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# Worked example: Using the first-order integrated rate law and half-life equations

AP.Chem:
TRA‑3 (EU)
,
TRA‑3.C (LO)
,
TRA‑3.C.4 (EK)
,
TRA‑3.C.5 (EK)

## Video transcript

we've already looked at the conversion of cyclopropane to propane and shown that it's a first-order reaction and we also found the rate constant at 500 degrees Celsius in an earlier video so the rate constant is six point seven times 10 to the negative 4 1 over seconds and so in Part A if the initial concentration of cyclopropane is point zero five molar what is the concentration of cyclopropane after 30 minutes well to solve for this concentration we can use the integrated rate law that we found in an earlier video since this is a first-order reaction the integrated rate law or one form of it is the natural log of the concentration of a at any time T is equal to the negative KT where K is the rate constant plus the natural log of the initial concentration of a and here cyclopropane or c3h6 is a so this is the natural log of the concentration of cyclopropane c3h6 is equal to K is six point seven times 10 to the negative 4 so this will be negative times six point seven times ten to the negative four and time we need to find the time and since we have K in seconds right we have 30 minutes here we need to convert 30 minutes into seconds so 30 30 minutes right there are 60 seconds in one minute right 60 seconds in one minute so if we multiply those two right minutes cancels out so 30 times 60 gives me 1800 seconds so that's 1800 seconds so I put that in here so 1800 seconds plus the natural log the natural log of the initial concentration of cyclopropane right the initial concentration is point zero five so this is that the natural log of point zero five so starting right now face there all right so let's let's think about what we would do to solve for the concentration of cyclopropane all right well we have this sum we have this natural log in here all right so we could exponentiate both sides to get rid of our natural log so if we exponentiate both sides that gets rid of our natural log here and that would give us the concentration of cyclopropane after 1800 seconds so let's do that math it's get out the calculator here let's do all that math on the right side this was this was let's see here we have we have negative six point seven times ten to the negative four I need to multiply that by 1800 seconds and to that we're going to add the natural log of point zero five and then we need to take e to our answer right and e to our answer gives us point zero one five molar and so that's our answer our answer our concentration of cyclopropane is point zero one five molar now you didn't have to use this form of the integrated rate law you could have used a different form the concentration as a function of time all right so you could have written the the concentration of cyclopropane right at a certain time is equal to the initial concentration of cyclopropane at the initial concentration e to the negative KT so we saw this in in the previous video so the concentration as a function of time so we can plug in our values here we're solving for the concentration of cyclopropane the initial concentration was point zero five so we plug in here point zero five e to the negative K was six point seven six point seven times ten to the negative four and our time was 1800 seconds so we could do the problem this way obviously we should get the same answer let's just go ahead and show that here so we have negative six point seven times ten to the negative four time's 1800 all right and then we would take e to our answer and then we need to multiply by the initial concentration of point zero five and so obviously we get point zero one five molar again so it doesn't matter how you do the problem all right doesn't matter which equation you start with you're going to get point zero one five molar for your concentration of cyclopropane all right let's let's look at Part B of this question how long does it take for the concentration of cyclopropane to reach point zero one molar all right so to reach point zero 1 molar once again doesn't really matter which form you use which equation you use I'll just take the first equation that we discussed right the natural log of the concentration of a is equal to negative KT plus the natural log of the initial concentration of a and let's plug in let's plug in what we need here so the concentration of cyclopropane to reach point zero one molar so that's our concentration here right that's our concentration at some time so that would be the natural log of this would be point zero one more is equal to negative KT so this is negative six point seven times ten to the negative four the time is what we don't know all right how long does it take so we're trying to find the time and we know the initial concentration of cyclopropane once again that's point zero five molar so this is the natural log of this is natural log of point zero five molar again all right so to solve for time this would be the natural log of 0.1 all right minus the natural log of point zero five and then you have to divide that by negative six point seven times ten to the negative four so just some algebra here gets you your time so we can do that on our calculator all right we can take let's make some room over to put it over here all right so this would be the natural log of points zero-one and then we're going to subtract the natural log of 0.5 and then divide that by negative six point seven times ten to the negative four and then we get as our time at 2400 and - all right that's in seconds right so two thousand four hundred and two seconds time is equal to two thousand four hundred and two seconds and we could leave it in seconds or we could we could put that in two minutes all right so how many minutes is that well two thousand four hundred and two seconds if we divide by a conversion factors v60 there are 60 seconds in every minute all right so seconds would cancel out we get 1 over 1 over minutes and so that gives us minutes of course and so we could do that really quickly on the calculator or you could do that in your head all right I'll go ahead and do it on the calculator just to show you all right let's take that and let's divide it by 60 obviously we're gonna get pretty close to 40 all right so once again that's something you could do in your head but 40 minutes so approximately 40 minutes is how long it would take to reach a final concentration of Oh point zero one molar all right let's look at Part C alright so in Part C they want us to find the half-life well from the previous video for a first-order reaction the half-life is equal to 0.693 divided by your rate constant K and so the rate constant for this reaction was six point seven times 10 to the negative 4 so the half-life is equal to 0.693 divided by six point seven times 10 to the negative 4 and this was 1 over seconds all right so we can do that on our calculator to solve for the half-life so we have 0.693 and divide that by six point seven times ten to the negative four and so we get 1000 and the four right so T 1/2 which is the half life which is 1034 and the units will be seconds all right this would be 1 over 1 over seconds or seconds so 1034 seconds if you wanted to convert that into minutes all right we could just we could just divide that by 60 seconds and that gives us 17 we'll just say approximately 17 minutes all right so we'll round that to approximately 17 minutes is the half-life and remember from the previous video what the half-life refers to so if we started with where we started with a concentration of point zero five molar right that was our initial concentration half-life is how long it takes right for for us to get half of our initial concentration so how long does it take to go from point zero five molar to half that which is point zero two five zero molar all right that's the half-life it takes 17 minutes all right take 17 minutes for our concentration of cyclopropane to go from point zero five molar to point zero two five molar and how long would it take for our concentration to reach point zero one to five molar all right so this concentration is half of this so therefore take another 17 it would take another 17 minutes so again watch the previous video for more on half-life
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