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Half-life of a first-order reaction

AP.Chem:
TRA‑3 (EU)
,
TRA‑3.C (LO)
,
TRA‑3.C.5 (EK)
The half-life of a reaction is the time required for a reactant to reach one-half its initial concentration or pressure. For a first-order reaction, the half-life is independent of concentration and constant over time. Created by Jay.

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  • leaf blue style avatar for user Chris
    So, to clarify, the main point here is that no matter the initial concentration of a reactant, it will take the same amount of time for half of the reactant to be disappear?
    (8 votes)
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  • male robot johnny style avatar for user ranaajrawat
    I have a little confusion about first order reactions that produce products being dependent upon the concentration of the reactant, (i.e. if you double the reactant in a first order reaction, you double the amount of product produced), while the half life decay of a first order reaction that produces a product (i.e. half the initial concentration of the reactant) is not dependent upon the initial concentration. Could you please explain these two differences when you get a chance? By the way, these videos have been extremely helpful and I appreciate all the hard work you and Khan Academy have put into making these materials available to anyone, anytime, anyplace.
    (8 votes)
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    • leafers sapling style avatar for user Erika Jensen
      In earlier videos we see the rate law for a first-order reaction R=k[A], where [A] is the concentration of the reactant. If we were to increase or decrease this value, we see that R (the rate of the reaction) would increase or decrease as well. When dealing with half-life, however, we are working with k (the rate constant). While the rate of reaction is measured in units molar/second, a rate constant for a first-order reaction is 1/(second).
      (4 votes)
  • male robot hal style avatar for user RandomDad
    Why would we need to know about the half-time ? and where does the symbol t 1/2 come from ?
    (4 votes)
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  • blobby green style avatar for user david.scalais
    Why is the function of half life exponential. And the function in the previous video is a straight line. If both are first-order reaction.
    (3 votes)
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    • mr pants purple style avatar for user Ryan W
      If you graph half life data you get an exponential decay curve. It’s kind of the definition of it. If you graph something that starts at 100 and decays by half every 1 minute, 50 by minute 2, 25 by 3, 12.5 by 4, 6.25 by 5 etc. you’ll see.

      Go back to the previous video and look at the label on the y axis, then compare with the y axis on this one. A quick method for working out the reaction order is to plot [A] vs t, ln[A] vs t and 1/[A] vs t, one of them will give a straight line which tells you the order
      See: https://www.chem.purdue.edu/gchelp/howtosolveit/Kinetics/IntegratedRateLaws.html
      (3 votes)
  • blobby green style avatar for user HamzamAfsar
    Does half life increase or decrease
    (1 vote)
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  • blobby purple style avatar for user Dj Korzelius
    Is the time you get from the half life always seconds?
    (1 vote)
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  • female robot amelia style avatar for user ayushkoul00
    why is there e in the half life formula, i thought that half life was calculated using .5 as the exponential base?
    (1 vote)
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    • male robot hal style avatar for user Andrew M
      any exponential decay can be written with any base you want (check your properties of logarithms)
      e is common but 1/2 is also good for half-life.
      e is common because it is much easier to deal with when you want to manipulate these equations with higher math, like calculus. When you study calculus you will see why.
      (3 votes)
  • blobby green style avatar for user RatuBS
    At he said that k is a constant. I know that k is a constant, but it has a different value depending on the ordo of the reaction (like if the ordo is one, the unit of k would be 1/s) Is that alright?

    Or am I get it wrong? Can someone help me?
    (1 vote)
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    • leaf red style avatar for user Richard
      So each chemical reaction has it's own rate constant, which indeed is constant, at a specific temperature. So a reaction like A -> B will have a rate constant associated with the rate law of the reaction, but we have to specify the temperature of the reaction. This is because the rate constant for the same reaction at say 300K will be different in value than the rate constant at 400K. Now this is juts talking about the value of a rate constant, but what I think what you mean are the units of the rate constant the same?

      In which case then no the units of the rate constant will be different depending on the overall order of the reaction. So a first order reaction's rate constant will indeed be using units of 1/s (or s^(-1)) while the rate for a second order reaction will be in units of 1/(M*s) (or s^(-1)*M^(-1)).

      So a first order reaction and a second order reaction can have the same numerical value for their rate constants, but it's inaccurate to say that they are the same since they are using different units.

      I think I might be getting hung up on your use of different value as opposed to different unit, but I hope that made sense.
      (3 votes)
  • male robot hal style avatar for user 𝐢ᴀɴᴅʏ_Qɪ
    How can we tell if a reaction is first order?
    (1 vote)
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    • leaf red style avatar for user Richard
      We can tell what order a reaction is graphically by plotting the reactant's concentration versus time and seeing if it produces a linear curve. We essentially rearrange a reaction order's integrated rate to resemble a linear equation of the form: y=mx+b.

      For a zeroth order reaction: [A] = -kt +[A]0, the y variable is the reactant concentration or [A], the x variable is time or t. If plotting [A] versus t yields a straight line then the reaction is zeroth order. Additionally the slope, m, will be -k and the y-intercept, b, will be the initial concentration of the reactant. So if the reaction is truly zeroth order then it'll form a straight line modeled on the integrated rate law. This same logic follows for first, second, etc. order reactions too.

      If we plot [A] versus t and we do not find a straight line, then it's not zeroth order and will follow another order's integrated rate law. For a first order reaction: ln([A]) = -kt + ln([A]0), the y variable is now ln([A]) and the x variable is still time. If we tried plotting ln([A]) versus time and get a straight line now, then it's first order. The slope will be -k and the y-intercept will be ln([A]0).

