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CATHERINE DRENNAN: All right.
So enzymes, of
course, are catalysts,
and that leads us to our
next unit, which is kinetics.
And it's our next
unit and, in fact,
our last unit of the semester.
So we're going to be talking
about kinetics for the rest
of the semester, which
sounds like a long time,
but it really isn't.
All right.
So we'll switch to
today's handout.
So kinetics and thermodynamics
is a sort of yin-yang.
You really have to think about
both of them at the same time.
So when you're thinking
about whether a reaction will
go forward spontaneously, you're
thinking about thermodynamics.
And you're thinking about
how fast that reaction is
going to go, you're
thinking about kinetics.
So let's just do
a little review.
We've talked about this before,
but it's a clicker challenge.
So today, let's think back
to thermodynamics and tell me
what stable/unstable refers to,
thinking about thermodynamics
and kinetics now.
OK, 10 more seconds.
OK.
So most people got stability.
When you're talking
about stable/unstable,
you're not talking about rate.
You are talking
about delta G. You're
talking about the
spontaneous tendency.
A chemist, if it's
around a long time,
would talk about it not being
stable, but being inert.
And in terms of this
one here, if you're
thinking about the delta
G for the decomposition
into its elements, that versus
the delta G for formation,
if that's negative then
it would be stable.
But we want to be positive
for the decomposition.
All right.
So let's just look
at this up here.
So again, it refers to delta
G. And labile and inert
refers to the rate.
And again, when you have a
negative delta G of formation,
then that makes
something stable.
OK.
So thermodynamics, kinetics.
We want to think about, is
the reaction spontaneous?
But we also need to think about
how long it's going to take.
So rate is important.
And so if we think about
the average rate of a Ferris
wheel-- I looked it up.
It is 31.4 seconds
to spin around
if you're at an amusement park.
And you can think about your
experience-- or sorry, that
should say imagine-- one
revolution for every 5 seconds
or every 5 hours would be a
very different experience.
Five seconds would
be way too fast.
You'd probably be
throwing up everywhere.
In 5 hours, you'd be
like, oh, my goodness,
get me off of this thing.
So rate matters.
And chemical kinetics
doesn't measure
the rate of Ferris wheels,
except maybe it does
if the ferrous is ferrous iron.
Then we might be
talking about its rate
if it's spinning around doing
things in a chemical reaction.
So we might think about the
concentration of ferrous iron
changing with time.
So chemical kinetics
experiments,
you measure concentrations and
think about how fast they're
changing with time.
So let's think about
some of the factors that
are going to affect the
rates of chemical reactions,
and we'll write some
of these on the board.
So what's one thing that
you may have observed
that will affect, either
speed up or slow down, say,
a chemical reaction?
What's one thing
you can think of?
AUDIENCE: Temperature.
CATHERINE DRENNAN: Temperature.
What's another thing
you can think of?
AUDIENCE: [INAUDIBLE].
AUDIENCE: Catalysts.
CATHERINE DRENNAN: Catalysts.
So right, so changing
the sort of environment.
So it depends a little.
It could be pressure.
It depends on what
you're talking about,
what kind of nature of
material you're talking about.
So possibly pressure
could have an effect.
And that brings us to the
nature of the material,
whether it's gas or
solid or other things.
And that also comes to
the other issue, which
is, what is its mechanism?
Oops, sorry, mechanism.
So does it have one step?
Does it have many steps?
Do the steps involve
changes in phase?
What's going on
in the mechanism?
And so those are all
some good things.
One other thing
that I'll put on,
which also only
applies in some cases,
depending on the type of
mechanism and type of reaction,
would be the concentration of
the material, which kind of
also gets to this pressure
idea, depending on what it is.
How much do you have in there?
All right.
So those are some of
the things, and we're
going to talk about most
of these in the next unit.
All right.
So let's now think about one
example of a chemical reaction,
and this chemical reaction is
called the oscillating clock.
And before we see
the demo, we're
going to think about
what's happening.
So this demo, as with
most chemical reactions,
involves thermodynamics,
chemical equilibrium, kinetics.
It also has some acid-base.
You have to have some acid
in there to get it to go.
It's also an
oxidation-reduction reaction.
And it has colors, and we'll
see that the colors change
depending on what the
oxidation state of the material
is and also what its
liganded state is.
So basically, in one
demo, you get every unit
that we've had in the
second half of the course.
