Energy Profiles and Catalysts

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. You add a little catalyst to a reaction, and it runs thousands of times faster. So at equilibrium, you must get more product, right? Today we will see why that answer is wrong. We will read an energy profile, see what a catalyst does to it, and find how much faster it goes. In JEE, questions on this come up often. So, let's begin.

0:27Reading an energy profile

So, here is an energy profile. Energy goes up the side, and the reaction coordinate runs along the bottom: it tracks the reaction from start to finish. Reactants sit on the left, products on the right, and in between, a hill. For now, look only at the solid dark blue curve: the reaction without a catalyst. Okay? To become products, the reactants must climb over this hill. The height of the climb is the activation energy, E a: capital E with a small a below it. So the threshold energy, the energy at the very top of the hill, equals the energy of the reactants plus E a. With me? Now, a reaction can also run backwards. The products climb back over the hill to become reactants again. So tell me, starting from the products, where does that backward climb end? At the same peak, exactly! E a f, with a small f for forward, starts at the reactants. And E a b, with a small b for backward, starts at the products. Their difference is the energy of the products minus the energy of the reactants: delta H, the enthalpy change. The little triangle, delta, means change. So delta H equals E a f minus E a b. Clear? Now, in the figure, the products sit lower than the reactants, so this reaction is exothermic. Which climb is taller, forward or backward? The backward one, correct! For an exothermic reaction, delta H is less than zero, so E a b is greater than E a f. The products have further to climb. Got it?

2:10What a catalyst does

Now bring in a catalyst. It opens a new path for the reaction, with a lower activation energy. That is the dashed orange curve. Its height is E a prime: E a with a small dash after it, which we read as prime. Okay? Look carefully at the two ends. The reactants are at the same level, and so are the products. Only the peak has come down. So what happens to the backward climb? It gets shorter too, and by exactly the same amount! Both climbs now end at the new, lower peak. So E a f and E a b fall by the same amount. Right? Delta H is E a f minus E a b. Take the same amount off both, and the difference does not change. In the figure, one delta H arrow serves both paths. Delta G, the Gibbs energy change, and capital K, the equilibrium constant, do not change either. So a catalyst can never make a non-spontaneous reaction spontaneous. With me? Now, our puzzle from the start. The catalyst makes the forward reaction thousands of times faster. So do we get more product at equilibrium? No, and this is the trap. The catalyst speeds up the backward reaction too. The rate constants, small k, of the forward and backward reactions rise by the same factor. Okay? So equilibrium is reached sooner, but its position does not shift. Same equilibrium, same amount of product, just reached faster. And that solves our puzzle. Got it?

3:50How much faster? Same A, lower Eₐ

Remember the Arrhenius equation: k equals capital A times e to the power minus E a over R T, okay? Here k is the rate constant, capital A is the frequency factor, R is the gas constant, and T is the temperature in kelvin. With the catalyst, k cat, written with cat below the k, equals capital A times e to the power minus E a prime over R T, with the same capital A for both. Right? So, how many times faster is the catalysed reaction? Divide one by the other. What do you get? Yes, exactly! Capital A cancels, and the powers subtract. k cat over k equals e to the power E a minus E a prime, over R T, okay? Look, only the drop in activation energy appears here, not E a itself. Since it sits in the power, a small drop gives a huge factor. Clear? One more useful result. Run the uncatalysed reaction at temperature T, and the catalysed one at another temperature, T prime. When are their rates equal? When the two powers are equal! Same capital A, so we need E a prime over R T prime to equal E a over R T, and R cancels. So E a prime over T prime equals E a over T, okay? Temperatures in kelvin, always.

5:16Example 1: backward Ea from forward Ea and delta H

Now, example one. The forward activation energy, E a f, is eighty kilojoules per mole, and delta H is minus thirty kilojoules per mole. In the units, the minus one up top means per. Find E a b, the backward activation energy. Pause the video and try it first. Done? Okay, let's check. Start from the energy profile rule: delta H equals E a f minus E a b. Right? Rearrange it: E a b equals E a f minus delta H, okay? So eighty minus thirty, that is fifty kilojoules per mole, right? Oh no, careful here, that is exactly the trap. Delta H is minus thirty, not thirty. So E a b equals eighty minus minus thirty, which is eighty plus thirty: one hundred and ten kilojoules per mole. With me? Always check with the picture. Delta H is negative, so the reaction is exothermic, and the backward climb must be taller. One hundred and ten is greater than eighty. Perfect!

6:29Example 2: rate rise when a catalyst lowers Ea by 20 kJ/mol

Now a slightly harder one. Here, a catalyst lowers E a by twenty kilojoules per mole at three hundred kelvin, with capital A unchanged. By what factor does the rate rise? Pause the video and try it first. Done? Okay, let's check. Same capital A, so k cat over k equals e to the power E a minus E a prime, over R T, okay? Only the drop is needed: twenty kilojoules per mole. Right? So, put in twenty, over eight point three one four times three hundred. That is tiny, so e to that power is about one. The wavy equals sign means approximately. So the catalyst does nothing at all? No, no, careful here, look at the units! R is eight point three one four joules per kelvin per mole: joules, not kilojoules. So the drop must be in joules too: twenty thousand joules per mole. Clear? Now the power: twenty thousand, over eight point three one four times three hundred. That gives eight point zero two. With me? So k cat over k equals e to the power eight point zero two, which is about three point zero times ten to the power three. About three thousand times faster, from a drop of just twenty kilojoules per mole! That is the thousands of times faster from our puzzle.

7:58Quick revision: delta H from Ea, catalyst rules, kcat over k

Got it? So, quick revision. Delta H equals E a f minus E a b; for an exothermic reaction, the backward climb is taller. The catalyst lowers both activation energies equally, so delta H, delta G and capital K do not change. Equilibrium is reached sooner, never shifted. And k cat over k equals e to the power E a minus E a prime, over R T, energies in joules; for equal rates, E a prime over T prime equals E a over T, okay? Now go and try the practice problems on this page. All the best!