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## MCAT

### 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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# First-order reaction (with calculus)

Deriving the integrated rate law for first-order reactions using calculus. How you can graph first-order rate data to see a linear relationship.

## Want to join the conversation?

- At1:26, I read below that "d" stands for derivative, but what exactly is a derivative and why do we need it in this equation? I would suggest that the person making this video explain for us non-calculus viewers what a derivative is and it's function or reference a video where we can find a simple explanation. Additionally, at2:25, what is integration? Why do we need to "integrate"?(11 votes)
- In this case we' re saying that the rate of consumption of the reagent d{A}/dt is directly proportional to the concentration {A}. Therefore when the concentration {A} decreases, the rate of consumption also decreases, it' s a differential equation and you have to integrate in order to come up with a function of {A} respect to time.(3 votes)

- 0:47, How did you know it was a first order reaction?(8 votes)
- He is saying, IF we know that the reaction is first order, this is how to do the math.(7 votes)

- how is the integration of d(A)/(A) = ln(A) at3:21(7 votes)
- Let's just say that A is equal to x for now.

d/dx (ln x) = 1/x

To derive the rate law, I'm taking the antiderivative of dx/x, or 1/x dx.

So then if I take the antiderivative of both sides:

In x = antiderivative of 1/x(5 votes)

- We have been taught first order formula as 0.693/k...can't we use that one??(1 vote)
- yes, it is exactly the same thing because the natural log of 2 is 0.693(5 votes)

- Why does taking the indefinate integral of

(d (A))/[A] = -kt

only leaves the left side of the equation with a constant and not also the right?

Int[ (d (A))/[A] = -kt ] =

ln[A]t + X = -kt + Y, where X & Y are both constants.

Is it that

Y - X = ln[A]o

after setting the general boundaries?

thnx in advance...(2 votes)- One of the constants is zero.

dA/A = -kdt

∫dA/A = -k∫dt

lnA] = -kt]

A runs from A₀ to A, and t runs from 0 to t.

lnA – lnA₀ = -kt + 0

lnA = -kt + lnA₀(4 votes)

- At 0408, he says this formula is on the AP chemistry equation sheet. Are these videos really made with the MCAT content in mind? Any thoughts?(2 votes)
- I dont think they were made for mcat review but they were chosen because they go over mcat content. The mcat doesnt use calculus so I'm assuming this is just for deeper understanding(3 votes)

- Don't we get an exponential graph if we solve the equation this way or am I making a mistake in my calculations?

ln(At) - ln(a0) = -kt

ln(At/A0) = -kt

e^-kt = At/A0

At = A0*e^-kt(2 votes) - At2:03, why can't you take the derivative of both sides of the equation?(1 vote)
- You could take the derivative of both sides, but it wouldn't achieve a meaningful result. The goal here is to get a function that allows us to calculate [A] at any given time, t. We start out with the rate of how [A] changes with time, and this rate is changing, so we need to integrate in order to be able to calculate [A] for a particular time, t. If we took a derivative, as you suggested, then we'd be able to determine that rate at which the rate is changing. This, however, is not what we're interested in here.(3 votes)

- What is the integrated rate law? I don't see any videos on that.(2 votes)
- Any place you see the concentration of a reactant as a function of time, it is an integrated rate law. Just to mention, the other type, the differential rate law, gives the rate of a reaction as a function of concentration of a reactant, assuming the effects of the reverse reaction are negligible.(1 vote)

- In this video are we assuming that the reaction is a first order reaction because there is only one reactant.(1 vote)
- Nope, we are assuming it is a first order reaction because this reaction only has one step. And this step only has one reactant with a coefficient of 1.

If this reaction has multiple steps, even if it has one reactant, the order of reaction may not necessarily be equal to 1.

Hence, the order of reaction depends on whether the reaction is one-step or multi-step.

