A rate law is a tool used to predict the rate of a reaction. So if we start with something generic, like A plus B turning into C plus D, the rate law for this process would take this form. The rate would be dependent on the concentrations of the reactants. So if we are going to calculate the rate of reaction, we would take K, which is a rate constant. Hopefully by now we see that the letter K is referring to a constant. And we would take K and multiply it by the concentrations of each of our reactants. So whenever you see these square brackets like this and like this, it refers to measuring the concentration of A and the concentration of B. As I said in the last video, we generally measure concentration in molarity. But we also have these exponents X and Y. So the rate is going to equal the rate constant times the concentration of A to some power times the concentration of B to some power. And these exponents for now have to be found experimentally. So let me show you how this works. I want to find the rate law for the following process. If I take hydrogen gas and combine it with chlorine gas, I'll get hydrogen chloride gas. So I know that the rate law will take this generic form. It equals K times the concentration of hydrogen to some power times the concentration of chlorine to some power. The first thing we want to do is find the exponents. And in order to do that, I need some data. We have to do this experimentally. We can use this table to find the exponents for our rate law. And then once we have our exponents, we can then figure out the rate constant. And we're going to take time to consider the units of the rate constant after we find it. And then finally, you'll see that for experiment number four, no initial rate is given. So once we complete the rate law, we can use the rate law to find that rate for experiment number four. Let's start with our generic rate law. I haven't written already, but let's start here. We're going to say that the rate equals the constant times the concentration of my first reactant, which is hydrogen, to some power times the concentration of chlorine to some power. And as I said, the exponents have to be found experimentally. Let's start with exponent x here, the exponent that goes with hydrogen. If I want to find out what's happening with hydrogen, I need to find two experiments where the chlorine is held constant. Well, if you look, you can see here the concentration of chlorine in experiments one and two are constant. Nothing's happening to the chlorine here. So the only thing that's affecting our rate of reaction here is the concentration of the hydrogen. If we look at our concentration of hydrogen, well, the hydrogen initially is 0.01 and at the end is 0.02. So my hydrogen has gone up by a factor of two. If I look at my rate, I see that my rate started at 2.1 times 10 to the negative four and has gone to 4.2 times 10 to the negative four. So my rate has also doubled. I've doubled my concentration and I've doubled my rate. If I'm looking specifically at my hydrogen, I can say that my rate is going to equal some constant times hydrogen to some power. And then in this case, we've kept the chlorine constant. So I can say that there is another constant, k2. Or sometimes you'll see this written as rate equals what's called a pseudo constant, k prime times the concentration of hydrogen to some power. Well, what have I done to the rate? I've taken the rate and I've doubled it. What have I done to the concentration? I've taken the concentration and doubled it. So you have to ask yourself, 2 to what power equals 2. In this case, the only exponent that works is B1. If I had an exponent of 2, if I doubled the concentration, then the rate would go up by 2 squared and you would see an increase of 4. So I know that the exponent for hydrogen, the x value in this rate law, equals 1. Let's go through the same process. To do this for the chlorine, we need to find two reactions where the hydrogen is held constant. So I see in reaction number 1 and reaction number 3, I have the same concentration of hydrogen. If I look at reaction number 1, my concentration of chlorine is 0.001 and in reaction number 3, my concentration chlorine is 0.002. So again, I've doubled my concentration. However, I've gone from initial rate of 2.1 times 10 to the negative 4 to 8.4 times 10 to the negative 4. So here, if I want to do rate equals a pseudo constant times the concentration of chlorine to some power, I increase my concentration of chlorine by 2 and I increase my rate by 4. So the only exponent that will do that is 2. So 2 squared will give me a rate increase of 4. So that tells me that my y exponent is 2. So I can take my rate law and say that my total rate law is going to equal my constant times the concentration of hydrogen to the first power, times the concentration of chlorine to the second power. That is what's meant when we say that the exponents have to be found experimentally. You have to look at a series of data, including their initial rates in order to figure out what the exponents are. Well, now that we have that, we can now find k. Now, k is a constant. So it doesn't actually matter which one of these reactions you use, because k will be the same for all of them. I'm going to take this data from reaction 1 and plug it in to that rate law that I've written. I'm going to say I know that if my concentration of hydrogen is 0.001, and my concentration of chlorine is 0.001, then my rate is going to be 2.1 times 10 to the negative 4. So I can say my rate, which is 2.1 times 10 to the negative 4, molarity per second, is going to equal this constant times my concentration of hydrogen, which we said was 0.001 molar. I am the concentration of chlorine, which is 0.001 molar squared. So with all of these known values, I can now solve for k. And when you do this, you get a value of k that's equal to 210,000, or 2.1 times 10 to the 5. But I also said we have to be careful with units. We want to end up with molarity per second. And you'll see that on the right side of the equation, I have molarity and molarity squared over here. I want to get this molarity cubed to equal molarity per second. And so how I'm going to do that is for my units of k, I'm going to introduce seconds, and they have to be on the bottom. And I already have three molarities on the right side, so I want to cancel out two of them. So if I do one over seconds times molarity squared, when I take that unit and multiply it by molarity, and then molarity squared again, I'll be left with molarity per second. Sometimes you will see this written as leaders squared over moles squared times seconds. It's the same thing. Remember, molarity is moles per liter. So the inverse of molarity is leaders over moles. So these units are the same. The problem is that k doesn't always have the same units. The units for k will depend on what exponents that you get in your rate law. Finally, coming to the end here. Now I have a full rate law with k. And here we can write that out. Rate equals 210,000, one over seconds times molarity squared times the concentration of hydrogen to the first power times the concentration of chlorine to the second power. That's my full rate law with the exponents, with the constant and the units of the constant. If I want to find the rate for experiment number four, all I have to do is plug in the values. My concentration of hydrogen, I have 0.0015, and my concentration of chlorine, I have 0.0018, and so we should be able to figure out what the rate is. My rate is going to equal 210,000, one over seconds times molarity squared times 0.0015 molar times 0.0018 molar squared. And so I'm getting a rate for reaction four equal to 0.00102 or 1.2 times 10 to the negative three. And then we can take a look at our units. I've got molarity and molarity squared here, so I could cancel out my molarity squared, cancel out my molarity, and then I'm going to have molarity divided by seconds, which are the correct units for rate. I would like to close this video by discussing the term reactant order. When we say what the order is of a reactant, we're talking about the exponent. So in our last example, the hydrogen was a first order reactant, and the chlorine is a second order reactant. Normally our reactant order is going to be whole numbers. First order, second order, we'll even see 0th order reactants. It is possible to have reaction orders that are fractions or decimals, but we'll stick to the whole numbers in our work. Thank you.