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Question: A hypothetical reaction \({{\text{A}}_{\text{2}}}{\text{ + }}{{\text{B}}_{\text{2}}} \to {\text{2AB}...

A hypothetical reaction A2 + B2→2AB{{\text{A}}_{\text{2}}}{\text{ + }}{{\text{B}}_{\text{2}}} \to {\text{2AB}} follows the mechanism as given below:
A2⇄A + A{{\text{A}}_{\text{2}}} \rightleftarrows {\text{A + A}}…… (Fast)
A + B2⇄AB + B{\text{A + }}{{\text{B}}_{\text{2}}} \rightleftarrows {\text{AB + B}}…….. (Slow)
A + B→AB{\text{A + B}} \to {\text{AB}}……… (Fast)
The order of the overall reaction is?
A.2
B.1
C.32\dfrac{3}{2}
D.0

Explanation

Solution

In chemical kinetics, the overall rate of the reaction is often approximately determined by the slowest step and the order of the reaction is also determined from the same. This step is called the Rate-Determining Step or the Rate-Limiting step.

Complete step by step answer:
The steps involved in the hypothetical reaction, A2 + B2→2AB{{\text{A}}_{\text{2}}}{\text{ + }}{{\text{B}}_{\text{2}}} \to {\text{2AB}} can be given as follows:
A2⇄A + A{{\text{A}}_{\text{2}}} \rightleftarrows {\text{A + A}}…… (Fast)
A + B2⇄AB + B{\text{A + }}{{\text{B}}_{\text{2}}} \rightleftarrows {\text{AB + B}}…….. (Slow)
A + B→AB{\text{A + B}} \to {\text{AB}}……… (Fast)
Now, for any chemical reaction, the prediction of the rate equation is simplified by the assumption of the Rate-Determining Step. In the above mechanism, the second step is the slowest step which can be considered as the rate determining step of the reaction. In this step, there are two species of reactants involved.
The rate of the reaction can be written as:
R = k[A][B]1{\text{k}}\left[ {\text{A}} \right]{\left[ {\text{B}} \right]^1}…………. (1), where k is the rate constant of the reaction.
The concentration of A{\text{A}} can be determined from the first equation.
A2⇄A + A{{\text{A}}_{\text{2}}} \rightleftarrows {\text{A + A}}, at equilibrium, Keq = [A]2[A2]{\text{Keq = }}\dfrac{{{{\left[ {\text{A}} \right]}^{\text{2}}}}}{{\left[ {{{\text{A}}_{\text{2}}}} \right]}}
Therefore, [A]=Keq[A2]\left[ {\text{A}} \right] = \sqrt {{\text{Keq}}\left[ {{{\text{A}}_{\text{2}}}} \right]}
Replacing the value ofA{\text{A}} in equation 1, we get,
Rate = k×(Keq)12 [A2]12[B]1{{k \times }}{\left( {{\text{Keq}}} \right)^{\dfrac{{\text{1}}}{{\text{2}}}}}{\text{ }}{\left[ {{{\text{A}}_{\text{2}}}} \right]^{\dfrac{{\text{1}}}{{\text{2}}}}}{\left[ {\text{B}} \right]^1}
From here, we get the order of the reaction as the sum of the stoichiometric coefficient of the reactants which is equal to,
Order of the reaction = 12+1=32\dfrac{1}{2} + 1 = \dfrac{3}{2}

Hence, the correct answer is option C.

Note:
The order of a chemical reaction or the reaction order can be defined as the sum of the exponents of the concentration terms of the reactants obtained from the rate-determining step of the reaction. The value of this order can be even zero or of fractional order and this symbolises the dependence of the rate of the reaction on the concentration of the reactants.