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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.

## 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.
• 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.
• 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.
• 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) • 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)
• 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)
• In most of the rate law lectures, the order of the reaction is mentioned as "1/2 with respect to A/B". Why the specification? Won't the order be the same for the entire reaction?
(1 vote) • Not necessarily, the order of the overall reaction can be different from the individual reactant’s orders. For example, both of these reactions are second order overall:

(1) A^(2) → B
(2) A + B → C
But in reaction 1 it is second order with respect to A, while in reaction 2 it is first order with respect to A and B. In both cases the orders of the reactants sum to 2, but the reactants are not always second order.

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 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.
(1 vote) • here the order of the reaction is not determined by the coefficient, while in rate law it is? : https://www.khanacademy.org/science/ap-chemistry-beta/x2eef969c74e0d802:kinetics/x2eef969c74e0d802:reaction-mechanisms/v/mechanisms-and-the-rate-determining-step
Why is this?
(1 vote) • The order of the reaction is determined by the coefficient of the slowest elementary step here. The slow step is: NO2 + NO2 → NO + NO3, but this can be written as 2NO2 → NO + NO3, since NO2 appears twice. It’s like combining like terms in algebra. The coefficient in the slowest elementary step is 2, which means the reaction is second order with respect to NO2, and also second order overall.

Hope that helps.
(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) 