Animated Solution for Chemistry - Chemical Kinetics: For a reaction scheme, $A\stackrel{k_1}{\to }B\stackrel{k_2}{\to }C$, if the rate of formation of $B$ is set to be zero then the concentration of $B$ is given by

Imagine a relay race, but instead of runners passing a baton, we have molecules transforming into one another. This is the essence of a consecutive reaction. In our problem, we are given a classic scheme: reactant $A$ transforms into an intermediate $B$ with a rate constant $k_1$, and then $B$ quickly transforms into the final product $C$ with a rate constant $k_2$.

This setup is incredibly common in complex chemical processes. The intermediate $B$ is often a highly reactive, fleeting species. Our goal is to understand how its concentration behaves over time.

Rate of Formation

Let's focus our analytical lens entirely on the intermediate, $B$. Its concentration is caught in a chemical tug-of-war. On one side, it is constantly being produced from $A$. On the other side, it is continuously being consumed to form $C$.

To express this mathematically, we write the differential rate equation for $B$. The net rate of change of its concentration, $\frac{d[B]}{dt}$, is the difference between its rate of production and its rate of consumption.

According to the rate law, the rate at which $B$ is produced from $A$ is $k_1[A]$. Similarly, the rate at which $B$ is consumed to form $C$ is $k_2[B]$. Therefore, our master equation becomes:

Here is where the magic happens. The problem explicitly states a crucial condition: the rate of formation of $B$ is set to be zero.

What does this mean physically? It implies that the intermediate $B$ is so reactive that the moment it is formed, it is immediately consumed. Its concentration remains very small and essentially constant throughout the major part of the reaction. This powerful assumption is known in chemical kinetics as the Steady-State Approximation.

By applying this approximation, we set the net rate of change to exactly zero:

With our equation perfectly balanced, finding the concentration of $B$ is a straightforward algebraic maneuver. We simply rearrange the terms to isolate $[B]$.

First, we move the consumption term to the other side of the equation, showing that the rate of production perfectly equals the rate of consumption:

Finally, we divide both sides by the rate constant $k_2$ to solve for the concentration of $B$:

And there we have it! This elegant result tells us that the concentration of the intermediate is directly proportional to the concentration of the reactant $A$, scaled by the ratio of the two rate constants. This perfectly matches option (b) and demonstrates the profound utility of the steady-state approximation in unraveling complex reaction mechanisms.