Question

The rate constant of a certain first order reaction is $200{s^{ - 1}}$. What is its half-life period?

Answer 1

Hint: Half-life is the term used in radioactivity to define the period of time in which the radioactive nuclei undergoes radioactive decay so that it is reduced to half of its initial amount. The formula used for calculating half-life is,
$N(t) = {N_0}{e^{ - \lambda t}}$

Complete step by step answer:
For a first order reaction, the decay does not take place exponentially, and in one half-life the amount of radioactive substance is reduced by half, that is, half of the initial amount is left.
The reduced formula for the radioactive decay following first order reaction, is given as,
${t_{1/2}} = \dfrac{{0.693}}{\lambda }$
Where, ${t_{1/2}}$ is the half-life of the substance
$\lambda $ is the rate constant of the reaction
Substituting the value of $\lambda $ in the above equation,
${t_{1/2}} = \dfrac{{0.693}}{{200}}s$
$ \Rightarrow {t_{1/2}} = 0.003465s$
$ \Rightarrow {t_{1/2}} = 3.465 \times {10^{ - 3}}s$
Therefore, the half-life of the radioactive substance is $3.465 \times {10^{ - 3}}s$.

Note:
The reduced formula for calculating the half-life is derived from the formula
$N(t) = {N_0}{e^{ - \lambda t}}$. For first order reactions, as the amount of substance is also reduced by half at half-life, therefore, $N(t) = \dfrac{{{N_0}}}{2}$. Substituting the value of N(t) in the equation,
$\dfrac{{{N_0}}}{2} = {N_0}{e^{ - \lambda t}}$
$\dfrac{1}{2} = {e^{ - \lambda {t_{1/2}}}}$
Taking log of both sides,
$\ln \dfrac{1}{2} = \ln {e^{ - \lambda {t_{1/2}}}}$
As, $\ln {e^{ - \lambda {t_{1/2}}}} = - \lambda {t_{1/2}}$
So, $\ln \dfrac{1}{2} = - \lambda {t_{1/2}}$
$ \Rightarrow \ln 2 = \lambda {t_{1/2}}$
$ \Rightarrow 0.693 = \lambda {t_{1/2}}$
$ \Rightarrow {t_{1/2}} = \dfrac{{0.693}}{\lambda }$