- 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
- Kinetic and thermodynamic enolates
The Arrhenius equation
The Arrhenius equation is k = Ae^(-Ea/RT), where A is the frequency or pre-exponential factor and e^(-Ea/RT) represents the fraction of collisions that have enough energy to overcome the activation barrier (i.e., have energy greater than or equal to the activation energy Ea) at temperature T. This equation can be used to understand how the rate of a chemical reaction depends on temperature. Created by Jay.
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- At2:49, why solve for f and not k?(5 votes)
- 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.(14 votes)
- 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!(3 votes)
- 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 :)(13 votes)
- how does we get this formula, I meant what is the derivation of this formula(8 votes)
- The derivation is too complex for this level of teaching.(3 votes)
- 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?
- THE WATCHER(4 votes)
- 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.(7 votes)
- This Arrhenius equation looks like the result of a differential equation. Is it?(6 votes)
- So what is the point of A (frequency factor) if you are only solving for f?(4 votes)
- we avoid A because it gets very complicated very quickly if we include it( it requires calculus and quantum mechanics)(5 votes)
- 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(2 votes)
- 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(4 votes)
- 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?(2 votes)
- 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.(2 votes)
- isn't R equal to 0.0821 from the gas laws? Or is this R different?(1 vote)
- 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.(3 votes)
- so what is 'A' exactly and what does it signify?
thank you(1 vote)
- 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.(3 votes)
- In the last video, we talked about collision theory, and we said that molecules must collide to react, and we also said those collisions must have the correct orientation in space to be effective collisions, and finally, those collisions must have enough energy for the reaction to occur. And these ideas of collision theory are contained in the Arrhenius equation. So down here is our equation, where k is our rate constant. So k is the rate constant, the one we talk about in our rate laws. A is called the frequency factor. So, A is the frequency factor. Also called the pre-exponential factor, and A includes things like the frequency of our collisions, and also the orientation of those collisions. And then over here on the right, this e to the negative Ea over RT, this is talking about the fraction of collisions with enough energy for a reaction to occur. So we symbolize this by lowercase f. So the fraction of collisions with enough energy for the reaction to occur. f depends on the activation energy, Ea, which needs to be in joules per mole. R is the gas constant, and T is the temperature in Kelvin. So let's see how changing the activation energy or changing the temperature for a reaction, we'll see how that affects the fraction of collisions with enough energy for our reaction to occur. So, let's start with an activation energy of 40 kJ/mol, and the temperature is 373 K. So, let's solve for f. So, f is equal to e to the negative of our activation energy in joules per mole. So we need to convert 40 kilojoules per mole into joules per mole, so that would be 40,000. So, 40,000 joules per mole. All right, this is over our gas constant, R, and R is equal to 8.314 joules over K times moles. All right, and then this is going to be multiplied by the temperature, which is 373 Kelvin. So, 373 K. So let's go ahead and do this calculation, and see what we get. So, let's take out the calculator. e, e to the, we have -40,000, one, two, three divided by 8.314 times 373. So, we get 2.5 times 10 to the -6. So this is equal to 2.5 times 10 to the -6. So what does this mean? All right, well, let's say we had one millions collisions. All right, so 1,000,000 collisions. What number divided by 1,000,000, is equal to 2.5 x 10 to the -6? So this number is 2.5. 2.5 divided by 1,000,000 is equal to 2.5 x 10 to the -6. So what this means is for every one million collisions in our reaction, only 2.5 collisions have enough energy to react. So obviously that's an extremely small number of collisions with enough energy. All right, let's see what happens when we change the activation energy. So we're going to change the activation energy from 40 kilojoules per mole to 10 kilojoules per mole. So, we're decreasing the activation energy. We're keeping the temperature the same. So let's see how that affects f. So let's plug in this time for f. So f is equal to e to the now we would have -10,000. So we've changed our activation energy, and we're going to divide that by 8.314 times 373. So let's do this calculation. So now we have e to the - 10,000 divided by 8.314 times 373. And here we get .04. So this is equal to .04. So .04. Notice what we've done, we've increased f. We've gone from f equal to 2.5 times 10 to the -6, to .04. So let's stick with this same idea of one million collisions. So let's say, once again, if we had one million collisions here. So 1,000,000 collisions. What number divided by 1,000,000 is equal to .04? So that number would be 40,000. 40,000 divided by 1,000,000 is equal to .04. So for every one million collisions that we have in our reaction this time 40,000 collisions have enough energy to react, and so that's a huge increase. Right, it's a huge increase in f. It's a huge increase in the number of collisions with enough energy to react, and we did that by decreasing the activation energy. So decreasing the activation energy increased the value for f. It increased the number of effective collisions. All right, let's do one more calculation. This time we're gonna change the temperature. So let's keep the same activation energy as the one we just did. So 10 kilojoules per mole. So 10 kilojoules per mole. This time, let's change the temperature. Here we had 373, let's increase the temperature to 473, and see how that affects the value for f. So f is equal to e to the negative this would be 10,000 again. e to the -10,000 divided by 8.314 times, this time it would 473. So times 473. So let's do this calculation. So e to the -10,000 divided by 8.314 times 473, this time. So we get, let's just say that's .08. So I'll round up to .08 here. So this is equal to .08. So we've increased the value for f, right, we went from .04 to .08, and let's keep our idea of one million collisions. Right, so it's a little bit easier to understand what this means. So what number divided by 1,000,000 is equal to .08. That must be 80,000. Right, so this must be 80,000. So for every 1,000,000 collisions that we have in our reaction, now we have 80,000 collisions with enough energy to react. So we've increased the temperature. Gone from 373 to 473. We increased the number of collisions with enough energy to react. We increased the value for f. Finally, let's think about what these things do to the rate constant. So we go back up here to our equation, right, and we've been talking about, well we talked about f. So we've made different calculations over here for f, and we said that to increase f, right, we could either decrease the activation energy, or we could increase the temperature. So decreasing the activation energy increased the value for f, and so did increasing the temperature, and if we increase f, we're going to increase k. So if we increase f, we increase the rate constant, and remember from our rate laws, right, R, the rate of our reaction is equal to our rate constant k, times the concentration of, you know, whatever we are working with for our reaction. Here I just want to remind you that when you write your rate laws, you see that rate of the reaction is directly proportional to the rate constant k. So if you increase the rate constant k, you're going to increase the rate of your reaction, and so over here, that's what we've been talking about. If we decrease the activation energy, or if we increase the temperature, we increase the fraction of collisions with enough energy to occur, therefore we increase the rate constant k, and since k is directly proportional to the rate of our reaction, we increase the rate of reaction. And this just makes logical sense, right? We know from experience that if we increase the temperature of a reaction, we increase the rate of that reaction. So, once again, the ideas of collision theory are contained in the Arrhenius equation, and so we'll go more into this equation in the next few videos.