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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:

$$[\ce{B}]\overset{\text{def}}{=}[\ce{B_0}]-[\ce{AB}]$$

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.

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  • $\begingroup$ Crossposted to physics.stackexchange.com/q/149712/2451 $\endgroup$ – Qmechanic Nov 30 '14 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 '14 at 23:23
  • $\begingroup$ Maybe, but this term doesn't influence the differential equation. $\endgroup$ – dgrat Nov 30 '14 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 '14 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 '14 at 0:37
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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}

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