Question

The half life period of a first order reaction is $6.0\;h$ . Calculate the rate constant.

Answer 1

To calculate the rate constant for the first order reaction, we must derive the relation between rate constant and time required to complete $50$ of the reaction from the differential rate law for the first order reaction. On substituting the value of half life period, we will get the value of rate constant.
$k = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}$
Where, k is the rate constant and ${t_{\dfrac{1}{2}}}$ is the half life period of the first order reaction.

Complete Step By Step Answer:
A first order reaction is a chemical reaction which proceeds at a rate that depends linearly on concentration of only one reactant. The differential rate law for the first order reaction can be represented as follows:
$- \dfrac{{d[A]}}{{dt}} = k[A]$
On rearranging terms, the expression will be as follows:
$\Rightarrow \dfrac{{d[A]}}{{[A]}} = - kdt$
Integrating both sides by applying proper limits:
$\Rightarrow \int_{{{[A]}_o}}^{{{[A]}_t}} {\dfrac{{d[A]}}{{[A]}} = - k\int_0^t {dt} }$
Where, ${[A]_o}$ is the initial concentration of the reactant and ${[A]_t}$ is the concentration of reactant left after time $t$ .
$\Rightarrow \ln {\left[ A \right]_t} - \ln {\left[ A \right]_o} = - kt$
On simplifying, the integrated rate law for the first order reaction will be as follows:
$k = \dfrac{1}{t}\ln \dfrac{{{{[A]}_o}}}{{{{[A]}_t}}}\;\;\;\;\;\;...(1)$
We know that when $50$ of the reaction is completed, ${[A]_t} = \dfrac{{{{[A]}_o}}}{2}$ and $t = {t_{\dfrac{1}{2}}}$ . Substituting values in equation (1):
$\Rightarrow k = \dfrac{1}{{{t_{\dfrac{1}{2}}}}}\ln \dfrac{{{{[A]}_o}}}{{\dfrac{{{{[A]}_o}}}{2}}}$
$\Rightarrow k = \dfrac{{\ln 2}}{{{t_{\dfrac{1}{2}}}}}$
Substituting the value of $\ln \;2 = 0.693$ :
$\Rightarrow k = \dfrac{{0.693}}{{{t_{\dfrac{1}{2}}}}}\;\;\;\;\;...(2)$
Now, as per question the half life period of first order reaction is $6.0\;h$ . Substituting value in equation (2), the value of rate constant will be:
$k = \dfrac{{0.693}}{6}$
$\Rightarrow k = 0.116\;{h^{ - 1}}$
Hence, the rate constant for the given first order reaction is $0.116\;{h^{ - 1}}$ .

Note:
It is important to note that for first order reactions, the half-life is independent of the initial concentration of the reactant, which is a unique aspect to the first order reactions. Always remember that, the ${[A]_t}$ is the amount of reactant left after time t but not the amount of reactant consumed in the reaction.