Steady State Approximation final integral proof $\frac{d[P]}{d t}=k_{a}[A]$ yielding $[P]=\left(1-e^{-k_{a} t}\right)[A]_{0}$

Question

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?

Answer 1

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.

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