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?


3 Answers 3


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


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.


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 ;)


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