- First-order reactions
- First-order reaction (with calculus)
- Plotting data for a first-order reaction
- Half-life of a first-order reaction
- Half-life and carbon dating
- Worked example: Using the first-order integrated rate law and half-life equations
- Second-order reactions
- Second-order reaction (with calculus)
- Half-life of a second-order reaction
- Zero-order reactions
- Zero-order reaction (with calculus)
- Kinetics of radioactive decay
- 2015 AP Chemistry free response 5
Plotting data for a first-order reaction
Example of graphing first-order rate data to see a linear relationship, and calculating rate constant k from the slope.
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- I got 7.66x10E -4, what did I do wrong?
I used -2.54+2.31 all over 300 seconds.(9 votes)
- The points provided won't produce a perfectly straight line. So the slope between points 1 and 2 may be slightly different between that of 2 and 3, etc. A best fit line is put in place and the slope of the best fit line is what would be used to determine K.(7 votes)
- wouldn't it be easier to take the concentration at 900s minus the initial concentration and divide by 900? Same as rise over run and you don't need to worry about the sloppy graph.(5 votes)
- It is if you know that the data is perfectly linear. This will happen with low level book problems however when you start doing calculations based on lab work there will be a degree of variance in your results. Even if you do flawless work there is systematic and instrument error to take into account. When this is the case you can't just take a shortcut and get accurate results. Even though you are able to do so at this point it is best to practice making calculations based on all the data points so you will be able to do the work later on.(5 votes)
- i did not very well grasp how the slope was calculated..are there any basics for this i should understand first. if yes from where please?(3 votes)
- i'm calculating the slope manually and I keep getting 7.67 x 10 ^-4; although i got the 6.7x10-4 via graphing in calculator as well. can you show me how to find slope manually (using the rise/run formula)(2 votes)
- you can get a pretty close estimate by using the data in the table:
change in [A] / change in time = R
(0.079 - 0.099) / (300s) = R
and then use R / [0.099] = k = 6.734 x 10^-4(3 votes)
- is there no other way to prove if a given reaction is first order or not and to find the rate constant other than plotting a graph?(2 votes)
- If you are given experimental data and an equation, you can plug in and solve for R=k[A]^x[B]^y. Khan Academy has another video on this: https://www.khanacademy.org/science/chemistry/ap-chemistry/kinetics-ap/reaction-rates-and-rate-laws-ap/v/experimental-determination-of-rate-laws?modal=1(2 votes)
- Wouldn't a concentration of cyclopropane vs. time graph be an exponential curve?(1 vote)
- Yes. If we let [A] = the concentration of cyclopropane, the integrated rate law is
[A] = [A]₀e^(- k t)
A plot of [A] vs. t is a curve that starts at [A]₀ and gradually approaches the horizontal axis asymptotically as t increases.
The shape of the curve is called an exponential decay curve.
If you take the natural logarithm of each side of the equation, you get
ln[A] = ln[A]₀ - kt
If you plot ln[A] vs t, you get a straight line with slope = -k and y-intercept =ln[A]₀.(4 votes)
- is the equation thingy only valid for the first-order reaction or is it valid for any(2 votes)
- only for the first-order reactions. For others, we would've to integrate their equation again.(1 vote)
- I get the graph for 1st order would be linear, but why not just plot concentration vs time...isn't that more practical?(2 votes)
- For part b of the problem, when calculating the slope of the plot line, I kept getting m= -6.333 x 10^-4. However, I didn't use a stats function on my calculator. I used the method of change in y/change in x. I did this with several of the points in the table, and I kept getting the same thing. Am I doing something wrong?(2 votes)
- What does ln[A]t represent and why didn't he write t with ln[C3H6] around0:51(2 votes)
- [Voiceover] Let's see how to plot data for a first-order reaction, so the conversion of cyclopropane into propene 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 propene. If we want to prove that 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 K T 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 cycolopropane, C three H six, 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 M X plus B, so if we put the natural log of the concentration of cyclopropane on the Y axis, and we put time on the X axis, if we do that, and 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 graph that, all right, 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 of 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 negative two point three one here. Next the natural log of point zero seven 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 four. Natural log of zero five four is equal to negative two point nine two, so we have negative two point nine two, and so now we're going to do the natural log of the concentration of cyclopropane on the Y axis, right? And we're going to time on the X axis, so let's go down here and look. I already have the axises labled, right? So on the X axis, all right, down here we have time, and on the Y axis we have the natural log of the concentration of cyclopropane, so let's figure out some points, let's figure out some points here on our graph, so when time is equal to zero, all 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, it's bery hard to graph something perfectly given what we're trying to do here on this video, so I'll just say that's approximately two point three one. Next point is that when time is equal to 300 seconds we have Y is equal to negative two point five four, so 300 seconds would be here. Two point five four, that's pretty close to here, all right? So we'll say that that's approximately that point. Next, time is equal to 600 seconds. Negative point two 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 900 seconds. Negative two point nine two, so 900 seconds, negative two point nine two would be pretty close to there. All right, so let's see if we can draw a straight line through those points, or pretty close to being through those points here, so we're putting a line of best fit, right? Let's see what we can do. Well, that looks pretty good, actually, all right, 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, all right? So this reaction is first-order, and we plotted everything, and we got Y is equal to M X 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. All right, 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, all 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, all 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 a delta X. Your units would be one over seconds, right? So your units for K are going be one over seconds, and we could have figured out the units 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 a molar. This is to the first power, so, obviously, molars would cancel, and you would get K is equal to one over seconds, so however you want to think about it. The units for K will be one over seconds. All right, so we could figure out some points and calculate K, but we wouldn't get super accurate value for K using this graph, which we just plotted by hand, so let's go ahead and get out the calculator, and let's find out what K is using the calculator, all right? So we can plug in our points. This isn't the calculator I'm used to using, so it's actually a little bit more difficult on this calculator than the one I usually use, but we can do it. We can go to Stats, and then go to F two 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, 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, Y is equal to negative two point seven three, and then, finally, one more point. When time is equal to 900, Y is equal negative two point nine two, so we have all of our data in there now. We can exit, we can go back into Stats, and then hit F one for 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 F two, 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 and put that in here. The slope is negative six point seven times ten to the negative four, and that's equal to negative K, so obviously K, or the rate constant, K is equal to six point seven times ten to the negative four, and this would be 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 of our best fit line.