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## Relationship between reaction concentrations and time

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# Rate constant k from half-life example

## Video transcript

- [Voiceover] In this video,
we're going to be talking about how to find the rate constant K for a first order reaction. We're going to work through this example, and we have a table of data,
so we know the concentration of our reactant, hydrogen
peroxide or H2O2, over time, so we have one, two, three,
four, five different data points and concentrations for
each of those data points. The overall reaction we're
looking at is the decomposition of hydrogen peroxide to go
to water or H20 and oxygen. We're looking at this reaction
at 40 degrees Celsius. The question we'd like to
answer here is, what is K? Based on the data we
have, this concentration over time data, we would like to know what is the rate constant K. So there are two pretty
common ways that you might go about trying to answer this question. The first way you might do it, and I think for a lot of people, this
might be your first instinct, is that you actually solve it graphically. You can graph, and since we know this is a first order reaction, you
might know that in order to look at a first order
reaction, you usually graph the natural log of your
concentration over time versus time. When you do this, you can
use the integrated rate law for a first order reaction,
so the natural log of the concentration at
any point in time, is equal to the natural log of the
initial concentration minus Kt. We can graph our data. We can take the natural log of this, graph it versus time, and then we know that the slope will give us minus K. That's a perfectly fine way to solve this particular problem. But I would suggest,
especially since we know that this particular reaction
is a first order reaction, we're going to solve this
a slightly different way. So what makes first
order reactions special, is that their half-life,
or t 1/2, is constant. Not only is it constant, it is related to the rate constant K. Our half-life does not change over time. And we have an equation: t 1/2 is equal to the
natural log of two over K. The natural log of two
we can calculate out, so it's actually 0.693
over the rate constant K. We have a separate video where we talk about deriving this particular equation. All you really need to do is,
you know that your half-life by definition, is the time it takes to go from your initial concentration to half the initial concentration,
so all you have to do is plug in 1/2 times your
initial concentration, and then solve for t. But if you're taking an exam, chances are you might already know this equation, or you've been given
this equation already, so we can actually use this
equation to solve for K, without actually graphing our data. This is useful too if you don't have a graphing calculator for an exam. We can look at our data and
then see if we can figure out our half-life from our table. Here, from our first data
point to our second data point, and I will just actually
number these so you know what I'm talking about, we go
from data point one to two, we've halved our
concentration, and the time that's gone by is 2.16 times
10 to the four seconds. Just to be thorough, we can look at some of the other data points, and make sure the half-life
indeed stays constant. If we look at points two and three, again, our concentration goes down by half, and three and four, the same is true. In fact, for all of these
data points, if you look at any two data points, the
concentration goes down by half, and it always takes 2.16
times 10 to the four seconds. So we know that our
half-life is equal to 2.16 times 10 to the four seconds. I just want to add here that this is a slightly contrived example. This data is very nice-looking data. If you were to do an experiment
with a first order reaction, and actually measure this kind of data, there's a really high
chance that your data wouldn't look quite this
clean, and that's totally okay. In fact, that's perfectly
normal, so don't feel bad if your experimental data doesn't look quite as nice as this. That's actually probably a good sign. Anyways, now that we know the half-life, from looking at our data table,
we can actually rearrange this equation here to solve for K. So if we solve for K, K is equal to 0.693 divided by our half-life. Now that we know our
half-life, we can plug that in and get K, so it's 0.693 divided by 2.16 times ten to the four seconds. If we plug this into
our calculator, we get that K is equal to 3.21
times ten to the minus five, and our units are one over seconds. The main things to remember here, are that for first order
reaction, since your half-life is constant, you can
use this handy equation to calculate K if you know
half-life and vice versa; if you know half-life, you
can always calculate K, and it should always be
really easy to recognize a first order reaction based
on a graph or table of data, by checking for constant half-life.