Question

A first order reaction takes 40 min for 30% decomposition. Calculate${{\text{t}}_{{}^{1}/{}_{2}}}$.

Answer 1

First order reaction: When rate of reaction depends on the first power of concentration of a reactant, it is first order reaction.The rate of first order reaction is directly proportional to the concentration of reactant. Its rate law is written as$-\dfrac{\text{d}\left[ {{\text{A}}_{o}} \right]}{\text{dt}}=\left[ {{\text{A}}_{o}} \right]$.

Complete step by step answer:
The formula of first order reaction to find rate is $\text{k = }\dfrac{2.303}{\text{t}}\log \left( \dfrac{\text{a}}{\text{a-x}} \right)$. Here, t is the time of the reaction, k is the rate of reaction,a is the initial amount of reactant present at t=0 and (a-x) is the amount of reactant left after dissociating x from it. Let us solve this question step wise
(1) Find the value of x by the decomposition of substance. The decomposition is 30%; so; x is equal to 30% of a. The value of x is equal to$\dfrac{30}{100}\times \text{a}$. Now, x is equal to$\dfrac{3}{10}\times \text{a equals 0}\text{.3a}$.
(2) Now, a-x will be$\text{a - 0}\text{.3a is 0}\text{.7a}$. The output of$\text{log}\left( \dfrac{\text{a}}{0.7\text{a}} \right)\text{will be log}\left( \dfrac{10}{7} \right).$ $\text{log}\left( \dfrac{10}{7} \right)$will give 0.1549 as its answer.
(3) Rate of reaction or k is $\dfrac{2.303}{\text{40}}\times 0.1549$equals 0.0089$\text{mi}{{\text{n}}^{-1}}$.
(4) Half-life of first order reaction is defined as the time needed by a reactant to reach half of its original value or amount. The formula of half-life is${{\text{t}}_{{1}/{2}\;}}=\dfrac{0.693}{\text{k}}$. Put the value of k found above to find${{\text{t}}_{{}^{1}/{}_{2}}}$of the reaction. The${{\text{t}}_{{}^{1}/{}_{2}}}$will be$\dfrac{0.693}{0.0089}$ which equals 77.86 min.
-The half-life of the reaction is 77.86 min.
Additional Information: Examples of first order reaction:
(1) Thermal decomposition of${{\text{N}}_{2}}{{\text{O}}_{5}}$; which occurs as${{\text{N}}_{2}}{{\text{O}}_{5}}\to {{\text{N}}_{2}}{{\text{O}}_{3}}+{{\text{O}}_{2}}$.
(2) Hydrolysis of hydrogen peroxide as${{\text{H}}_{2}}{{\text{O}}_{2}}\to {{\text{H}}_{2}}\text{O}+\dfrac{1}{2}{{\text{O}}_{2}}$.
(3) Decomposition of calcium carbide as$\text{CaC}{{\text{O}}_{3}}\to \text{CaO}+\text{C}{{\text{O}}_{2}}$.

Note: First order reaction and pseudo first order reaction are two different terms. Pseudo first order reaction is found in cases where one reacting substance is present excess and other is maintained at constant concentration. It seems as if a bimolecular reaction is made to work like a first order reaction. The rate is determined by one reactant when two are present.