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Course: Chemistry library > Unit 15
Lesson 2: Relationship between reaction concentrations and time- First-order reactions
- First-order reaction (with calculus)
- Plotting data for a first-order reaction
- Half-life of a first-order reaction
- 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
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Second-order reactions
The integrated rate law for the second-order reaction A → products is 1/[A]_t = kt + 1/[A]_0. Because this equation has the form y = mx + b, a plot of the inverse of [A] as a function of time yields a straight line. The rate constant for the reaction can be determined from the slope of the line, which is equal to k. Created by Jay.
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Hello, on the last part you mention that the gradient for second order could be either k or 2k. In an exam situation which would you recommend using? Thank you(5 votes)
Well, that depends on the reaction. Like Jay said, if the coefficient in front of your reactant A is 1, then 1k is the slope. But if the coefficient in front of the reactant A is 2, then 2k is the slope.
At the end where he says a lot of textbooks just regard the slope as 1k for all situations even with reactions like the one used as an example with a coefficient of 2 in front of the reactant A, he's basically just saying most textbooks are incorrect on that point. In an exam setting, especially with more modern ones, they would prefer using the correct multiple of k for the slope. But of course, always show your work so that test graders can see your logic for using either 1k or 2k for the slope.
Hope that helps.(9 votes)
Can someone explain how the formula was derived at the beginning of the video using the two equations?(2 votes)- You need an understanding of calculus and separable differential equations to really know how its done.
We start by setting the two rate equations equal to each other since they equal the same rate, R. Also since this is calculus instead of using deltas, ∆, we use 'd's to show that we're using a more particular derivate definition of change.
So we get: -d[A]/dt = k[A]^(2)
This is a separable differentiable equation, a relatively straightforward problem in differential equations. We multiply the dt over the right and divide the [A]^(2) over the left.
This gives us: -d[A]/[A]^(2) = -kdt
So now we integrate both sides. The left integral with with concentration of A will have bounds from [A]0 (starting concentration) to [A]t (final concentration of some variable end time). The right integral will have bounds from 0 (initial time) to t (some final time). The bounds are kept as general as possible to allow for any times we want.
After integrating and evaluating the bounds we get: 1/[A]t - 1/[A]0 = kt
We can simply add the 1/[A]0 term over the right side to get the form used in the video.
Hope that helps.(7 votes)
- So, for a second order reaction the square root of the rate of the reaction is proportional to the concentration of the reaction? Is that right?(2 votes)
- This is true, but only if it is second order to a single reactant. So if the rate law is Rate = k[A]^(2) then yes what you said is true. But if the rate law is Rate = k[A][B] which is also second order overall then it is the product of the reactant's concentrations which are directly proportional to the rate of the reaction.
Hope that helps.(5 votes)
- This isn't a question, but a comment to those who are wondering how to do the integration of that equation at0:38. First the equation is -dA/dt=k[A]^2. This is equivalent to -dA*(1/[A]^2)=k*dt. Performing integration:
1/[A]=kt, but remember, there are bounds to the integration, so it is actually 1/[A] at time t minus 1/[A] at time 0 which is equal to kt and we are done.(2 votes) 
But what if there were multiple reactants? For a reaction like A+B -> C, if the rate law was R=k[A][B], would that also be a second order reaction...?(1 vote)- Yes that would be a second order overall reaction. We would say also that the reaction is first order with respect to A and B. The integrated rate law however is different from a second order reaction of the form Rate = k[A]^(2).(2 votes)
- In the First-order Reactions video, he says if the graph is a straight line, we know it is a first order reaction. Then the same thing in this video. Is it both?(1 vote)
- It depends on what you plot on the y-axis.
For example, assume we have a reaction which is truly first-order. If you plot the natural log of the concentration of your reactant, ln([A]), on the y-axis versus time on the x-axis and you create a linear curve, then it’s first-order. If you plot the same data, but with [A] or 1/[A] on the y-axis, it’ll produce a nonlinear curve which will also tell us the reaction is not zeroth or second-order.
Now assume you have a reaction which is truly second-order. If you plot the reciprocal of the concentration of your reactant, 1/[A], on the y-axis versus time and you create a linear curve, then it’s second-order. If you plot the same data, but with [A] or ln([A]) on the y-axis, it’ll produce a nonlinear curve which will tell us the reaction is not zeroth or first-order.
So you plot the same data, but with different variations of the y-axis variable, and you’ll produced one curve which is noticeable more linear than the others. That’ll tell you what order it is. So using the integrated rate laws to determine order, you really need several plots to determine know for sure.
Hope that helps.(1 vote)
Why is the reaction rate divided by two if the reactant is increasing.(1 vote)- We want a single reaction rate which is equivalent if we were to write the rate in terms of any of the participating chemicals. This means we have to write the reciprocal of the chemical's stoichiometric coefficient in the reaction rate. This idea being is that if the reaction is consuming twice as many moles of reactant as everything else we need to multiple by a factor of 1/2.
