Question

The reaction \( \text{A}_2 + \text{B}_2 \to 2\text{AB} \) follows the mechanism: \[ \text{A}_2 \xrightarrow{k_1} \text{A} + \text{A} \, (\text{fast}), \quad \text{A} + \text{B}_2 \xrightarrow{k_2} \text{AB} + \text{B} \, (\text{slow}), \quad \text{A} + \text{B} \to \text{AB} \, (\text{fast}). \] The overall order of the reaction is:

• A. 3
• B. 1.5
• C. 2.5
• D. 2

Hint:

The overall order of a reaction is determined by the rate-determining step and the exponents of the reactants in that step.

Answer 1

The Correct Option is B

The reaction mechanism comprises three stages:

• \(\text{A}_2 \xrightarrow{k_1} \text{A} + \text{A}\) (rapid)
• \(\text{A} + \text{B}_2 \xrightarrow{k_2} \text{AB} + \text{B}\) (slow, rate-determining)
• \(\text{A} + \text{B} \to \text{AB}\) (rapid)

The overall order is determined by the rate-determining step. The rate law for the slow step is: \[\text{Rate} = k_2[\text{A}][\text{B}_2]\] The rapid initial step reaches equilibrium, where \( [\text{A}] = K_1^{0.5} [\text{A}_2]^{0.5} \). Substituting this into the rate law yields: \[\text{Rate} = k_2(K_1^{0.5}[\text{A}_2]^{0.5})[\text{B}_2]\] \[\text{Rate} = k'[\text{A}_2]^{0.5}[\text{B}_2]\] The overall order is the sum of the exponents of the reactant concentrations: \[0.5 (\text{for } [\text{A}_2]) + 1 (\text{for } [\text{B}_2]) = 1.5\] Therefore, the overall reaction order is 1.5.