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I want to ask a question about the Steady State Approximation integral.

I understand the steady state integral works in Leyman's terms as the intermediate is formed as fast as it is consumed so the intermediate's concentration is really low and nearly always the same.

If I had a reaction:

$$\ce{A->I->P}$$

Then essentially,

$$\frac{d[I]}{d t} \approx 0$$

and therefore using this information and previous knowledge of rate

$$\begin{array}{l}{\frac{d[I]}{d t}=k_{a}[A]-k_{b}[I]} \\ {\frac{d[I]}{d t}=k_{a}[A]-k_{b}[I] \approx 0}\end{array}$$

which results after rearrangement:

$$[I] \approx \frac{k_{a}[A]}{k_{b}}$$

and hence $$\frac{d[P]}{d t}=k_{b}[I] \approx k_{b} \frac{k_{a}[A]}{k_{b}} \approx k_{a}[A]$$

The trouble is, I couldn't understand how:

$$\frac{d[P]}{d t}=k_{a}[A]$$

can be solved to give

$$[P]=\left(1-e^{-k_{a} t}\right)[A]_{0}$$

I recognise from previous work that:

$$\frac{d[A]}{d t}=-k_{a}[A]$$ gives

$$A =[A]_{0}e^{-k_{a} t}$$

from my final year at school, but I couldn't seem to apply this understanding to the final step of the integral above.

How can this be shown?

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    $\begingroup$ One molecule of A makes one molecule of P, so $[\ce{A}] + [\ce{P}] = $ some constant $c$, if you ignore the intermediate $\ce{I}$. At time $t = 0$, you know that $[\ce{A}] = [\ce{A}]_0$ and $[\ce{P}] = 0$... $\endgroup$ – orthocresol Jun 4 at 22:03
  • $\begingroup$ When you make the steady state approximation species I is removed from any time dependence effectively making the problem $A \to P$ as $A=A_0e^{-kt}$ then $P=A_0(1-e^{-kt} )$ because A+P is constant. $\endgroup$ – porphyrin Jun 5 at 9:06
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There are a couple of ways to approach this problem. Bear in mind this problem requires some calculus.

$\boldsymbol{1^{st}}$ Way (repeated integration): To start, you were able to recognize the first order integrated rate law.

$$\frac{d[A]}{dt} = -k_a[A] \implies [A] = [A]_0e^{-k_at}$$

This result may be plugged into the rate equation for $P$ to obtain

$$\frac{d[P]}{dt} = k_a[A] = k_a[A]_0e^{-k_at}$$

This equation is directly integrable.

$$\int_0^t\frac{d[P]}{dt}\,dt = \int_0^tk_a[A]_0e^{-k_at}\,dt$$ $$[P] - [P]_0 = -\frac{k_a[A]_0}{k_a}\left(e^{-k_at} - 1\right)$$ $$\boxed{[P] = [A]_0\left(1-e^{-k_at}\right)}$$

The last step applies the knowledge from the problem that $[P]_0=0$.

$\boldsymbol{2^{nd}}$ Way (conservation law): orthocrestol in the comments above pointed out that the sum of the concentrations of all chemical species should equal to some constant in this problem. Call this constant $C$ and ignore $I$ due to the steady state approximation. Thus,

$$[A] + [P] = C$$

Now substitute this into your rate law for $P$.

$$\frac{d[P]}{dt} = k_a[A] = k_a(C - [P])$$

This differential equation is separable so we can use the neat tricks to integrate this equation.

$$\int_0^{[P]}\frac{d[P]}{C-[P]} = \int_0^tk_a\,dt$$ $$-\ln\left(1-\frac{[P]}{C}\right) = k_at$$ $$[P] = C\left(1-e^{-k_at}\right)$$

Almost done. Now recognize that at time zero, we can write the conservation law as

$$C = [A]_0 + [P]_0 = [A]_0 + 0 = [A]_0$$

This makes physical sense because since $C$ is a constant, this value cannot change at any time. Moreover, the total number of molecules at any time should be the same as the amount of molecules that you start with due to stoichiometry (everything is 1:1 in this particular problem). So make the substitution to get your answer.

$$\boxed{[P] = [A]_0\left(1-e^{-k_at}\right)}$$

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  • $\begingroup$ Hi there! I fully understand every element of your working and I'm super grateful, but I just have a question about the 2nd to last step of your first way. You wrote $\int_{0}^{t} \frac{d[P]}{d t} d t=\int_{0}^{t} k_{a}[A]_{0} e^{-k_{a} t} d t$ and then gave $[P]-[P]_{0}=-\frac{k_{a}[A]_{0}}{k_{a}}\left(e^{-k_{a} t}-1\right)$ which all makes sense apart from the $-1$ inside the brackets ... how did you get that term? I might be missing something but apart from that it's a perfect answer that I can tick to confirm I understood the solution! $\endgroup$ – vik1245 Jun 5 at 1:16
  • $\begingroup$ @vik1245 You might have forgotten to apply the limits of integration $\endgroup$ – William R. Ebenezer Jun 5 at 4:11
  • $\begingroup$ @vik1245 The left hand side is the fundamental theorem of calculus which more or less states the definite integral of the derivative is the original function evaluated at the limits. As for the right hand side, William realized you may have missed it by forgetting to apply the limits of integration, here is some step-by-step for the integral:$$\int_0^tk_a[A]_0e^{-k_at}=k_a[A]_0\int_0^te^{-k_at}=k_a[A]_0\left(-\left.\frac{1}{k_a}e^{-k_at}\right|_0^t\right) = k_a[A]_0\left(-\frac{1}{k_a}\right)\left(e^{-k_at} - 1\right)$$ $\endgroup$ – MasterYoda Jun 5 at 4:52
  • $\begingroup$ oh of course, $e^{0} = 1$ how can I be so silly?! Many thanks tho anyway for your help! @MasterYoda $\endgroup$ – vik1245 Jun 5 at 16:48

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