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(3 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.(6 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)
- Can someone explain how the formula was derived at the beginning of the video using the two equations?(1 vote)
- 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.(3 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)
- 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)
- 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.(1 vote)
- If you were just given a data table, without the reaction equation, would it be possible to determine the coefficient in front of k (i.e. the how many moles of your reactant are in the balanced reaction equation)?(1 vote)
- If it was an elementary reaction, yes. But not the coefficients of an overall reaction with several elementary reactions.(1 vote)
- [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.