# Integrating the differential rate equation

1. Apr 23, 2012

### mycotheology

Heres the differential rate equation for a 0 order reaction in chemistry:
$Rate = {{-d[A]} / {dt}} = k$
which can be rearranged to this:
$-d[A] = dt k$
and when you integrate this you get the integrated rate equation but I don't understand how this works. The site I'm reading says you integrate both sides of the equation and get this:
http://www.uni-regensburg.de/Fakultaeten/nat_Fak_IV/Organische_Chemie/Didaktik/Keusch/Grafik/ord_z5.gif
is that supposed to be a definite integral or something? Anyhow, after integrating they get this:
http://www.uni-regensburg.de/Fakultaeten/nat_Fak_IV/Organische_Chemie/Didaktik/Keusch/Grafik/ord_z3.gif
I don't get it. Firstly [A] and t are variables so shouldn't they be raised a power and get divided by 2? In other words, shouldn't they become $[A]^{2} / 2$ and $t^{2} / 2$? Secondly, whats going on with the d. d means the change in something but I've never seen it been integrated before so I don't know what happens to it. Why did it disappear?

The 1st order equation has me equally confused:
http://www.uni-regensburg.de/Fakultaeten/nat_Fak_IV/Organische_Chemie/Didaktik/Keusch/Grafik/ord1_ip1.gif
integratin that, they get this:
http://www.uni-regensburg.de/Fakultaeten/nat_Fak_IV/Organische_Chemie/Didaktik/Keusch/Grafik/ord1-3.gif
has does d[A]/[A] become ln[A]?

Last edited: Apr 23, 2012
2. Apr 23, 2012

### scurty

Any time there are limits of integration it is a definite integral.

You need to remember back to your Calc I/II days. The d means an infitessimal change, you covered this when going over Riemann sums in Calc I. In more general terms, the d term lets you know for what term you are integrating with respect to. dx means you are integrating with respect to x and all other variables are held constant.

$\int x \ dx = \frac{x^2}{2} + C$ because the variable you are integrating with respect you is x. However, $\int y \ dx = xy + C$ because y is held constant.

So, in this problem we have $\displaystyle \int_{[A]_0}^{[A]} d[A] = -k \int_{t_o}^{t} dt$ We are simply integrating the constant 1 in both of these equations, so the integral yields $[A]$ and t respectively, evaluated at the proper limits of integration.

$\displaystyle \int_{[A]_0}^{[A]} d[A] = -k \int_{t_o}^{t} dt$ yields $[A] - [A]_0 = -k (t - t_0)$. In these types of problems, $t_0$ is usually 0 so the equation simplifies to $[A] = -kt + C$. I believe the + C term is the initial amount of the concentration of A, we never wrote it like this, we kept the $[A]_0$ term in the equation.

Remember $\displaystyle \int \frac{1}{x} dx = ln|x| + C$.

3. Apr 24, 2012

### HallsofIvy

Staff Emeritus
$$\int_{[A]_0}^{[A]} \frac{d[A]}{[A]}$$
is just bad notation. Although it is, unfortunately, often done you should NOT use the same symbol for the "variable of integration" and the limits of integration. Better would be
$$\int_{[A]_0}^{[A]} \frac{dt}{t}$$