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.

Main content

Current time:0:00Total duration:7:14

let's say that our reaction here is second order in a so a turns into our products and what time is equal to zero we're starting with our initial concentration of a after some time T we would have the concentration of a at that time T so if we're expressing the rate of our reaction we know we can do that in a couple different ways so we can say that the rate is equal to the negative change in the concentration of a over the change in time so we've done this in earlier videos we can also say that our rate is equal to opting what the rate law the rate is equal to the rate constant K times the concentration of a and since I said this is second order in a this is a second order reaction this is the concentration of a to the second power now we can set these two rates equal we can set this equal to this so let's do that we have the negative change in the concentration of a over the change in time and that's equal to the rate constant K times the concentration of a squared next we need to think about our calculus alright so instead of instead of expressing this as an average rate change in a over change in time we can think about the instantaneous rate right so the rates of change of the concentration of a with respect to time so we can write negative D a D concentration of a over DT so the rate of change in the concentration of a with respect to time and this is equal to K a squared and we have our differential equation right we we could solve our differential equation and get a function and the first thing that we do to solve a differential equation is to separate our variables all right we need to put the A's on one side and we need to put the T on the other side the DT on the other side so we divide both sides by a squared so now on the Left we have negative da over concentration of a squared and we multiply both sides by DT so we get K DT on the right side next we're ready to integrate alright so we're ready to integrate and let's let's rewrite what we have on the left here so this would be the negative I'm going to say this is the concentration of a to the negative two power D a just to make the integral a little bit easier to follow and on the right here we still have we still have K DT all right so we're ready to integrate so we're going to integrate the left here and we're going to integrate the right K is a constant so we can pull it out of our integral and let's go back up here and remind ourselves what we will be integrating from so for time all right let's look right here we're going from time is equal to zero to time is equal to T and for our concentration we'd be going from our initial concentration to our concentration at time a T so let's go ahead and put those in so we have we're integrating from zero to T so from zero to T and we're integrating from our initial concentration our initial concentration to our concentration at some time T all right so on the left side and the left side what do we have here so what's the integral this is like X to the negative two DX right so it's like the integral of x to the negative two DX here so that would give us let me give us the concentration of a to the negative first over negative one so obviously you have to add some calculus if you're watching the video at this point all right and then we still had our negative sign all right so let me go ahead and put this in here this negative sign right here is still there and we would be evaluating this this would be equal to let me just rewrite this here so this is the same as this would be the same as 1 over the concentration of a so you have two negative signs and so now you have a positive so this would be 1 over the concentration of a since a to the negative first is 1 over the concentration of a we're evaluating this from our initial concentration to our concentration at time T and on the right side the integral of DT would just be T all right so we have K T evaluated from 0 to T next we use the fundamental theorem of calculus right so we go ahead and plug in what we have right as we plug in this first so we have one over the concentration of a T so B minus one over the constant the initial concentration of a and then on the right side this of course would just be K T so we've found we found our integrated rate law for a second-order reaction all right so this is this is our integrated our integrated rate law some people call this the integrated rate equation right it doesn't really matter which one you're thinking about but this this is very useful because we can rearrange this right we can we can move the one over initial concentration of H to the right side so we get 1 over the concentration of a right is equal to K T plus 1 over the initial concentration of a and now this looks very familiar right this is the equation for a straight line we can see that this would be y is equal to MX plus B all right so if you're graphing if you're graphing time on your x axis and on your Y axis you have 1 over the concentration of a you're going to get you're going to get a straight line right and the slope of that straight line the slope which is M the slope should be K your rate constant and the y-intercept right the y-intercept here which is B should be 1 over the initial concentration of a so let's let's really quickly sketch out sketch out this graph here so we have our axes alright and on our x-axis we're going to put time alright so on our x-axis we put time and on the y-axis we put 1 over the concentration of a so we have 1 over the concentration of a and if this is a second-order reaction your graph should be a straight line so I'm going to go ahead and draw a straight line in here or attempt to draw a straight line so you get the idea and the slope of this line all right the slope of our line slope over line is equal to K it's equal to the rate constant all right that's what we saw up here the slope is equal to the rate constant and the y-intercept so this point right here right this should be 1 over the initial concentration of a alright so this point right here is 1 over the initial concentration of a so that's the idea of the integrated rate law or the integrated rate equation for a second-order reaction