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### Course: AP®︎/College Chemistry>Unit 9

Lesson 2: Relationship between reaction concentrations and time

# Half-life of a first-order reaction

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.

## Want to join the conversation?

• 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?
• You got it! That's how half-life works, whether you're talking reactions or nuclear decay.
• 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.
• 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).
• Why would we need to know about the half-time ? and where does the symbol `t 1/2` come from ?
• when you know the half life of a rxn, then you can determine the amount of product formed and we all know time is 't' 1/2 is half, so t1/2 is just a symbol for half life.
• 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.
• 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
• Does half life increase or decrease
(1 vote)
• For a given first-order reaction, the half-life is a constant.
• How can we tell if a reaction is first order?
(1 vote)
• 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.
• Is the time you get from the half life always seconds?
(1 vote)
• No. It depends on the substance. Eg. The half life of Carbon 14 is 5730 years
• why is there e in the half life formula, i thought that half life was calculated using .5 as the exponential base?
(1 vote)
• 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.
• 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)
• 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.