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- [Instructor] Let's say we
have a hypothetical reaction

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where reactant A turns into products

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and that the reaction is
first-order with respect to A.

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If the reaction is first-order
with respect to reactant A,

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for the rate law we can write
the rate of the reaction

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is equal to the rate constant K

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times the concentration
of A to the first power.

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We can also write that
the rate of the reaction

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is equal to the negative of
the change in the concentration

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of A over the change in time.

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By setting both of these
equal to each other,

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and by doing some calculus,

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including the concept of integration,

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we arrive at the integrated rate law

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for a first-order reaction,

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which says that the natural
log of the concentration of A

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at some time T, is equal to negative KT,

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where K is the rate constant

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plus the natural log of the
initial concentration of A.

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Notice how the integrated rate law

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has the form of Y is equal to mx plus b,

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which is the equation for a straight line.

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So if we were to graph the
natural log of the concentration

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of A on the Y axis, so let's
go ahead and put that in here,

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the natural log of the concentration of A,

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and on the X axis we put the time,

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we would get a straight line

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and the slope of that straight line

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would be equal to negative K.

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So the slope of this line,

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the slope would be equal to the negative

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of the rate constant K,

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and the Y intercept would
be equal to the natural log

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of the initial concentration of A.

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So right where this line meets the Y axis,

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that point is equal to the natural log

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of the initial concentration of A.

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The conversion of methyl
isonitrile to acetonitrile

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is a first-order reaction.

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And these two molecules
are isomers of each other.

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Let's use the data that's
provided to us in this data table

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to show that this conversion
is a first-order reaction.

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Since the coefficient in
front of methyl isonitrile

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is a one, we can use this form
of the integrated rate law

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where the slope is equal to the negative

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of the rate constant K.

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If our balanced equation
had a two as a coefficient

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in front of our reactant, we
would have had to include 1/2

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as a stoichiometric coefficient.

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And when we set our two
rates equal to each other now

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and go through the calculus,
instead of getting negative KT,

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we have gotten negative two KT.

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However for our reaction we
don't have a coefficient of two.

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We have a coefficient of one and therefore

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we can use this form of
the integrated rate law.

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Also notice that this form
of the integrated rate law

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is in terms of the concentration of A

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but we don't have the
concentration of methyl isonitrile

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in our data table,

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we have the pressure of methyl isonitrile.

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But pressure is related to concentration

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from the ideal gas law,
so PV is equal to nRT.

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If we divide both sides by V,

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then we can see that pressure is equal to,

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n is moles and V is volumes,

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so moles divided by
volume would be molarity,

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so molarity times R times T.

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And therefore pressure
is directly proportional

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to concentration, and for a
gas it's easier to measure

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the pressure than to
get the concentration.

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And so you'll often see data

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for gases in terms of the pressure.

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Therefore, we can imagine this form of the

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integrated rate law as the
natural log of the pressure

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of our gas at time T
is equal to negative KT

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plus the natural log of the
initial pressure of the gas.

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Therefore, to show that this reaction

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is a first-order reaction we
need to graph the natural log

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of the pressure of methyl
isonitrile on the Y axis

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and time on the X axis.

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So we need a new column in our data table.

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We need to put in the natural log

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of the pressure of methyl isonitrile.

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So for example, when time is equal to zero

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the pressure of methyl
isonitrile is 502 torrs.

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So we need to take the natural log of 502.

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And the natural log of
502 is equal to 6.219.

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To save time, I've gone ahead

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and filled in this last column here,

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the natural log of the
pressure methyl isonitrile.

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Notice what happens as
time increases, right,

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as time increases the
pressure of methyl isonitrile

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decreases since it's being
turned into acetonitrile.

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So for our graph, we're
gonna have the natural log

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of the pressure of methyl
isonitrile on the y-axis.

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And we're gonna have time on the X axis.

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So notice our first point here

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when time is equal to zero seconds,

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the natural log of the
pressure as equal to 6.219.

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So let's go down and
let's look at the graph.

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All right, so I've
already graphed it here.

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And we just saw when time
is equal to zero seconds,

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the first point is equal to 6.219.

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And here I have the other data
points already on the graph.

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Here's the integrated rate
law for a first-order reaction

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and I put pressures in there
instead of concentrations.

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And so we have the natural
log of the pressure

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of methyl isonitrile on the y-axis

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and we have time on the X axis,

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and the slope of this line should be equal

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to the negative of the rate constant K.

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So there are many ways to
find the slope of this line,

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one way would be to use
a graphing calculator.

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So I used a graphing calculator

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and I put in the data from the data table

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and I found that the slope of this line

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is equal to negative 2.08

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times 10 to the negative fourth.

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And since if I go ahead and
write y is equal to mx plus b,

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I need to remember to take
the negative of that slope

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to find the rate constant K.

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Therefore K is equal to positive 2.08

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times 10 to the negative fourth.

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To get the units for the rate constant,

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we can remember that slope is equal to

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change in Y over change in X.

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So change in Y would be the
natural log of the pressure,

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which has no unit, and X
the unit is in seconds.

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So we would have one over
seconds for the units for K.

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And finally, since we got a straight line

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when we graphed the natural log
of the pressure versus time,

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we know that this data is
for a first-order reaction.

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And therefore we've proved
that the transformation

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of methyl isonitrile to acetonitrile

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is a first-order reaction.