Question

The rate constant for a first order reaction is 60 $s^{−1}$. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?

Answer 1

Hint: A first-order reaction refers to a reaction which proceeds at a rate that depends only on one reactant concentration linearly. You should have at-least basic knowledge about differential rate laws which are usually utilised to explain what is actually occurring on a molecular level while a reaction is occurring. On the other hand, integrated rate laws are employed for identifying the reaction order as well as the value of the rate constant from the experimental measurements.

Complete step by step answer:
We are given the value of rate constant for a first order reaction as 60 $s^{-1}$.
We know that, for the first order reaction:

$t = \dfrac{{2.303}}{k}\dfrac{{\log a}}{{a - x}}$

Let us suppose that ‘a’ M be the initial concentration, and final concentration (a – x) = $\dfrac{a}{{16}}, k = 60{s^{ - 1}}$
Now:
$
t = \dfrac{{2.303}}{60}\dfrac{{\log a}}{{\dfrac{a}{{16}}}} = \dfrac{{2.303}}{60}\log 16 \\

= \dfrac{{2.303 \times 1.2041}}{60} = 0.0462 = 4.62 \times {10^{ - 2}}s \\
$
As a result, it will take $4.62 \times {10^{ - 2}}s$ to reduce the initial concentration of the reactant to its 1/16th value.

Note: The differential equation explaining the first-order kinetics is written below:
$Rate = - \dfrac{{d\left[ A \right]}}{{dt}} = k{\left[ A \right]^1} = k\left[ A \right]$
Here, "rate" refers to the reaction rate (units are molar/time) and k corresponds to the reaction rate coefficient (units are 1/time). However, the units of k vary in case of non-first-order reactions. $\left[ A \right] = {\left[ A \right]_0}{e^{ - kt}}$
The integrated form of rate law can be written in the following manner:
$\left[ A \right] = {\left[ A \right]_0}{e^{ - kt}}$
The integrated form of the rate law can even be utilised to determine the population of reactants at any time after the initiation of the reaction.