\text{A} + \text{B}_2 \xrightarrow{k_2} \text{AB} + \text{B}$$

(slow, rate-determining step)

$$\text{A} + \text{B} \to \text{AB}$$

(fast)

Find: The overall order of the reaction.

The slow step controls the rate, so

$$\text{Rate} = k_2[\text{A}][\text{B}_2]$$

Here $[\text{A}]$ is an intermediate, so express it in terms of stable reactants. From the fast step, as given in the solution,

$$[\text{A}] = K_1^{0.5}[\text{A}_2]^{0.5}$$

Substituting into the rate law,

$$\text{Rate} = k_2\left(K_1^{0.5}[\text{A}_2]^{0.5}\right)[\text{B}_2]$$ $$\text{Rate} = k'[\text{A}_2]^{0.5}[\text{B}_2]$$

Therefore, the overall order is

$$0.5 + 1 = 1.5$$

Hence, the overall order of the reaction is $1.5$. The correct option is B.

Given:

$$\text{A}_2 + \text{B}_2 \rightarrow 2\text{AB}$$

Mechanism:

1. $\text{A}_2 \xrightarrow{k_1} 2\text{A}$ (fast)
2. $\text{A} + \text{B}_2 \xrightarrow{k_2} \text{AB} + \text{B}$ (slow)
3. $\text{A} + \text{B} \xrightarrow{k_3} \text{AB}$ (fast)

Find: The overall order.

The second step is the rate-determining step, so

$$\text{Rate} = k_2[\text{A}][\text{B}_2]$$

Using the steady-state treatment shown in the solution for intermediate $\text{A}$:

$$2k_1[\text{A}_2] = k_2[\text{A}][\text{B}_2] + k_3[\text{A}][\text{B}]$$ $$[\text{A}] = \frac{2k_1[\text{A}_2]}{k_2[\text{B}_2] + k_3[\text{B}]}$$

The solution further states that

$$[\text{B}] \propto [\text{B}_2]^{1/2}$$

so the effective dependence on $[\text{B}_2]$ becomes $[\text{B}_2]^{1/2}$. Hence the overall rate law is approximately

$$\text{Rate} \propto [\text{A}_2]^1[\text{B}_2]^{1/2}$$

Therefore, the overall order is

$$1 + \frac{1}{2} = 1.5$$

So, the overall order of the reaction is $1.5$.