A hypothetical reaction A2+B2→ 2AB follows the mechanism as given below, A2A+A(fast) A+B2 xrightarrowAB+B(slow) A+B xrightarrowAB(fast) The order of the overall reaction is

Question

A hypothetical reaction A2+B2→ 2AB follows the mechanism as given below, A2A+A(fast) A+B2 xrightarrowAB+B(slow) A+B xrightarrowAB(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 Keq=([A]2/[A2]) or [A]=√Keq[A2]=Keq1/2A21/2 Hence, rate =k[B2]Keq1/2[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, $A_{2} A + A (f a s t)$ $A + B_{2} A B + B (s l o w)$ $A + B A B (f a s t)$ 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, $A_{2} A + A (f a s t)$ $A + B_{2} A B + B (s l o w)$ $A + B A B (f a s t)$ 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 $= 1 \frac{1}{2}$