# Chemical kinetics using Laplace transformation

I have a simple chemical reaction $A\leftrightarrow B$ with forward rate $k_1$ and backward rate $k_2$. I can now write the differential equation of this system as following.

$\frac{dA}{dt} = -k_1A + k_2B, \quad \frac{dB}{dt} = k_1A - k_2B$

Assuming that reactant A initial concentration is $A_0$, I took the Laplace transform:

$sA(s) - A_0 = -k_1 A(s) +k_2B(s), \quad sB(s) - 0 = k_1A(s) - k_2B(s)$

So far so good. I was hoping to solve for $B(t)$ using any of these two equations, with the initial condition, $A(s) = \frac{A_0}{s}$.

Now, for the first equation,

$$sA(s) - A_0 = -k_1 A(s) + k_2B(s) \\ A_0 - A_0 = -k_1\frac{A_0}{s} + k_2 B(s) \implies B(s) = \frac{k_1A_0}{k_2s}\\$$ This is not correct!

If I use the second equation,

$$B(s) = \frac{k_1A(s)}{s+k_2} = \frac{k_1A_0}{s(s+k_2)}$$ This gives me what I was expecting.

Certainly, I missed something somewhere but I can't figure out what!

• If the first equation would hold the rate $\frac{d[A]}{d[T]}=0$ and the reaction would not proceed, ever. Apr 23, 2015 at 19:51

You may simplify the derivation considerably by noticing that

$$A(t) + B(t) = A_0 \implies \hat{A}(s) + \hat{B}(s) = \frac{A_0}{s}$$

Then

$$s \hat{B}(s) = k_1 \hat{A}(s) - k_2 \hat{B}(s) = \frac{k_1 A_0}{s} - (k_1+k_2) \hat{B}(s)$$

Therefore

$$\hat{B}(s) = \frac{k_1 A_0}{s (s+k_1+k_2)}$$

• Thanks. I was making the terrible mistake of considering $A(s)$ as step-function input to the system. I should have started with state-space representation of the system, and then computing the transfer function. Apr 23, 2015 at 20:23