If it is one-step, you can take the coefficient of reactant as the order

If it is multi-step, there is a video later on that will tell you how to derive the order(3 votes)

## Video transcript

- [Voiceover] Let's say we
have a first order reaction where A turns into our products, and when time is equal to zero we have our initial concentration of A, and after some time T, we have the concentration
of A at that time T, and let's go ahead and write
out the rate for our reaction. In an early video, we
said that we could express the rate of our reaction in
terms of the disappearance of A, so we said that's the change in the concentration of A
over the change in time, and we put a negative in sign in here to give us a positive value for the rate. We could also write out the rate law, so for the rate law, the rate is equal to the rate constant K times the concentration of A, and since this is a first order reaction, this would be to the first power, so we talked about this in an early video. We can set these equal
to each other, right? Since they're both equal to the rates, we can say that the rate
of disappearance of A, so the negative change
in the concentration of A over the change in time is
equal to the rate constant K times the concentration
of A to the first power. This is the average rate of
reaction over here on the left, and so if we wanted to write
this as the instantaneous rate, we need to think about
calculus, all right? So the instantaneous rate would
be at negative D A, right? D A, the negative rate of change
of A with respect to time, so this would be D T, and this is equal to K
A to the first power, and now we have a differential equation, and when you're solving
a differential equation eventually you get to a function, and our function would be the concentration as a function of time, so we will eventually get there, but your first step for
solving a differential equation is to separate your variables, so you need to put all
the A's on one side, and we'll put the T on the opposite side, so we need to divide through by A, so we get on the left side... We divide both sides by A, and we get D A over the
concentration of A here. We're going to multiply both sides by D T to get the T on the right side, so we would get K D T
over here on the right, and I'll go ahead and put the negative sign on the right as well, so we just rearranged a few things to get us ready for integration, right? After you separate your variables you need to integrate, so we're going to integrate the left, and since K is a constant I can pull it out of my integral, and I can go like that, and let's see, what would be integrating from? Well, for time, let's go back up to here. All right, time we would
be integrating from time is equal to zero
to time is equal to T, and for concentration we'd be integrating from the initial concentration to the concentration at some time T, so let's go ahead and plug those in, so we'd be integrating from
zero to T on the right side, and the left side we'd be integrating from the initial concentration of A to the final concentration, or I should say the
concentration of A at any time T. All right, so on the left, right, D A over A, that's natural
log of A, all right? So that would be equal
to the natural log of A, and we would be evaluating this from the initial concentration to the concentration at some time T, and, on the right side,
integral of D T is just T, so we have negative K T
evaluated from zero to T. Next, we use the fundamental
theorem of calculus, so on the left side we
would have natural log of the concentration of A at some time T minus the natural log of A... Minus the natural log of the
initial concentration of A, and then on the right side
we would have negative K T, and so this is one way to write the integrated rate law, so on the current AP Chem
formula sheet, right, this is your equation for
a first order reaction, so this is your integrated rate law. Your integrated rate law, and this is one way to write it. You could keep on going and express this in a different way, which we'll do in a later video, but this is one equation that you can use to solve some problems here, and let's go ahead and
rearrange that a little bit. All right, let's add the natural log of the initial concentration
to the right side, so let's rearrange that, and, therefore, we get natural
log of the concentration of A at some time T is equal to negative K T plus the natural log of the
initial concentration of A, and the reason why I
wanted to rearrange this is so you can see that this follows the graph of a straight line. Think about Y is equal to M X plus B. All right, so we know that
Y is equal to M X plus B is the graph of a straight line, and if you think about what
you would need to graph here, all right, you would put
the natural log of A, all right, on your Y axis, and on your X axis you
would put time, right? And then we can see that
M, which is the slope, would be equal to negative K, and B, of course, is your Y intercept, and so that should be the natural log of the initial concentration of A, and so if we put together
a little graph here... Let me just real quickly
sketch in a graph, so we would have... Let's say that's our Y axis here, right? So here's our Y axis, and then we would have our X
axis right here, all right? And let's go ahead and label on axises, so for our Y axis we would be
putting over here on the left this would be the natural log of the concentration of A at any time T, and then on our X axis we
would put time, all right? So here's what we're graphing, and then the graph of this should be a straight line, all right? So the graph is going
to be a straight line. Let me go ahead and draw it in right here, so I'm just going to attempt to sketch a straight line here like that, and this would be our
Y intercept, all right? And our Y intercept
should be the natural log of the initial concentration
of A, all right? So this point right here, our Y intercept, should be the natural log of
the initial concentration of A, and the slope of our line, all right, let's go back over here... The slope of our line, which is M, should be equal to negative K, so if you get the slope of this line, this slope is equal to negative K, so from the slope you can
find the rate constant, and the graph of the
natural log of A verses time gives a straight line with
a slope of negative K.