Integration of the first order rate equation

Question

Given the simple first order reaction $\ce{A -> P}$ derive the integrated rate law.

$$\frac{\mathrm d[\ce{A}]}{\mathrm dt} = -k[\ce{A}]$$ Collect terms: $$\frac{\mathrm d[\ce{A}]}{[\ce{A}]} = -k\,\mathrm dt$$

Now for the bit I need help with, the integration: Apparently the integrated form of $\frac{\mathrm d[\ce{A}]}{[\ce{A}]}= \ln[\ce{A}]$ but I'm struggling to see exactly how.

The next step (usually omitted)I think should be to split the fraction before integration $$\frac{\mathrm d[\ce{A}]}{1}\frac{1}{[\ce{A}]}=\mathrm d[\ce{A}]\frac{1}{[\ce{A}]}=-kt$$

$$\int{\mathrm d[\ce{A}]\frac{1}{[\ce{A}]}}=\int{-kt}$$

$$\int{\mathrm d[\ce{A}]}\int{\frac{1}{[\ce{A}]}}=\int{-kt}$$

Now, to evaluate this I use the general integration law for a constant $\int{[\ce{A}]\,\mathrm dt} = [\ce{A}]t + C$ and reciprocal $\int{\frac{1}{[\ce{A}]}=\ln[\ce{A}]+C}$. Therefore $$\int{\mathrm d[\ce{A}]}\int{\frac{1}{[\ce{A}]}}=[\ce{A}]t + C_1+ \ln[\ce{A}]+C_2=\int{\frac{\mathrm d[\ce{A}]}{[\ce{A}]}}$$

I know this is wrong because the textbooks say something different but I don't know how to correct it. Can anyone help?

Answer 1

Klaus and Binary Geek have already made the main points but I think it would be helpful to see a full explanation.

The differential equation we are trying to solve is: $$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -k[\ce{A}]$$

Separating the variables: $$\int{\frac{\mathrm{d}[\ce{A}]}{[\ce{A}]}} = \int{-k\,\mathrm{d}t}$$

Substituting in the limits; initial concentration $[\ce{A}]_0$ at time $t=0$ and final concentration $[\ce{A}]$ at time $t$: $$\int_{[\ce{A}]_0}^{[\ce{A}]}{\frac{\mathrm{d}[\ce{A}]}{[\ce{A}]}} = \int_0^t{-k\,\mathrm{d}t}$$

Integrating both sides: $$\bigg[{\ln[\ce{A}]}\bigg]_{[\ce{A}]_0}^{[\ce{A}]} = -k\bigg[{t}\bigg]_0^t$$

Evaluating: $$\ln[\ce{A}] - \ln[\ce{A}]_0 = -k(t-0)$$

Condensing the logarithms: $$\ln{\left(\frac{[\ce{A}]}{[\ce{A}]_0}\right)} = -kt$$

Solving for $[\ce{A}]$: $$[\ce{A}] = [\ce{A}]_0 \mathrm{e}^{-kt}$$

Answer 2

You can't write $\int \frac{\mathrm{d}[A]}{[A]} $ as $\int \mathrm{d}[A] \int\frac{1}{[A]}$.

$\int \frac{1}{[A]}$ doesn't make any sense. Integration is essentially summation of infinite infinitely small terms and so each term must be infinitely small. $\frac{1}{[A]}$ isn't infinitely small. $\frac{\mathrm{d}[A]}{[A]}$ is.

$\int \frac{1}{[A]} \neq \ln([A]) + C$ instead $\int \frac{\mathrm{d}[a]}{[A]} = \ln([A]) + C$

Also as Klaus Warzecha has stated, you forgot to put the limits.

Answer 3

You omitted the limits for the integration and therefore miss the practical (and most important) aspect of the whole exercise!

Remember why all this integration is done:

It is about measuring $[\ce{A}_t]$ over time, starting with $[\ce{A_0}]$ at $t = 0$, in order to confirm a hypothesis on the rate law.

If you work the limits for the integration in, you'll reach a useful functional relation between concentration of the starting material and time - useful for a chemist and a biologist ;)