Question

A first order reaction is 50% completed after 30 minutes. This implies that the time required to complete 90% of the reaction is?

Answer 1

Hint: We are given a first order reaction and consider it to be $ A \to P $ . For solving this question we need to know the integrated rate law for first order reactions. First order reactions are those where the rate of the reaction only depends on the rate of the reactant and not the substrate.

Complete answer:
Consider a first order reaction $ A \to P $ where A is the reactant and P is the product. The rate law for the reaction is given as: $ Rate = k{[A]^1} $ where the exponent 1 is the order of the reaction; First order. Consider the initial concentration of the reactant to be $ {A_0} $ and that after time ‘t’ be $ {[A]_t} $ .
The integrated rate law for the first order reaction hence can be given as: $ - kt = \ln \left( {\dfrac{{{A_t}}}{{{A_0}}}} \right) $
Where k is the rate constant usually expressed in $ {s^{ - 1}} $ or $ {\min ^{ - 1}} $ for first order reaction.
The information given to us is the time taken for 50% reaction to compete is 30mins, i.e. $ t = 30 \to {[A]_t} = \dfrac{{{{[A]}_0}}}{2} $
This time required to complete 50% of the reaction is known as the half life of the reaction. Substituting the values in the above equation we get: $ - k(30) = \ln \left( {\dfrac{{\dfrac{{{{[A]}_0}}}{2}}}{{{{[A]}_0}}}} \right) $
 $ - k(30) = \ln \left( {\dfrac{1}{2}} \right) $
 $ k = \dfrac{1}{{30}}\ln \left( {\dfrac{1}{2}} \right) = 2.31 \times {10^{ - 2}}{\min ^{ - 1}} $
Now that we know the value for rate constant, k, we can find out the time required for the reaction to be 90% complete. If the reaction is 90% complete it means that 10% reactant is still remaining. Hence the concentration of the reactant after time t can be given as: $ {[A]_t} = \dfrac{1}{{10}}{[A]_0} $
Substituting the values in the equation to find the time required:
 $ - (2.31 \times {10^{ - 2}})t = \ln \left( {\dfrac{{\dfrac{1}{{10}}{{[A]}_0}}}{{{{[A]}_0}}}} \right) $
 $ t = \dfrac{{ - \ln (10)}}{{ - 2.31 \times {{10}^{ - 2}}}} = 99.68\min $
Rounding off the value for t, we get the time required to complete 90% of the reaction as $ = 100\min $ .

Note:
The half life of a first order reaction and the rate constant can be related by the formula: $ {t_{1/2}} = \dfrac{{0.693}}{k} $ Where k is the rate constant and $ {t_{1/2}} $ is the time required for the 50% reaction to complete.