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## Relationship between reaction concentrations and time

# Second-order reactions

AP.Chem:

TRA‑3 (EU)

, TRA‑3.C (LO)

, TRA‑3.C.1 (EK)

, TRA‑3.C.3 (EK)

, TRA‑3.C.4 (EK)

## Video transcript

- [Instructor] Let's say we have a hypothetical reaction where reactant A turns into products. And let's say the reaction is second order with respect to A. If the reaction is second
order with respect to A, then we can write the rate of the reaction is equal to the rate constant
k times the concentration of A to the second power since this is a second order reaction. We can also write that
the rate of the reaction is equal to the negative of the change in the concentration of A
over the change in time. If we set these two
ways of writing the rate of reaction equal to each other, and use of calculus, including the concept of integration, we will arrive at the integrated rate law for a second order reaction. The integrated rate law
for a second order reaction says that one over the concentration of reactant A at some time t, is equal to the rate
constant k times the time plus one over the initial
concentration of A. Notice how the integrated rate law has the form of y is equal to mx plus b, which is the equation for a straight line. So if we graph one over
the concentration of A on the y axis, and time on the x axis, so let's go ahead and put that in here, one over the concentration
of A on the y axis, and time on the x axis, we
will get a straight line and the slope of that line is equal to the rate constant k. So slope is equal to K and the y intercept is equal to one over the
initial concentration of A. So the point where our
line intersects the y axis is equal to one over the
initial concentration of A. Let's look at an example
of a second order reaction. C5H6 is cyclopentadiene, and two molecules of cyclopentadiene will
react with each other to form dicyclopentadiene. Our goal is to use the
data from this data table to prove that this
reaction is second order. However, we have to be careful because in our balanced equation, we have a two as a coefficient in front of cyclopentadiene. Going back to our hypothetical reaction where reactant A turned into products, there's also one as a
coefficient in front of the A. And if there's a one as a
coefficient in front of the A, we can use this form of
the integrated rate law for a second order reaction. However, for our problem, we have a two as a coefficient in
front of cyclopentadiene. And that means we need to have a stoichiometric
coefficient of 1/2 in here, and that changes the math. Now when we set these two
rates of reactions equal to each other, and we use
calculus and we integrate to get our integrated rate law, because of this one half, we end up with a two in front of the K. So thinking about y is equal to mx plus b, now the slope of the
line is equal to two K. So now for our reaction, we can write our integrated rate law as one over the concentration
of cyclopentadiene at some time t, is
equal to two kt plus one over the initial concentration
of cyclopentadiene. So if we look at our data table, we have time in seconds and
we have the concentration of cyclopentadiene,
but we need to have one over the concentration of cyclopentadiene so we need a new column. So we're going to calculate
one over the concentration of cyclopentadiene. So if the concentration of cyclopentadiene when time is equal to zero
seconds is 0.0400 molar, if we take one divided by 0.0400, we would get 25.0. To save some time I've gone ahead and filled in the rest of this column. Notice as time increases so
as we go from zero seconds to 50 to 100 to 150 to 200, the concentration of
cyclopentadiene decreases because cyclopentadiene is
turning into dicyclopentadiene. Next, we need to graph our data. So we're gonna have one
over the concentration of cyclopentadiene on the y axis and time on the x axis. Our first point, so when
time is equal to zero, one over the concentration
of cyclopentadiene is 25.0. So if we go down to our graph, we can see that when
time is equal to zero, our first point here is 25.0. And plotting the other points
gives us a straight line. Next, we need to find the
slope of this straight line. And there are many ways to do that. One way to do it is to
use a graphing calculator. And when I used a graphing calculator to find the slope of this line, I found that the slope is equal to 0.1634. Thinking about y is equal to mx plus b, our slope should be equal to two k. So to find the rate constant k, we need to divide the slope by two, which gives us 0.0817. To find the units for
K, remember that slope is equal to change in y over change in x, and on our Y axis, our
units are one over molar, and the x axis the units are seconds. So therefore, we can
write the rate constant k is equal to 0.0817. It's to be one over
molar divided by seconds which is the same thing as
one over molar times seconds. It's important to point
out that most textbooks don't cover how the two
as a coefficient changes the integrated rate law. And so a lot of textbooks will simply say that the slope of the line for the second order integrated rate law, is equal to K. So a lot of books would
just say the final answer for the rate constant is 0.163. So you'll see you'll see
a lot of textbooks say that the rate constant would be 0.163, one over molar times seconds. However, since the coefficient in front of cyclopentadiene is a two, technically this rate
constant is the correct one. Finally, since we got a straight line, when we graphed one over the concentration of cyclopentadiene versus time, we know that this reaction
has second order kinetics.