Sequential first order reaction graph in chemical kinetics

In graph given below is a sequential first order reaction


$$\ce{A ->[$k\_1$] B ->[$k\_2$] C}$$


For $k\_2 \gg k\_1$ the graph of concentration of $\ce{A}$, $\ce{B}$ and $\ce{C}$ is as follows.


[![enter image description here](https://i.stack.imgur.com/OuQ76.jpg)](https://i.stack.imgur.com/OuQ76.jpg)


Won't the graph of the concentrations of $\ce{B}$ and $\ce{C}$ be the same as if $\ce{B}$ is formed then only $\ce{C}$ will be formed simultaneously? Then why is $\ce{C}$'s graph different?




This question is about the **rate-determining step** of multi-step reactions. Bear in mind that the overall rate of reaction is equal to the slowest step of the reaction, in which case, from $\ce{A}$ to $\ce{B}$ ($k\_2 \gg k\_1$). Since it is a first-order reaction, the overall reaction rate = $k\_1 [\ce{A}]$. The graph of concentration vs time for a first-order reaction should be like $y = \frac1x$.


On the other hand, $\ce{B}$ is the **reaction intermediate**. $[\ce{B}]$ initially increases when the reactant $\ce{A}$ is gradually consumed. But soon, $[\ce{B}]$ reduces to form $\ce{C}$.


The graph of $[\ce{C}]$ must be symmetrical about the horizontal line $y = \frac12[\ce{A}]$ because as stated earlier, the overall reaction rate $= k\_1 [\ce{A}]$.


 1

This is a pseudo first-order reaction. The rate of product formation is:


$$
\mathrm{R} = \dfrac{\mathrm{d}}{\mathrm{d}t}\ce{[C]}= k\_2\ce{[B]} \tag{1}\\
$$


Applying steady state approximation:


$$
\dfrac{\mathrm{d}}{\mathrm{d}t}\ce{[B]}= k\_1\ce{[A]} - k\_2\ce{[B]} \approx 0 \tag{2}
$$


$$
\implies k\_1\ce{[A]} \approx k\_2\ce{[B]}\\
\implies k\_1\ce{[A]} \approx k\_2\ce{[B]}
$$


Substituting in Equation (1):


$$
R \approx k\_1\ce{[A]} 
$$


For the overall reaction $\ce{A -> C}$:


$$
-\dfrac{\mathrm{d}}{\mathrm{d}t}\ce{[A]} = k\_1\ce{[A]}\\
\implies \ce{[A]} = \ce{[A]\_o}\mathrm{e}^{-k\_1t} \tag{3}
$$


 1