Question 1

how does we get this formula, I meant what is the derivation of this formula

Accepted Answer

The derivation is too complex for this level of teaching.

Question 2

At 2:49, why solve for f and not k?

Accepted Answer

So that you don't need to deal with the frequency factor, it's a strategy to avoid explaining more advanced topics. f is what describes how the rate of the reaction changes due to temperature and activation energy. As with most of "General chemistry" if you want to understand these kinds of equations and the mechanics that they describe any further, then you'll need to have a basic understanding of multivariable calculus, physical chemistry and quantum mechanics.

Question 3

Two questions :

1. Why are none of the equations derived? 
2. Why must the collisions have enough energy to overcome an activation barrier? What would happen if they didn't?

Thank you.

- THE WATCHER

Accepted Answer

For students to be able to perform the calculations like most general chemistry problems are concerned with, it's not necessary to derive the equations, just to simply know how to use them. But if you really need it, I'll supply the derivation for the Arrhenius equation here.

So Svante Arrhenius, after whom the equation is named for, was a Swedish chemist who performed much of the work for this equation in the late 19th century. In his work he was concerned with the effect temperature had with the kinetic rate constant, k, of chemical reactions. So essentially he plotted several variables and observed their relationship to find a connection. Eventually he found that plotting the natural log of the rate constant, ln(k), on the y-axis and reciprocal temperature on the x-axis, 1/T, resulted in a straight line. Straight lines are nice to discover in science because you can fit them to the slope-intercept equation: y= mx+b, where x an y are the variables, m is the slope or gradient, and b is the y-intercept. Using this straight line, which is referred to as an Arrhenius plot, Arrhenius created the equation: ln(k) = (-Ea/R)(1/T)+ln(A), where ln(k) is the y variable, 1/T is the x variable, -Ea/R is the slope, and ln(A) is the y-intercept. Then through some algebra you can arrive at the more recognizable equation.

* ln(k) = (-Ea/R)(1/T)+ln(A)
* ln(k) = (-Ea/RT)+ln(A)
* ln(k) = ln(A) - (Ea/RT)
* e^(ln(k)) = e^(ln(A) - (Ea/RT))
* k = e^(ln(A))e^(-Ea/RT)
* k = Ae^(-Ea/RT)

The physical meaning of the activation barrier is essentially the collective amount of energy required to break the bonds of the reactants and begin the reaction. The breaking of bonds requires an input of energy, while the formation of bonds results in the release of energy. Essentially if you can't overcome the activation energy barrier you can't break the initial reactant's bonds and being the reaction and so nothing happens. This is why you can place two chemicals in contact with each other and not observe a reaction because there isn't sufficient energy to begin the reaction and they remain unreacted.

Hope that helps.

Question 4

so if f = e^-Ea/RT, can we take the ln of both side to get rid of the e? I am trying to do that to see the proportionality between Ea and f and T and f. But I am confused.

ln f = - Ea/RT

Can someone explain how to do it this way? Thanks!

Accepted Answer

Yes you can! When you do, you will get: ln(k) = -Ea/RT + ln(A). This is the y= mx + c format of a straight line. When it is graphed, you can rearrange the equation to make it clear what m (slope) and x (input) are. So it will be: ln(k) = -Ea/R (1/T) + ln(A). So then, -Ea/R is the slope, 1/T is x, and ln(A) is the y-intercept. So the graph will be a straight line with a negative slope and will cross the y-axis at (0, y-intercept).

I hope this answered your question :)

I hope this answered your question :)

Question 5

So what is the point of A (frequency factor) if you are only solving for f?

Accepted Answer

we avoid A because it gets very complicated very quickly if we include it( it requires calculus and quantum mechanics)

Question 6

Hi, the part that did not  make sense to me was, if we increased the activation energy, we decreased the number of "successful" collisions (collision frequency) however if we increased the temperature, we increased the collision frequency.
In theory, wouldn't increasing the temperature, increase kinetic energy hence the activation energy - so shouldn't they both have the same effect on the "successful collision frequency"? would really appreciate any feedback, thanks

Accepted Answer

The activation energy is the amount of energy required to have the reaction occur. If you have more kinetic energy, that wouldn't affect activation energy. They are independent. Imagine climbing up a slide. If you climb up the slide faster, that does not make the slide get shorter. Hope this helped. :D

Question 7

So f has no units, and is simply a ratio, correct?  So does that mean A has the same units as k?  What are those units?

Accepted Answer

I believe it varies depending on the order of the rxn such as 1st order k is 1/s, 2nd order is L/mol*s, and 0 order is M/s.

Question 8

isn't R equal to 0.0821 from the gas laws? Or is this R different?

Accepted Answer

R can take on many different numerical values, depending on the units you use.  All such values of R are equal to each other (you can test this by doing unit conversions).  The value you've quoted, 0.0821 is in units of (L atm)/(K mol).  This R is very common in the ideal gas law, since the pressure of gases is usually measured in atm, the volume in L and the temperature in K.  However, in other aspects of physical chemistry we are often dealing with energy, which is measured in J.  Thus, it makes our calculations easier if we convert 0.0821 (L atm)/(K mol) into units of J/(mol K), so that the J in our energy values cancel out.  Through the unit conversion, we find that R = 0.0821 (L atm)/(K mol) = 8.314 J/(K mol).  That is, these R's are equivalent, even though they have different numerical values.

Question 9

so what is 'A' exactly and what does it signify?
thank you

Accepted Answer

In the Arrhenius equation, k = Ae^(-Ea/RT), A is often called the *per-exponential factor*.
It comes more from experiment than from some theoretical calculations.
It is also often called the *frequency factor*.
A represents the number of collisions per second with the proper orientation to react, and the e^(-Ea/RT) term represents the probability that a given collision will result in a reaction.

Course: Chemistry library > Unit 15

The Arrhenius equation

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