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## Relationship between reaction concentrations and time

Current time:0:00Total duration:8:09

# Second-order reaction example

## Video transcript

- [Voiceover] We've seen
this reaction before, this is the decomposition of nitrogen dioxide to nitric oxide and oxygen. And we've already proved
it's a second-order reaction, and we also found the rate constant at 300 degrees C. So in part a, if the
initial concentration of nitrogen dioxide is .008 Molar, what is the concentration
of nitrogen dioxide after 20 minutes? Well, we know this is a
second-order reaction. So we can use our integrated rate law for a second-order reaction. And that was one over
the concentration of A minus one over the
initial concentration of A is equal to the rate constant K times the time. So this is one form of our
integrated rate equation. Now let's think about what
we're gonna plug in here. We're trying to solve for
the final concentration of NO2. So we're gonna put in here one over the concentration
of nitrogen dioxide at 20 minutes, right this would be minus one over the initial
concentration of nitrogen dioxide, which is .008. So we plug that into here. So one over .008, this is equal to our rate constant which is .54, so we have .54, one over Molar times seconds. And then we're gonna
multiply that by the time. Notice we have seconds here, for our rate constant
we were given minutes. So we need to convert minutes to seconds. So if you have 20 minutes, all right we know that
there are 60 seconds for every one minute. So if we multiply, minutes cancels, and that gives us seconds. So 20 times 60 is 1200. So we have 1200 seconds. All right, so let's go
ahead and let's do the math. So let's get out the calculator here, and we'll multiply .54 by 1200. And so on the right side we have 648. Notice your units, your
seconds would cancel here. And the left, what's one over .008? So let's do that. What's one divided by .008? That gives us 125. So we have 125 here, and we have our minus, this would be one over
the concentration of NO2. All right so now of
course we're just solving for the concentration of nitrogen dioxide. So one over the concentration of NO2 is equal to 648 plus 125. And that of course, is 773. So to solve for the concentration of nitrogen dioxide, that would be one over 773. So let's get our calculator out, and let's do one divided by 773. And so the concentration would be .0013. So this is equal to .0013 Molar. So if you start with a
concentration of .008 Molar, the concentration after 20 minutes goes down to .0013 Molar. All right, let's move on to part b. So in part b, if the initial concentration of nitrogen dioxide is .008 Molar again, what is the half-life? So in the previous video, we talked about the
equation for the half-life of a second-order reaction. And we found that the half life
for a second-order reaction is equal to one over the rate constant K, times the initial concentration of A. And so for this case, this would be one over the
rate constant which is .54, one over Molar times seconds. And then we need to multiply that by the initial concentration of A, which of course would be
the initial concentration of NO2. So that's times .00800 Molar. All right, so let's do that math. So we get out our calculator here, and we take .54 and we
multiply that by .008. So we get .00432, and so we take one divided by that, one divided by our answer here will get us our half-life. So we get 2.3 times ten to the second, seconds right? So this is equal to, t 1/2 is equal to 2.3 times 10 to the second seconds. Or we could say approximately 230 seconds. So if we're going from
a concentration of .008, if we're going from a
concentration of .008 Molar, and we wait approximately 230 seconds, the concentration goes down to .004. So we have half of our
initial concentration after one half-life. All right, let's compare what we got in b to what we're gonna get in c. Because in c we're changing the initial concentration
of N02 to .004 Molar. So now what's the half-life? So the calculation is the same, right? We're gonna plug in t 1/2
is equal to one over K, times the initial concentration of A. The difference is we're changing
the initial concentration. All right so K stays the same, so .54 one over Molar times seconds. But now we're gonna plug
in .004 instead of .008. So we have .004 Molar and we're solving for the half-life. So once again we can get
out our calculator here, take .54, multiply that by .004, and then if we take one over our answer then we should get our half-life. So that's 4.6 times ten to the second. So this would be equal to, so our half life would be equal to 4.6 times ten to the second... Seconds. All right? So this is seconds. So I think I forgot to point that out in the previous example. If you're looking at your units, the Molar would cancel here, and so you'd have one
over one over seconds. Which of course is equal to seconds. So the half-life is measured in seconds. It'd be the same thing for up here, right? The Molars would cancel, we'd have one over one over seconds, so we'd get seconds here. So this half-life, 4.6, let me just make this look more like a 6, 4.6 times ten to the second, that's twice our first half-life. So if we're going from a
concentration of .004 Molar, and we wait approximately 460 seconds, we're gonna end up with a
concentration of .002 Molar. So remember, for a second-order reaction, the half-life is not constant. So we talked about this
in the previous video. The half-life for a second order reaction depends on the concentration
that you're starting with. And so we can see here, we're starting with a concentration that's half of the first one, and we can see that our
half-life is twice as long. So this second half-life
here is twice as long as the first half-life that
we talked about for part b.