If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

## AP®︎/College Chemistry

### Unit 5: Lesson 3

Concentration changes over time

# Second-order reactions

AP.Chem:
TRA‑3 (EU)
,
TRA‑3.C (LO)
,
TRA‑3.C.1 (EK)
,
TRA‑3.C.3 (EK)
,
TRA‑3.C.4 (EK)
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? • 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)
• 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) • 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) 