In this experiment we are going to be determining the rate law of the reaction is giving at the top. A plus B, yields 2C. Substance D is included as a catalyst and we see that data in the table as including D. So, D could be part of the rate law and we need to find out if it is. So, the rate law will have the form, Oops. [LAUGH] D D raised to to the Z power. So there is our rate law. Now we need to choose experiment to figure out the x, the y, and the z. We're going to begin by looking at x. x is the order of the substance A. So we need to choose experiments in which B and D are held constant. Now experiment one, three hold B and D constant. So I want to choose experiments one and three. For those two experiment,s I want to take a ratio of rates for those two experiments and that would be three times to the minus 2 molarity per second divided by 1.5 times 10 to the minus 2. That's going to give me a ratio of 2 and what that is telling me is between those two experiments, the rate is doubling. Now I chose to do it in order of the large over the small, so it was experiment 3 over experiment 1. Want to make sure I keep that same ratio, as I do a ratio of concentration of a. For experiment three, the value is 0.6 molar. For experiment one, it's 0.3 molar. And that gives me a two as well. So in those two experiment I double the a's concentration and that doubled the rate. So 2 to what power gives you a 2? Concentration of what power is going to give me a doubling of rate. Well, x would have to be equal to 1 and we now know it's first order in A. Next we want to figure out why. To figure out why we need to find two experiments in which the other two substances are being held constant. Now, why is this associated with this substances B, right? So, we need to find the experiments in which A and D are being held constant. Let's look over here. A and D well in experiment number one and experiment number two A is held constant and D is held constant so those are the two that we're going to choose to determine the power y. I've got an empty space over here for me to work on for y I'm going to look at experiments, one, and two, I, again, am going to do a ratio of rates for those two experiments. Let's look at the rates. Well, they're both 1.5, for the two experiments. See, 1.5, 1.5. So, that's easy enough, ratio rates at 1.5, times to the minus two 1.5 times 10 minus 2, that's a value of 1. We're also going to do a ratio of the concentration of B, that's what we're trying to find a power for is B. So we come over here and we see that B is going from, 0.1 to 0.05 or the other way round really won't matter but I would larger over small. 0.1, 0.05 and we see we have a double in. So we say to ourselves, the concentration raised to what power will give me a rate change of 1. Well, Y would have to be equal to 0. Any number raised to the 0 power gives me 1. Now know that y is 0 and that B's concentration has no affect on the rate. So now we're ready to look at Z. Now for Z we're going to need to look at A and B and figure out experiments in which A and B are being held constant. Now let's see here looks like experiment number one In experiment number no, A and No, no, sorry, A and B. Experiment 1 definitely with experiment 4, A is not changing. In experiment 1, experiment 4, B is not changing. So those two experiments that I'm going to want to use. We're going to do a ratio of rates. So for z, we're looking at experiments One and Four, ratio of rates. I'm going to choose large over small 3.0 x 10 to the -2 /1.5 x 10 to the -2 And then I'm going to do a ratio of concentrations. Ratio of a concentration or D. So let's look at those two experiments. I did 4/1 for that, so I need to do 4/1 for the D and that is going to give me the values of 0.10 over 0.05 and that is 2. So again, we say, the concentration raised to what power gives me this rate and Z would have to be 1. So now we know the rate law. Rate is equal to K. It was A to the first power. We won't include B, because it's to the 0th power and d to the first power. So that's the rate law. Let's see what else is asked. Determine the rate law, we've done that. What is the order of the reaction? Well the order of the reaction, the overall order would be 1 plus 1. We don't see those written in, but we know they're there. So it's an overall order of 2, and we'd say it's second order overall. Lastly, it wants us to determine the value of the rate constant K. Go take this experiment and I'm going to solve it for k. Rate divided by a times d, would give me k. The rate I can choose anyone of these experiments. So I want to use experiment one. The rate is 1.5 times 10 to the minus 2. 1.5 x 10 to the -2. Make sure you include the units because you not only want to know the value of k but you want to know the unit for k. In experiment one, A was 0.30 and D was 0.05. So we'll put 0.30, and this is in concentration in molarity, and d is 0.05. Now if I take those numbers I will come up with a K of 1 and the molarity on top will cancel with the one of these in bottom and that will leave me with 1/M. So the right constant value is one and the unit is one over molarity per second, otherwise written, sometimes written as molarity to the minus one seconds to the minus one. This unit that we see for K is the unit for K every time you have a second or reaction You always have one over molarity times seconds as your units.