Hope that helps.(1 vote)
- can we find the k without graph?(1 vote)
- hello, how do I find the slope?(1 vote)
- wait but how do you determine the value of k when calculating 1/[C5H6](1 vote)
- Well we’re not getting the rate constant, k, from just taking the reciprocal of cyclopentadiene’s concentration, 1/[C5H6]. We’re plotting 1/[C5H6] versus time, t, so that it forms a straight line. A line has a general equation of y = mx+b which we fit to the integrated rate law of a second-order reaction, 1/[A]t = kt + 1/[A]0. This means the slope of that 1/[C5H6] versus t graph is equal to k for a general second-order equation. Jay explains this at1:09.
In the example though cyclopentadiene has a coefficient of 2 so the integrated rate law has the form 1/[A]t = 2kt + 1/[A]0 which means the slope of the graph is equal to 2k. When we do plot the data, the slope of the line is 0.1634 which is equal to 2k. Jay explains this at5:55.
So, 2k = 0.1634
k = 0.0817 after dividing both sides of the equation by 2.
Hope that helps.(1 vote)
0:38Video transcript
- [Instructor] Let's say we have a hypothetical reaction where reactant A turns into products. And let's say the reaction is second order with respect to A. If the reaction is second
order with respect to A, then we can write the rate of the reaction is equal to the rate constant
k times the concentration of A to the second power since this is a second order reaction. We can also write that
the rate of the reaction is equal to the negative of the change in the concentration of A
over the change in time. If we set these two
ways of writing the rate of reaction equal to each other, and use of calculus, including the concept of integration, we will arrive at the integrated rate law for a second order reaction. The integrated rate law
for a second order reaction says that one over the concentration of reactant A at some time t, is equal to the rate
constant k times the time plus one over the initial
concentration of A. Notice how the integrated rate law has the form of y is equal to mx plus b, which is the equation for a straight line. So if we graph one over
the concentration of A on the y axis, and time on the x axis, so let's go ahead and put that in here, one over the concentration
of A on the y axis, and time on the x axis, we
will get a straight line and the slope of that line is equal to the rate constant k. So slope is equal to K and the y intercept is equal to one over the
initial concentration of A. So the point where our
line intersects the y axis is equal to one over the
initial concentration of A. Let's look at an example
of a second order reaction. C5H6 is cyclopentadiene, and two molecules of cyclopentadiene will
react with each other to form dicyclopentadiene. Our goal is to use the
data from this data table to prove that this
reaction is second order. However, we have to be careful because in our balanced equation, we have a two as a coefficient in front of cyclopentadiene. Going back to our hypothetical reaction where reactant A turned into products, there's also one as a
coefficient in front of the A. And if there's a one as a
coefficient in front of the A, we can use this form of
the integrated rate law for a second order reaction. However, for our problem, we have a two as a coefficient in
front of cyclopentadiene. And that means we need to have a stoichiometric
coefficient of 1/2 in here, and that changes the math. Now when we set these two
rates of reactions equal to each other, and we use
calculus and we integrate to get our integrated rate law, because of this one half, we end up with a two in front of the K. So thinking about y is equal to mx plus b, now the slope of the
line is equal to two K. So now for our reaction, we can write our integrated rate law as one over the concentration
of cyclopentadiene at some time t, is
equal to two kt plus one over the initial concentration
of cyclopentadiene. So if we look at our data table, we have time in seconds and
we have the concentration of cyclopentadiene,
but we need to have one over the concentration of cyclopentadiene so we need a new column. So we're going to calculate
one over the concentration of cyclopentadiene. So if the concentration of cyclopentadiene when time is equal to zero
seconds is 0.0400 molar, if we take one divided by 0.0400, we would get 25.0. To save some time I've gone ahead and filled in the rest of this column. Notice as time increases so
as we go from zero seconds to 50 to 100 to 150 to 200, the concentration of
cyclopentadiene decreases because cyclopentadiene is
turning into dicyclopentadiene. Next, we need to graph our data. So we're gonna have one
over the concentration of cyclopentadiene on the y axis and time on the x axis. Our first point, so when
time is equal to zero, one over the concentration
of cyclopentadiene is 25.0. So if we go down to our graph, we can see that when
time is equal to zero, our first point here is 25.0. And plotting the other points
gives us a straight line. Next, we need to find the
slope of this straight line. And there are many ways to do that. One way to do it is to
use a graphing calculator. And when I used a graphing calculator to find the slope of this line, I found that the slope is equal to 0.1634. Thinking about y is equal to mx plus b, our slope should be equal to two k. So to find the rate constant k, we need to divide the slope by two, which gives us 0.0817. To find the units for
K, remember that slope is equal to change in y over change in x, and on our Y axis, our
units are one over molar, and the x axis the units are seconds. So therefore, we can
write the rate constant k is equal to 0.0817. It's to be one over
molar divided by seconds which is the same thing as
one over molar times seconds. It's important to point
out that most textbooks don't cover how the two
as a coefficient changes the integrated rate law. And so a lot of textbooks will simply say that the slope of the line for the second order integrated rate law, is equal to K. So a lot of books would
just say the final answer for the rate constant is 0.163. So you'll see you'll see
a lot of textbooks say that the rate constant would be 0.163, one over molar times seconds. However, since the coefficient in front of cyclopentadiene is a two, technically this rate
constant is the correct one. Finally, since we got a straight line, when we graphed one over the concentration of cyclopentadiene versus time, we know that this reaction
has second order kinetics.