Question

A reaction A2 + B2 → 2AB occurs by the following mechanism -
A2 → A + A ......... (slow)
A + B2 → AB + B .......(fast)
A + B → AB .........(fast) its order would be -
(A) 3/2
(B) 1
(C) 0
(D) 2​

Answer 1

Answer: (B) 1

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

Order of the reaction is defined as the sum of the concentration of terms on which the rate of the reaction actually depends. It is the sum of the exponents of the molar concentration in the rate law expression.

For reactions which takes place in multiple steps are complex reactions and the order is given by the slowest step.

For the given reaction, the slowest step is

$A_2\rightarrow A+A$

$Rate=k[A_2]^1$

Order of the reaction = 1

Thus order of the reaction is 1.

Answer 2

Order of the given reaction will be 1

Explanation:

In a mechanism of the reaction, the slow step in the mechanism determines the rate of the reaction.

For the given chemical reaction:

$A_2+B_2\rightarrow 2AB$

The intermediate reaction of the mechanism follows:

Step 1:  $A_2\rightleftharpoons A+A;\text{ (slow)}$

Step 2:  $A+B_2\rightarrow AB+B;\text{ (fast)}$

Step 3:  $A+B\rightarrow AB;\text{ (fast)}$

As, step 1 is the slow step. It is the rate determining step

Rate law for the reaction follows:

$\text{Rate}=k[A_2]$

Order of the reaction is defined as the sum of the concentration of terms on which the rate of the reaction actually depends. It is the sum of the exponents of the molar concentration in the rate law expression.

Order of the reaction = 1

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