Now that we understand how to write rate laws, let's apply this to a reaction. So here we have the reaction of nitric oxide, which is NO, and hydrogen to give us nitrogen and water at 1280 degrees C. In part A, our goal is to determine the rate law.
And from the last video, we know that the rate of the reaction is equal to K, which is the rate constant, times the concentration of nitric oxide. So nitric oxide is one of our reactants. So the rate of the reaction is proportional to the concentration of nitric oxide to some power, x.
We don't know what x is yet. We also know the rate of the reaction is proportional to the concentration of our other reactant, which is hydrogen. So we put hydrogen in here, but we don't know what the power is. So we put a y for now. We can figure out what x and y are by looking at the data in our experiments.
So let's say we wanted to first figure out what x is. So to figure out what x is, we need to know how the concentration of nitric oxide affects the rate of our reaction. So we're gonna look at experiments one and two here. And the reason why we chose those two experiments is because the concentration of hydrogen is constant in those two experiments. The concentration of hydrogen is.002 molar in both.
So if we look at what we did to the concentration of nitric oxide, we went from a concentration of.005 to a concentration of.010. We increased the concentration of nitric oxide. We increased the concentration of nitric oxide by a factor of two.
We doubled the concentration. What happened to the initial rate of reaction? Well, the rate went from 1.25 times 10 to the negative five.
to five times 10 to the negative five. So the rate increased by a factor of four. We increased the rate by a factor of four.
And if you have trouble doing that math in your head, you could just use a calculator and say five times 10 to the negative five, and if you divide that by 1.25 times 10 to the negative five, this would be four over one, or four. So this rate is four times this rate up here. Now we know enough to figure out the order for nitric oxide, because remember from the previous video what we did is we said two to the x, two to the x is equal to four.
So over here, two to the x is equal to four. Obviously x is equal to two. Two squared is equal to four. equal to four. So we can go ahead and put that in for our rate loss.
And now we know the rate is equal to k times the concentration of nitric oxide. This would be to the second power. So the reaction is second order in nitric oxide.
Next, let's figure out the order with respect to hydrogen. So this time we want to choose two experiments where the concentration of nitric oxide is constant. So that would be experiments two and three, where we can see the concentration of nitric oxide has not changed. It's.01 molar for both of those experiments.
But the concentration of hydrogen has changed. It goes from.002 to.004. 2.004, so we've increased the concentration of hydrogen by a factor of two. And what happened to the rate of reaction?
We went from five times 10 to the negative five to one times 10 to the negative four. So we've doubled the rate. The rate has increased by a factor of two.
And sometimes the exponents bother students, right? How is this doubling the rate? Well, once again, if you can't do that in your head, you could take out your calculator and take 1 times 10 to the negative 4 and divide that by 5 times 10 to the negative 5. And you'll see that's twice that. So the rate goes up by a factor of 2. So now we have 2 to what power is equal to 2?
So 2 to the y is equal to 2. Obviously, y is equal to 1. 2 to the to the first power is equal to two. So now we know that our reaction is first order in hydrogen. So we can go ahead and put that in here.
So we can put in hydrogen, and we know that it's first order in hydrogen. So we've now determined our rate law. In part B, they want us to find the overall order of the reaction, and that's pretty easy to do because we've already determined the rate law in part A. We know that the reaction is second order in nitric oxide and first order in hydrogen.
So to find the overall order, all we have to do is add our exponents. Two plus one is equal to three, so the overall order of the reaction is three. Let's compare our exponents to the coefficients in our balanced equation for a minute here.
So it's very tempting for students to say, oh, we have a two here for our coefficient for nitric oxide. Is that why we have a two down here for the exponent and the rate law? But if you look at hydrogen, hydrogen has a coefficient of two, and we determined that the exponent was a one down here in the rate law.
So you can't just take your coefficients in your balanced chemical equation and put them in for your exponents in your rate law. Alright, so you need to look at your experimental data to determine what your exponents are in your rate law. And later we'll get more into mechanisms and we'll talk about that a little bit more. Alright, let's move on to part C. In part C they want us to find or calculate the rate constant K.
