Question

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

Answer 1

Hint: 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.