# 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.

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?

• The answer to your doubt is directly on your graph. First some B has to be created. Initially, the rate of C formation is zero. Jun 24, 2023 at 16:06
• My doubt is if B is only not formed then will C form? Then why the graph of concentration is different? Jun 24, 2023 at 17:00
• Neither of questions is clear. Try to reformulate. // C is formed only from B. If no B is present, no C is being formed. Jun 24, 2023 at 17:07
• The concentration profiles of B and C are not the same because B is formed only from A and then B only forms C so the concentration of B rises and falls but C only rises as it does not react to form anything else. Jun 25, 2023 at 12:21
• @porphyrin thanku sir Jun 25, 2023 at 14:33

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 $$= k_1 [\ce{A}]$$.

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

$$\mathrm{R} = \dfrac{\mathrm{d}}{\mathrm{d}t}\ce{[C]}= k_2\ce{[B]} \tag{1}\\$$

$$\dfrac{\mathrm{d}}{\mathrm{d}t}\ce{[B]}= k_1\ce{[A]} - k_2\ce{[B]} \approx 0 \tag{2}$$

$$\implies k_1\ce{[A]} \approx k_2\ce{[B]}\\ \implies k_1\ce{[A]} \approx k_2\ce{[B]}$$

Substituting in Equation (1):

$$R \approx k_1\ce{[A]}$$

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

$$-\dfrac{\mathrm{d}}{\mathrm{d}t}\ce{[A]} = k_1\ce{[A]}\\ \implies \ce{[A]} = \ce{[A]_o}\mathrm{e}^{-k_1t} \tag{3}$$

• Are you assuming mass action kinetics? Jun 27, 2023 at 8:06
• @RodrigodeAzevedo it is applicable for low concentrations. Jun 27, 2023 at 8:11
• Do you mean (1) is applicable? Jun 27, 2023 at 8:12
• @RodrigodeAzevedo (1), (2), and (3). Jun 27, 2023 at 8:14