And that's kind of true
about any chemical reaction.
Every chemical reaction,
if you're interested in it,
involves thinking about
all of these things
that we've been talking about.
All right.
So oscillating clock reaction.
It's a fairly
complicated reaction.
It has many steps,
and we're just
going to break the
overall reaction shown
here down into two steps today.
As with many
reaction mechanisms,
there are multiple steps.
But here is step 1
that we'll talk about,
and here is step 2
that we'll talk about.
And step 1 occurs when your
I2 concentration is low.
I2 here is a product
of the reaction,
and when it builds up too much,
then that reaction will stop.
And so in this reaction,
we have acidic conditions.
You'll see our H+.
And under those
acidic conditions,
we're going to change
a clear solution--
so I, iodide, in its certain
oxidation states is clear.
But when it goes to I2,
it turns an amber color.
In the second reaction,
which will only
occur when I2 is high--
so this will build up,
shutting off reaction a
and starting reaction b.
In this reaction, you have
a complex being formed,
and that's going to
change the color.
So we're going to go from
amber to a blue complex.
All right.
So before we see
this happen-- and I
should say that once
you get over here,
you find the blue complex.
This will consume the
I2, which will cause
that to drop to be low again.
And then you're going to
start reaction a again,
which, when your concentration's
built up too much,
you'll start reaction b again.
And so reaction a, then reaction
b, then reaction a, reaction b.
And you'll notice that the
product of one of the reactions
is a substrate here, product
here, substrate here.
So this goes back and
forth, hence the name
oscillating clock.
All right.
So before I show you
something pretty,
I'm going to make
you do work and think
about the oxidation-reduction
reactions that are occurring
that you're going
to see that lead
to these specular color changes.
All right.
So here we go, clicker question.
All right, let's just
do 10 more seconds.
All right.
So somebody asked
once, like, how
will I know that
it's a peroxide?
So I tried to help you
out and said reaction
with hydrogen peroxide.
Let's see how many people-- some
people definitely noticed that.
All right.
So let's just take a look
over here at what's going on
and fill in our chart.
So here the oxidation
state is plus 5.
So we have now this oxygen
is its normal minus 2.
So minus 2 times 3 minus
6 equals the whole thing.
Charge needs to equal minus 1.
So we need plus 5 there.
And I2 is 0.
The next one, I-, is being
oxidized to I2, minus 1 to 0.
The oxygen in hydrogen
peroxide is being oxidized.
So it's minus 1
here because we have
plus 2 for H and two oxygens.
So that's minus 1 each to 0.
And then, again, minus 1, and
oxygen and water is minus 2.
So there's lots of
oxidation-reduction reactions
going on.
And in fact, this occurs in
multiple different steps.
So there's even more
different reactions going on
that I didn't even show you.
All right.
So we have one more
clicker question
to answer before we do the demo,
which is that if I told you--
you want to think about
iodide you're using
because it's changing color.
Why are you using the
hydrogen peroxide?
So tell me what is true
about hydrogen peroxide
using the information given.
All right, 10 more seconds.
All right.
So if we look over here,
the large positive value.
And again, it's a large positive
value of the standard reduction
potential is a big value,
which makes it a good oxidizing
agent.
The reduction is spontaneous.
So if you have a large positive
E for the reaction written
as a reduction, that means
that delta G for that reduction
is going to be negative,
which means it's spontaneous.
So things that
like to be reduced
are good oxidizing agents.
Actually, hydrogen peroxide is
a really good oxidizing agent,
and that's why it's
used in a lot of things.
All right.
So let's look at this demo now.
So again, clear to
amber, amber to blue.
And the reaction rate is also
sensitive to temperature,
and you told me temperature
does affect reaction rates,
and it does in this case.
All right.
Let's see it.
AUDIENCE: OK.
So we've got the
hydrogen peroxide.
Eric's got that in his hand.
He's going to take that off
before he actually unscrews it.
Yay.
OK.
Hydrogen peroxide, as we said,
it gets reduced really easily,
which is actually the reason
why we kept it in a plastic bag,
because even the presence of
oxygen can cause it to react.
So he's going to add solution b.
That was solution a.
Solution b contains
the iodate, which
is going to provide the-- huh?
CATHERINE DRENNAN: Oh, sorry.
AUDIENCE: Yeah, I
got it stirring.
CATHERINE DRENNAN: All right.
