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:9:21

let's see how to plot data for a first-order reaction so the conversion of cyclopropane into propane is a first-order reaction and in Part A they want us to use the experimental data to show that it's first-order so we look at the data over here and we can see as time increases right the concentration of cyclopropane decreases which makes sense because cyclopropane is turning into propane if we want to prove this is first-order we need to use the integrated rate law from the previous video so in the previous video we showed that the natural log of the concentration of a is equal to negative KT plus the natural log of the initial concentration of a and for this reaction a is cyclopropane so we could write this as the natural log of the concentration of cyclopropane c3h6 is equal to negative K T plus the natural log of the initial concentration of cyclopropane which we also talked about in the previous video follows the form y is equal to MX plus B so if we put if we put the natural log of the concentration of cyclopropane on the y axis let me put time on the x axis if we do that if we get a straight line or close to a straight line then we know that the reaction is first-order and the slope of that line right M should be equal to negative K where K is your rate constant and the y-intercept right should be equal to the natural log of the initial concentration of cyclopropane so if we're going to if we're going to graph that alright we need to figure out the natural log of the concentration of cyclopropane right so right now we have only the concentration of cyclopropane we need to take the natural log of all these numbers and before we graph something right so we need to take the natural log of point zero nine nine so we get out the calculator here so natural log of point zero nine nine gives us negative two point three one so we put in negative two point three one here next the natural log of 0.7 nine so the natural log of point zero seven nine and we get negative two point five four so negative two point five four next natural log of point zero six five so the natural log of point zero six five gives us negative two point seven three so this is negative two point seven three and then one more so the natural log of point zero five for natural log of point zero five four is equal to negative two point nine two so we have negative two point nine two and so now now we're going to do the natural log of the concentration of cyclopropane on the y-axis all right and we're going to do time on the x-axis let's go down here and look I already have the axes labeled alright so on the x-axis right down here we have time and on the y-axis we have the natural log of the concentration of cyclopropane so let's let's figure out some points let's figure out some some points here on our graph so when time is equal to zero right y is equal to negative two point three one so when time is equal to zero we have negative two point three one this is negative two so this is negative two point one negative two point two and so this would be negative two point three so negative two point three one would be pretty close to there and obviously I it's very hard to graph something perfectly given what we're trying to do here on this video so I'll just say that's approximately negative two point three one next point is that when x is equal to three hundred seconds we have y is equal to negative two point five four so three hundred seconds would be here two point five four that's pretty close to you here all right so we'll say that's approximately that point next time is equal to 600 seconds negative two point seven three so we have 600 seconds two point seven three this would be negative two point six negative two point seven so negative two point seven three close to there all right certainly not perfect but close enough all right and then finally time is equal to nine hundred seconds negative two point nine two so nine hundred seconds negative two point nine two would be pretty close pretty close to you there alright so let's see if we can draw a straight line through those points are pretty close to being through those points here so we're putting a line of best fit alright let's see if we can do well that looks pretty good actually alright so we put our line and our line is pretty close to passing through our points so the points fall on our straight line so we can say that the reaction is first-order alright so this reaction is first-order we plotted everything and we got y is equal to MX plus B so we're done with Part A because we got a straight line here all right we can say that this is a first-order reaction alright for Part B our job is to calculate the value of the rate constant and the rate constant remember let's go back up here the rate constant is K and we know the slope is M all right so the slope of that line is M and the slope is equal to negative K so let's go back down to here so the slope of this line the slope of this line is equal to negative K so we can find the slope a few different ways one way would be to do Delta Y over Delta X right so your slope is equal to change in Y over change in X so if you picked a point here and you picked a point here right you could figure out your slope that way all right so I'm just showing you what the slope would be this would be Delta Y right here and then this would be Delta X your units would be 1 over seconds all right so your units for K are going to be 1 over second and we could have figured this the unit's out a different way by using the rate law all right so this is for writing the rate law for our reaction the rate of our reaction is equal to the rate constant K times the concentration of cyclopropane to the first power because this is a first-order reaction all right so the rate of a reaction would be in molar per second and so this would be times K and then we have the concentration would be molar and this is to the first power so obviously right molars would cancel and you'd get K is equal to 1 over second so however you want to think about it the units for K will be 1 over seconds all right so we could we could figure out some points and calculate K but we wouldn't get super accurate value for K using using this graph which we just plotted by hand so let's go ahead and get out the calculator and let's find what K is using the calculator all right so we can we can plug in our points this isn't the calculator I'm used to using so it's actually a bit more difficult on this calculator than then not the one I usually use but we can do it we can go to stat and then go to f2 for edit and then hit enter twice and then we can start putting in our data points all right so when X is equal to zero all right y is equal to negative two point three one all right next when time is equal to 300 seconds when X is equal to 300 Y is equal to negative two point five four when time is equal to 600 all right Y is equal to negative two point seven three and then finally one more point when time is equal to 900 y is equal to negative two point nine two so we have all of our data in there now we can exit we can go back into stat and then hit f1 for a calc all right so that's what we want hit enter twice and then we want a linear regression so that's right here we want a linear regression so we hit f2 and B all right B is M B is the slope so the slope of that line is negative six point seven times ten to the negative four all right so let's go ahead put that in here the slope is negative six point six point seven times ten to the negative four and that's equal to negative K so obviously K or the rate inte K is equal to six point seven times ten to the negative four and this would be one over seconds all right so we've now figured out the rate constant all right so we've proved that this is a first-order reaction all right by graphing our data and then we found the value of the rate constant by finding the slope by finding the slope of our best fit line