Question

For the reaction scheme $A\xrightarrow{{{k}_{1}}}B\xrightarrow{{{k}_{2}}}C$, if the rate of formation of B is set to be zero then the concentration of B is given by:
(A) $\left( \dfrac{{{k}_{1}}}{{{k}_{2}}} \right)\left[ A \right]$
(B) $\left( {{k}_{1}}+{{k}_{2}} \right)\left[ A \right]$
(C) ${{k}_{1}}{{k}_{2}}\left[ A \right]$
(D) $\left( {{k}_{1}}-{{k}_{2}} \right)\left[ A \right]$

Answer 1

Hint: In the multi-step reaction, the rate of formation of the intermediate B taken to be zero, that is, as soon as it is formed, it gets consumed. The Steady state approximation can be applied to determine the concentration.

Complete answer:
In the given complex reaction, compound A proceeds in the forward direction with the rate constant ${{k}_{1}}$, forming an intermediate compound B, which further reacts with rate constant ${{k}_{2}}$ to give compound C.
Given the rate of formation of intermediate B to be zero, the Steady state approximation is applied. That is, the rate of formation of the intermediate is equal to its rate of consumption.
So, we have the rate of formation of intermediate B, $\dfrac{d\left[ B \right]}{dt}={{k}_{1}}\left[ A \right]$ and,
The rate of consumption of B is $\dfrac{d\left[ B \right]}{dt}=-{{k}_{2}}\left[ B \right]$ , where the negative sign indicates the decrease in the concentration of the intermediate.
Combining the above two equations for the rate of formation of intermediate B, we get,
> $\dfrac{d\left[ B \right]}{dt}={{k}_{1}}\left[ A \right]-{{k}_{2}}\left[ B \right]=0$
> ${{k}_{1}}\left[ A \right]={{k}_{2}}\left[ B \right]$
> $\left[ B \right]=\dfrac{{{k}_{1}}}{{{k}_{2}}}\left[ A \right]$
Therefore, the concentration of the intermediate with respect to the reactant in terms of the rate constants is obtained.

So, the concentration of B is given by option (A), $\left( \dfrac{{{k}_{1}}}{{{k}_{2}}} \right)\left[ A \right]$.

Note: The given reaction is a consecutive reaction, as the product formed in one step is the reactant for the other step. Also, both the reactions are first order reactions and with compound B being the intermediate, the first step is slow, that is the rate determining step. The approximation is applied to this consecutive reaction having a slow reaction in the first step compared to the subsequent steps and can also determine the overall reaction rate.