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- [Voiceover] Here we have the balance equation for the decomposition of nitrogen dioxide into nitric oxide and oxygen. In part A they want us to use the data to show that the decomposition of nitrogen dioxide is a second order reaction. So over here we have the time, and we also have the concentration of NO2. To show this is a second order reaction we need to remember the integrated rate law from the previous video. So we found that 1 over the concentration of A is equal to the rate constant, k times the time plus 1 over the initial concentration of A. So that's the integrated rate law for a second order reaction. For our reaction, A=NO2, so we can plug that in. So now we have 1over the concentration of NO2 is equal to the rate constant, k times the time plus 1 over the initial concentration of NO2. This is in the form of y=mx+b. So in the previous video, we talked about if you put time on the x axis and 1 over the concentration of NO2 on the y axis, if the reaction is second order you're going to get a straight line. And the slope of that straight line, m is equal to the rate constant, k. And the y intercept is equal to the initial... Oh, 1 over the initial concentration of NO2. So if we're going to show that this is a second order reaction we already have the time, but we have the concentration of NO2. We need 1 over the concentration of NO2. So let's plug that in here... So we need to get 1 over the concentration of NO2. So that would be molar to the negative first. So let's go ahead and do that. Here's the concentration of NO2, so we can take 1 over .008. So we get out the calculator, and we take 1 divided by .008 and we get 125, so this is 125. Next, we have 1 over .0066. So 1 divided by .0066 gives us 152. So we have 152 here, next. 1 over .0056 and that gives us 179, so rounding to 179. And then finally we have our last one which is 1 divided by .00485 and we get 206. So we have 206, so we know this is going to go on our y axis, time is going to go on our x axis. So let's look at the graph here. Lets's find some of our points. So we have time on our x axis, and we have 1 over the concentration of NO2 on the y axis. Our first point would be when time=0, y=125. So when time=0, 125 is right about there. So that's our first point. Our next point would be time=50, y=152. So when time=50, 152 would be pretty close to there. Next we have time=100, y=179. So when the time=100, 179 would be somewhere about there. And then finally our last point, when time=150, y=206. So when we are talking about 150, 206 should be somewhere in there. Let's see if we have a straight line. So let's go ahead and draw a line in here. We can see that these points are pretty much on a straight line. So we've shown that this is a second order reaction. Obviously I didn't do a perfect job of graphing, but you can see that these points are on our straight line and we know the slope of this line, right, the slope... Is equal to the rate constant, k. So, we're done with A, right? We've taken our data and we've shown that we get a straight line when we put time in the x axis and 1 over the concentration of NO2 in the y axis. So we're done with A, move on to B where we're trying to figure out the value of the rate constant. This reaction was at 300 degrees C. So all we have to do is find the slope of our line and we'll find our rate constant. So of course, one way to find the slope would be to take delta y over delta x. So the slope is equal to the change in y over the change in x. So you could find a point here and a point here, on your line and you could find the slope that way. So that's one way to do it. This would be the change in y, and this would be the change in x. And the units, for y we have molar to the negative first so, molar to the negative first. and for the x we have second, so the unit would be... I could re-write this as 1 over molar times seconds. So that's our unit for our second order reaction. We could have also figured out our unit using the rate law. Rate is equal to the rate constant, k times the concentration of, we had, NO2 and since this is a second order reaction this would be squared. So rate would be molar per second, and then we have the rate constant, k times the concentration squared. So this would be molar squared. So let's go down and get a little more room. So this would be molar over seconds is equal to k times molar squared. One of those molars would cancel so now we have 1 over seconds is equal to k times molar, and so once again we find that k is equal to 1 over molar times seconds. So it doesn't matter how you get your units here. We know that's going to be the units for k, we could find it out by looking at those two points that I marked on our graph but to get an exact answer, let's use our calculator. So let's take out our calculator and let's plug everything in. So let's go to stats and we'll go to edit here and we'll go ahead and plug in our points. So when x=0, y=125. When x=50, y=152. Next one, when x=100, y=179. Then finally, we have when time=150 seconds, y=206. So we have our points in there now and we can go back to stats and we can hit calc, so F1 right here and we can hit enter twice and then we can do a linear regression which should be right here for F2. And when we do our linear regression we find that b on our calculator here, b is the slope. So b=.54, so that's the slope of our line and therefore that's the value of our rate constant. So k, or our rate constant= .54 and this was 1 over molar times seconds. So we have figured out the value for k using the data that was given to us.