What are the steps to integrate the common rate law to find the integrated rate law for any order

Question

I cannot find any sources that actually list the steps for integrating a rate law, and it’s been driving me crazy. My Ap chem 2 teacher doesn’t know how to do it and just wants me to memorize the common rate laws, however I don’t memorize equations unless I know how to write them out and where they come from. I have the equation

$$ \text{rate} = \dfrac{\mathrm d[A]}{\mathrm dt} = k \cdot [A_0]^{n} $$

where $n$ is the order of the equation
$[A_0]$ is the concentration at time $0$
$[A]$ is the concentration with respect to time
$k$ is the rate constant

I know that for a first order ($n=1$) reaction the equation becomes

$$ [A] = [A_0] \cdot \mathrm e^{kt} $$

and that a second order ($n=2$) reaction the equation is

$$ [A] = \dfrac{[A_0]}{1 + kt \cdot [A_0]} $$

But I don't know what the steps are to find these equations. If I find that the order of my equation needs to be 1.5 ($n=1.5$), I will have no Idea what the equation will be.

From these equations I will know most of the time $[A_0]$, $[A]$ and sometimes Rate.

Edit:

After following the links I learned a portion of the process to solve the equation for any $n$. However after testing it out I am either getting a incorrect answer of a incomplete answer

$$ [A] =\int_{[A_0]}^{[A]}{\frac{1}{[A_0]^2}\,\mathrm d[A]} = \frac{[A_0]^{1}}{1} = \int_{0}^{t}{k\,\mathrm dt}$$

the answer should be

$$ [A] = \dfrac{[A_0]}{1 + kt \cdot [A_0]} $$

what parts am I missing?

Answer 1

The integrated rate laws can be derived by equating two expressions for the rate of the reaction. This gives us a first order differential equation which can be manipulated to get the integrated rate law. The steps that you need to follow are actually quite simple and structured. Here is an example for deriving the first order integrated rate law:

Step 1: Equate the two expressions for the rate of the reaction

For first order reactions, the two expressions for the rate of the reaction are: $$ r = -\frac{\mathrm{d}[A]}{\mathrm{d}t}$$ and $$r = k[A]$$ Equating them we get: $$-\frac{\mathrm{d}[A]}{\mathrm{d}t} = k[A]$$

Step 2: Rearrange so that all the same variables are on the same side to form an first order differential equation

So we rearrange this expression to get all of the same variables on the same side (note that k is a constant, not a variable, so it doesn't matter which side it is on) to get: $$\frac{\mathrm{d}[A]}{[A]} = -k\mathrm{d}t$$

Step 3: Integrate both sides

Now we integrate both sides: $$\int_{[A_0]}^{[A]}{\frac{\mathrm{d}[A]}{[A]}} = -\int_{t_i}^{t_f}k\mathrm{d}t$$ $$\mathrm{ln}[A] - \mathrm{ln}[A_0] = -kt$$

Step 4: Using some algebraic manipulation, rearrange expression to get an equation for [A] $$\mathrm{ln}\frac{[A]}{[A_0]} = -kt$$ $$\frac{[A]}{[A_0]} = \mathrm{e}^{-kt}$$ Therefore we get as required: $$[A] = [A_0]\cdot e^{-kt}$$

Using this guide, you should be able to derive the other integrated rate laws. If you have any trouble, feel free to ask.

Edit:

Since you have also mentioned that you are struggling with deriving the second order integrated rate law, I will show you how to derive it, using the steps that I have outlined above.

Step 1:

$$r = - \frac{\mathrm{d}[A]}{\mathrm{d}t} = k[A]^2$$

Step 2:

$$\frac{\mathrm{d}[A]}{[A]^2} = -k\mathrm{d}t$$

Step 3:

$$\int_{[A]_0}^{[A]}\frac{\mathrm{d}[A]}{[A]^2} = \int_{t_i}^{t_f}k\mathrm{d}t$$

$$\frac{-1}{[A]} + \frac{1}{[A]_0} = -kt$$

Step 4:

Okay now everything up to here has been exactly the same thing we did for deriving the first order integrated rate law. We have gotten an equation relating $[A], [A]_0$ and $t$. To make $[A]$ the subject, we multiply by $[A]$ and put all the terms containing $[A]$ on one side: $$\frac{[A]}{[A]_0} + kt[A] = 1$$ Now we factorise $[A]$ out on the left hand side and divide both sides so that the left hand side equals to $[A]$: $$[A]\left(\frac{1}{[A]_0} + kt\right) = 1$$ $$[A] = \frac{1}{\frac{1}{[A]_0} + kt}$$ Now we multiply the right hand side by $[A]_0$/$[A]_0$ and we get the required equation: $$[A] = \frac{[A]_0}{1 + kt\dot~ [A]_0}$$

Answer 2

If you have questions about maths, ask them in MSE.. well it's not to difficult:

$$\begin{align}\frac{\mathrm{d}[A]}{\mathrm{d}t}=k\cdot[A]_0^2 &\Longrightarrow \frac{\mathrm{d}[A]}{[A]_0^2}=k\cdot \mathrm{d}t\\ &\Longrightarrow \int_{[A]_0}^{[A]}\frac{\mathrm{d}[A']}{[A']_0^2}=k\cdot\int_0^t \mathrm{d}t'\\ &\Longrightarrow \left[-\frac{1}{[A']}\right]_{[A]_0}^{[A]}=k\cdot t\end{align}$$

And now you can finish :)