Question

$A(aq)\to B(aq)+C(aq)$ is a first order reaction:
                    $Time\,\,\,\,\,\,\,\,\,\,\,t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\infty $
$\text{moles of reagent }\,\,\,\,\,\,\,{{n}_{1}}\,\,\,\,\,\,\,\,\,\,\,\,{{n}_{2}}$
Reaction progress is measured with the help of titration of reagent $'R'$. If all $A,B\,and\,C$ reacted with reagent and have $'n'$ factors $\left[ \text{n}\,\text{factors;eq}\text{.wt}=\dfrac{mol.wt.}{n} \right]$ in the ratio of $1:2:3$ 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) $k=\dfrac{1}{t}\ln \left( \dfrac{4{{n}_{2}}}{5({{n}_{2}}-{{n}_{1}})} \right)$

Answer 1

Hint: In the first-order reaction, the rate is dependent on the concentration of only one reactant. As the reaction progresses, the concentration of the reactant decreases, and so does the rate. At a certain time interval, the concentration of the reactant left can be used along with the initial concentration to determine the rate.

Complete step by step solution:
As we know, the integrated form of first order reaction is given as:
$k=\dfrac{1}{t}\ln \dfrac{a}{a-x}$ -------- (a)
where $(a-x)$ is the concentration of the reactant after time = t, $(a)$ is the initial concentration at time = 0 and $k$ is the rate constant of first order reaction.
So, in the given reaction, when the reagent R is titrated against the reaction, then it measures the rate of reaction. The moles of the reagent ‘R’ which reacted in the given reaction during titration, is given by the n-factor, that is, in the ratio of $1:2:3$for A, B, C respectively.
So, let the moles of reagent R, at time t = 0 be n. Then,
$A\,\,\,\,\,\,\,\to \,\,\,\,B\,\,\,+\,\,\,\,\,C$
At t = 0, at the beginning of the reaction, the moles of reactant are unchanged. So, moles present is:
$n\,\,\,\,\,\,\,\,\,\,\,\,0 \,\,\,\,\,\,\,\, 0$
At t = t, when ‘x’ moles of reactant has reacted, then moles left is:
${n – x}\,\,\,\,\,\,\,\,{2x}\,\,\,\,\,\,\,\,{3x}$
At t = $\infty $, when all the reactant is used up, then moles left is:
$0\,\,\,\,\,\,\,\,\,\,\,{2n}\,\,\,\,\,\,\,\,{3n}$
Thus, the moles of the reagent which reacted at time t = t, is ${{n}_{1}}=(n-x)+2x+3x=n-4x$ and the moles of reagent reacts at t = $infinity $, ${{n}_{2}}=2n+3n=5n$.
Then, we get the value of $n=\dfrac{{{n}_{2}}}{5}$ and $x=\dfrac{{{n}_{1}}-n}{4}=\dfrac{{{n}_{1}}-\left( {{n}_{2}}/5 \right)}{4}$ , that is, the moles of reactant A at time t = 0 and moles of reactant that reacted at t = t respectively.
Substituting the value of n and n-x in equation (a), where $a=n$ and $a-x=n-x$. We get,
\[k=\dfrac{1}{t}\ln \left( \dfrac{n}{n-x} \right)\]
\[k=\dfrac{1}{t}\ln \left( \dfrac{4{{n}_{2}}}{5\left( {{n}_{2}}-{{n}_{1}} \right)} \right)\]

Therefore, the rate constant of the first order reaction is option (D)- $k=\dfrac{1}{t}\ln \left( \dfrac{4{{n}_{2}}}{5({{n}_{2}}-{{n}_{1}})} \right)$.

Note: The first order rate reaction is also expressed in the differential form as: $Rate=-\dfrac{d\left[ A \right]}{dt}=k\left[ A \right]$
where $\left[ A \right]$ is the concentration of the reactant and the negative sign indicates the decrease in the concentration of the reactant with time.
For the first order reaction, the unit of rate is $mole\,\,{{\sec }^{-1}}$ and the unit of rate constant is ${{\sec }^{-1}}$.