Collision Theory

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. In a gas, the molecules rush around and collide with each other again and again, a huge number of times every second. So if every collision made products, every gas reaction would be over in a flash, right? But many reactions take minutes, hours, even days. So where do all those collisions go? Today, collision theory gives the answer. It also tells us what capital A in the Arrhenius equation really means, and questions on it come up often in JEE, so let's begin.

0:37Not every collision gives products

So, picture every collision in a gas inside this box; the capital U in its corner stands for all of them. Say molecules of A collide with molecules of B, again and again. How often they collide is the collision frequency, ZAB: capital Z, with a small A B written below it. The two circles inside, we will fill in a moment. Okay? Now tell me, does every one of these collisions turn reactants into products? No, not every collision works! First condition: energy. The colliding molecules must bring energy at least equal to E a, the activation energy: capital E with a small a below it. The sign you see is greater than or equal to. So circle E, E for energy, holds the collisions with enough energy. Right? So, if the energy is enough, the reaction must happen, right? Not yet, careful here! The molecules must also hit each other the right way round: the proper orientation. Hit the wrong way, and even a collision with plenty of energy gives nothing. So circle O, O for orientation, holds the collisions with the proper orientation. Clear? So, only the collisions in both circles, this overlap, actually give products. Enough energy and the proper orientation, both together. We call these effective collisions. This is the heart of collision theory. With me?

2:06Collision theory: the rate

Now let us turn this picture into a formula. Start with all the collisions: ZAB, the collision frequency. Okay? Next, the energy filter. Remember from the Arrhenius lesson: e to the power minus E a over R T is the fraction of collisions with energy at least E a. Here e is the base of natural logs, R is the gas constant, and T is the temperature in kelvin. With me? Then the orientation filter: capital P, the steric factor, also called the orientation factor. It allows for the collisions with the wrong orientation. The sign says P is less than or equal to one, so it can only cut the number down. Right? So, all the collisions, the energy fraction, and the orientation factor. How do we combine these three to get the rate? We multiply them, exactly! Rate equals P, times ZAB, times e to the power minus E a over R T, that's it! Each factor keeps only the collisions that pass its filter. Got it?

3:13Comparing with Arrhenius

Now put this next to the Arrhenius equation: k, the rate constant, equals capital A, times e to the power minus E a over R T, okay? And collision theory says: rate equals PZAB, times the same exponential. Compare the two lines. So what must capital A be? Yes, exactly! Capital A equals P times ZAB, the steric factor times the collision frequency. So the frequency factor is really about collisions: how often they happen, times the orientation factor. Clear? Now, here is a trap. Does capital A count only the collisions with enough energy? No, no, careful here! Look, P times ZAB has no E a in it anywhere. Capital A counts collisions whatever their energy; the energy part sits entirely in the exponential, e to the power minus E a over R T, nowhere else. JEE has asked about this, so remember it. Got it?

4:18Example 1: why the rate is far below the collision frequency

Now, example one, and it is our puzzle from the start. Why is the rate of a reaction far smaller than the collision frequency? Pause the video and try it first. Done? Okay, let's check. Only effective collisions give products: energy at least E a, and the proper orientation, both together. Only this overlap counts. Right? Now, how big is this overlap? Is it most of the collisions, or only a tiny part of them? Only a tiny part, absolutely! The energy factor, e to the power minus E a over R T, is tiny at ordinary temperatures. Okay? To feel how tiny, take the numbers from a solved example in the Arrhenius lesson: E a of fifty kilojoules per mole, at three hundred kelvin. In the unit, k J means kilojoules, the minus one on mol means per mole, and capital K is kelvin. With me? Then e to the power minus E a over R T is e to the power minus twenty point zero five, which is about two point zero times ten to the power minus nine; the wavy equals sign means about. Only about two collisions in every billion have enough energy! Clear? And P, at most one, can cut it down further. So the rate is much less than ZAB: the double less than sign means much less than. That solves our puzzle: most collisions lack the energy, or hit the wrong way. Perfect!

5:50Example 2: why order 3 or more is rare in one step

Now a harder one, example two. Why do reactions of order three or more rarely occur as single steps? Hint: count the particles that must collide. Pause the video and try it first. Done? Okay, let's check. Remember from chapter one: for a single step, called an elementary step, the order equals the molecularity, which is the number of particles colliding in that step. Okay? So, a single step of order three needs three particles to collide at the same instant. Not one after another: all three at once. Right? Now think. Even with just two particles, we saw that only a tiny part of the collisions work. So what happens with three? It gets much harder, of course! Now all three must arrive with the proper orientation, and the collision must still bring enough energy, all at the same moment. With me? So such a simultaneous collision is very improbable, and with four or more particles it is the same story. That is why reactions of order three or more rarely occur as single steps. Clear? So, a single step with three particles colliding can never happen, right? Oh no, careful here! The word is rarely, not never. Molecularity is one, two, or rarely three. In the exam, read words like never and rarely very carefully. Got it?

7:19Quick revision: effective collisions, steric factor P

So, quick revision. An effective collision needs energy at least E a, and the proper orientation. Rate equals PZAB, times e to the power minus E a over R T, with P at most one, so capital A equals PZAB, okay? And three particles colliding at once is very improbable, so a single step of order three or more is rare. Now go and try the practice problem on this page. All the best!