Well we could calculate the rate constant k by using the rate law that we determined in part a and by choosing one of the experiments and plugging in the numbers into the rate law. So it doesn't matter which experiment you choose. You could choose one, two, or three. I'm just gonna choose one here, so experiment one.
And we're gonna plug all of our information into the rate law that we just determined. For example, in our rate law we have the rate of reaction. over here?
Well for experiment one, the initial rate of reaction was 1.25 times 10 to the negative five. And it was molar per second. So we're gonna plug this in, we're gonna plug this in to our rate law. So let's go down here and plug that value in.
1.25, 1.25 times 10 to the negative five. And this was molar per second. Alright, so that takes care of the rate of the reaction, next we have that equal to the rate constant K, so we're trying to solve for K, times the concentration of nitric oxide, times the concentration of nitric oxide squared.
So let's go back up here and find the concentration of nitric oxide in the first experiment. The concentration is.005 molar. Alright, so we're gonna plug in.005 molar in here, so we have.005 molar.
0.005 molar. And next, we're gonna multiply that by the concentration of hydrogen, right? The concentration of hydrogen to the first power. So we go back up to experiment one, and we find the concentration of hydrogen, which is 0.002 molar. So we plug that in.
So we have 0.002 molar. 0.002 molar. Next, all we have to do is is solve for k. So let's go ahead and do that. So let's get out the calculator here.
And we could say.005 squared gives us 2.5 times 10 to the negative five. We need to multiply that by.002. So on the right side, we would have five times 10 to the negative eight. So we have five times 10 to the negative eight. Think about your units.
This would be molar, squared times molar over here. So molar squared times molar, and all of this times... our rate constant k is equal to 1.25 times 10 to the negative five molar per second. Let's go ahead and find the number first and then we'll worry about our units here. So to find what k is, we just need to take 1.25 times 10 to the negative five and if we divide that, if we divide that by five times 10 to the negative eight, then we get that k is equal to 250. Alright, so k is equal to 250. We can go ahead and put that in here.
K is 250. What would the units be? Well, we have molar on the left, we have molar on the right, so we could cancel one of those molars out. And so on the left we have one over seconds, and on the right we have molar squared.
So we divide both sides by molar squared, and so we get for our units, this would be one over molar squared, this would be one over molar squared times seconds. And so we've found the rate constant for our reaction. And notice this was for a specific temperature, right? So our reaction was at 1280 degrees C. This is the rate constant at 1280 degrees C.
Finally, let's do part D. So what is the rate of the reaction when the concentration of nitric oxide is.012 molar and the concentration of hydrogen is.007? six molar? Well, we can use our rate law.
So our rate law is equal to what we found in A. Our rate law is equal to K times the concentration of nitric oxide squared times the concentration of hydrogen to the first power. So our goal is to find the rate of the reaction. So we're solving for R here, and we know what K is now. K is 250, one over molar squared times seconds.
So we have that. The concentration of nitric oxide is.012. So we have.012.
This would be molar, and we need to square that. And then, so all I did was take this, I took this and I plugged it into here. And now we're gonna take the concentration of hydrogen, which is.006 molar, and plug that into here. So let's go ahead and do that. So that would be times.006 molar.
Let me go ahead and put in the molar there. So.006 molar to the first power. And we solve for our rate.
So the rate is equal to, let's do the numbers first. So we have 0.012 squared, 0.012 squared, and we're going to multiply that by point, so times 0.006, and then we also need to multiply that by our rate constant, k, so times 250. This gives us our answer of 2.16 times 10 to the negative 4. Let's round that to 2.2. We have 2.2, 2.2 times, times 10 to the negative four. And in terms of our units, if we think about what happens to the units here, we would have molarity squared, right here, molarity squared, molarity squared, so we end up with molar per second, which we know is our units for the rate of reaction, so molar per second. So we found the rate of our reaction.