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### Course: MCAT > Unit 10

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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# Using the Arrhenius equation

How to use the Arrhenius equation to calculate the activation energy.

## Want to join the conversation?

- Is there a specific EQUATION to find A so we do not have to plot in case we don't have a graphing calc??(9 votes)
- k = A e^(-Ea/RT)(2 votes)

- For the first problem, How did you know it was a first order rxn?(3 votes)
- In this problem, the unit of the rate constants show that it is a 1st-order reaction. The only reactions that have the unit 1/s for k are 1st-order reactions.(4 votes)

- why the slope is -E/R why it is not -E/T or 1/T(3 votes)
- R is a constant while temperature is not.(5 votes)

- is y=mx+b the same as y=mx+c in the UK?(3 votes)
- Yes, I thought the same when I saw him write "b" as the intercept. In the UK, we always use "c" :-)(2 votes)

- For T1 and T2, would it be the same as saying Ti and Tf?

Ti meaning Initial Temperature

Tf meaning Final Temperature(2 votes)- Yes, of corse it is same. You can write whatever you want ,but provide the correct value(3 votes)

- Shouldn't the Ea be negative? -19149=-Ea/8.314

-19149*8.314=-Ea

Did you multipy by -8.314(2 votes)- The negatives cancel. Also, think about activation energy (Ea) being a hill that has to be climbed (positive) versus a ditch (negative)(3 votes)

- how do you find ln A without the calculator?(2 votes)
- You can't do it easily without a calculator.

If you don't have a personal calculator, you can use an on-line calculator like this:

http://www.eeweb.com/toolbox/calculator

or simply enter an expression like ln 2 in the search box of your search engine.(0 votes)

- what does inK=lnA-Ea/RT mean?(1 vote)
- It is ARRHENIUS EQUATION used to find activating energy or complex of the reaction when rate constant and frequency factor and temperature are given . It can also be used to find any of the 4 date if other 3are provided(2 votes)

- How would you know that you are using the right formula?(1 vote)
- See the given data an what you have to find and according to that one judge which formula you have to use(2 votes)

- What are the units of the slope if we're just looking for the slope before solving for Ea?(1 vote)

