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, right, so we have one reactant, A, turning into one product, B.
Now let's say at time is equal to zero, we're starting with an initial concentration of A of one molar. And A hasn't turned into B yet, so at time is equal to zero, the concentration of B is zero. 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 0.98 molar. So we lost 0.02 molar for the concentration of A.
So that turns into, since A turns into B, after two seconds, the concentration of B is 0.02 molar, right, because A turned into B. So this is our concentration of B after two seconds. If we want to know the average rate of reaction here, we can plug into our definition for rate of reaction. change in concentration, let's do 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.02, so we write in here.02, and from that we subtract the initial concentration of our product, which is zero. So.02 minus zero, that's all over the change in time.
So we're done with the first time. That's the final time minus the initial time. So that's two minus zero. So the rate of reaction, the average rate of reaction would be equal to.02 divided by two, which is.01 molar per second.
So that's our average rate of reaction. from time is equal to zero to time is equal to two seconds. We could do the same thing for A.
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. of A. So the rate would be equal to, alright, the change in the concentration of A, that's the final concentration of A, which is.98, minus the initial concentration of A. the initial concentration of A is one.
So 0.98 minus one, and this is all over the final time minus the initial times. This is over two minus zero. Now this would give us a negative 0.02, negative 0.02 here, over two.
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.02 divided by two, which of course is.01. molar per second, so we get a positive value for the rate of reaction. Alright, 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. Alright, 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, right? 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 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, 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 nine times 10 to the negative six molar per second. 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 3.6 times 10 to the negative five 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 two, and since we're losing dinitrogen pentoxide, we put a negative sign here. So this gives us negative 1.8 times 10 to the negative five.
molar per second. So dinitrogen pentoxide disappears at twice the rate that oxygen appears. Alright, 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 nine times 10 to the negative six 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. Alright, 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 times 10 to the negative five. Alright, so that's 3.6 times 10 to the negative five.
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 1 fourth. So 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 1 fourth.
Finally, let's think about dinitrogen pentoxide. Alright, so we said that that was disappearing at negative 1.8 times 10 to the negative five. So once again, what do I need to multiply this number by in order to get nine times 10 to the negative six?
Well, this number, right, in terms of magnitude was twice this number, so I need to multiply by 1 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. Alright, so that would give me, that gives me nine times 10 to the negative six. 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 1 here as well.
Alright, so now that we've figured out how to express our rate, we can look at our balanced equation. So over here we had a two for dinitrogen pentoxide, and notice where the two goes here for expressing our rate. For nitrogen dioxide, we had a four for our coefficient.
So the four goes in here. And for oxygen, over here, let's use green, we had a one. So I could have written one over one just to show you the pattern of how to express your rate.