If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:8:16

- [Voiceover] Here we have one form of the integrated rate law for
a first order reaction. And we're gonna keep
going with the math here, so we eventually will
talk about the half-life. So over here on the left, the natural log of the concentration of A at time t minus the natural log of the
initial concentration of A. That's the same thing as the natural log of the concentration of A over the initial concentration of A. So that's just the log property. And this is equal to negative kt, where k is your rate constant. Next, we need to get
rid of our natural logs. So we're going to exponentiate both sides. So we're gonna take e to both sides here and that's gonna get
rid of our natural log. So now, on the left side
we have our concentration over the initial concentration. On the right side we have
e to the negative kt. So we're gonna multiply both sides by the initial concentration of A. So we get that our
concentration of A at time t is equal to the initial concentration of A times e to the negative kt. And now, it's a little bit
easier to think about the graph. We can put the concentration
of A on the y-axis and we can put time on the x-axis. And this is in the form
of an exponential decay. So down here I've graphed
an exponential decay graph, just to show you what it looks like here. Let's think about this point on our graph. So that's when time is equal to zero. So when time is equal to zero, what is the concentration? So you would just plug in time
is equal to zero into here. So you would have your
concentration is equal to the initial concentration
times e to the zero. And e to the 0 is of course one. So this is one. So our initial concentration is obviously this point right here,
time is equal to zero. This is obviously our
initial concentration. So I'll write that in here. So that's this point. And as time approaches infinity, as time goes to infinity, your concentration of A goes to zero. So as you go out here, obviously your concentration of A is going to approach zero. So that's the idea of an
exponential decay graph. Next, let's think about half-life. So over here is our
definition of half-life. It's the time it takes
for the concentration of a reactant to decrease to half of its initial concentration. So if the initial concentration, if this is the initial concentration here, what would be the concentration after half of it has reacted? We would get our initial
concentration divided by two. So we're gonna plug this
in for our concentration and then the symbol
for half-life is t 1/2. So t 1/2. So we're gonna plug this in for time. So when the time is
equal to the half-life, your concentration is half of
your initial concentration. So let's plug those in and
solve for the half-life. So on the left side we would have our initial concentration divided by two. And then this would be equal to the initial concentration of A times e to the negative k and then this would be the half life, so we plug in t 1/2 here. And so now we're just
gonna solve for t 1/2. We're gonna find the half-life
for a first order reaction. Let's get some more space down here. And we can immediately cancel out our initial concentration of A. So now we have 1/2 is equal to this is e to the negative kt 1/2. Alright, next, we need
to get rid of our e here. So we can take the
natural log of both sides. So we can take the natural log of 1/2 this is equal to the natural log of e to the negative kt to the 1/2. And so that gets rid of our e. So now we have the natural log of 1/2 is equal to negative kt 1/2. So we're just solving for t 1/2, cause t 1/2 is our half-life. So our half-life, t 1/2, would be equal to this would be negative natural log of 1/2 divided by k. Let's get out the calculator
and let's find out what natural log of 1/2 is. So let's get some space over here. So natural log of .5 is
equal to negative .693. So we have the negative of that, so we get a positive value
here for our half life. So our half-life is equal
to, let me rewrite this here, so our half-life, t 1/2, is equal to .693 divided by k, where k is our rate constant. So here is your half-life
for a first order reaction. Now let's think about this. If k is a constant, obviously .693 is a constant. And so your half-life is constant. Your half-life of a first order reaction is independent of the
initial concentration of A. So you're gonna get the same half-life. And let's think about that for an example. Let's go back up here to our graph and let's think about half-life. So lets' say we're starting
with some initial concentration, let me go ahead and change colors here, so we can think about it. I'm just going to represent
our initial concentration here with eight dots. So that's our, let's say
we have eight particles we're starting out here. So obviously this is a
theoretical reaction. So we're gonna wait until we've
lost half of our reactant. Alright, so we've lost
half of our reactant, obviously we'd be left with four. We'd be left with four here. So, where would that be on our graph? Well this point right here
is out initial concentration. This is concentration
here, so we'd go half that. So that would be right here on our graph. So we'd go over here
and we find this point. Then we drop down to here on our x-axis. And let's just put in some times. Let's say that this is 10 seconds and 20 seconds and 30 seconds and 40 and so on. So we can see that it took, this took 10 seconds for us to decrease the concentration
of our reactant by half. And so the first half-life is 10 seconds. So let me write that in here, so t 1/2 is equal to 10 seconds. Again, a made up reaction, just to think about the idea of half-life. And then, let's say, now we have four. How long does it take for
half of that to react? So if half of that reacts,
we're left with two particles. And on our graph, let's see,
that would be right here. This point would be half of this, so we go over to here, and then we drop down. And how long did it take for us to decrease the concentration
of our reactant by half? Once again, 10 seconds. So this time here would be 10 seconds. So this half-life is 10 seconds. We could do it again. So we lose half of our reactant again. And we go over to here on our graph and we drop down to here. How long did it take to go from two particles to one particle? Once again, it took 10 seconds. So the half-life is once again 10 seconds. So your half-life is independent of the initial concentration. So it didn't matter if we started with eight particles or four or two. Our half-life was always 10 seconds. And so, this is the idea of half-life for a first order reaction.