Question

Two first order reactions have half-lives in the ratio 8 :1, calculate the ratio of time intervals ${T_1}:{T_2}$ . The time ${T_1},{T_2}$ are the time period for ${\left( {\dfrac{1}{4}} \right)^{th}}$ and ${\left( {\dfrac{3}{4}} \right)^{th}}$ completion respectively
(A) 1:0.0301
(B) 0.125:0.602
(C) 1: 0.602
(D) none of these

Answer 1

Hint : A first-order reaction is one in which the rate of the reaction is directly proportional to the concentration of only one ingredient. In other terms, a first-order reaction is a chemical reaction whose pace is determined by changes in just one of the reactants' concentration. As a result, the order of these reactions is 1.

Complete Step By Step Answer:
The half-life of a quantity is the time it takes for it to decrease to half of its original value. In nuclear physics, the phrase is used to explain how rapidly unstable atoms disintegrate radioactively and how long stable atoms survive. The word is also used to describe any sort of exponential or non-exponential decay in general. The biological half-life of medicines and other substances in the human body, for example, is discussed in medical research. Doubling time is the inverse of half-life.
In first-order reactions, the reaction's concentration decreases over time until it approaches zero, and the length of the half-life is constant regardless of concentration. A first order reaction's half-life is independent of its starting concentration and is exclusively determined by the reaction rate constant, k.
${t_{1/2}} = \dfrac{{\ln 2}}{k}$
For first order reaction,
${{\text{k}}_1} = \dfrac{{\ln 2}}{{{{\text{t}}_{1/2}}}}$
Given from the question $\dfrac{{{{\left( {{t_{1/2}}} \right)}_1}}}{{{{\left( {{t_{1/2}}} \right)}_2}}} = \dfrac{{{k_2}}}{{{k_1}}} = \dfrac{8}{1} \ldots (i)$
${{\text{T}}_1} =$ time interval for $\dfrac{1}{4}$ the completion of reaction 1
Now we know that
${\mathbf{N}} = {{\mathbf{N}}_0}{{\text{e}}^{ - {\text{kt}}}}$
Hence
$\dfrac{3}{4} = {{\text{e}}^{ - {{\text{k}}_1}\;{{\text{T}}_1}}}$
$\Rightarrow {{\text{k}}_1}\;{{\text{T}}_1} = \ln \left( {\dfrac{4}{3}} \right)$
${{\text{T}}_2} =$ time period of $\dfrac{3}{4}$ the completion of reaction (2):
$\dfrac{1}{4} = {{\text{e}}^{ - {{\text{k}}_2}\;{{\text{T}}_2} = }}{{\text{k}}_2}\;{{\text{T}}_2} = \ln (4)$
Hence we obtain
$\dfrac{{{{\text{T}}_1}}}{{\;{{\text{T}}_2}}} = \left( {\dfrac{{{{\text{k}}_2}}}{{{{\text{k}}_1}}}} \right) \cdot \dfrac{{\ln 4}}{{\dfrac{3}{{\ln 4}}}} = \dfrac{{8 \times \ln \dfrac{4}{3}}}{{\ln 4}} = 1:0.0301$
So option A is correct.

Note :
The word "half-life" is almost exclusively applied to exponential (radioactive decay, for example) or roughly exponential decay processes (such as biological half-life discussed below). The half-life of a degradation process that is not even close to exponential will fluctuate drastically while it is occurring. In this situation, it is uncommon to speak of half-life in the first place, but people will occasionally refer to the decay as having a "first half-life," "second half-life," and so on, where the first half-life is defined as the time required to decay from the initial value to 50%, the second half-life is defined as the time required to decay from 50% to 25%, and so on.