# For a 1st order reaction, how do you derive the integrated rate law for a product versus time?

For a chemical reaction in which some stoichiometric quantity of arbitrary chemical $$A$$ decomposes into products $$B$$ and $$C$$,

$$a \ce{A} \ce{->} b \ce{B} + c \ce{C}$$

or more simply:

$$a\ce{A} \ce{->} \text{Products}$$

The depletion of substrate A at a concentration [A] can be modeled:

$$\frac{{\rm d}[\ce{A}]}{{\rm d} t} = - k[\ce{A}]^n$$

If this is a first order reaction then n = 1:

d[A] / dt = -k * [A]

This is a separable differential equation. When integrated from time 0 to time t and initial concentration [A]0 to [A]t which is the concentration at time t, the integrated form is this:

ln([A]t) = ln([A]0) - k * t

or in terms of [A]:

[A]t = [A]0 * e-k * t

This is useful if you want to know how the concentration of A as [A] changes with respect to time. But if you want the concentration of products, I think you could consider the total concentration:

[A]0 = [A]t + [P]t

Substitution into the the last equation yields a form in terms of products:

[P]t = [A]0 * (1 - e -k * t)

This kind of makes sense to me. The substrate is being exponentially depleted and the products are being exponentially produced to a maximum concentration which is [A]0. The following assumptions were made that the reverse reaction is negligible and that the reaction proceeds to completion which isn't always true. In fact, the time coordinate where these two curves intersect is actually the half-life!

If [P]t = [B]t + [C]t

Solving the equation for [B]t yields:

[B]t = [A]0 * (1 - e -k * t) - [C]t

This doesn't make sense because at the start of the experiment (time = 0), negative concentration of [B] would be observed. Am I doing something wrong?

I have never seen integrated rate laws modeling the concentration of a single product versus time, only the depletion of reagent concentration versus time.

• What is wrong with $$[\ce{B}]=[\ce{B}]_0 + \frac ba \cdot [\ce{A}]_0 \cdot (1 - \exp{(-kt)} )$$ $$[\ce{C}]=[\ce{C}]_0 + \frac ca \cdot [\ce{A}]_0 \cdot (1 - \exp{(-kt)} )$$ Jan 18, 2020 at 6:58
• The flaw in your reasoning is [A]$_0$ = [A]$_t$ + [P]$_t$. Look at your reaction scheme. If a=b=c=1, for example, [P]$_t$=2[A]$_0$. Jan 18, 2020 at 20:42

If you are doing this on a molar basis, you have to be careful about assumptions like: $$[A_o] = [A]_t + [P]_t$$ In the same way that you have defined a component balance for the reactant system, you can also do so for the product system. Let's look at the concentration of B in your reaction equation $$\ce{aA -> bB + cC}$$.
Defining the component balance, $$\frac{d[B]}{dt}=k[A]$$ $$[B]_o = 0, t = 0$$ This is essentially saying that the generation of $$B$$ is proportional to the inverse depletion of $$A$$ (as defined in your component balance). We know that from the solution of the A component balance that, $$[A](t) = [A]_oe^{-kt}$$ Plugging in this expression to the differential equation for $$[B](t)$$ yields, $$\frac{d[B]}{dt} = k[A]_oe^{-kt}$$ Solving this differential equation will yield the following expression for $$[B]$$, $$[B](t)=-[A]_oe^{-kt}+C_1$$ Using our intial condition that $$[B]=0, t = 0$$ we solve for our constant $$C_1$$ and end up with, $$[B](t)=[A]_o(1-e^{-kt})$$ Note that this is similar to the form you had already derived, however... this is only for $$B$$. Since the kinetic expression will be the same assuming no initial mols of $$B$$ or $$C$$ and same stoiciometric coefficients (i.e $$b = c$$) then we can reasonably assume that $$[P]_t = 2[A]_o(1-e^{-kt})$$