Question

The decomposition of ${{N}_{2}}{{O}_{5}}$ occurs as:
\[2{{N}_{2}}{{O}_{5}}\to 4N{{O}_{2}}+{{O}_{2}}\], and follows first order kinetics hence.
(A) The reaction is bimolecular.
(B) The reaction is unimolecular
(C) ${{t}_{\dfrac{1}{2}}}\propto {{\left[ {{a}_{0}} \right]}^{0}}$
(D) None of the above

Answer 1

Hint: Molecularity of the reaction is the number of species that will collide with each other to give products. Half life of any reaction can be given from its order of reaction by following formula.
     ${{t}_{\dfrac{1}{2}}}\propto {{\left[ {{a}_{0}} \right]}^{1-n}}$
Where n is the order of the reaction.

Complete step by step answer:
Let’s know about first order kinetics.
We should know about first order kinetics that in this type of reaction, the rate of the reaction depends on the concentration of only one reactant, and is proportional to the amount of the reactant.
It may be represented by the equation, rate = kA, where k is the reaction rate constant, and A is the concentration of the reactant. The differential equation describing the first-order reaction’s rate is given below:
     \[Rate=\dfrac{-d\left[ A \right]}{dt}=k{{\left[ A \right]}^{1}}=k\left[ A \right]\]
The "rate" is the reaction rate (in units of moles/time) and k is the reaction rate coefficient (in units of 1/time).
     \[2{{N}_{2}}{{O}_{5}}\to 4N{{O}_{2}}+{{O}_{2}}\]

- In the above reaction, we should note that molecularity of reaction is 2. We should note that the molecularity of a reaction is the number of reactant molecules taking part in a single step of the reaction. So, in the above reaction molecularity should be 2. As the molecularity of this reaction is 2, we can call it a bimolecular reaction.
We will now use the half-life equation:
${{t}_{\dfrac{1}{2}}}$ = half-life of first order
$\left[ {{a}_{0}} \right]$ =initial concentration
n=1, for first order reaction
     $\begin{align}
 & {{t}_{\dfrac{1}{2}}}\propto {{\left[ {{a}_{0}} \right]}^{1-n}} \\
 & n=1 \\
 & {{t}_{\dfrac{1}{2}}}\propto {{\left[ {{a}_{0}} \right]}^{1-1}} \\
 & {{t}_{\dfrac{1}{2}}}\propto {{\left[ {{a}_{0}} \right]}^{0}} \\
\end{align}$

Thus, we can say from above discussion that this reaction is bimolecular and also it follows that ${{t}_{\dfrac{1}{2}}}\propto {{\left[ {{a}_{0}} \right]}^{0}}$ .
So, the correct answer is “Option A and C”.

Note: We should note that, the half-life of a reaction is the amount of time needed for a reactant concentration to decrease by half compared to its initial concentration. Do not get confused between molecularity of the reaction and order of the reaction. So, whenever unimolecular, bimolecular, etc terms are used, it shows molecularity of the reaction and it does not have any relation with the order of the reaction.