Question

Given that for the reaction of nth order, the integrated rate equation is:
$K = \dfrac{1}{{t(n - 1)}}[\dfrac{1}{{{C^{n - 1}}}} - \dfrac{1}{{{C_0}^{n - 1}}}]$
Where C and ${C_0}$are the concentration of reactant at time t and initially respectively. The ${t_{\dfrac{3}{4}}}$and ${t_{\dfrac{1}{2}}}$are related as (${t_{\dfrac{3}{4}}}$is time required for C to become${C_{\dfrac{1}{4}}}$)
A. ${t_{\dfrac{3}{4}}} = {t_{\dfrac{1}{2}}}[{2^{2n - 1}} + 1]$
B. ${t_{\dfrac{3}{4}}} = {t_{\dfrac{1}{2}}}[{2^{2n - 1}} - 1]$
C. ${t_{\dfrac{3}{4}}} = {t_{\dfrac{1}{2}}}[{2^{2n + 1}} + 1]$
D. ${t_{\dfrac{3}{4}}} = {t_{\dfrac{1}{2}}}[{2^{2n + 1}} - 1]$

Answer 1

Hint: The equation which tends to represent the dependence of the rate of the reaction on the concentration of the species of the reactant is called the differential rate equation. The instantaneous rate of the reaction is expressed as the slope of the tangent at any instant of time in the graph of the concentration time graph. The unit for the rate constant changes with the order of the reaction.

Complete step-by-step answer: So from the question we get to know that at time ${t_{\dfrac{3}{4}}}$the concentration is equal to${C_{\dfrac{1}{4}}}$. If the time is equal to ${t_{\dfrac{1}{2}}}$then the concentration is equal to ${C_{\dfrac{1}{2}}}$.
Now the integrated rate equation given to us is $K = \dfrac{1}{{t(n - 1)}}[\dfrac{1}{{{C^{n - 1}}}} - \dfrac{1}{{{C_0}^{n - 1}}}]$. Here we will equate it in terms of t so the equation will be :
$t = \dfrac{1}{{K(n - 1)}}[\dfrac{1}{{{C^{n - 1}}}} - \dfrac{1}{{{C_0}^{n - 1}}}]$
Now we will find the integrated rate equation for time $${t_{\frac{3}{4}}}$$:
${t_{\dfrac{3}{4}}} = \dfrac{1}{{K(n - 1)}}[\dfrac{1}{{{{(\dfrac{{{C_o}}}{4})}^{n - 1}}}} - \dfrac{1}{{{C_0}^{n - 1}}}]$………………equation 1
Now we will find the integrated rate equation for the time $${t_{\frac{1}{2}}}$$:
${t_{\dfrac{3}{4}}} = \dfrac{1}{{K(n - 1)}}[\dfrac{1}{{{{(\dfrac{{{C_o}}}{2})}^{n - 1}}}} - \dfrac{1}{{{C_0}^{n - 1}}}]$…………………equation 2
So now dividing the equation 1 by equation 2, we get,
$\dfrac{{{t_{\dfrac{3}{4}}}}}{{{t_{\dfrac{1}{2}}}}} = \dfrac{{[\dfrac{{{4^{n - 1}}}}{{{C_o}^{n - 1}}} - \dfrac{1}{{{C_o}^{n - 1}}}]}}{{[\dfrac{{{2^{n - 1}}}}{{{C_o}^{n - 1}}} - \dfrac{1}{{{C_o}^{n - 1}}}]}}$
On solving this we get,
$\dfrac{{{t_{\dfrac{3}{4}}}}}{{{t_{\dfrac{1}{2}}}}} = \dfrac{{{4^{n - 1}} - 1}}{{{2^{n - 1}} - 1}} = \dfrac{{{{({2^{n - 1}})}^2} - 1}}{{({2^{n - 1}}) - 1}} = \dfrac{{({2^{n - 1}} - 1)({2^{n - 1}} + 1)}}{{{2^{n - 1}} - 1}}$
On further solving this we will get,
$\dfrac{{{t_{\dfrac{3}{4}}}}}{{{t_{\dfrac{1}{2}}}}} = ({2^{n - 1}} + 1)$
So now writing the answer is terms of ${t_{\dfrac{3}{4}}}$we get,
${t_{\dfrac{3}{4}}} = {t_{\dfrac{1}{2}}}[{2^{2n - 1}} + 1]$

So the correct answer is option A.

Note: The order of the reaction is defined as the relationship between the concentration of the substance taking part in it and the rate of the chemical reaction. the first order of the reaction is dependent on the concentration of the single species. Whereas the second order of the reaction is dependent on the concentration of the two species taking part in the reaction.