0:03

In our last couple of lessons we learned about rate laws and

Â we learned about rate constants and the orders and

Â how to determine those orders from experimental data.

Â This lesson is a little bit different

Â in that we're gonna be talking about integrated rate laws.

Â Integrated rate laws incorporate time.

Â The others was just rate versus concentration.

Â This will incorporate time.

Â So we're gonna be able to see how the concentration changes as time goes by.

Â 0:34

This is kind of like the difference between saying this is how fast a train is

Â traveling in miles per hour.

Â Or, this is how much time it takes for

Â the train to go from one point to some other point down the line.

Â Our integrated rate laws are like that latter one.

Â How much time does it take to go from

Â this amount of reactant to having it drop down to this amount?

Â Or, if I know I'm starting with this amount and I wanna know how much

Â time it's gonna take to end up with another amount, we can calculate the time.

Â 1:08

We will be focusing on the simplest cases of 1st order overall reactions.

Â Then we'll look at 2nd order overall reactions.

Â Then we'll back up a little bit and

Â spend a little bit of time talking about 0th order

Â because there are reactions in which the concentration has no effect on the rate.

Â 1:31

So, in our lesson today, we're going to be focusing on how to derive

Â the first order integrated rate law, and

Â the interdependence between time and concentration for that.

Â 1:46

So for first-order kinetics, we know a couple of things already.

Â If we had a generic reaction like this one, A turning into products,

Â we could write this rate law, rate is equal to K times A,

Â we know that simply because we've been told it is first order.

Â If we back up a few more lessons and remember what we learned about integrated,

Â no not integrated, but writing a rate expression for reaction,

Â we could take that reaction up here and we could write this rate expression.

Â We remember that the minus sign is because that's a reactant.

Â And the reactant's concentration is decreasing as time goes by.

Â We want the rate of reaction to be positive, so

Â we put that minus sign in there.

Â 2:45

Now, taking that equation and

Â doing a little calculus will give us the integrated rate law.

Â Before we look at that integrated rate law, let me say that calculus is

Â not a prerequisite for this course, so you don't have to be able to derive this.

Â But if you've had calculus and you take it and you look at it like this,

Â because that would be a little bit more familiar symbols for you.

Â 3:24

It's easy to remember if you can derive it using calculus, but that is the equation.

Â The parts and pieces that we see here is the bottom,

Â which is the concentration of your reactant, we are calling it A,

Â at the beginning or at the initial point of the reaction.

Â And up here would be the concentration of A,

Â that reactant, at some time down the line.

Â 3:52

And that time would be that time that's in the equation when you solve it.

Â Now I'm gonna do a little rearranging.

Â The purpose for this rearranging is I want to get it into a nice,

Â linear expression, something that we can plot a straight line using.

Â The first thing I've done is, when you've got the natural log

Â of a quotient as we see here, that can be translated as this portion.

Â You subtract numerator and not denominator.

Â 4:23

Then I'm gonna do a little rearranging, to rearrange it into this form.

Â Now, all I've done is I have moved the natural log of A over here,

Â and I've grouped the minus k out.

Â This is gonna give me the slope intercept form of a line, m stands for

Â slope, b stands for intercept.

Â 4:44

And what this tells me is that this I could plot along the y-axis,

Â t I could plot along the x-axis, and when I did that,

Â I would obtain a straight line, and the slope of that line would be equal to -k.

Â Because that's m and

Â of course y intercept would be natural log of the initial concentration.

Â So that's what that says.

Â A plot of natural log of A, that would be the y-axis versus time,

Â gives a straight line with slope equal to a -k.

Â 5:29

But if you take the natural log of that concentration,

Â you suddenly have a straight line.

Â The straight line will always point in the down direction, okay?

Â It'll be a negative slope because if you take the negative of a negative slope,

Â you will get a positive value for k, okay?

Â 6:11

Looks like we've got four data plots that have been posted and

Â the best straight line through those data points.

Â Now, if you wanna know the slope of that line,

Â what you would do is you would take two points on the line.

Â Not two data points, but two points on the line,

Â and then you would do a change in y over a change in x.

Â Let me erase that dot because it's in the way now.

Â So the slope of that line would be obtained by

Â 6:51

Keep consistent with your data points, so that you're doing the same order.

Â That the two y values, which are these two here, are plugged in.

Â The two x values, which are these here and here,

Â are plugged in and you will obtain a value for k,

Â in this instance, of being 0.034 minutes to the -1.

Â Like I said, it doesn't always have to be units of seconds.

Â We could convert this to 1 over seconds if we wanted using that there's

Â 60 seconds in a minute.

Â But any time you have 1 over a time unit,

Â whether it be minutes, seconds or hours for your k value,

Â that's also an indication that it is a first-order reaction.

Â 7:44

Now, if we remember, we have done this in a previous lesson.

Â If we're dealing with gases, and

Â we know that PV=nrt, we could put our n and

Â our V together and get P equals n over V times RT.

Â This that I'm circling are the units of concentration.

Â So there is a relationship between concentration and pressure.

Â So all the same plots, linear expressions,

Â could be obtained using pressure instead of concentration.

Â So we could plot natural log of pressure versus time and

Â also get a straight line if it's first order kinetics.

Â Here is a depiction of this, natural log of pressure versus time for

Â this generic reaction where I've got an A that's a gas.

Â It would also give me a nice straight line.

Â Well that's the end of our learning objective five,

Â where we're looking at only first order kinetics.

Â We're looking at that integrated rate law for it, we haven't yet

Â done any calculations with it, but

Â you might have to do some, get some information from graphs.

Â And if you ever see a straight line plot, or natural log of pressure or

Â concentration is plot along with the Y-axis and time along the X-axis.

Â Whenever you see a straight line,

Â you will automatically know that's a first-order reaction.

Â