Question

When the concentration of the reactant is $A \to B$ is increased by 8 times but rate increases only 2 times, the order of the reaction would be $\dfrac{1}{n}$. Value of n is?

Answer 1

Hint: The common equation for the rate of the reaction for the given reaction can be given as
\[{{\text{r}}_1}{\text{ = k[A}}{{\text{]}}^n}{\text{ }}\]
where ‘n’ is the order of the reaction.

Complete Step-by-Step Solution:
We know that the rate of the reaction is the increase or decrease in the concentration of reactants or products. There are many factors which influence the rate of the reaction.
- The rate of the reaction can depend upon the concentration of one or more reactants. We know that the representation of rate of the reaction in terms of concentration is called rate law expression.
- The given reaction and the number of moles of components are as below.
\[nA \to mB\]
 - Here, we can see that ‘n’ moles of A react to give m moles of B.
Here, we can say that the rate of the reaction can be given by the following equation.
\[{{\text{r}}_1}{\text{ = k[A}}{{\text{]}}^n}{\text{ }}...{\text{(1)}}\]
Here, ${r_1}$ is the rate of the reaction.
- Now, we are given that the rate of the reaction increases two times as we increase the concentration of the reactant by 8 times. So, we can write that
\[{r_2} = 2{r_1} = k{[8A]^n}{\text{ }}...{\text{(2)}}\]
Here, ‘n’ is the order of the reaction.
From equation (1) and (2), we can write that
\[\dfrac{{{r_1}}}{{2{r_1}}} = \dfrac{{k{{[A]}^n}}}{{k{{[8A]}^n}}}\]
So,
\[\dfrac{1}{2} = \dfrac{1}{{{{(8)}^n}}}\]
Thus, we can write that 2 = ${(8)^n}$
So, we can write that n = $\dfrac{1}{3}$ .

Thus, we obtained that the order of the reaction is $\dfrac{1}{3}$ .

Note: Remember that the sum of the power of the concentration of the reactants in the rate law expression is called order of the reaction. Order of the reaction can be any number. It can also be a fraction.