Question

The reaction \( {A}_2 + {B}_2 \to 2{AB} \) follows the mechanism: \[ {A}_2 \xrightarrow{k_1} {A} + {A} \ ({fast}) \quad {A} + {B}_2 \xrightarrow{k_2} {AB} + {B} \ ({slow}) \quad {A} + {B} \to {AB} \ ({fast}) \] 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 (slow step). If there is equilibrium in one of the steps, the concentration of that reactant can be substituted with the equilibrium expression.

Answer 1

The Correct Option is B

The overall rate law is determined by the slow step of the mechanism, which is: \[ {Rate} = k_2 [{A}][{B}_2] \] Since the fast step \( {A}_2 \to {A} + {A} \) is equilibrium, we can use the equilibrium constant \( k_1 \) to express \( [{A}] \) in terms of \( [{A}_2] \): \[ [{A}] = \sqrt{k_1 [{A}_2]} \] Substituting this into the rate law: \[ {Rate} = k_2 \sqrt{k_1 [{A}_2]} [{B}_2] \] Thus, the overall rate law is: \[ {Rate} = k [{A}_2]^{1/2} [{B}_2]^1 \] The overall order of the reaction is \( 1/2 + 1 = 1.5 \).