November 8, 2019May 17, 2024 by gary_mail@hotmail.com

1st order consecutive reaction

A 1st order consecutive reaction of the type A → B → C is composed of the reactions:

$A\rightarrow B\; \; \; \; \; \; \; v_1=k_1[A]$

$B\rightarrow C\; \; \; \; \; \; \; v_2=k_2[B]$

The rate laws are:

$\frac{d[A]}{dt}=-k_1[A]\; \; \; \; \; \; \; \; 20$

$\frac{d[B]}{dt}=k_1[A]-k_2[B]$

$\frac{d[C]}{dt}=k_2[B]$

To understand how a 1^st order consecutive reaction proceeds over time, we need to develop equations for $[A]$ , $[B]$ and $[C]$ . The expression for $[A]$ is the solution for eq20, i.e. $[A]=[A_0]e^{-k_1t}$ . Substituting this in the 2nd rate law above and rearranging:

$\frac{d[B]}{dt}+k_2[B]=k_1[A_0]e^{-k_1t}\; \; \; \; \; \; \; \; 21$

Eq21 is a linear first order differential equation of the form y’ + P(t)y = f(t). Multiplying eq21 with the integrating factor $e^{k_2t}$ , we have

$e^{k_2t}\frac{d[B]}{dt}+k_2e^{k_2t}[B]=k_1e^{k_2t}[A_0]e^{-k_1t}\; \; \; \; \; \; \; \; 22$

The LHS of eq22 is the derivative of the product of $e^{k_2t}$ and [B], i.e. $\frac{d\left ( e^{k_2t}[B] \right )}{dt}$ . So,

$\frac{d\left ( e^{k_2t}[B] \right )}{dt}=k_1e^{k_2t}[A_0]e^{-k_1t}$

Integrating both sides with respect to time, noting that [B] = 0 at t = 0, and rearranging, we have

$[B]=\frac{k_1}{k_2-k_1}\left ( e^{-k_1t}-e^{-k_2t} \right )[A_0]\; \; \; \; \; \; \; \; eq23$

As t → ∞, [B] = 0.

At all times, [A] + [B] + [C] = [A₀], so from eq23,

$[A_0]-[A]-[C]=\frac{k_1}{k_2-k_1}\left ( e^{-k_1t}-e^{-k_2t} \right )[A_0]$

Substitute $[A]=[A_0]e^{-k_1t}$ in the above equation and rearranging,

$[C]=\left( 1+\frac{k_1e^{-k_2t}-k_2e^{-k_1t}}{k_2-k_1} \right )[A_0]$

As t → ∞, [C] = [A₀].

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