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### Course: MCAT > Unit 9

Lesson 18: Kinetics- Kinetics questions
- Introduction to reaction rates
- Rate law and reaction order
- Worked example: Determining a rate law using initial rates data
- First-order reaction (with calculus)
- Plotting data for a first-order reaction
- Half-life of a first-order reaction
- Worked example: Using the first-order integrated rate law and half-life equations
- Second-order reaction (with calculus)
- Half-life of a second-order reaction
- Zero-order reaction (with calculus)
- Collision theory
- The Arrhenius equation
- Forms of the Arrhenius equation
- Using the Arrhenius equation
- Elementary reactions
- Reaction mechanism and rate law
- Catalysts
- Kinetic and thermodynamic enolates

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

In this video, we'll use the first-order integrated rate law to calculate the concentration of a reactant after a given amount of time. We'll also calculate the amount of time it takes for the concentration to decrease to a certain value. Finally, we'll use the first-order half-life equation to calculate the half-life of the reaction. Created by Jay.

## Want to join the conversation?

- Sir ,isn't there any other method to find the concentration other than using ln and exponential? Because during our calculations we cannot use a calculator which is pretty tough.(5 votes)
- If you are referring to the MCAT exam, from what I have heard, the MCAT tests more conceptual knowledge about rate laws and simpler math, such as finding rate laws rather than complex problems like this. I think it is important to understand the main ideas behind this problem, such as half-life, but I don't think we will be expected to do this much math.(18 votes)

- is the half life always t1/2 = .693/K ?(2 votes)
- Only if the reaction is a first order reaction.(13 votes)

- At6:22, why don't you divide the two ln's? I thought you were supposed to divide when you subtract.(3 votes)
- well, he could have written it as 'ln(0.01/0.05)' but his method is more straightforward(1 vote)

- how would you find the starting concentration at time= zero if you were given a set of data (time vs concentration)?(1 vote)
- I would plot ln(c) vs. t and extrapolate the straight line back to t=0.(4 votes)

- How much math knowledge is needed for AP Chemistry? Do I need to know Pre-Calc?(2 votes)
- At a minimum, I would say algebra since it can get you through a majority of chemistry problems. But the more math you know, the better your problem solving capabilities.(1 vote)

- is it that the average rate of a first-order reaction always gets slower with time?(2 votes)
- In the worked example video: Are significant figures of any importance?(1 vote)
- Yes, they're important. You can't do science, at least professionally, without using correct significant figures.(2 votes)

- Shouldn't we round 2402 seconds to 2400 seconds, and 1034 seconds to 1000 seconds, because we only have two sig-figs in the rate constant?7:02,9:00(1 vote)
- You're correct, both those answers should be correctly reported with only two sig figs because of the rate constant. However writing 2400 and 1000 seconds can be misconstrued as having differnt amounts of sig figs and are considered ambigious answers. For example one person could argue that 2400 has two sig figs (the 2 & 4) while another person could argue that all four digits are significant.

To avoid this issue we should write them in scientific notation. So 2400 s becomes 2.4 x 10^(3) s, and 1000 becomes 1.0 x 10^(3) s. And now everyone is clear that both numbers have two sig figs each.

Hope that helps.(2 votes)

- do you have to convert the rate constant to seconds when determining the half live, if the rate constant is given in hours? or would you find the half live using the hours k and then convert the units to seconds?(1 vote)
- What happened to the units? I imagine that the logarithm only makes sense if the quantity is dimensionless, but I don't understand how we just make something unitless.(1 vote)
- Taking a logarithm of something doesn’t make it unitless. Taking the ln([A]) still results in a unit of molarity (M) since that is the unit of [A]. Only canceling units through division makes numbers dimensionless (unitless).

Jay doesn’t include the units throughout the problem, only the numerical values since it makes the calculations more compact. However, he does include units for the answers which is where they are truly needed.

If you follow the first-order integrated rate law using units, it works out.