AUDIENCE: It's
going to provide--
CATHERINE DRENNAN: I
forgot to turn this on.
AUDIENCE: --the iodine
for the solution.
And as you can see, it's
turning kind of yellowish, which
is what we were expecting.
So now the iodate is starting
to produce the iodine.
And when he adds solution
d, we can see it changing.
Yay, it works.
It's always a struggle whether
these things will actually
work, but this one
usually works very nicely.
I'm going to turn this
up a little bit more.
And it will actually
continue to do this
for quite a while,
I think, right?
So--
CATHERINE DRENNAN: Yeah.
AUDIENCE: Yay.
CATHERINE DRENNAN: Awesome.
AUDIENCE: Awesome.
CATHERINE DRENNAN: [LAUGHS].
[APPLAUSE]
CATHERINE DRENNAN: Yeah.
So we can move this
off to the side,
and now we're going to see
the effect of temperature.
AUDIENCE: Actually, we'll
move that in a second.
So as Cathy mentioned,
affecting temperature
will affect the
rates of reaction.
So this guy will continue to go.
And what we have here is
we have all three solutions
again, except they've
been put in ice.
So we're going to
see if we've kept
this a little bit cold again.
You should have
taken the thing off.
All right.
So that's the peroxide.
We've got solution b.
You got it?
OK.
And yeah, we're going to
stir in a second, after b.
So as before, you saw the
temperature change kind
of quickly.
This time it's not really
changing that much.
So before, you saw temperatures
change pretty quickly.
This is probably going to take
a while to actually change,
and it might not change at all.
It's because the reaction
was so cold that you
won't have that second step.
And also I can't-- there we go.
So it happens really, really
slowly, as you can tell,
whereas this one happened
very, very quickly.
So yay.
CATHERINE DRENNAN: Yeah.
So example of many things.
So we had chemical equilibrium.
We had thermodynamics,
talking about which
reactions were spontaneous.
So spontaneous until
making the amber.
And then when it's made too
much of the iodide, the I2,
then we see the
next reaction going
until we now have too little.
And then we switch
back and forth.
And yeah, this, I
think, very clearly
demonstrates that the
temperature is very different,
has a very different effect.
It's a much, much
slower reaction.
All right.
Thank you.
We can leave them-- do you want
to just leave them up here?
AUDIENCE: I"m going to
leave this one up here.
CATHERINE DRENNAN: Yeah, OK.
We'll leave the
fast one up there.
It will eventually stop.
So it's good probably, I
don't know, maybe five minutes
or more or a little longer.
And then eventually it just kind
of stays this more dark brown.
Maybe it's already happened.
We'll see if it switches again.
All right.
So let's talk about measuring
reaction rates then,
because we just saw
this reaction going
in another reaction rate.
We saw the effect
of temperature.
And let's think about
how we would actually go
about measuring some of these.
So we can think about
measuring two different kinds
of rates-- an average rate
and an instantaneous rate.
And so first let's talk
about measuring average rate.
And to do that, I should
switch back over here.
OK.
So let's consider a
different reaction now,
one that's a little
less complicated
and kind of made up,
but it's a good example.
NO2 plus CO goes to NO plus CO2.
And so if you're
measuring a rate,
you could think about
measuring the decrease
in either of the reactants or
the increase in concentration
of either of the products.
So in this case, let's
consider that we're
measuring the change in NO.
And usually people
pick the thing
to measure based on
what's easiest to measure,
what kind of handle you are.
Does it have a spectroscopic
signal or something else
that you can easily measure?
So here we have
concentration versus time.
And so we should see
none in the beginning,
and then we should see it
increase and then level off.
And we can calculate
an average rate,
which is just going to be
the change in concentration
over the change in time.
We can also express
the average rate
as delta the
concentration of NO over
delta t, the change in time.
And we can calculate
what that would be.
We can pick an interval,
say, from 50 to 150.
We can measure some
concentrations there and then
calculate the average
rate-- the change
in concentration
from the time 150
to the time 50 concentration
over that time interval.
And it will give us an
answer, in this case,
1.3 times 10 to the
minus 4 molar per second.
All right.
Well, that could
be useful to know.
But the average rate depends
on the time interval I picked.
If I picked a different
time interval,
I might have gotten
a different rate.
So instead of calculating
average rate, a lot of people
want to calculate instantaneous
rate, where you're asking,
what is the rate at a
particular time point?
