I was reading about the integrated rate law. However I have problems to follow the solution. I have an equilibrium reaction:

$$\ce{A + B<=>[{k_{on}}][{k_{off}}]AB}$$

with forward and back reaction. I approximate the measurement to a pseudo first order reaction. The complex can be measured. B can be calculated:


Therefore the equilibrium is

$$0 = k_\mathrm{on} \cdot [\ce{A_0}] \cdot ([\ce{B_0}]-[\ce{AB}])-k_\mathrm{off} \cdot [\ce{AB}]$$

which can be expressed also like this:

$$0 = k_\mathrm{on} \cdot \ce{[A_0]} \cdot \ce{[B_0]} - (k_\mathrm{on} \cdot A_0 - k_\mathrm{off})F$$

To solve the differential equation I simplified to this: \begin{align} \frac{\mathrm{d}F(t)}{\mathrm{d}t} &= c_1 - c_2 \cdot F(t)\\ c_1 &= k_\mathrm{on} \cdot A_0 \cdot B_0\\ c_2 &= k_\mathrm{on} \cdot A_0 - k_\mathrm{off}\\ \end{align}

So far the theory was easy and I am sure everything is correct, but now I am stuck. The solution of this differential equation would be straightforward:

$$F(t) = \frac{c_1}{c_2}+k\cdot \mathrm{e}^{-(c_2\cdot t)}$$

However, the solution to fit and simulate the kinetic should be (in the simplified writing). This equation is published in http://afm1.pharm.utah.edu/pnscourse/Anal_Biochem_1995.pdf.

$$F(t) = \frac{c_1\cdot (1-\mathrm{e}^{-(c_2\cdot t)})}{c_2}$$

I really would like to know, where is my mistake and how the derivations of this formula has to be.

  • $\begingroup$ Crossposted to physics.stackexchange.com/q/149712/2451 $\endgroup$
    – Qmechanic
    Nov 30, 2014 at 22:43
  • $\begingroup$ In your equation 0=k_onA_0*B_0-(k_onA_0-k_off )F : If I understand you correctly then F = [AB], right? If so, then I think, you have a sign mistake there: it should be 0=k_onA_0*B_0-(k_onA_0 + k_off )F instead. $\endgroup$
    – Philipp
    Nov 30, 2014 at 23:23
  • $\begingroup$ Maybe, but this term doesn't influence the differential equation. $\endgroup$
    – dgrat
    Nov 30, 2014 at 23:29
  • $\begingroup$ Looking at your equations again, aren't your solution and the solution from the paper identical? I mean the solution from the paper is $F(t) = c_1 \frac{1 - e^{-(c_2 t)}}{c_2} = \frac{c_1}{c_2} - \frac{c_1}{c_2} e^{-(c_2 t)}$ which is exactly the form your equation has, i.e $F(t) = \frac{c_1}{c_2} + k e^{-(c_2 t)}$ with $k = -\frac{c_1}{c_2}$, right? $\endgroup$
    – Philipp
    Dec 1, 2014 at 0:22
  • $\begingroup$ Yes the equation is nearly okay, and k should be -c1/c2. But k came from the integration step. Unfortunately, I don't know why it is okay to define k=-c1/c2. $\endgroup$
    – dgrat
    Dec 1, 2014 at 0:37

1 Answer 1


You have made no mistake. Your solution and the solution from the paper are practically identical. The solution from the paper is

\begin{align} F(t)=c_1 \frac{1−e^{−c_2 t}}{c_2}=\frac{c_1}{c_2} - \frac{c_1}{c_2} e^{−c_2 t} \end{align}

which is exactly the form your equation has, i.e.

\begin{align} F(t)=\frac{c_1}{c_2} + K e^{−c_2 t} \qquad \text{with} \qquad K = - \frac{c_1}{c_2} \ . \end{align}

The integration constant $K$ is not defined the way it is but fixed via the boundary conditions of the reaction. In this case you get $K = - \frac{c_1}{c_2}$ by requiring that at the start of the reaction, i.e. $t=0$, the reactants have not yet reacted with each other and there is no product $\ce{AB}$ present initially, i.e. $F(t\!=\!0) = 0$. This leads to the desired result:

\begin{align} F(t\!=\!0) \overset{!}{=} 0 &=\frac{c_1}{c_2} + K e^{0} \qquad \Rightarrow \quad K=- \frac{c_1}{c_2} \ . \end{align}


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.