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AP Chem: TRA‑3 (EU), TRA‑3.A (LO), TRA‑3.A.1 (EK), TRA‑3.A.2 (EK)

- The rate of a chemical reaction is defined as the change
in the concentration of a reactant or a product over the change in time, and concentration is in
moles per liter, or molar, and time is in seconds. So we express the rate
of a chemical reaction in molar per second. Molar per second sounds a lot like meters per second, and that, if you remember your physics is our unit for velocity. So, average velocity is equal to the change in x over the change in time, and so thinking about average velocity helps you understand the definition for rate
of reaction in chemistry. If we look at this applied to a very, very simple reaction. So we have one reactant, A, turning into one product, B. Now, let's say at time is equal to 0 we're starting with an
initial concentration of A of 1.00 M, and A hasn't turned into B yet. So at time is equal to 0, the concentration of B is 0.0. Let's say we wait two seconds. So, we wait two seconds, and then we measure
the concentration of A. Obviously the concentration of A is going to go down because A is turning into B. Let's say the concentration of A turns out to be .98 M. So we lost .02 M for
the concentration of A. So that turns into, since A turns into B after two seconds, the concentration of B is .02 M. Right, because A turned into B. So this is our concentration
of B after two seconds. If I want to know the average
rate of reaction here, we could plug into our definition for rate of reaction. Change in concentration, let's do a change in
concentration of our product, over the change in time. So, the Rate is equal to the change in the concentration of our product, that's final concentration
minus initial concentration. So the final concentration is 0.02. So, we write in here 0.02, and from that we subtract
the initial concentration of our product, which is 0.0. So, 0.02 - 0.0, that's all over the change in time. That's the final time
minus the initial time, so that's 2 - 0. So the rate of reaction, the average rate of reaction, would be equal to 0.02 divided by 2, which is 0.01 molar per second. So that's our average rate of reaction from time is equal to 0 to time is equal to 2 seconds. We could do the same thing for A, right, so we could, instead of defining our rate of reaction as the appearance of B, we could define our rate of reaction as the disappearance of A. So the rate would be equal to, right, the change in the concentration of A, that's the final concentration of A, which is 0.98 minus the initial concentration of A, and the initial
concentration of A is 1.00. So 0.98 - 1.00, and this is all over the final
time minus the initial time, so this is over 2 - 0. Now this would give us -0.02. - 0.02 here, over 2, and that would give us a
negative rate of reaction, but in chemistry, the rate
of reaction is defined as a positive quantity. So we need a negative sign. We need to put a negative sign in here because a negative sign gives us a positive value for the rate. So, now we get 0.02 divided by 2, which of course is 0.01 molar per second. So we get a positive value
for the rate of reaction. All right, so we calculated
the average rate of reaction using the disappearance of A and the formation of B, and we could make this a
little bit more general. We could say that our rate is equal to, this would be the change
in the concentration of A over the change in time, but we need to make sure to
put in our negative sign. We put in our negative sign to give us a positive value for the rate. So the rate is equal to the negative change in the concentration of A over the change of time, and that's equal to, right, the change in the concentration of B over the change in time, and we don't need a negative sign because we already saw in
the calculation, right, we get a positive value for the rate. So, here's two different ways to express the rate of our reaction. So here, I just wrote it in a
little bit more general terms. Let's look at a more complicated reaction. Here, we have the balanced equation for the decomposition
of dinitrogen pentoxide into nitrogen dioxide and oxygen. And let's say that oxygen forms at a rate of 9 x 10 to the -6 M/s. So what is the rate of formation of nitrogen dioxide? Well, if you look at
the balanced equation, for every one mole of oxygen that forms four moles of nitrogen dioxide form. So we just need to multiply the rate of formation of oxygen by four, and so that gives us, that gives us 3.6 x 10 to the -5 Molar per second. So, NO2 forms at four times the rate of O2. What about dinitrogen pentoxide? So, N2O5. Look at your mole ratios. For every one mole of oxygen that forms we're losing two moles
of dinitrogen pentoxide. So if we're starting with the rate of formation of oxygen, because our mole ratio is one to two here, we need to multiply this by 2, and since we're losing
dinitrogen pentoxide, we put a negative sign here. So this gives us - 1.8 x 10 to the -5 molar per second. So, dinitrogen pentoxide disappears at twice the rate that oxygen appears. All right, let's think about
the rate of our reaction. So the rate of our reaction is equal to, well, we could just say it's equal to the appearance of oxygen, right. We could say it's equal to 9.0 x 10 to the -6 molar per second, so we could write that down here. The rate is equal to the change in the concentration of oxygen over the change in time. All right, what about if
we wanted to express this in terms of the formation
of nitrogen dioxide. Well, the formation of nitrogen dioxide was 3.6 x 10 to the -5. All right, so that's 3.6 x 10 to the -5. So you need to think to yourself, what do I need to multiply this number by in order to get this number? Since this number is four
times the number on the left, I need to multiply by one fourth. Right, so down here, down here if we're
talking about the change in the concentration of nitrogen dioxide over the change in time, to get the rate to be the same, we'd have to multiply this by one fourth. All right, finally, let's think about, let's think about dinitrogen pentoxide. So, we said that that was disappearing at -1.8 x 10 to the -5. So once again, what do I need to multiply this number by in order to get 9.0 x 10 to the -6? Well, this number, right, in terms of magnitude was twice this number so I need to multiply it by one half. I need to get rid of the negative sign because rates of reaction are defined as a positive quantity. So I need a negative here. So that would give me, right, that gives me 9.0 x 10 to the -6. So for, I could express my rate, if I want to express my rate in terms of the disappearance
of dinitrogen pentoxide, I'd write the change in N2, this would be the change in N2O5 over the change in time, and I need to put a negative
one half here as well. All right, so now that we figured out how to express our rate, we can look at our balanced equation. So, over here we had a 2
for dinitrogen pentoxide, and notice where the 2 goes here for expressing our rate. For nitrogen dioxide, right, we had a 4 for our coefficient. So, the 4 goes in here, and for oxygen, for oxygen over here, let's use green, we had a 1. So I could've written 1 over 1, just to show you the pattern of how to express your rate.

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