Question

Derive the relation between half-life and rate constant for a first-order reaction.

Answer 1

Hint: First order reaction is a reaction whose reaction rate depends upon the concentration of only one reactant. The half-life can be defined as the time required to react to the half amount of the reactant’s initial amount. The rate constant is the coefficient of proportionality at the given temperature to the product of the concentrations of the reactants. This information can be applied to establish a mathematical equation and make the derivation of relation.

Complete step by step answer:
1) The derivation for the relation between half-life and rate constant for a first-order reaction will be as follows:
First, we will use the equation of integrated rate law,
$k = \dfrac{{2.303}}{t}{\log _{10}}\dfrac{{{{\left[ A \right]}_0}}}{{{{\left[ A \right]}_t}}}$
In the above equation, the notations are described as,
For a half-life reaction time, $t = {t_{1/2}}$
${\left[ A \right]_0} = $ the initial concentration of a reactant at the initial time where $t = 0$
${\left[ A \right]_t} = $ the final concentration of a reactant of the half-life reaction where $t = \dfrac{1}{2}$
In a half-life reaction, The value of ${\left[ A \right]_t}$ which is the final concentration of the reactant is half of the initial concentration of the reactant ${\left[ A \right]_0}$.
Therefore, The above value ${\left[ A \right]_t}$ can also be written as,
${\left[ A \right]_t} = \dfrac{{{{\left[ A \right]}_0}}}{2}$
By putting the value of the ${\left[ A \right]_t}$ in the equation of integrated rate law for the first-order reaction we get,
$k = \dfrac{{2.303}}{{{t_{1/2}}}}{\log _{10}}\dfrac{{{{\left[ A \right]}_0}}}{{\dfrac{{{{\left[ A \right]}_0}}}{2}}}$
By taking denominator value to numerator we get,
$k = \dfrac{{2.303}}{{{t_{1/2}}}}{\log _{10}}\dfrac{{{{\left[ A \right]}_0}}}{1} \times \dfrac{2}{{{{\left[ A \right]}_0}}}$
By canceling the same values we get,
$k = \dfrac{{2.303}}{{{t_{1/2}}}}{\log _{10}}\left( 2 \right)$
By calculating the logarithmic value of ${\log _{10}}\left( 2 \right) = 0.301$ and putting it in the following equation we get,
$k = \dfrac{{2.303}}{{{t_{1/2}}}} \times 0.301$
By doing multiplication of the multiplying factors we get,
$2.303 \times 0.301 = 0.693$
By putting the multiplication result in the above equation we get,
$k = \dfrac{{0.693}}{{{t_{1/2}}}}$
By changing sides we get,
${t_{1/2}} = \dfrac{{0.693}}{k}$
This is the derived equation showing the relation between half-life and rate constant for the first-order reaction.

Note:
While taking the value of final concentration ${\left[ A \right]_t}$ one must take it half of the initial concentration ${\left[ A \right]_0}$ as the final reactant will be exactly half of the amount of initial reactant amount. After the whole derivation, we concluded that the half-life and rate constant are inversely proportional to each other.