- First-order reactions
- First-order reaction (with calculus)
- Plotting data for a first-order reaction
- Half-life of a first-order reaction
- Half-life and carbon dating
- Worked example: Using the first-order integrated rate law and half-life equations
- Second-order reactions
- Second-order reaction (with calculus)
- Half-life of a second-order reaction
- Zero-order reactions
- Zero-order reaction (with calculus)
- Kinetics of radioactive decay
- 2015 AP Chemistry free response 5
Worked example: Using the first-order integrated rate law and half-life equations
In this video, we'll use the first-order integrated rate law to calculate the concentration of a reactant after a given amount of time. We'll also calculate the amount of time it takes for the concentration to decrease to a certain value. Finally, we'll use the first-order half-life equation to calculate the half-life of the reaction. Created by Jay.
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- Sir ,isn't there any other method to find the concentration other than using ln and exponential? Because during our calculations we cannot use a calculator which is pretty tough.(5 votes)
- If you are referring to the MCAT exam, from what I have heard, the MCAT tests more conceptual knowledge about rate laws and simpler math, such as finding rate laws rather than complex problems like this. I think it is important to understand the main ideas behind this problem, such as half-life, but I don't think we will be expected to do this much math.(18 votes)
- is the half life always t1/2 = .693/K ?(2 votes)
- Only if the reaction is a first order reaction.(10 votes)
- At6:22, why don't you divide the two ln's? I thought you were supposed to divide when you subtract.(3 votes)
- well, he could have written it as 'ln(0.01/0.05)' but his method is more straightforward(1 vote)
- how would you find the starting concentration at time= zero if you were given a set of data (time vs concentration)?(1 vote)
- I would plot ln(c) vs. t and extrapolate the straight line back to t=0.(4 votes)
- is it that the average rate of a first-order reaction always gets slower with time?(2 votes)
- In the worked example video: Are significant figures of any importance?(1 vote)
- Yes, they're important. You can't do science, at least professionally, without using correct significant figures.(2 votes)
- Shouldn't we round 2402 seconds to 2400 seconds, and 1034 seconds to 1000 seconds, because we only have two sig-figs in the rate constant?7:02,9:00(1 vote)
- You're correct, both those answers should be correctly reported with only two sig figs because of the rate constant. However writing 2400 and 1000 seconds can be misconstrued as having differnt amounts of sig figs and are considered ambigious answers. For example one person could argue that 2400 has two sig figs (the 2 & 4) while another person could argue that all four digits are significant.
To avoid this issue we should write them in scientific notation. So 2400 s becomes 2.4 x 10^(3) s, and 1000 becomes 1.0 x 10^(3) s. And now everyone is clear that both numbers have two sig figs each.
Hope that helps.(2 votes)
- do you have to convert the rate constant to seconds when determining the half live, if the rate constant is given in hours? or would you find the half live using the hours k and then convert the units to seconds?(1 vote)
- Can we find the half life without the initial concentration?(1 vote)
- We can, but only for a first order reaction. Each order has its own half-life equation.
Zeroth order: ([A0]/2k)
First order: (ln(2)/k)
Second order: (1/k[A0])
So the zeroth and second order require us to know the rate constant and the initial concentration, while the first order only requires the rate constant.
Hope that helps.(1 vote)
- Why do you need the s^-1 instead of just s. I have seen this for mole^-1 as well(1 vote)
- Do you mean the unit for the rate constant? The unit for the rate constant differs depending on the order of the reaction. This is because the units have the be equivalent on either side of a rate equation.
So for a first order reaction the rate law is: Rate = k[A], where k is the rate constant and A is the reactant to the first power. The rate's unit is given as molarity per second, or M/s, and the concentration uses just units of molarity, M. So writing the same rate law with just the units: M/s = k*M, we see that for both sides of the equation to have the same unit the rate constant must be 1/s or s^(-1). This is because M*1/s = M/s, the same unit as the rate's unit.
The same kind of logic goes into figuring out the other rate law constant's units.