So now let's talk about
instantaneous rate.
We'll have the same
equation up there.
Now, instantaneous
rate is defined
as the rate where you have a
limit of delta t, the change
in time, going to 0.
You're comparing concentrations
at time t and time
t plus the interval dt
over this time change,
again, as you approach 0.
You can also express
this as d change
in concentration of NO dt.
And so as that change
in time approaches 0,
then the rate becomes the
slope of the line tangent
to the curve at time t.
So if we wanted, say, find
the instantaneous rate
at 150-- here's 150-- we
can draw a line tangent
to the curve at that time.
So this would be
the point at 150,
and then we can
calculate the slope
to tell us the
instantaneous rate.
So let's do that.
We will calculate the slope.
We have to look what
the concentrations are,
what the times are.
And then the
instantaneous rate at 150
would be the slope of the line,
this change in concentration,
over that time interval.
And to the correct
significant figures--
there should be an
extra 0 in there--
7.70 times 10 to the
minus 5 molar per second.
So that would be the
instantaneous rate,
again, the slope of the
line tangent to the curve
at time 150.
So you could ask, what
is the instantaneous rate
at 100 seconds or
200 seconds, and get
that instantaneous rate.
People are often talking
about initial rate.
And by initial rate, they mean
the instantaneous rate at time
equals 0.
So again, we have
two different rates.
We have average rate
and instantaneous rate.
So now let's think about rate
expressions and then rate laws.
All right.
So keeping with
that same reaction,
again, we could monitor
NO forming or CO2 forming,
or we could monitor either of
our reactants disappearing.
And if we assume that there's
no intermediates or very slow
steps, then we could say
that the rate of the reaction
should be equal to
the disappearance
of the concentration of
NO2, or decrease in NO2, one
of our reactants, over time.
It should also be equal to
the disappearance of CO,
our other reactant
over time, and also
equal to the appearance
of NO with time, and also
equal to the appearance
of CO2 with time.
And again, for these all
to be equal to each other,
we are assuming here there are
no intermediates being formed,
or that if we are
forming them, then
they're independent of time.
In other words, if, say,
there was a large, slow step
somewhere in here, we might see
all the reactants disappear.
But there'd be a lag before
the products would appear,
because more steps and
things are going on.
So these are all
equal to each other
if you have no intermediates
or the intermediates are not
affecting the overall rate.
So we could think about
this for a generic equation.
So if we consider A plus
B going to C plus D,
when we're talking about
the rates of disappearance
or appearance, we have to
remember the stoichiometry
of the reaction.
So if we have a
stoichiometry of little a,
then we would say minus 1/a dA
dt, stoichiometry of little b
for reactant B, minus 1/b dB dt.
Then formation of products, 1/c
dC dt, and our other product,
1/d dD dt.
I got through it.
OK.
So let's look at an
example reaction,
and why don't you tell me
what the rate is equal to.
I didn't announce whether this
would be in the program or not.
Why don't you look at the
results and make a decision?
OK, 10 seconds.
All right.
Still haven't gotten much into
the 90s, but still I'm happy.
All right.
So here, again, we're talking
about the disappearance
of our reactant.
So it will be minus
1/2, and the formation
of one of our products
and the formation
of the other product as well.
All right.
So don't forget about the
stoichiometry of the reaction.
So these are called
rate expressions,
and you don't want to get
that confused with rate laws.
So a problem might ask,
write the rate expression.
This is what it's looking for.
If it says write
the rate law, it's
going to be looking for
the following thing.
So a rate law is a relationship
between rate and concentration,
and they are related by a
proportionality constant,
little k, which is
called the rate constant.
So big K is what?
AUDIENCE: Equilibrium.
CATHERINE DRENNAN:
Equilibrium constant.
Little k is a rate constant.
All right.
So let's look at a reaction.
And so reaction of A
plus B going to C plus D,
the rate law could be
expressed, rate equals
rate constant k times
the concentration of A
to the m and B to the
n, where m and n are
the order of the reaction
in A and B respectively,
and k, again, is
our rate constant.
All right.
So let's look at what I call
the truth about rate laws.
So the first thing, you cannot
just look at the stoichiometry
of the equation and say, oh,
it's going to have this m for A
based on its stoichiometry,
unless it's what's called
an elementary reaction.
