Question

In a reaction ${\text{A}} \to $ product, when the start is made from $8 \times {10^{ - 2}}M$ of $A$, half-life is found to be $120$ minutes. For the initial concentration $4 \times {10^{ - 2}}M$, half-life of the reaction becomes $240$ minutes. The order of the reaction is:
A.$0$
B.$1$
C.$2$
D.$0.5$

Answer 1

Hint: Rate of reaction: The rate of a reaction is the speed at which a chemical reaction happens.
Order of the reaction: It is defined as the power dependence of the rate of reaction on the concentration of the reactants.

Complete step by step solution:
Let us first talk about the rate of reaction and order of reaction.
Rate of reaction: The rate of a reaction is the speed at which a chemical reaction happens.
Order of the reaction: It is defined as the power dependence of the rate of reaction on the concentration of the reactants. For example: if order of reaction is one then rate of reaction depends linearly on the concentration of one reactant. The unit of first order of reaction is ${s^{ - 1}}$. The unit of second order of reaction is $1/Ms$.
Half-time: It is defined as the time duration in which the concentration of a reactant drops to one-half of its initial concentration. It is represented by ${t_{\dfrac{1}{2}}}$.
The relation between the half-time and concentration of reactant is as follows:
Half-time of a reaction is inversely proportional to the concentration of the reactant raised to the power of its order of reaction minus one.
Here in the question we are given with the initial concentration and half-time for the reaction two times. And as the relation between the half-time and concentration is inverse relation so we can say that if ${t_{\dfrac{1}{2}}}^1$ is the half-life for case one and ${a_1}$ is its initial concentration and for case two half-life is as ${t_{\dfrac{1}{2}}}^2$ and its initial concentration as ${a_2}$ and let the order of reaction as $n$ then the relation will be:
$\dfrac{{{t_{\dfrac{1}{2}}}^1}}{{{t_{\dfrac{1}{2}}}^2}} = {(\dfrac{{{a_2}}}{{{a_1}}})^{n - 1}}$.
Here, we are given with ${t_{\dfrac{1}{2}}}^1 = 120,{\text{ }}{t_{\dfrac{1}{2}}}^1 = 240,{\text{ }}{{\text{a}}_1} = 8M,{\text{ }}{{\text{a}}_2} = 4M$
Now these values in the relation and we will get,
$
\Rightarrow \dfrac{{120}}{{240}} = {(\dfrac{4}{8})^{n - 1}} \\
\Rightarrow \dfrac{1}{2} = {\dfrac{1}{2}^{n - 1}} \\
 $
By comparing the powers we get,
 $
\Rightarrow n - 1 = 1 \\
\Rightarrow n = 2 \\
 $.
So the order of the reaction will be $2$.

Hence option C is correct.

Note: Activation energy: It is defined as the least required energy for a chemical reaction to happen. It is represented by ${E_a}$.
Frequency factor is defined as the rate of molecular collisions that occur during the chemical reaction. It is also known as pre-exponential factor and is represented as A.