      And if it's not first order, then it could second order which uses: 1/[A] = kt + 1/[A]0, with y being 1/[A] and x being time again. The slope would be k and the y-intercept would be 1/[A]0. If 1/[A] versus time produces a straight line, it's second order. If not we keep repeating this process for other reaction orders until we find a straight line.

      It's a nice trick in chemistry of fitting data to a straight line to see if there's a relationship between variables.

      Hope that helps.
      (2 votes)
  • leaf yellow style avatar for user Kshitij D G
    How can half life be achieved, while a chemical reaction is actually going on?
    (1 vote)
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Video transcript

- [Voiceover] Here we have one form of the integrated rate law for a first order reaction. And we're gonna keep going with the math here, so we eventually will talk about the half-life. So over here on the left, the natural log of the concentration of A at time t minus the natural log of the initial concentration of A. That's the same thing as the natural log of the concentration of A over the initial concentration of A. So that's just the log property. And this is equal to negative kt, where k is your rate constant. Next, we need to get rid of our natural logs. So we're going to exponentiate both sides. So we're gonna take e to both sides here and that's gonna get rid of our natural log. So now, on the left side we have our concentration over the initial concentration. On the right side we have e to the negative kt. So we're gonna multiply both sides by the initial concentration of A. So we get that our concentration of A at time t is equal to the initial concentration of A times e to the negative kt. And now, it's a little bit easier to think about the graph. We can put the concentration of A on the y-axis and we can put time on the x-axis. And this is in the form of an exponential decay. So down here I've graphed an exponential decay graph, just to show you what it looks like here. Let's think about this point on our graph. So that's when time is equal to zero. So when time is equal to zero, what is the concentration? So you would just plug in time is equal to zero into here. So you would have your concentration is equal to the initial concentration times e to the zero. And e to the 0 is of course one. So this is one. So our initial concentration is obviously this point right here, time is equal to zero. This is obviously our initial concentration. So I'll write that in here. So that's this point. And as time approaches infinity, as time goes to infinity, your concentration of A goes to zero. So as you go out here, obviously your concentration of A is going to approach zero. So that's the idea of an exponential decay graph. Next, let's think about half-life. So over here is our definition of half-life. It's the time it takes for the concentration of a reactant to decrease to half of its initial concentration. So if the initial concentration, if this is the initial concentration here, what would be the concentration after half of it has reacted? We would get our initial concentration divided by two. So we're gonna plug this in for our concentration and then the symbol for half-life is t 1/2. So t 1/2. So we're gonna plug this in for time. So when the time is equal to the half-life, your concentration is half of your initial concentration. So let's plug those in and solve for the half-life. So on the left side we would have our initial concentration divided by two. And then this would be equal to the initial concentration of A times e to the negative k and then this would be the half life, so we plug in t 1/2 here. And so now we're just gonna solve for t 1/2. We're gonna find the half-life for a first order reaction. Let's get some more space down here. And we can immediately cancel out our initial concentration of A. So now we have 1/2 is equal to this is e to the negative kt 1/2. Alright, next, we need to get rid of our e here. So we can take the natural log of both sides. So we can take the natural log of 1/2 this is equal to the natural log of e to the negative kt to the 1/2. And so that gets rid of our e. So now we have the natural log of 1/2 is equal to negative kt 1/2. So we're just solving for t 1/2, cause t 1/2 is our half-life. So our half-life, t 1/2, would be equal to this would be negative natural log of 1/2 divided by k. Let's get out the calculator and let's find out what natural log of 1/2 is. So let's get some space over here. So natural log of .5 is equal to negative .693. So we have the negative of that, so we get a positive value here for our half life. So our half-life is equal to, let me rewrite this here, so our half-life, t 1/2, is equal to .693 divided by k, where k is our rate constant. So here is your half-life for a first order reaction. Now let's think about this. If k is a constant, obviously .693 is a constant. And so your half-life is constant. Your half-life of a first order reaction is independent of the initial concentration of A. So you're gonna get the same half-life. And let's think about that for an example. Let's go back up here to our graph and let's think about half-life. So lets' say we're starting with some initial concentration, let me go ahead and change colors here, so we can think about it. I'm just going to represent our initial concentration here with eight dots. So that's our, let's say we have eight particles we're starting out here. So obviously this is a theoretical reaction. So we're gonna wait until we've lost half of our reactant. Alright, so we've lost half of our reactant, obviously we'd be left with four. We'd be left with four here. So, where would that be on our graph? Well this point right here is out initial concentration. This is concentration here, so we'd go half that. So that would be right here on our graph. So we'd go over here and we find this point. Then we drop down to here on our x-axis. And let's just put in some times. Let's say that this is 10 seconds and 20 seconds and 30 seconds and 40 and so on. So we can see that it took, this took 10 seconds for us to decrease the concentration of our reactant by half. And so the first half-life is 10 seconds. So let me write that in here, so t 1/2 is equal to 10 seconds. Again, a made up reaction, just to think about the idea of half-life. And then, let's say, now we have four. How long does it take for half of that to react? So if half of that reacts, we're left with two particles. And on our graph, let's see, that would be right here. This point would be half of this, so we go over to here, and then we drop down. And how long did it take for us to decrease the concentration of our reactant by half? Once again, 10 seconds. So this time here would be 10 seconds. So this half-life is 10 seconds. We could do it again. So we lose half of our reactant again. And we go over to here on our graph and we drop down to here. How long did it take to go from two particles to one particle? Once again, it took 10 seconds. So the half-life is once again 10 seconds. So your half-life is independent of the initial concentration. So it didn't matter if we started with eight particles or four or two. Our half-life was always 10 seconds. And so, this is the idea of half-life for a first order reaction.