A hypothetical reaction A2+B2 arrow 2 A B follows the mechanism as given below, A2 leftharpoons A+A text (fast ) A+B2 longrightarrow A B+B text (slow) A+B longrightarrow A B text (fast) The order of the overall reaction is

Question

A hypothetical reaction A2+B2 arrow 2 A B follows the mechanism as given below, A2 leftharpoons A+A text (fast ) A+B2 longrightarrow A B+B text (slow) A+B longrightarrow A B text (fast) The order of the overall reaction is (A) 2 (B) 1 (C)  1(1/2)  (D) 0

Tardigrade Admin · Accepted Answer

From slow step, rate =k[B2][A] From 1st equation K eq =([A]2/[A2]) or [A]=√K eq [A2]=K eq 1 / 2 A21 / 2 Hence, rate =k[B2] Ke q1 / 2[A2]1 / 2 =k'[A2]1 / 2[B2] Hence, order =1 (1/2)

Q. A hypothetical reaction $A_{2} + B_{2} \to 2 A B$ follows the mechanism as given below,
$A_{2} ⇌ A + A (fast )$
$A + B_{2} ⟶ A B + B (slow)$
$A + B ⟶ A B (fast)$
The order of the overall reaction is

2

1

$1 \frac{1}{2}$

0

From slow step, rate $= k [B_{2}] [A]$
From 1st equation $K_{e q} = \frac{[ A ] ^{2}}{[ A _{2} ]}$
or $[A] = K_{e q} [A_{2}] = K_{e q}^{1/2} A_{2}^{1/2}$
Hence, rate $= k [B_{2}] K_{e q}^{1/2} [A_{2}]^{1/2}$
$= k^{'} [A_{2}]^{1/2} [B_{2}]$
Hence, order $= 1 \frac{1}{2}$

Q. A hypothetical reaction A2​+B2​→2AB follows the mechanism as given below, A2​⇌A+A (fast ) A+B2​⟶AB+B (slow) A+B⟶AB (fast) The order of the overall reaction is

2

1

121​

0

From slow step, rate =k[B2​][A] From 1st equation Keq​=[A2​][A]2​ or [A]=Keq​[A2​]​=Keq1/2​A21/2​ Hence, rate =k[B2​]Keq1/2​[A2​]1/2 =k′[A2​]1/2[B2​] Hence, order =121​

Q. A hypothetical reaction $A_{2} + B_{2} \to 2 A B$ follows the mechanism as given below,
$A_{2} ⇌ A + A (fast )$
$A + B_{2} ⟶ A B + B (slow)$
$A + B ⟶ A B (fast)$
The order of the overall reaction is

$1 \frac{1}{2}$

From slow step, rate $= k [B_{2}] [A]$
From 1st equation $K_{e q} = \frac{[ A ] ^{2}}{[ A _{2} ]}$
or $[A] = K_{e q} [A_{2}] = K_{e q}^{1/2} A_{2}^{1/2}$
Hence, rate $= k [B_{2}] K_{e q}^{1/2} [A_{2}]^{1/2}$
$= k^{'} [A_{2}]^{1/2} [B_{2}]$
Hence, order $= 1 \frac{1}{2}$