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### Course: Chemistry archive>Unit 10

Lesson 2: Relationship between reaction concentrations and time

# 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.

## Want to join the conversation?

• 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
• 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.
• Can someone explain how the formula was derived at the beginning of the video using the two equations?
• 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.
• 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?
• 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.
• 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...?
• 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).
• This isn't a question, but a comment to those who are wondering how to do the integration of that equation at . 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.
• 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 at .

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 at .

So, 2k = 0.1634
k = 0.0817 after dividing both sides of the equation by 2.

Hope that helps.
(1 vote)