Catalysis

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. In the Haber process, nitrogen and hydrogen make ammonia on the surface of iron, and the iron makes the reaction much faster. But at the end, the iron is still there, completely unchanged. So tell me, does the iron really take part in the reaction, or does it just sit and watch? Keep that question in mind. You will answer it yourself soon. Today: the two kinds of catalysts, how a surface catalyst works, enzymes, and one thing a catalyst can never change. In JEE, quick theory questions on catalysts come up often. So, let's begin.

0:40Two kinds of catalysts

First, catalysts come in two kinds. Homogeneous: the catalyst is in the same phase as the reactants. Okay? Two examples to remember. N O, nitric oxide, in the lead-chamber process. And H plus, the hydrogen ion, written H with a small plus sign at the top, in ester hydrolysis. Right? Now, our iron in the Haber process. The iron is a solid, but nitrogen and hydrogen are gases. So, is that homogeneous catalysis? No, it is not, because the phases are different. So it is heterogeneous: the catalyst is in a different phase from the reactants. Clear? The examples: F e, iron, in the Haber process, P t, platinum, in the contact process, and N i, nickel, in hydrogenation. Each one is a solid metal, working on reactants in another phase. With me?

1:35Adsorption theory: how a surface catalyst works

So now, let us see how a solid surface speeds up a reaction. The adsorption theory explains it in four steps. Step one: the reactant molecules are adsorbed, which means they stick, at special spots on the catalyst surface called active sites. Okay? Now think. Once a molecule is held on the metal surface, do the bonds inside it become stronger, or weaker? They become weaker, yes, exactly! The bonds in the reactants are weakened, so they are easier to break. Some questions say adsorption strengthens these bonds. That statement is false. Got it? Step three: the reaction occurs on the surface itself, through an intermediate. Right? And step four: the products desorb, which means they leave the surface, and the surface is free again. Now, our puzzle from the start. Does the iron take part? Yes, it does: the reaction happens on its surface. But then the products leave, and the iron is free again, so at the end it is unchanged. Clear?

2:38Promoters, poisons and transition metals

Now, other substances can change how active a catalyst is. First, a promoter makes the catalyst more active. M o, molybdenum, is the promoter for the iron in the Haber process. So iron is the catalyst, molybdenum is the promoter. Right? And a poison does the opposite: it makes the catalyst less active. Arsenic, written A s, is a poison in the contact process. With me? Now look at our catalysts: iron, platinum, nickel. All of them are transition metals. So why are transition metals such good catalysts? There are two reasons. First, they have variable oxidation states: the same metal can exist in more than one oxidation state. Second, to bond with the reactants, they use only their three d electrons, right? No, no, careful here, that is a trap! They use both the three d and the four s electrons to bond with the reactants. True or false questions on exactly this point do come up. Got it?

3:45Enzymes: catalysts of living things

Now, enzymes: the catalysts of living things. They are highly specific: an enzyme works only on its own particular reactant, which we call its substrate. And other species can activate it, or inhibit it. Okay? Now, for most reactions, a higher temperature means a faster rate, right? So if we keep heating an enzyme reaction, will it keep getting faster? No, it will not keep getting faster! Enzymes are most active at an optimum temperature, about three hundred and ten kelvin, the capital K on screen, and at an optimum p H as well. On either side of the optimum, the enzyme is less active. Clear?

4:27Michaelis–Menten: rate and substrate

Now the rate of an enzyme reaction: the Michaelis Menten equation. Rate equals V max times S, over K m plus S, okay? The square brackets mean concentration, so this is the substrate concentration. V max, capital V with a small max below it, is the maximum rate, and K m, capital K with a small m below it, is a constant for that enzyme and its substrate. With me? Now suppose the substrate concentration is very small, much smaller than K m. What does the bottom of this fraction become? It becomes just K m, yes! So the rate is approximately V max over K m, times the substrate concentration. The wavy equals sign means approximately. Rate proportional to concentration: first order. Okay? Now the opposite case. The substrate concentration is huge, much bigger than K m. What happens to the rate now? Now K m is negligible, S cancels, and the rate is approximately V max, a constant. It no longer depends on the substrate at all, so it is zero order. This is called saturation. The reason: all the active sites are occupied, so extra substrate cannot bind. Got it?

5:42Example 1: does a catalyst change the equilibrium constant

Now, example one, a quick theory question. Does a catalyst change the equilibrium constant, or only how fast equilibrium is reached? Pause the video and think it through first. Done? Okay, let's check. Many students argue like this: the catalyst makes the forward reaction faster, so more product forms, so the equilibrium constant, capital K, gets bigger, right? No, no, careful here, that is the trap! Remember from the lesson on energy profiles: a catalyst gives a new path. It lowers the activation energy, capital E with a small a, for the forward and the backward reactions by the same amount. So both speed up by the same factor. Okay? And the equilibrium constant, capital K, equals k f over k b: the forward rate constant, with a small f, over the backward one, with a small b. Both are multiplied by the same factor, so the ratio does not change. So a catalyst never changes the equilibrium constant. It only helps the reaction reach equilibrium sooner. Perfect!

6:51Example 2: Km 2 mM, fraction of Vmax at 2 mM and 18 mM

Now, example two. An enzyme has K m equal to two millimolar: the m M means millimoles per litre. What fraction of V max is reached when the substrate concentration is two millimolar, and when it is eighteen millimolar? Pause the video and try it first. Done? Okay, let's check. We want a fraction of V max, so divide the Michaelis Menten equation by V max. Rate over V max equals S, over K m plus S, okay? First, S equals two millimolar. Two, over two plus two, that is two over four: zero point five. Half of V max! See, when S equals K m, the rate is exactly half the maximum. Nice, isn't it? Now, S equals eighteen millimolar. That is nine times as much substrate as before. So the rate becomes nine times bigger too, right? Oh no, careful here, that cannot be right! Nine times zero point five is four point five times V max, more than the maximum itself. Use the formula: eighteen, over two plus eighteen, that is eighteen over twenty: zero point nine. Okay? So nine times the substrate gives only one point eight times the rate: zero point nine over zero point five. The enzyme is getting close to saturation, with most of its active sites already busy.

8:13Quick revision: phases, enzyme orders, K never changes

Perfect! So, quick revision. Same phase, homogeneous. Different phase, heterogeneous, and a surface catalyst works by adsorption. Enzymes: first order at low substrate concentration, zero order at high, because of saturation. And a catalyst never changes the equilibrium constant; equilibrium is only reached sooner. Now go and try the practice problems on this page. All the best!