Question

For the given reaction,
$A\mathop \rightleftharpoons \limits_{{K_2}}^{{K_1}} B$
Where ${K_1}$ is zero order rate constant and ${K_2}$ is first rate constant. The concentration of B at any time can be given by the expression:
A.${K_1} = \dfrac{1}{t}{\log _{10}}\dfrac{{[{B_0}]}}{{[B]}}$
B.${K_2} = \dfrac{{2.303}}{t}{\log _{10}}\dfrac{{[{B_0}]}}{{[B]}}$
C.$\dfrac{1}{{{K_2}}}\left[ {\log \dfrac{{\log {K_1} - {K_2}[{B_0}]}}{{{K_1} - {K_2}[B]}}} \right] = t$
D.Can’t be expressed

Answer 1

Hint: For a zero order reaction, the rate of reaction is independent of the concentration of reactants. That is, any change in the concentration of the reactants does not affect the rate of reaction. While in a first order reaction, the rate of reaction is directly proportional to the concentration of one of the reactants.

Complete step by step answer:
For the reaction,
$A\mathop \rightleftharpoons \limits_{{K_2}}^{{K_1}} B$
The conversion of A into B is a zero order reaction which means rate is independent of the concentration of A.
So, $rate = {K_1}{[A]^0}$
$ \Rightarrow rate = {K_1}$
$ \Rightarrow - \dfrac{{dA}}{{dt}} = {K_1}$
The conversion of B into A is a first order reaction having rate constant ${K_2}$. This means rate is dependent linearly on the concentration of reactant B.
Let the initial concentration of B is $({B_0})$ and after time ‘t’ the concentration of B left unreacted be (B).
So, $rate = {K_2}[B]$
$ \Rightarrow - \dfrac{{dB}}{{dt}} = {K_2}[B]$
$ \Rightarrow - \dfrac{{dB}}{{[B]}} = {K_2}dt$
$ \Rightarrow - \int {\dfrac{{dB}}{{[B]}}} = {K_2}\int {dt} $
$ \Rightarrow - \ln [B] = {K_2}t + c$ …$(1)$
At $t = 0$, $[B] = [{B_0}]$
So, $ - \ln [{B_0}] = c$ …$(2)$
At $t = t$, putting equation $2$ into $1$.
$ - \ln [B] = {K_2}t - \ln [{B_0}]$
$ \Rightarrow - \ln [B] + \ln [{B_0}] = {K_2}t$
$ \Rightarrow \ln \dfrac{{[{B_0}]}}{{[B]}} = {K_2}t$
$ \Rightarrow {K_2} = \dfrac{1}{t}\ln \dfrac{{[{B_0}]}}{{[B]}}$
As we know, on converting $\ln $ to $\log $ we multiply with $2.303$. So,
$ \Rightarrow {K_2} = \dfrac{{2.303}}{t}\log \dfrac{{[{B_0}]}}{{[B]}}$

Thus, the answer is option B.

Note:
In a first order reaction if there are two reactants, then the rate of reaction is linearly dependent on the concentration of one of the reactants and independent of the concentration of the other reactant. Or the rate of reaction is partially dependent on the concentration of both the reactants such that the overall reaction rate is one.