Question

The rate of decomposition of a substance increases by a factor $2.25$ for $1.5$ times increase in the concentration of the substance at the same temperature. Find out the order of the reaction?

Answer 1

To answer this question, you should recall the concept of kinetics of a reaction. Using rate law and the values given in the question we shall find the answer to this question. We shall substitute the values in the general rate law equation (given below).

Formula used: $\dfrac{{d\left[ A \right]}}{{dt}} = - k{\left[ A \right]^n}$
where $n$is the order of reaction, k is the rate constant and A is the concentration in molarity.

Complete step by step solution:
The order of reaction can be defined as the power dependence of rate on the concentration of all reactants. Many factors affect the rate of reactions involving different phases proceed more rapidly when there is greater surface area contact. Generally, increase in temperature or reactant concentration increases the rate of a given reaction. Now, for decomposition reaction rate law can be written as:
$\dfrac{{d\left[ A \right]}}{{dt}} = - k{\left[ A \right]^n}$----1
where $n$is the order of reaction.
As per the given condition:
$2.25\dfrac{{d\left[ A \right]}}{{dt}} = - k{\left[ {1.5A} \right]^n}\;$----2.
Solving these 1 and 2 equations by comparing we get
$n = 2$

Hence, the order of the reaction is 2.

Note: The overall order of any reaction is the sum of the partial components of a chemical reaction.
Rate$= {\text{ }}k{\left[ A \right]^x}{\left[ B \right]^y}\;$ then the overall order can be written as $x + y$. As we said that the order provides relationship between rate and concentration of components:
1.If the reaction is a zero-order reaction, change in the reactant concentration will have no effect on the reaction rate.
2.In case of first order, doubling the reactant concentration will double the reaction rate.
3.In case of second-order, doubling the concentration of the reactants will quadruple the overall reaction rate.
4.In case of third order, the overall rate increases by eight times when the reactant concentration is doubled.
The important formulas for the half-life of a reaction which varies with the order of the reaction:
a)In the case of a zero-order reaction, the expression for the half-life is: ${t_{1/2}}\; = {\text{ }}\dfrac{{{{\left[ R \right]}_0}}}{{2k}}$
b)In the case of the first-order reaction, the expression for the half-life is given by: ${t_{1/2}}\; = \dfrac{{{\text{ }}0.693}}{k}$
c)In case of a second-order reaction, the expression for the half-life of the reaction is: ${t_{1/2}} = \;\dfrac{1}{{k{{[R]}_0}}}$