2
$\begingroup$

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!

$\endgroup$
1
  • $\begingroup$ If the first equation would hold the rate $\frac{d[A]}{d[T]}=0$ and the reaction would not proceed, ever. $\endgroup$ – rightskewed Apr 23 '15 at 19:51
3
$\begingroup$

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

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

$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$ – Dilawar Apr 23 '15 at 20:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.