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theme of first order kinetics.
In our last lesson, we talked about the integrated
rate law for first order kinetics.
Now we are going to half about the concept of half-life.
You are going to employ the concept of half-life
and see the relationship between a rate constant
and the value for half-life
and be able to utilize those in calculations.
So what is half-life?
It is the amount of time, if you start with a
certain amount of reactant to drop that
down to half the amount you started with.
We have these symbols to represent half-life.
The half-life symbol is 't 1/2'
and in the statement above
in symbols the amount of time it takes
to get you concentration down to
half the amount you started with.
So lets derive the half-life formula.
This is our integrated first order rate law.
We are going to do some substitution.
What are we going to substitute? We are going to take
what this represents here
so we are ging to substitute in
half-life of 'A naught' there.
So this is going to give us, this equation
now of course the 'A naughts will cancel and we have
natural log of 1/2 is equal to minus
kt.
Once we bring the minus to the other sire
That actually gives us the natural log of two
equals kt.
Then we will
calculate the natural log
2, and that is .693.
So this equation right here
is our half-life equation
for first order kinetics.
Though, make sure as you commit
this to memory that you realize it only Though, make sure as you commit
this to memory that you realize it only
applies to first order kinetics.
It is helpful
to learn it from this stage
if you can derive it quickly.
It is helpful to look at it in this
derivation, to help you remember this equation.
Now for my class, at the University of Kentucky,
I require them to memorize
the equations that are boxed
there, so this is one I recommend
that you commit to memory.
So this is only for
first order kinetics.
It is independent of how much you start with.
It doesn't matter if you start with a lot
or a little, it is the amount of time
of however much you start with
and drop it to 1/2
it original value.
That is what that statement says.
It doesn't matter how much time it takes.
So lets look at this
representation here.
In the top sphere we see
16 molecules
and lets say each one of those
represents a tenth of a mole.
Then we could have
a representation of a certain amount.
As one time interval, whatever
the half-life is, whatever that interval is,
the amount of times it takes to drop it down
1/2 its value, is its half-life. the amount of times it takes to drop it down
1/2 its value, is its half-life.
What we see with first order kinetics
is that time interval is exactly the same
to drop it down to its next
half as much.
That time interval is exactly
the same, to drop it down to its next
half as much.
That is unique to first order kinetics.
Of course in the next time interval
will we have remaining?
There would only be one blue dot,
in my circle.
So lets think about this problem
and I will have you stop the video and ponder it.
I have a first order reaction
starting with an 8 molar solution so
there is its initial concentration.
The half-life
is a minute
so I want you to think
is going to remain after 3 minutes.
Did you say, 1 molar?
Well, if you did you would be correct.
What are we doing here if its
three minutes we have
three half-lives,
that have gone by.
In the first half-life it is
going to go from 8 to 4.
In the second half-life it is going to go
from 4 to 2.
The third half-life is going to go 2 to 1.
So we are going down in those,
as we go through those half-lives.
If we look at our picture
here we see that
we have said that these
time intervals are in minutes.
We are starting here with 8,
and going one minute down to 4.
Going the second minute down to 2.
Then the third minutes goes down to 1.
Lets do a calculation,
we have a first order reaction,
giving me the rate constant. We know we
can obtain a half-life from that rate
constant 0.693
over k which is
3.0
times 10 ^ -3 seconds to the ^ -1.
Now remember that is our unit
for first order kinetics
always for the rate constant. for first order kinetics
always for the rate constant.
We can obtain our half-life from that formula.
Let think about this, we want to know the time
required for the reaction to be 75% complete.
What happens after the first half-life?
First half-life it goes from
100% there to
50% there and
then it going to go from 50%
there to 25% there
cutting it in half again.
That would be 75% complete.
So this is asking how much time
is it going to take to go through 2
half-lives.
Well, if it we knew how much time it took to go
through one half, we would be set.
So what is one half-life?
When we do this calculation
we find it is 231 seconds.
So we will take 231 seconds
to go through that half-life. Now we really only know
it to 2 significant figures
so we could say it is 2.3 times
10 ^ 2 seconds.
That is one half-life,
two half-lifes would be twice that.
So 4.6
times 10 ^ 2
seconds, or around 460 seconds.
This problem is a little bit different
in that it does not have a nice
multiple of half-lives.
So it is 30% complete.
Well, if it is 30% complete
what we are going to want to do if we cannot
figure out by a certain number of half-lives
is we are going to want to go to our first
order kinetics problem.
Our integrated first law.
This is the really the first time we are seeing
this equation being used.
30% complete
is 70% still there.
So what we know is if we started with
all of it, that would be 100%
and can that not a concentration but this ratio
would be the same, it works for percentages as well.
If I started a 100%
if it 30% complete
70% is still there.
So we have the 70 over 100 [70/100]
or we have .7 .
We figured out, that this is the same
rate constant, so we know
it 3.0 x 10 ^ -3
seconds to the minus 1 [-1] here.
We can solve for
the time it takes for us to do that
so lets see here
obtain the natural log [ln] of .7
and that is a -0.357.
We will divide both sides by
-3.0 x 10 ^ -3
and that will give me
a t value [time]
of 119 seconds
we only know that to
two significant figures
so that would be 1.2 x 10 ^ 2 seconds.
Should it be let then the half-life?
The half-life was 230 seconds.
Yes, because
the half-life is to get o 50% complete.
This is only 30% complete
so that number makes sense.
I want to go back to the previous
problem for just a moment
because if you look at this problem
and you could not think
70% is two half-lives
you could work this problem in a similar
fashion as the one before
If you started with 100 and at
75% complete, there is a 25% there
and then we have a negative k
times t
and we could solve for t
and we would find that it would be
the same value, so you did not have
to notice that was two half-lives
but you know the half-life it is sometimes
very convenient to look for that.
That finishes
our learning objective. We have learned
between half-life and k
we can utilize that to go between the two.
Then we can use it
in a calculation.