## Video transcript

- [Voiceover] Let's see how we can use the Arrhenius equation to find the activation energy for a reaction. And so let's say our reaction is the isomerization of methyl isocyanide. So on the left here we
have methyl isocyanide and it's going to turn into its isomer over here for our product. This is a first-order reaction and we have the different rate constants for this reaction at
different temperatures. Let's just say we don't have anything on the right side of the
line I just drew yet. We only have the rate constants
at different temperatures. And in part a, they want us to find the activation energy for
the reaction in kJ/mol. One way to do that is to remember one form of the Arrhenius equation we talked about in the previous video, which was the natural log
of the rate constant k is equal to -Ea over R where Ea is the activation energy and R is the gas constant, times one over the temperature plus the natural log of A,
which is the frequency factor. And this is in the form of y=mx+b, right? So if you graph the natural
log of the rate constant on the y axis and one over
the temperature on the x axis, you're going to get a straight line. And the slope of that straight line m is equal to -Ea over R. And so if you get the slope of this line, you can then solve for
the activation energy. If you wanted to solve
for the frequency factor, the y-intercept is equal
to the natural log of A which is your frequency factor. So you could solve for
that if you wanted to. If you put the natural
log of the rate constant on the y axis, so up here
this would be on the y axis, and then one over the
temperature on the x axis, this would be your x axis here. I went ahead and did the math
just to save us some time. If you took the natural log
of this rate constant here, you would get this value. And if you took one over this temperature, you would get this value. And so now we have some data points. We have x and y, and we have
these different data points which we could put into the calculator to find the slope of this line. So let's do that, let's
plug those values in. So we go to Stat and we go to Edit, and we hit Enter twice
and then start inputting. So x, that would be 0.00213. So when x is equal to 0.00213, y is equal to -9.757. Let's put in our next data point. So that's when x is equal to 0.00208, and y would be equal to -8.903. Our third data point is when x is equal to 0.00204, and y is equal to - 8.079. Next we have 0.002 and we have - 7.292. And then finally our last data point would be 0.00196 and then -6.536. Alright, so we have everything inputted now in our calculator. Let's exit out of here, go back
into Stat, and go into Calc. And we hit Enter twice. We want a linear regression, so we hit this and we get
our linear regression. So we can see right
here on the calculator, b is the slope. So that's -19149, and then the y-intercept would be 30.989 here. So let's go ahead and write that down. So we have, from our calculator, y is equal to, m was - 19149x and b was 30.989. Alright, we're trying to
find the activation energy so we are interested in the slope. So the slope is -19149. And those five data points, I've actually graphed them down here. You can see that I have the natural log of the rate constant k on the y axis, and I have one over the
temperature here on the x axis. And here are those five data points that we just inputted into the calculator. And so the slope of our line is equal to - 19149, so that's what we just calculated. So to find the activation energy, we know that the slope m is equal to-- Let me change colors here to emphasize. The slope is equal to -Ea over R. So the slope is -19149, and that's equal to negative
of the activation energy over the gas constant. And R, as we've seen
in the previous videos, is 8.314. So we can solve for the activation energy. So let's get out the calculator
here, exit out of that. This would be 19149 times 8.314. And so we get an activation energy of, this would be 159205 approximately J/mol. Our answer needs to be in kJ/mol, so that's approximately 159 kJ/mol. So let's write that down. Activation energy is equal to 159 kJ/mol. And so we've used all that
data that was given to us to calculate the activation
energy in kJ/mol. In part b they want us to
find the activation energy, once again in kJ/mol. But this time they only want us to use the rate constants at two
different temperatures, at 470 and 510 Kelvin. And so we need to use the other form of the Arrhenius equation
that we talked about in the previous video. So this one was the natural log of the second rate constant k2 over the first rate constant k1 is equal to -Ea over R, once again where Ea is
your activation energy, times one over T2 minus one over T1. And so for our temperatures, 510, that would be T2 and then 470 would be T1. Let's go ahead and plug
in what we know so far. So the natural log, we have to look up these rate constants, we will look those up in a minute, what k1 and k2 are equal to. And that would be equal to
negative of the activation energy which is what we're trying to find, over the gas constant
which we know is 8.314. This would be times one over T2, when T2 was 510. So one over 510, minus one over T1 which was 470. So one over 470. Now let's go and look up those values for the rate constants. So we're looking for the rate constants at two different temperatures. And our temperatures are 510 K. Let me go ahead and change colors here. So we're looking for k1 and k2 at 470 and 510. So let's go back up here to the table. So 470, that was T1. And so this would be the value
for the first rate constant, 5.79 times 10 to the -5. And then T2 was 510, and so this would be our
second rate constant here. So 1.45 times 10 to the -3. And so let's plug those values back into our equation. So it would be k2 over k1, so 1.45 times 10 to the -3 over 5.79 times 10 to the -5. So let's plug that in. So this is the natural log of 1.45 times 10 to the -3 over 5.79 times 10 to the -5. So now we just have to solve
for the activation energy. So let's find the stuff on the left first. So the natural log of 1.45 times 10 to the -3, and we're going to divide that by 5.79 times 10 to the -5, and we get, let's round that up to 3.221. So we get 3.221 on the left side. On the right side we'd have - Ea over 8.314. And let's solve for this. So let's get the calculator out again. And let's do one divided by 510. From that we're going to subtract one divided by 470. So let's see what we get. We get, let's round that to - 1.67 times 10 to the -4. So just solve for the activation energy. So we have 3.221 times 8.314 and then we need to divide that by 1.67 times 10 to the -4. And so we get an activation energy of approximately, that would be 160 kJ/mol. We need our answer in
kJ/mol and not J/mol, so we'll say approximately
160 kJ/mol here. So the activation energy is equal to about 160 kJ/mol, which is almost the same value that we got using the other form of
the Arrhenius equation. So the other form we
ended up with 159 kJ/mol, so close enough. So you can use either version
of the Arrhenius equation depending on what you're
given in the problem.