Question

$A(aq) \to B(aq) + C(aq)$ is a first order reaction

Time	\[t\]	\[\infty \]
moles of reagent	\[{n_1}\]	\[{n_2}\]

Reaction progress is measured with the help of titration ’\[R\]’. If all \[A\], \[B\] and \[C\] reacted with reagent and have ‘\[n\]’ factors [\[n\] factor; \[eq.{\text{ }}wt = \dfrac{{mol.wt.}}{n}\] ] in the ratio \[1:2:3\] with the reagent. The \[k\] in terms of \[t\], \[{n_1}\] and \[{n_2}\] is:
A. $k = \dfrac{1}{t}\ln \left( {\dfrac{{{n_2}}}{{{n_2} - {n_1}}}} \right)$
B. $k = \dfrac{1}{t}\ln \left( {\dfrac{{2{n_2}}}{{{n_2} - {n_1}}}} \right)$
C. \[k = \dfrac{1}{t}\ln \left( {\dfrac{{4{n_2}}}{{{n_2} - {n_1}}}} \right)\]
D. \[k = \dfrac{1}{t}\ln \left( {\dfrac{{4{n_2}}}{{5\left( {{n_2} - {n_1}} \right)}}} \right)\]

Answer 1

Hint: A first order reaction is a reaction in which the overall rate of the reaction only depends on the concentration of reactant. However at half time the rate becomes independent of concentration of reactant.
Complete step by step answer:
As for first order reaction the rate law depends on the concentration of reactants, so the differential form of first order kinetics is:
Rate = \[\dfrac{{ - d\left[ A \right]}}{{dt}}\] =\[k\left[ A \right]\],
where k is the reaction rate coefficient and \[\left[ A \right]\] is the concentration of reactant \[\left[ A \right]\].
The integrated form of rate equation becomes
$k = \dfrac{{2.303}}{t}\log \left( {\dfrac{{{A_o}}}{A}} \right)$ where ${A_o}$ is the concentration at time \[0\] and $A$ is the concentration at time \[t\].
Considering the equation which follow the first order,
$A(aq) \to B(aq) + C(aq)$
The moles of $A$ present initially is \[n\]. Let \[x\] moles of reagent \[R\] reacted with substrate \[A\], then the concentration of \[A\], \[B\] and \[C\] at time \[t = 0\] , \[t = t\] and time \[t = \infty \] will be:
$A(aq) \to B(aq) + C(aq)$

\[n\]	\[0\]	\[0\]	\[t = 0\]
\[n - x\]	\[2x\]	\[3x\]	\[t = t\]
\[0\]	\[2n\]	\[3n\]	\[t = \infty \]

At time \[t\], \[{n_1}\] moles of reagent react with A, so \[{n_1} = n - x + 2x + 3x = n + 4x\].
\[x{\text{ }} = {\text{ }}\dfrac{{{n_1} - n}}{4}\]
At time \[\infty \], \[{n_2}\] moles of reagent reacts with A, so \[{n_2} = 0 + 2n + 3n = 5n\]
\[n = \dfrac{{{n_2}}}{5}.\]
Substituting in the equation of rate of first order,
$k = \dfrac{{2.303}}{t}\log \left( {\dfrac{n}{{n - x}}} \right)$
\[k = \dfrac{1}{t}\ln \left( {\dfrac{{4{n_2}}}{{5\left( {{n_2} - {n_1}} \right)}}} \right)\]
Thus option D is the correct answer.

Note:
When the rate of reaction is independent of concentration of reactants the rate is zero order reaction. But in first order reaction only at half time the rate is independent of concentration of reactants.