We said that reactions are normally zeroth-order, first-order and second-order.
What we should realize is with each one of these orders, we're going to have what's called
an i-rate equation or an integrated rate law equation.
We're going to say for zeroth-order, first-order and second-order, they have the same variables.
We're going to say At represents the final concentration or final amount of our reactant.
We're going to say Ao equals the initial concentration of our reactant.
We’re going to say K is the same K we've been dealing with.
It’s the rate constant and then t is just time.
One thing we should realize here is that when it comes to K and t, they have to have the
same units of time.
What does that really mean?
That means let's say our time was in hours.
K and t have to have the same units of time.
K would have to be hours inverse.
Always keep an eye out for that.
Make sure that both have same units of time.
If not, you have to convert one so that they both match.
We're going to say with each one of these orders, there comes a graph.
We’re going to say anytime we plot a graph, it’s of y versus x.
Our y-axis is this and our x-axis is this.
The x-axis will always give us time.
We're going to say here that the y can either be concentration of our reactant, ln of our
reactant or one over our reactant.
What you should realize is that each one of these integrated rate law equations can be
used to write out this plot.
We'll see how that works.
We’re going to say these are all the integrated rate law questions.
All these equations are equal to y equals mx plus b.
What you should realize, we're talking about plotting y versus x.
This first one is zeroth order.
Here the y is just the concentration of our reactant, which is right here, the y slope,
and x is time.
We're going to say m, remember m is your slope.
Here, remember, your slope is equal to your variable k.
Just remember that.
It's going to become important later on when we're doing certain types of calculations.
Here on the y-axis, it’s ln of my concentration.
Here, y is ln of our concentration.
Here again, x is time.
Then finally for second-order, one thing you should realize first is that second-order
is the only one with a positive slope.
Why is it the only with the positive slope?
Because here, m is a positive k. for zeroth and first order, k was negative.
It's negative here and it’s negative here, which means the slopes are decreasing over
time.
Here y is 1 over A and of course, time again is on the x-axis.
Just remember, we can manipulate these integrated rate laws to basically match them up with
y equals mx plus b.
In that way, we can plot y versus x.
Remember, your slope which is m is equal to your rate constant.
Finally, first-order, zeroth-order, second-order – they each have what we call a half-life
equation.
Remember, half-lie is basically the time it takes for about half of a substance to decompose
over a period of time.
For zeroth order, it is half-life equals initial concentration of my reactant divided by 2K.
For first, it is half-life equals ln2 over k.
Remember, ln 2 is the same thing as 0.693 over K. Then finally, half-life of a second-order
is 1 over K times the initial concentration.
On your exams, your professor will most likely give you these equations but they’ll just
mix up the orders.
I’d recommend just memorizing because if they're going to mix up the orders anyway,
you should know which is which.
These are the questions you need to remember when it comes to zeroth order, first order
and second order.
We’ll learn when exactly do we use them and how do we have to manipulate them sometimes
to solve for a given question.
Just remember, try your best to memorize each one of them because you most likely will need
all of them for your upcoming exams.