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Main content
Current time:0:00Total duration:9:24

The Arrhenius equation

AP.Chem:
TRA‑4 (EU)
,
TRA‑4.C (LO)
,
TRA‑4.C.4 (EK)

Video transcript

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 and 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 Iranian equation so down here is our equation where K is our rate constant so K is the rate constant the one we talked 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 eetu 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 e a 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 kilojoules per mole and the temperature is 373 Kelvin so let's let's solve for F right so f is equal to e to the negative of our activation energy in joules per mole so we need to convert 40 kilojoules role 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 / k x moles all right and then this is going to be multiplied by the temperature which is 373 Kelvin so 373 Kelvin 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 negative 40000 1 2 3 divided by 8 point 3 1 4 times 373 so we get 2.5 times 10 to the negative 6 so this is equal to 2.5 times 10 to the negative 6 so what does this mean all right well let's say we had 1 million collisions all right so 1 million collisions what number divided by 1 million is equal to 2.5 times 10 to the negative 6 so this number is 2.5 2.5 divided by 1 million is equal to 2.5 times 10 to the negative 6 so what this means is for every 1 million collisions in our reaction only 2.5 collisions have enough energy to react so obviously that's extremely small number of collisions with enough energy all right let's see what happens when we change the activation energy all right 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 or T 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 negative 10,000 right so we've changed our activation energy and we're going to divide that by 8.314 times 373 so let's do this calculations now we have e to the negative 10,000 divided by 8.31 four times 373 and here we get point zero four all right so this is equal to point zero four so point zero four notice what we've done we've increased F right we've gone from F equal to two point five times ten to the negative six two point zero four so let's stick with this same idea of one-million collision so let's say once again if we had 1 million collisions here so 1 million collisions what number divided by 1 million is equal to point zero 4 so that number would be 40,000 40,000 divided by 1 million is equal to point zero 4 so for every 1 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 alright so decreasing the activation energy increase the value for F it increased the number of effective collisions alright let's let's do one more calculation this time we're going to change the temperature so let's keep 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 negative 10,000 divided by 8.314 times this time it would be 473 so times 473 so let's do this calculation so e to the negative 10,000 divided by 8.314 times 473 this time so we get let's just say that's point 0h let's all round up 2.08 here so this is equal to point zero eight so we've increased the value for f right we went from point zero four two point zero eight and let's keep our idea of 1 million collisions right so it's a little bit easier to understand what this means so what number divided by 1 million is equal to point zero eight that must be 80,000 right so this must be 80,000 so for every 1 million collisions that we have in our reaction now we have 80,000 collisions with enough energy to react all right so we've increased the temperature we've gone from 373 to 473 we increase the number of collisions with enough energy to react we increase 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 alright and we've been talking about when we talk about F right 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 right or we could increase the temperature so decreasing the activation energy increase the value for F right and so did increasing the temperature and if we increase F right we're going to increase K so if we increase F we increase the rate constant and remember from our rate laws right are the rate of our reaction is equal to our rate constant K times the concentration of you know whatever we are working with for a reaction here I just want to remind you that when you write your rate laws you see that the rate of the reaction is directly proportional to the rate constant K so if you increase 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 right we if we decrease the activation energy or if we increase the temperature we increase 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 makes this just makes logical sense right we know from experience that if we increase the temperature of a reaction all right we increase the rate of that reaction so once again the ideas of collision theory are contained in the Irenaeus equation and so we'll go more into this equation in the next few videos