Ln([A]t) = -kt + ln([A]0)

M = -(1/s)(s) + M

M = M

Hope that helps.(1 vote)

## Video transcript

- [Voiceover] We've already looked at the conversion of cyclopropane to propene and shown that it's a
first-order reaction. And we also found the rate constant at 500 degrees Celsius
in an earlier video. And so the rate constant is 6.7 times 10 to the negative 4 one over seconds. And so in part a, if the
initial concentration of cyclopropane is .05 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 concentraion 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 C three H six, is A. So this is the natural
log of the concentration of cyclopropane, C three H six. Is equal to k, is 6.7 times 10 to the negative four. So this would be negative times 6.7 times 10 to the negative four. And time. We need to find the time. And since we have k in seconds, we have 30 minutes here. We need to convert 30
minutes into seconds. So 30, 30 minutes. There are 60 seconds in one minute. 60 seconds in one minute. So if we multiply those two, 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. The initial concentration is .05. So this is the natural log of .05. So starting to run out of space there. Alright, so, let's think about what we would do to solve for the concentration of cyclopropane. Well we have this natural log in here. So we could exponentiate both sides to get rid of our natural logs. 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. Let's get out the calculator here. Let's do all that math on the right side. This was, let's see here, we have We have negative 6.7 times 10 to the negative four. I need to multiply that by 1800 seconds. And to that, we're going to
add the natural log of .05. And then we need to take e to our answer. And e to our answer gives us .015 molar. And so that's our answer. Our answer, our
concentration of cyclopropane is .015 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. So you could have written, the concentration of cyclopropane at a certain time is equal to the initial concentration of cyclopropane, the initial concentration
e to the negative kt. So we saw this in the previous videos. 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 .05 So we plug in here .05. e to the negative k was 6.7, 6.7 times 10 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 6.7 times 10 to the negative four times 1800, and then we would take e to our answer. And then we to multiply by
the initial concentration of .05. And so obviously we get .015 molar again. So it doesn't matter
how you do the problem, it doesn't matter which
equation you start with, you're gonna get .015 molar for your concentration of cyclopropane. Alright, let's look at
part b of this question. How long does it take
for the concentration of cyclopropane to reach .01 molar? Alright, so to reach .01 molar, once again, it doesn't really matter which form you use,
which equation you use. I'll just take the first
equation that we discussed. Write 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 what we need here. So the concentration of cyclopropane to reach .01 molar, so that's
our concentration here. That's our concentration at some time. So that would be the natural log of this would be .01 molar is equal to negative kt. So this is negative 6.7 times
10 to the negative four. The time is what we don't know. How long does it take? So we're trying to find the time. And we know the initial concentration of cyclopropane, once
again that .05 molar. So this is the natural log of .05 molar again. So to solve for time, this would be the natural log of .01 minus the natural log of .05 and then you have to divide that by negative 6.7 times
10 to the negative four. So just some algebra here gets you your time. So we can do that on our calculator. We can take, let's make some room over, let's put it over here. So this would be the natural log of .01 and then we're gonna subtract
the natural log of .05 and then divide that by negative 6.7 times 10 to the negative four. And then we get as our time 2,402. That's in seconds. So 2,402 seconds. Time is equal to 2,402 seconds. And we could leave it
in seconds, or we could we could put that into minutes. So how many minutes is that? Well, 2,402 seconds, if we divide by a conversion factor. So it'd be 60, there are
60 seconds in every minute. So seconds would cancel out. We get one over, one over minutes. So that gives us minutes, of course. And so we can do that really
quickly on the calculator, or you could do that in your head. I'll go ahead and do
that on the calculator just to show you. Let's take that and and
let's divide it by 60. Obviously we're gonna
get pretty close to 40. So once again, that's something that 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 .01 molar. Alright, let's look at part c. 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 .693 divided by your rate constant k. So the rate constant for this reaction was 6.7 times 10 to the negative four. So the half-life is equal to .693 divided by 6.7 times 10 to the negative four. And this was one over seconds. So we can do that on our calculator to solve for the half-life. So we have .693, divide that by 6.7 times 10 to the negative four. And so we get 1,034. So t 1/2, which is the half-life, which is 1,034 and the
units would be seconds. This would be one over one
over seconds, or seconds. So 1,034 seconds. If you wanted to convert
that into minutes, we could just divide that by 60 seconds. And that gives us 17, eh we'll just say approximately 17 minutes. 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, remember, we started with a
concentration of .05 molar. That was our initial concentration. A half-life is how long it takes for us to get half of our
initial concentration. So how long does it take
to get from .05 molar to half that, which is .0250 molar. That's the half life. It takes 17 minutes. It takes 17 minutes for our concentration of cyclopropane to go from .05 molar to .025 molar. And how long would it take for our concentration to reach .0125 molar? So this concentration is half of this, so therefore it would take another 17, it would take another 17 minutes. Again, watch the previous
video for more on half-life.