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Current time:0:00Total duration:12:28

AP.Chem:

TRA‑3 (EU)

, TRA‑3.B (LO)

, TRA‑3.B.2 (EK)

, TRA‑3.B.3 (EK)

, TRA‑3.B.4 (EK)

, TRA‑3.B.5 (EK)

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 n O and hydrogen to give us nitrogen and water at 1280 degrees C and 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 right 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 right 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 all right 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 right 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 going to look at experiments 1 & 2 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 point zero zero two molar in both so if we look at what we did to the concentration of nitric oxide right we went from a concentration of point zero zero five to a concentration of Oh point zero one zero we increased the concentration of nitric oxide we increase 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 one point to five times 10 to the negative 5 to 5 times 10 to the negative 5 so the rate increased by a factor of four right we increase the rate by a factor of four and if you have trouble doing that math in your head you could just use a calculator right and say 5 times 10 to the negative 5 and if you divide that by one point - five times ten to the negative five this would be 4 over 1 or 4 so this rate is 4 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 2 to the X 2 to the X is equal to 4 so over here 2 to the X is equal to 4 obviously X is equal to 2 2 squared is equal to 4 so we can go ahead and put that in for our rate law so now we know the rate is equal to K times the concentration of nitric oxide all right 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 2 & 3 where we can see the concentration of nitric oxide has not changed its point 0 1 molar for both of those experiments but the concentration of hydrogen has changed it goes from point zero zero to two point zero zero four so we've increased we've increased the concentration of hydrogen by a factor of two and what happened to the rate of reaction we went from 5 times 10 to the negative 5 to 1 times 10 to the negative 4 so we've doubled the rate the rate has increased by a factor of 2 and sometimes the exponents bother students right how is this 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 right 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 first power is equal to two so now we know that our reaction is first order in hydrogen's 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 2 plus 1 is equal to 3 so the overall order of the reaction is 3 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 2 here for our coefficient for nitric oxide is that why we have a 2 down here for the exponent in the rate law but if you look at hydrogen hydrogen has a coefficient of 2 and we determined that the exponent was a 1 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 all right 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 you into mechanisms and we'll talk about that a little bit more all right 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 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 1 2 or 3 I'm just going to choose one here so experiment 1 and we're going to plug all of our information into the rate law that we just determined for example right in our rate law we have the rate of reaction over here well for experiment 1 the initial rate of reaction was 1.25 times 10 to the negative 5 and it was molar per second so we're going to plug this in we're going to 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 5 and this was molar per second all right 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 point zero zero five molar right so we're going to plug in point zero zero five molar in here so we have point zero zero five molar and next we're going to 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 point zero zero two molar so we plug that in so we have point zero so we have point zero zero two molar next all we have to do is solve for K so let's let's go ahead and do that so let's get out the calculator here and we could say point zero zero five squared right gives us two point five times ten to the negative five we need to multiply that by point zero zero two so on the right side we'd have five times ten to the negative eight all right so we have five times 10 to the negative eight think about your units right this would be molar squared times molar over here so molar squared times molar right in all of this times our rate constant K is equal to one point two five times ten to the negative five all right 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 one point to five times ten to the negative five and if we divide that right if we divide that by five times 10 to the negative H then we get that K is equal to 250 all right 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 1 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 for K this would be one over molar squared this would be one over molar squared times seconds and so we found we found the rate constant for our reaction and notice this was for a specific temperature right so our reaction was at 1280 degrees C so 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 point zero one two molar and the concentration of hydrogen is point zero zero six molar well we can use our rate law right so our rate law our rate law is equal to what we found in a write 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 251 over molar squared times seconds right so we have that the concentration of nitric oxide is point zero one two so we have point zero one two all right this would be molar and we need to square that and then so all I did was take this right this and I plugged it into here and now we're going to take the concentration of hydrogen which is point zero zero six molar and plug that into here so let's go ahead and do that so that would be times point zero zero six all right molar let me go ahead and put in the molar there so point zero zero six molar to the first power and we solve for our rate so the rate is equal to let's do let's do the numbers first so we have point zero one two squared point zero one two squared all right and we're going to multiply that by point so times point zero zero six and then we also need to multiply that by our rate constant K so times 250 and this gives us our answer of two point one six times ten to the negative four let's round that to two point two so we have two point two two point two times ten to the negative four in terms of our units right if we think about what happens to the units here we have molarity squared right right here molarity squared molarity squared so we end up with molar per second right which we know is our units for the rate of reaction so molar per second so we found we found the rate of our reaction

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