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### Course: Chemistry archive > Unit 10

Lesson 3: Arrhenius equation and reaction mechanisms- Collision theory
- The Arrhenius equation
- Forms of the Arrhenius equation
- Using the Arrhenius equation
- Collision theory and the Maxwell–Boltzmann distribution
- Elementary reactions
- Reaction mechanism and rate law
- Reaction mechanism and rate law
- The pre-equilibrium approximation
- Multistep reaction energy profiles
- Catalysts
- Types of catalysts
- Types of catalysts

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# Forms of the Arrhenius equation

How to write different forms of the Arrhenius equation. Using the Arrhenius equation to look at how changing temperature and activation energy affects collisions.

## Want to join the conversation?

- I really enjoyed by the derivation of this equation, but I still wonder; why did he start by natural log ?
`In`

? why not any other step ?!

I hope you understand my question.(6 votes)- What other step might he have taken? The natural log is the most reasonable way to pull down the exponents of an exponential function.(21 votes)

- At2:23, Mathematically, how could we have y-intercept of ln(a) (because t = 0 is undefined)?

Thank you. I love this.(2 votes)- The y-intercept is the value of
**y**when

. In this case**x**= 0**x**is not**T**, but 1/**T**.

Of course, this is still not really defined since it requires**T**to approach ∞.

However, we can still draw a line using real values for the temperature**T**and extend that line to where it intercepts the y-axis without any mathematical problems.

EDIT: Note the next lecture helps make this clear (https://www.khanacademy.org/science/chemistry/chem-kinetics/arrhenius-equation/v/using-the-arrhenius-equation)(3 votes)

- Isnt the equation supposed to be ln(k1/k2)= -Ea/r (1/T2-1/T1) instead of ln(k2/k1)...

??(2 votes)- This is another form that you can use relating two reactions at different temperatures.(2 votes)

- I am assuming that to go from -Ea/R * (1/T2) + Ea/R * (1/T1) you factorize. So it turns into the end equation. Am I right?(2 votes)
- In the second form of arrhenius equation ,,the frequency factor gets canceled ..so what does it mean ...that there is no frequency for collision(1 vote)
- In this particular example where k is equal to -Ea/R, what would be the units of k? Or is it unitless?(1 vote)
- According to the equation @2:30we cannot find frequency factor exactly as because if we were to find it, then 1/t must equal 0, which is impossible.So how do we precisely calculate the frequency factor (A) ?(1 vote)
- Yes it is practically impossible to get T--> inf but you can always extend the graph and calculate value of frequency factor by the use of mathematics. Since The equation is of the form y=mx+b since it is a straight line it will have intercept at y axis (it doesn't matter if 1/T will reach 0 or not) which will be equal ln(A). So, A=e^b(1 vote)

- Why is this so similar to the Clausius-Clapeyron equation?(1 vote)
- The Clausius-Clapeyron equation can be derived from the Arrhenius equation by using the rate constants for the forward and back steps for vaporisation of a liquid. The derivation is explained here in this short paper - https://www.researchgate.net/publication/240904514_From_the_Arrhenius_to_the_Clausius-Clapeyron_Equation(1 vote)

- Will this be something you have to know for the AP Chemistry exam?(1 vote)
- I'm not sure but it is probably something good to know.(1 vote)

- Why does it have to we have to subtract to find the other form for the Arrhenius equation?(1 vote)
- it is basically mathematical manipulation so as to give the equation another useful forms(1 vote)

## Video transcript

- [Voiceover] We've already seen one form of the Arrhenius equation. Which says that the rate constant k is equal to the frequency factor A times e to the negative ea over RT where ea is the activation energy, R is the gas constant, and T is the temperature. There are other forms of
the Arrhenius equation, which you might want to use, depending on the problem. So let's go ahead and
find those other forms. And we start by taking the
natural log of both sides. So we would get the natural log of k is equal to the natural log of A times e to the negative ea over RT. So the left side, we still
have the natural log of k. And on the right side, we
can use a log property. The natural log of A times
e to the negative ea over RT is equal to the natural log of A plus the natural log of e to the negative ea over RT. And the natural log of e
to the negative ea over RT would just be equal to
negative ea over RT. So we can write that in here. The natural log of the rate constant k is equal to the natural log of A minus ea over RT. And I'm just going to rewrite this as natural log of k is equal to negative ea over R times one over T plus the natural log of A. And the reason I wrote it this way, is it's easier to see the form y is equal to mx plus b. So y is equal to mx plus b. So if we graph the natural
log of k on the y-axis and one over T on the x-axis, we're going to get a straight line and the slope of that
line, which of course is m, the slope of that line is
equal to negative ea over R. So we can find the activation energy from the slope of the line. And if we wanted to find
the frequency factor, we know that the y intercept is equal to the natural log of A. So we could find the frequency
factor if we wanted to. So this is another for of
the Arrhenius equation. So I'll go ahead and box it here. So sometimes you might
want to use this form. And we can figure out another form. So let's start with this equation
that we just figured out. And let's write this for
one specific temperature. So at one specific temperature, we're going to have a
specific rate constant. So the natural log of k one, so I'll call this rate constant k one. Is equal to negative ea over R times one over a specific temperature, T one. So when the temperature is T one, we have rate constant k one. And so let's also put a
natural log of A in here. Alright, let's use a
different temperature. Let's say this is T two. Well at a different temperature, we're going to have a
different rate constant. So we'll call the rate constant k two. So the natural log of k two is equal to negative ea over R times one over T two this time plus natural log of A. Alright, so now we have two
different equations here for two different temperatures with two different rate constants. Let's take the natural log of k two minus the natural log of k one. So let's go ahead and write that. So the natural log of k two
would be all of this right here. So let's write that in. Negative ea over R times one over T two plus natural log of A. Next, from that we're
going to subtract, right? So subtract the natural log of k one. The natural log of k one is all of this. So this would be minus, and let's put all this in parentheses, negative ea over R, one over T one plus the natural log of A. Next, on the left side, we can use a log property. So over here on the left, natural log of k two
minus natural log of k one is equal to the natural log of k two over k one. Again, that's a log property. And on the right we would have, this would be negative ea over R, one over T two plus natural log of A. And then we have these
two negative signs here, so that would be plus ea over R, one over T one. And then this would be a
negative natural log of A. So a negative natural log of A. So minus natural log of A. Notice the natural logs of A go away here. And let's write the form, this next form of our Arrhenius equation. So on the left side we
have natural log of k two over k one is equal to, let's pull out a negative ea over R. So if we pull out a negative ea over R, then we would have one over T two minus, we need a minus sign here, one over T one. So that's just some algebra. And we've arrived at another
form of the Arrhenius equation. So the nice thing about this form is there's no longer an A in there. So if we know, if we know, the rate constants at two
different temperatures, right? If we know rate constants at
two different temperatures, we can find the activation energy. So here are some more forms
of the Arrhenius equation. And in the next video we'll
see when to use which form. And you might even see a
different form than this. So this is how I wrote it,
but different textbooks might have different things on the right side of this equation. So use whichever one you like the best.