And we're going to get
to that on Wednesday
after Thanksgiving.
Make sure you come Wednesday
after Thanksgiving.
That's the lecture on mechanism.
It's a really important
lecture in kinetics.
So we'll learn about
elementary reactions.
So you can't just look
at the stoichiometry
and write the rate law.
You need experiment.
So the rate law is
not just limited
to reactants, although
largely it's true
that the rate laws will
just have reactants.
It can also have a
product term, like C
could appear in the rate law.
How would you know this?
Well, it would be
experiment would tell you
whether this was true or not.
So for this rate law,
we saw this already,
the order of the
reaction in A is m,
and the order of the
reaction in B is n.
And m and n can be integers.
They can be fractions.
They can be positive,
and they can be negative.
So now we're going
to think about all
these different orders of
reaction and think about what--
and we'll do it just for
A-- what m is equal to.
So we're going to fill
in this table here.
And some of these things are
in your notes, some of them
you need to fill in.
So we're going to
start with m equals 1,
and this is called a
first-order reaction.
The rate law for this, the rate
would equal the rate constant
times the concentration
of A. And now
let's just think, if this is a
first-order reaction that just
depends on A, if you double the
concentration of A, what do you
expect will happen to the rate?
AUDIENCE: It'll double.
CATHERINE DRENNAN: What?
AUDIENCE: Double.
CATHERINE DRENNAN: It
will double, right.
So in a first-order
reaction, you
double the concentration
of something,
and the rate will double.
And this is how you figure out
what the order of the reaction
is.
You double the
concentration of something,
leaving everything
else the same,
and see the effect on rate.
So again, these are
experimentally determined.
Now let's consider m equals 2.
And notice the blue went away
here, but it's still there.
So you can see it.
All right.
Second order-- what
does the rate equal?
So it would equal k,
our rate constant,
times the concentration
of A, and now we have a 2.
So it's raised to the
power of m, and that's 2.
So that's how we would write
the rate law for something that
was second order in A.
So now let's think about
if you double the
concentration of A,
what you would you
observe for the rate?
What would happen?
AUDIENCE: Quadruple.
CATHERINE DRENNAN: It
would quadruple, right.
Now, what about if it triples?
Why don't you tell me
what would happen then.
All right, 10
seconds, very fast.
90%, yes!
OK.
So yes, 9 times.
All right.
And in problem set 9,
which is already posted,
there's all sorts of problems
where you see the effects,
and you can figure out what
the order of the reaction is.
All right.
So now let's consider
minus 1 over here,
which is often not
referred to by any name.
The rate here would be equal
to k concentration of A
to the minus 1.
And if we double the
concentration of this--
you can just yell out-- what
would happen to the rate?
AUDIENCE: Half.
AUDIENCE: Half.
CATHERINE DRENNAN: Half, yes.
All right.
Now, to m equals minus 1/2.
So the rate here would
be k equals concentration
of A raised to the minus 1/2.
And now if we double, that's
our last clicker question
on this sheet.
So why don't you
tell me what happens,
and this should be fast too.
All right, 10 more seconds.
All right.
So going back over
here, right, we
have 0.7 times, which
is 2 to the minus 1/2.
And when you're
looking at the effects
of the rate on
concentration, you'll
see the concentration double.
And you're like, what on
earth happened to the rate?
Keep in mind this minus 1/2,
because it's very hard to think
about what the relationship is.
So again, they can be fractions,
positive, or negative.
All right.
So back up here now, this
actually is called half order.
So we would have a
rate that is equal to k
times the concentration
of A raised to the 1/2.
And if we double
that-- you can just
yell out-- what would happen?
AUDIENCE: [INAUDIBLE].
CATHERINE DRENNAN: So
1.4 times the rate.
And finally, m equals 0.
What do you think that
is likely to be called?
AUDIENCE: Zero order.
CATHERINE DRENNAN:
That is zero order.
So at least some things
are pretty easy to guess.
And if we think
about the rate, what
would the rate be equal to?
AUDIENCE: k.
CATHERINE DRENNAN:
Just k, right.
It's just going
to be equal to k.
So zero order, the
A term is not there.
So if we double the
concentration of something
that is zero order, what
happens to the rate?
AUDIENCE: Nothing.
CATHERINE DRENNAN:
Nothing, right.
No effect on rate.
The concentration term
isn't part of the equation.
