KVPY 2020-21 SB SX Exam Previous Papers

Question : 105 of 120

Marks: +1, -0

Consider a reaction that is first order in both directions

A⇌↖{{k}_f}↙{{k}_b}B

Initially only A is present, and its concentration is

{A}_{0} .

Assume

{A}_{{t}}

and

{A}_{{eq}}

are the concentrations of

{A}

at time "t" and at equilibrium, respectively. The time "t" at which

{A}_{{t}}=({A}_{0}+{A}_{{eq}}) \/ 2

is :

Solution:

\table , A,⇌,B ;t=0,[A_0],,0;t=t,[A_0]-x=A_t,,x;{t}={t}_{{eq}} ,[{A}_{0}]-{x}_{e q}={A}_{e q} ,,{x}_{{eq}}

Given at time

{t}={t} {A}_{{t}}={({A}_{0}+{A}_{{aq}})}/{2}

and

x_{\text "eq. "}=A_{0}-A_{\text "eq "}

Now,

{t}={1}/{{k}_{{f}}+{k}_{{b}}} \ln ({{x}_{{c}}}/{{x}_{{c}-{x}}})

=({l{n} 2}/{{k}_{{f}}+{k}_{{b}}})

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