4.3: Integrated Rate lAws
Calculus! (If you don't know any calculus yet, that's fine. You can still use the results of this without knowing their exact derivation).
1. A rate is just a derivative (or slope) of a curve, or a "speed". 2. In this case, our rate is how fast the concentration is changing. 3. Therefore, if you integrate the rate equation, you will get an equation that finds the concentration as a function of time. 4. Since we have three basic rate laws, each with a different equation, we'll end up with three different integrated rate laws: 

0^{th} order: Rate = k integrates to [A]_{t} = [A]_{0}  kt
1^{st} order: Rate = k[A] integrates to ln[A]_{t} = ln[A]_{0}  kt
2^{nd} order: Rate = k[A]^{2} integrates to 1/[A]_{t} = 1/[A]_{0} + kt
In all cases, [A]_{t} is the concentration at time t (our y variable, essentially) and [A]_{0} is the original concentration (a constant). t is time (our x variable).
1^{st} order: Rate = k[A] integrates to ln[A]_{t} = ln[A]_{0}  kt
2^{nd} order: Rate = k[A]^{2} integrates to 1/[A]_{t} = 1/[A]_{0} + kt
In all cases, [A]_{t} is the concentration at time t (our y variable, essentially) and [A]_{0} is the original concentration (a constant). t is time (our x variable).
Second big problem: your data often has some error in it. You could probably use a computer to do a curve fitting to perfect data, but once you have some error, it's going to be nighimpossible to get a good comparison.

Here are the concentration versus time curves that result from the same starting concentration, using the same rate constant, for each order.
While they are very different when you have them all sidebyside, it can be very difficult to tell the difference for an individual data set: do you have a 1st order with a low rate constant, or a 0th order with a higher rate constant? Example: what is the order of the graph shown below? It looks very linear, but it might not be; the curvature is often very subtle. 
So how do you figure out the order of the reaction?
1. Use the method of initial rates (see 4.2). The problem with this is that it requires doing a bunch of trials, which may not be possible.
2. Take your concentration and time data and graph it three ways: [A] vs time, ln[A] vs time, and 1/[A] vs time. Whichever one is the best line will tell you what the order was:
1. Use the method of initial rates (see 4.2). The problem with this is that it requires doing a bunch of trials, which may not be possible.
2. Take your concentration and time data and graph it three ways: [A] vs time, ln[A] vs time, and 1/[A] vs time. Whichever one is the best line will tell you what the order was:
The thing we want to look at is the Rsquared value; the closer to one it is, the better the fit. In this case, the best fit is a perfect one for ln[A] vs time, so the reaction was first order.
The slope and intercept of the best fit line can also be used: ln[A]_{t} = ln[A]_{0}  kt or, in terms of our graph:
y = a  mx The slope is the opposite of k (k is always positive). The intercept is ln[A]_{0}, so [A]_{0} = e^{intercept} 
HalfLives:
For any reaction, the halflife is defined as the amount of time it takes half the current material to go away.
However, this is a basically useless concept for everything except first order reaction. For first order, it's great!
Mathematically:
For any reaction, the halflife is defined as the amount of time it takes half the current material to go away.
However, this is a basically useless concept for everything except first order reaction. For first order, it's great!
Mathematically:
Let 't_{1/2} be the time at which half the material is gone. Therefore, at that time, [A]_{t} = ½ [A]_{0}
Substituting back into the integrated first order rate law:
ln(½ [A]_{0}) = ln[A]_{0}  kt_{1/2} which rearranges to (using properties of logs)
kt_{1/2} = ln([A]_{0} / ½ [A]_{0}) which simplifies to
t_{1/2} = ln(2)/k AND THIS IS A CONSTANT NUMBER!
Substituting back into the integrated first order rate law:
ln(½ [A]_{0}) = ln[A]_{0}  kt_{1/2} which rearranges to (using properties of logs)
kt_{1/2} = ln([A]_{0} / ½ [A]_{0}) which simplifies to
t_{1/2} = ln(2)/k AND THIS IS A CONSTANT NUMBER!
So what does that mean? It means that if you have 100 g of something with a halflife of 30 minutes:
Graphically:
 After 30 minutes, there will be 50 g left
 After 60 minutes, there will be 25 g left (half of the remaining 50 g)
 After 90 minutes, there will be 12.5 g left (half of the remaining 25 g)
 Etc
Graphically:
Every 5 seconds, half the previous amount of material goes away.
You can use this to work backwards to the rate constant, too: 5 = ln(2)/k k = 0.1386 s^1 Why only first order? Both mathematically and graphically, it can be shown that zeroth and second order rate laws don't simplify down to a nice number, so your halflife keeps changing over the course of the reaction...making it a completely useless concept. 