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Question: \(A(aq)\to B(aq)+C(aq)\) is a first order reaction: \(Time\,\,\,\,\,\,\,\,\,\,...

A(aq)B(aq)+C(aq)A(aq)\to B(aq)+C(aq) is a first order reaction:
TimetTime\,\,\,\,\,\,\,\,\,\,\,t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\infty
moles of reagent n1n2\text{moles of reagent }\,\,\,\,\,\,\,{{n}_{1}}\,\,\,\,\,\,\,\,\,\,\,\,{{n}_{2}}
Reaction progress is measured with the help of titration of reagent R'R'. If all A,BandCA,B\,and\,C reacted with reagent and have n'n' factors [nfactors;eq.wt=mol.wt.n]\left[ \text{n}\,\text{factors;eq}\text{.wt}=\dfrac{mol.wt.}{n} \right] in the ratio of 1:2:31:2:3 with the reagent. The kk in terms of t,n1andn2t,{{n}_{1}}\,and\,{{n}_{2}} is:
(A) k=1tln(n2n2n1)k=\dfrac{1}{t}\ln \left( \dfrac{{{n}_{2}}}{{{n}_{2}}-{{n}_{1}}} \right)
(B) k=1tln(2n2n2n1)k=\dfrac{1}{t}\ln \left( \dfrac{2{{n}_{2}}}{{{n}_{2}}-{{n}_{1}}} \right)
(C) k=1tln(4n2n2n1)k=\dfrac{1}{t}\ln \left( \dfrac{4{{n}_{2}}}{{{n}_{2}}-{{n}_{1}}} \right)
(D) k=1tln(4n25(n2n1))k=\dfrac{1}{t}\ln \left( \dfrac{4{{n}_{2}}}{5({{n}_{2}}-{{n}_{1}})} \right)

Explanation

Solution

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=1tlnaaxk=\dfrac{1}{t}\ln \dfrac{a}{a-x} -------- (a)
where (ax)(a-x) is the concentration of the reactant after time = t, (a)(a) is the initial concentration at time = 0 and kk 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 1:2:31:2:3for A, B, C respectively.
So, let the moles of reagent R, at time t = 0 be n. Then,
AB+CA\,\,\,\,\,\,\,\to \,\,\,\,B\,\,\,+\,\,\,\,\,C
At t = 0, at the beginning of the reaction, the moles of reactant are unchanged. So, moles present is:
n00n\,\,\,\,\,\,\,\,\,\,\,\,0 \,\,\,\,\,\,\,\, 0
At t = t, when ‘x’ moles of reactant has reacted, then moles left is:
nx2x3x{n – x}\,\,\,\,\,\,\,\,{2x}\,\,\,\,\,\,\,\,{3x}
At t = \infty , when all the reactant is used up, then moles left is:
02n3n0\,\,\,\,\,\,\,\,\,\,\,{2n}\,\,\,\,\,\,\,\,{3n}
Thus, the moles of the reagent which reacted at time t = t, is n1=(nx)+2x+3x=n4x{{n}_{1}}=(n-x)+2x+3x=n-4x and the moles of reagent reacts at t = infinityinfinity , n2=2n+3n=5n{{n}_{2}}=2n+3n=5n.
Then, we get the value of n=n25n=\dfrac{{{n}_{2}}}{5} and x=n1n4=n1(n2/5)4x=\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=na=n and ax=nxa-x=n-x. We get,
k=1tln(nnx)k=\dfrac{1}{t}\ln \left( \dfrac{n}{n-x} \right)
k=1tln(4n25(n2n1))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=1tln(4n25(n2n1))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=d[A]dt=k[A]Rate=-\dfrac{d\left[ A \right]}{dt}=k\left[ A \right]
where [A]\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 molesec1mole\,\,{{\sec }^{-1}} and the unit of rate constant is sec1{{\sec }^{-1}}.