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## Activation energy and reaction rate

# The Arrhenius equation

AP.Chem:

TRA‑4 (EU)

, TRA‑4.C (LO)

, TRA‑4.C.4 (EK)

## Video transcript

- In the last video, we
talked about collision theory, and we said that molecules
must collide to react, and we also said those
collisions must have the correct orientation in space to
be effective collisions, and finally, those collisions
must have enough energy for the reaction to occur. And these ideas of collision theory are contained in the Arrhenius equation. So down here is our equation, where k is our rate constant. So k is the rate constant, the one we talk about in our rate laws. A is called the frequency factor. So, A is the frequency factor. Also called the pre-exponential factor, and A includes things like the frequency of our collisions, and also the orientation
of those collisions. And then over here on the right, this e to the negative Ea over RT, this is talking about the
fraction of collisions with enough energy for
a reaction to occur. So we symbolize this by lowercase f. So the fraction of collisions with enough energy for
the reaction to occur. f depends on the activation energy, Ea, which needs to be in joules per mole. R is the gas constant, and T is the temperature in Kelvin. So let's see how changing
the activation energy or changing the
temperature for a reaction, we'll see how that affects the fraction of collisions
with enough energy for our reaction to occur. So, let's start with an activation energy of 40 kJ/mol, and the temperature is 373 K. So, let's solve for f. So, f is equal to e to the negative of our activation energy in joules per mole. So we need to convert
40 kilojoules per mole into joules per mole, so that would be 40,000. So, 40,000 joules per mole. All right, this is over
our gas constant, R, and R is equal to 8.314 joules over K times moles. All right, and then this is going to be multiplied by the temperature, which is 373 Kelvin. So, 373 K. So let's go ahead and do this calculation, and see what we get. So, let's take out the calculator. e, e to the, we have -40,000, one, two, three divided by 8.314 times 373. So, we get 2.5 times 10 to the -6. So this is equal to 2.5 times 10 to the -6. So what does this mean? All right, well, let's say we
had one millions collisions. All right, so 1,000,000 collisions. What number divided by 1,000,000, is equal to 2.5 x 10 to the -6? So this number is 2.5. 2.5 divided by 1,000,000 is equal to 2.5 x 10 to the -6. So what this means is for every one million
collisions in our reaction, only 2.5 collisions have
enough energy to react. So obviously that's an
extremely small number of collisions with enough energy. All right, let's see what happens when we change the activation energy. So we're going to change
the activation energy from 40 kilojoules per mole to 10 kilojoules per mole. So, we're decreasing
the activation energy. We're keeping the temperature the same. So let's see how that affects f. So let's plug in this time for f. So f is equal to e to the now we would have -10,000. So we've changed our activation energy, and we're going to divide that by 8.314 times 373. So let's do this calculation. So now we have e to the - 10,000 divided by 8.314 times 373. And here we get .04. So this is equal to .04. So .04. Notice what we've done, we've increased f. We've gone from f equal
to 2.5 times 10 to the -6, to .04. So let's stick with this same idea of one million collisions. So let's say, once again, if we had one million collisions here. So 1,000,000 collisions. What number divided by 1,000,000 is equal to .04? So that number would be 40,000. 40,000 divided by 1,000,000 is equal to .04. So for every one million collisions that we have in our reaction this time 40,000 collisions have enough energy to react, and so that's a huge increase. Right, it's a huge increase in f. It's a huge increase in
the number of collisions with enough energy to react, and we did that by decreasing
the activation energy. So decreasing the activation energy increased the value for f. It increased the number
of effective collisions. All right, let's do one more calculation. This time we're gonna
change the temperature. So let's keep the same activation energy as the one we just did. So 10 kilojoules per mole. So 10 kilojoules per mole. This time, let's change the temperature. Here we had 373, let's increase
the temperature to 473, and see how that affects the value for f. So f is equal to e to the negative this would be 10,000 again. e to the -10,000 divided by 8.314 times, this time it would 473. So times 473. So let's do this calculation. So e to the -10,000 divided by 8.314 times 473, this time. So we get, let's just say that's .08. So I'll round up to .08 here. So this is equal to .08. So we've increased the value for f, right, we went from .04 to .08, and let's keep our idea
of one million collisions. Right, so it's a little bit easier to understand what this means. So what number divided by 1,000,000 is equal to .08. That must be 80,000. Right, so this must be 80,000. So for every 1,000,000 collisions that we have in our reaction, now we have 80,000 collisions with enough energy to react. So we've increased the temperature. Gone from 373 to 473. We increased the number of collisions with enough energy to react. We increased the value for f. Finally, let's think
about what these things do to the rate constant. So we go back up here to our equation, right, and we've been talking about, well we talked about f. So we've made different
calculations over here for f, and we said that to increase f, right, we could either decrease
the activation energy, or we could increase the temperature. So decreasing the activation energy increased the value for f, and so did increasing the temperature, and if we increase f, we're going to increase k. So if we increase f, we
increase the rate constant, and remember from our rate laws, right, R, the rate of our reaction is equal to our rate constant k, times the concentration of, you know, whatever we are working
with for our reaction. Here I just want to remind you that when you write your rate laws, you see that rate of the reaction is directly proportional
to the rate constant k. So if you increase the rate constant k, you're going to increase
the rate of your reaction, and so over here, that's what
we've been talking about. If we decrease the activation energy, or if we increase the temperature, we increase the fraction of collisions with enough energy to occur, therefore we increase the rate constant k, and since k is directly proportional to the rate of our reaction, we increase the rate of reaction. And this just makes logical sense, right? We know from experience that if we increase the
temperature of a reaction, we increase the rate of that reaction. So, once again, the
ideas of collision theory are contained in the Arrhenius equation, and so we'll go more into this equation in the next few videos.

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