Hope that helps.(1 vote)
- [Voiceover] We've already looked at the conversion of cyclopropane to propene and shown that it's a first-order reaction. And we also found the rate constant at 500 degrees Celsius in an earlier video. And so the rate constant is 6.7 times 10 to the negative 4 one over seconds. And so in part a, if the initial concentration of cyclopropane is .05 molar, what is the concentration of cyclopropane after 30 minutes? Well, to solve for this concentration, we can use the integrated rate law that we found in an earlier video. Since this is a first-order reaction, the integrated rate law, or one form of it, is the natural log of the concentraion of A at any time t is equal to the negative kt. Where k is the rate constant plus the natural log of the initial concentration of A. And here cyclopropane, or C three H six, is A. So this is the natural log of the concentration of cyclopropane, C three H six. Is equal to k, is 6.7 times 10 to the negative four. So this would be negative times 6.7 times 10 to the negative four. And time. We need to find the time. And since we have k in seconds, we have 30 minutes here. We need to convert 30 minutes into seconds. So 30, 30 minutes. There are 60 seconds in one minute. 60 seconds in one minute. So if we multiply those two, minutes cancels out, so 30 times 60 gives me 1800 seconds. So that's 1800 seconds, so I put that in here. So 1800 seconds plus the natural log, the natural log of the initial concentration of cyclopropane. The initial concentration is .05. So this is the natural log of .05. So starting to run out of space there. Alright, so, let's think about what we would do to solve for the concentration of cyclopropane. Well we have this natural log in here. So we could exponentiate both sides to get rid of our natural logs. So if we exponentiate both sides, that gets rid of our natural log here. And that would give us the concentration of cyclopropane after 1800 seconds. So let's do that math. Let's get out the calculator here. Let's do all that math on the right side. This was, let's see here, we have We have negative 6.7 times 10 to the negative four. I need to multiply that by 1800 seconds. And to that, we're going to add the natural log of .05. And then we need to take e to our answer. And e to our answer gives us .015 molar. And so that's our answer. Our answer, our concentration of cyclopropane is .015 molar. Now, you didn't have to use this form of the integrated rate law. You could have used a different form. The concentration as a function of time. So you could have written, the concentration of cyclopropane at a certain time is equal to the initial concentration of cyclopropane, the initial concentration e to the negative kt. So we saw this in the previous videos. So the concentration as a function of time. So we can plug in our values here, we're solving for the concentration of cyclopropane. The initial concentration was .05 So we plug in here .05. e to the negative k was 6.7, 6.7 times 10 to the negative four. And our time was 1800 seconds. So we could do the problem this way. Obviously we should get the same answer. Let's just go ahead and show that here. So we have negative 6.7 times 10 to the negative four times 1800, and then we would take e to our answer. And then we to multiply by the initial concentration of .05. And so obviously we get .015 molar again. So it doesn't matter how you do the problem, it doesn't matter which equation you start with, you're gonna get .015 molar for your concentration of cyclopropane. Alright, let's look at part b of this question. How long does it take for the concentration of cyclopropane to reach .01 molar? Alright, so to reach .01 molar, once again, it doesn't really matter which form you use, which equation you use. I'll just take the first equation that we discussed. Write the natural log of the concentration of A is equal to negative kt plus the natural log of the initial concentration of A. And let's plug in what we need here. So the concentration of cyclopropane to reach .01 molar, so that's our concentration here. That's our concentration at some time. So that would be the natural log of this would be .01 molar is equal to negative kt. So this is negative 6.7 times 10 to the negative four. The time is what we don't know. How long does it take? So we're trying to find the time. And we know the initial concentration of cyclopropane, once again that .05 molar. So this is the natural log of .05 molar again. So to solve for time, this would be the natural log of .01 minus the natural log of .05 and then you have to divide that by negative 6.7 times 10 to the negative four. So just some algebra here gets you your time. So we can do that on our calculator. We can take, let's make some room over, let's put it over here. So this would be the natural log of .01 and then we're gonna subtract the natural log of .05 and then divide that by negative 6.7 times 10 to the negative four. And then we get as our time 2,402. That's in seconds. So 2,402 seconds. Time is equal to 2,402 seconds. And we could leave it in seconds, or we could we could put that into minutes. So how many minutes is that? Well, 2,402 seconds, if we divide by a conversion factor. So it'd be 60, there are 60 seconds in every minute. So seconds would cancel out. We get one over, one over minutes. So that gives us minutes, of course. And so we can do that really quickly on the calculator, or you could do that in your head. I'll go ahead and do that on the calculator just to show you. Let's take that and and let's divide it by 60. Obviously we're gonna get pretty close to 40. So once again, that's something that you could do in your head, but 40 minutes. So approximately 40 minutes is how long it would take to reach a final concentration of .01 molar. Alright, let's look at part c. So in part c they want us to find the half-life. Well, from the previous video, for a first order reaction, the half-life is equal to .693 divided by your rate constant k. So the rate constant for this reaction was 6.7 times 10 to the negative four. So the half-life is equal to .693 divided by 6.7 times 10 to the negative four. And this was one over seconds. So we can do that on our calculator to solve for the half-life. So we have .693, divide that by 6.7 times 10 to the negative four. And so we get 1,034. So t 1/2, which is the half-life, which is 1,034 and the units would be seconds. This would be one over one over seconds, or seconds. So 1,034 seconds. If you wanted to convert that into minutes, we could just divide that by 60 seconds. And that gives us 17, eh we'll just say approximately 17 minutes. So we'll round that to approximately 17 minutes is the half-life. And remember from the previous video, what the half-life refers to. So if we started with, remember, we started with a concentration of .05 molar. That was our initial concentration. A half-life is how long it takes for us to get half of our initial concentration. So how long does it take to get from .05 molar to half that, which is .0250 molar. That's the half life. It takes 17 minutes. It takes 17 minutes for our concentration of cyclopropane to go from .05 molar to .025 molar. And how long would it take for our concentration to reach .0125 molar? So this concentration is half of this, so therefore it would take another 17, it would take another 17 minutes. Again, watch the previous video for more on half-life.