So the problem set 9 will
have a lot of experiments,
and you need to
figure out by looking
at what is happening what
the order of the reaction is.
Is it doubling when
you double the rate?
Is it quadrupling?
Is it doing some
weird thing that
seems to be a negative--
an inverse fraction?
And that will allow you to
figure out what the order is.
OK.
So once you figure out
the order of the reaction,
then the next truth
about rate laws
is that the overall
order of the reaction
is just the sum of the
exponents in the rate law.
So for example, if
we had this reaction,
what would be the overall order?
You can just yell it out.
AUDIENCE: 3.
CATHERINE DRENNAN: Yep.
It would be a third order.
The order would be 3.
Now, some people get
this wrong on the test.
It's like, no, you want
to save your points
for something that's hard.
2 plus 1 is 3.
So remember that.
OK.
And this would be then
second order in A.
You would say that's first
order in B, overall third order.
So if you're going
to lose points
on an exam on this
material, I recommend
losing them determining
the units for the rate
constant, because
that's much more
complicated than figuring
out the overall order
of the reaction.
The units, it
depends on the order.
You can have squared.
You can have molar
squared, quadrupled,
molar to the minus fractions.
All sorts of crazy things
happen in your units
of rate constants.
So save your points if
you're going to lose some.
And I think it would be
great if everyone got 100,
but if you're going to lose
some, lose some on that.
That's harder.
OK.
So we'll end there for today.
And Wednesday, we're going to
be talking about integrated rate
laws, half-life for first
order, nuclear chemistry.
It's going to be very exciting
and slightly radioactive.
So where were we?
We were talking about
the fact about how
you measure instantaneous
rates and average rates.
And one of the issues of
doing kinetics experiments,
it's all experimental.
You want to measure things.
But it can be very challenging
to measure initial rates,
because you're often talking
about a very small change
in concentration.
And so it can be
helpful sometimes
to use integrated
rate laws, which
allow you to measure a lot
of different concentrations
as times elapses and plot that
data to get out rate constants
and things like that.
So we're going to talk
about integrated rate laws.
And so the alternative, then,
is to use this integrated rate
law.
Again, expresses concentrations
directly as a function of time
and gets at this
small changes problem.
All right.
So we're going to talk
about first order now,
and then we'll talk about
second order in a little bit.
So for a first-order equation,
we have A going to B.
And we learned last time
that we can write a rate
expression for a first-order
process, or for this process
here.
So we could talk about
the rate expression
as the disappearance of A
minus d concentration of A dt.
We can also write the rate law
for a first-order equation,
and that would be k,
our rate constant,
times our concentration.
So the rate expression
has the d dt,
and the rate law has k,
our rate constant, in it.
All right.
So using these two things
that we learned last time,
we can do a derivation to get
our integrated first-order rate
law.
So in this derivation,
we're going
to separate our concentration
terms on one side and our time
terms on the other side.
So we're going to bring over our
concentration of A over here.
So we'll divide by the
concentration of A.
We have our d concentration
of A term from here.
We're going to take our minus
sign, put it on the other side.
And we're going to take
our dt, our time term,
and put it over there as
well with our rate constant.
So we have the terms that have
concentration of A on one side,
and we have our terms with rate
constant and time on the other.
Now we can integrate
both sides, and we
will integrate from our
original concentration of A.
So that's A to the O, for
our original concentration,
up to concentration at whatever
time t we stop the experiment
and through all the
times in between, over 1
over the concentration of A dA.
And on the other side, we have
our minus rate constant k,
and we're looking at the
time from time 0 to time t,
when we stop the experiment.
All right.
So we can take this
expression now and move it up,
because I have more
derivations to go.
So I'm going to put it
on the top of the screen.
And now I am going to solve it.
So this integral solves
to the natural log
of the concentration
of A at time t
minus the natural log of our
original concentration of A,
on the other side
minus kt, our time.
So there are two ways we
can rewrite this equation.
We can write it as the
equation for a straight line.
And all I did was take
this natural log of A0,
or A original over here.
And we could also
rearrange this term.
Instead of minus,
we have natural log
of the concentration
of A at time t
over our original concentration.
And now we can take the inverse
natural log of both sides.
And so we get rid of natural
log here, and on the other side
we have e to the minus kt,
again, the rate constant times
the time.
And then I can write
it-- break this out here,
moving the original
concentration of A
to this side.
