November 8, 2019September 26, 2024 by gary_mail@hotmail.com

1st order reversible reaction

A 1st order reversible reaction of the type $A\rightleftharpoons B$ is composed of the reactions:

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

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

The rate law is:

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

Substitute [B] = [A₀] – [A], where [A₀] is the initial concentration of A, in the above equation and rearranging, we have.

$\frac{d[A]}{dt}=-(k_1+k_2)[A]+k_2[A_0]\; \; \; \; \; \; \; \; 16$

Let

$x=(k_1+k_2)[A]-k_2[A_0]\; \; \; \; \; \; \; \; 17$

Eq16 becomes $\frac{d[A]}{dt}=-x$ . Differentiating eq17 with respect to [A], we have $d[A]=\frac{dx}{k_1+k_2}$ , and differentiating this expression with respect to time, we have $\frac{d[A]}{dt}=\frac{1}{k_1+k_2}\frac{dx}{dt}$ , which is equivalent to

$-x=\frac{1}{k_1+k_2}\frac{dx}{dt}\; \; \Rightarrow \; \; -(k_1+k_2)dt=\frac{dx}{x}$

Let x = x₀when t =0 and integrate the above expression. We have

$-(k_1+k_2)t=lnx-lnx_0\; \; \; \; \; \; \; \; 18$

Substitute eq17 in eq18, noting that the 2nd term on RHS of eq18 refers to concentrations at t = 0, where [A] = [A₀],

$-(k_1+k_2)t=ln\frac{(k_1+k_2)[A]-k_2[A_0]}{(k_1+k_2)[A_0]-k_2[A_0]}$

which rearranges to

$[A]=\frac{k_2+k_1e^{-(k_1+k_2)t}}{k_1+k_2}[A_0]\; \; \; \; \; \; \; \; 19$

When t = 0, eq19 becomes [A] = [A₀]. As t → ∞,

$[A_\infty ]=[A_{eqm}]=\frac{k_2}{k_1+k_2}[A_0]$

Since $[B_{eqm}]=[A_0]-[A_\infty ]$ ,

$[B_{eqm}]=[A_0]-\frac{k_2}{k_1+k_2}[A_0]=[A_0]\frac{k_1}{k_1+k_2}$

Therefore, the equilibrium constant for the reaction is:

$K=\frac{[B_{eqm}]}{[A_{eqm}]}=\frac{k_1}{k_2}$

This is the link between chemical kinetics and thermodynamics.

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