Divergent First-order Kinetics

Scheme:

$Bstackrel(k_1) larrA stackrel(k_2)rarrC$

Differential Equation:

$A'(t)=-k_1A-k_2A$

$B'(t)=k_1A$

$C'(t)=k_2A$

Mass balance equation:

$A_0=A+B+C$

Initial conditions:

$A(0)=A_0$ , $B(0)=0$ and $C(0)=0$

Solution:

$A(t)=A_0exp(-(k_1+k_2)t)$

$B=(k_1A_0)/(-(k_1+k_2) ) exp(-(k_1+k_2)t) + (k_1A_0)/(k_1+k_2)$

$C=(k_2A_0)/(-(k_1+k_2) ) exp(-(k_1+k_2)t) + (k_2A_0)/(k_1+k_2)$

Graph:

Plotting A₀=1Mol, B₀=0, C₀=0, k₁=1e-2, k₂=2e-2

Derivation:

The solution can be obtained directly by integration of the separable equation. First $A'(t),$

$A'(t)=-k_1A-k_2A=-(k_1+k_2)A$

$int_(A_0)^(A_t)(dA)/A=-(k_1+k_2)t$

$lnA_t=lnA_0-(k_1+k_2)t$

$A(t)=A_0exp(-(k_1+k_2)t)$

With the solution for $A(t)$ in terms of t only, we can solve $B'(t)$ ,

$B'(t)=k_1A=k_1A_0exp(-(k_1+k_2)t)$

We solve this integral with u-substitution. With $u=-(k_1+k_2)t$ , then $(du)/(dt)=-(k_1+k_2)$ ; applying the chain rule we have,

$(dB)/(dt)=-(k_1+k_2) (dB)/(du)=k_1A_0exp(u)$

which becomes,

$-(k_1+k_2) dB=k_1A_0exp(u) du$

We then integrate the expression,

$-(k_1+k_2) int dB=k_1A_0 int exp(u) du$

...and obtain,

$-(k_1+k_2) B=k_1A_0 int exp(u) + c_1$

Reversing the substitution for u,

$-(k_1+k_2) B=k_1A_0 exp(-(k_1+k_2)t) + c_1$

Making B the subject,

$B=(k_1A_0)/(-(k_1+k_2) ) exp(-(k_1+k_2)t) + c_1$

We then solve the initial value problem by noting that at t=0, B=0,

$0=(k_1A_0)/(-(k_1+k_2) ) + c_1$

Hence we can solve for c₁,

$c_1=(k_1A_0)/(k_1+k_2)$

The solution for B, becomes,

$B=(k_1A_0)/(-(k_1+k_2) ) exp(-(k_1+k_2)t) + (k_1A_0)/(k_1+k_2)$

We then perform the same procedure for $C'(t)$ , however we note that this differential equation differs from that of $B'(t)$ only by replacement of the rate constant k₁ with k₂. Hence the solution is,

$C=(k_2A_0)/(-(k_1+k_2) ) exp(-(k_1+k_2)t) + (k_2A_0)/(k_1+k_2)$