Answer
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Hint: The order of reaction refers to the relationship between the rate of a chemical reaction and the concentration of the species taking part in it. This can be defined as the power dependence of rate on the concentration of all reactants. The reaction order applicable in all chemical reactions.
Complete step by step solution:
Given reaction, \[{{A}_{2}}+{{B}_{2}}\to 2AB\]
From the given reaction mechanism steps, observe that [A] is the intermediate reactive species where concentration is determined from the equilibrium step.
In mechanism reactions, there are two fast reaction steps with one slow step reaction which is the rate-determining step.
\[A+{{B}_{2}}\to AB+B(slow)\]
The rate of the above reaction, $r=k[A][{{B}_{2}}]--(1)$
From equilibrium steps in the given mechanism, ${{A}_{2}}\rightleftharpoons A+A(fast)$
\[\begin{align}
& {{k}_{eq}}=\dfrac{[A][A]}{[{{A}_{2}}]} \\
& [A]={{({{k}_{eq}}[{{A}_{2}}])}^{\dfrac{1}{2}}}--(2) \\
\end{align}\]
From equation (1) and (2),
\[r=k.{{k}_{eq}}^{\dfrac{1}{2}}{{[{{A}_{2}}]}^{\dfrac{1}{2}}}[{{B}_{2}}]--(3)\]
Thus, the above equation represents the order of the given reaction \[{{A}_{2}}+{{B}_{2}}\to 2AB\]
Hence, from equation (1),
\[r={{k}_{1}}{{[{{A}_{2}}]}^{\dfrac{1}{2}}}{{[{{B}_{2}}]}^{1}};(\because {{k}_{1}}=k.{{k}_{eq}})\]
From the definition order of a reaction is the sum of the order of with respective reactants $[{{A}_{2}}]\And [{{B}_{2}}]$ .
Hence the order of the given reaction is n=$1+\dfrac{1}{2}=\dfrac{3}{2}$
The order of a reaction, not necessarily an integer. Zero-order, negative order, positive integer, non-integer order of reactions are possible. So the given reaction belongs to mixed order or higher order reactions. i.e, fractional order
Therefore the correct answer is option C.
Note: The overall order of the reaction is found by adding up the individual orders of the reaction of reactants taking part in a chemical reaction. The order of reaction of a reactant indicates how much the rate of reaction changes if the concentration of the reactant is changed.
Complete step by step solution:
Given reaction, \[{{A}_{2}}+{{B}_{2}}\to 2AB\]
From the given reaction mechanism steps, observe that [A] is the intermediate reactive species where concentration is determined from the equilibrium step.
In mechanism reactions, there are two fast reaction steps with one slow step reaction which is the rate-determining step.
\[A+{{B}_{2}}\to AB+B(slow)\]
The rate of the above reaction, $r=k[A][{{B}_{2}}]--(1)$
From equilibrium steps in the given mechanism, ${{A}_{2}}\rightleftharpoons A+A(fast)$
\[\begin{align}
& {{k}_{eq}}=\dfrac{[A][A]}{[{{A}_{2}}]} \\
& [A]={{({{k}_{eq}}[{{A}_{2}}])}^{\dfrac{1}{2}}}--(2) \\
\end{align}\]
From equation (1) and (2),
\[r=k.{{k}_{eq}}^{\dfrac{1}{2}}{{[{{A}_{2}}]}^{\dfrac{1}{2}}}[{{B}_{2}}]--(3)\]
Thus, the above equation represents the order of the given reaction \[{{A}_{2}}+{{B}_{2}}\to 2AB\]
Hence, from equation (1),
\[r={{k}_{1}}{{[{{A}_{2}}]}^{\dfrac{1}{2}}}{{[{{B}_{2}}]}^{1}};(\because {{k}_{1}}=k.{{k}_{eq}})\]
From the definition order of a reaction is the sum of the order of with respective reactants $[{{A}_{2}}]\And [{{B}_{2}}]$ .
Hence the order of the given reaction is n=$1+\dfrac{1}{2}=\dfrac{3}{2}$
The order of a reaction, not necessarily an integer. Zero-order, negative order, positive integer, non-integer order of reactions are possible. So the given reaction belongs to mixed order or higher order reactions. i.e, fractional order
Therefore the correct answer is option C.
Note: The overall order of the reaction is found by adding up the individual orders of the reaction of reactants taking part in a chemical reaction. The order of reaction of a reactant indicates how much the rate of reaction changes if the concentration of the reactant is changed.
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