And now this is the equation for
the integrated first-order rate
law, where you have
the concentration of A
at some point t equals
its original concentration
times e to the minus k, the
rate constant, times the time.
So if you know a rate constant
and how much time has elapsed,
and you know how much you
have of something originally,
you can figure out how
much you should have now.
Or if you figure out how the
concentration changes over time
and how much you
had originally, you
can calculate the rate constant
for that particular material.
And you can do those
calculations for rate constants
by plotting, using this
equation for a straight line.
So here I'm going to
plot natural log of A.
So I measured the
concentration of A
at various different
times against time.
And we have an equation
for a straight line.
So we should get a
straight line if this
is a first-order process.
And so what would
this be up here?
What is the y-intercept?
What is it?
AUDIENCE: Natural log of
the original concentration.
CATHERINE DRENNAN: Yeah.
So that's the natural log of
our original concentration.
And this you can yell more
loudly-- what is our slope?
AUDIENCE: Minus k.
CATHERINE DRENNAN: Yes.
It's a little bit
easier to yell loudly.
Minus k.
So you can
experimentally determine
the rate constant from this.
So if you measure
how the concentration
of A changes with
time, plot your data,
natural log of those
concentrations versus time,
get a lot of data
points, from the slope
you can measure
the rate constant.
And so rate constants for a
lot of different materials
have already been measured
in this kind of way.
All right.
So for a first-order process,
these are important equations.
But also for first
order, we spend
a lot of time talking
about half-life.
So half-life is
the time it takes
for half of the original
material to go away.
So it's a really easy thing.
I like it when things
are called what they are.
And here the time involved has
a little special abbreviation,
t 1/2.
So whenever you
see t 1/2, that's
talking about a half-life.
So we can derive this
expression as well
for the first-order half-life
using the expressions that we
just looked at, so
from the expression
we had above, where
we had the natural log
of the concentration
of A at time
t over our original
concentration
equals minus the rate
constant k times time.
And now we can substitute in.
So we're not interested
in just any old time.
We're interested in
time 1/2, so we're going
to want to put a 1/2 in there.
And that is the time it
takes for the original amount
to go to half, so
to be divided by 2.
So we can put that in.
So now our At, our
concentration of A at time t,
is our original
concentration divided by 2,
and our t is t 1/2.
So you'll see that the A0
terms are going to cancel out.
And you're going to end
up with natural log of 1/2
equals minus k times
our half-life, t 1/2.
And so then we can put in
our value for the natural log
of 1/2 and get rid of all those
minus signs and rearrange it.
And so then our half-life
for this first-order process
is going to be equal
to 0.9631 divided
by k, the rate constant.
So you'll note a couple of
perhaps important things
about this expression.
And one of the
important things is
that half-life doesn't
depend on the concentration.
So the concentration
term has dropped out.
So what does half-life
depend on then
if it doesn't depend
on the concentration?
Again, this is for first order.
So it's going to depend on k.
That's all that's in there.
There's only one
thing that's in there.
There's a number, a
constant, and there's k.
So half-life depends on
k, this rate constant,
and the rate constant depends
on the material in question.
So different materials
will have different values
of rate constant.
And we just saw how you can
calculate a rate constant.
You can measure it and
get the slope of the line
and tells you about what k is.
All right.
So let's use this
expression now.
And why don't you tell
me for the same material,
which of these events
will take longer.
All right, 10 more seconds.
So it takes, in fact,
the same amount of time,
because the concentration
doesn't show up in there,
and it was the same material.
All right.
So before we move
away from this,
you can write same amount
of time in your notes.
Let's take a look at this
plot on the bottom of the page
and just think about
what's happening.
Because you can talk
about a first half-life
and a second half-life
and a third half-life.
So we can just
fill this right in.
So for a concentration at
your first half-life, how much
is left?
So what's this number?
The concentration is what?
Half.
Second half-life,
what do we got?
AUDIENCE: 0.25.
CATHERINE DRENNAN: 0.25.
Third?
AUDIENCE: 0.125.
CATHERINE DRENNAN: 0.125, right.
So this is pretty
easy to think about.
But sometimes when you have
data and you're looking at it,
you have to remember what the
sort of possibilities are.
OK.
So that's first-order
half-life, and that's
the end of this lecture.
But we're not really
moving away from the topic,
because now we're going
to talk about an example
of a first-order process,
which is radioactive decay.
