Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Okay, one quick statement before we start. An elementary reaction can be of zero order. True or false? Now, zero order reactions are real; we have already met them in this unit. So the statement must be true, right? Hold that thought. We will settle it in Example two. This is Part C of the unit problem bank: five concept checks on ideas that are often tested in statement form, where a single word can change the answer. So, let's begin.

0:34Check 1: molecularity of a complex reaction

Check one. First, two words that students often mix up. The first is order. In the rate law, small r, the rate, equals k, the rate constant, times the concentration of A to the power m, times the concentration of B to the power n; the square brackets mean concentration. The powers m and n are the orders, and they are found by experiment. Okay? And orders can be zero, fractional, or even negative. Keep this line in mind; it comes back in Example two. Right? The second word is molecularity. An elementary step is a single step, in which the reacting species collide together. Molecularity is the number of species colliding in that one step. With me? Now the check. Some reactions are complex: they go through several elementary steps. Is molecularity defined for the overall complex reaction? No, it is not defined for the overall reaction. Molecularity is defined only for an elementary step. The overall equation is just the sum of the steps, so there is no single collision to count. Got it? So what do we say for a complex reaction? We speak of the molecularity of its slowest step. So if a statement gives a molecularity for a whole complex reaction, read it very carefully. Clear?

1:58Check 2: why is ester hydrolysis pseudo first order?

Check two. Here is the acid hydrolysis of an ester. Ester plus H two O, water, gives an acid plus an alcohol; the arrow means gives, and the small two counts the hydrogen atoms. The H plus written above the arrow is the acid, which acts as the catalyst. Okay? Two molecules react, the ester and water. And the rate law has both concentrations: the rate equals k, times the concentration of ester, times the concentration of water. Two concentrations, so it looks second order. Right? And yet, this reaction is called pseudo first order. Why? Here is a hint: think about how much water there is. Lots of it, yes, exactly! Water is in large excess, so as the reaction runs, its concentration hardly changes. Only a tiny part of all that water is used up. With me? So we absorb it into the constant. The rate equals k prime times the concentration of ester, where k prime, the k with a little dash, equals k times the concentration of water. Now the rate depends on the ester alone, so it behaves as first order. Got it? But see the word pseudo: it only behaves as first order. Two molecules still react, so the reaction is bimolecular, and the true rate law still has both concentrations. The water is simply hidden inside k prime. Clear?

3:25Check 3: does a catalyst change ΔG?

Check three: the catalyst. On screen is an energy profile. Energy goes up the side, and the reaction coordinate runs along the bottom, from reactants to products. The solid dark blue curve is the path without a catalyst, and the dashed orange curve is the path with one. Okay? The catalyst gives a new path with a lower activation energy. E a, capital E with a small a below it, is the climb without the catalyst: the taller two-headed arrow. E a prime, with the little dash, is the shorter arrow: the smaller climb with the catalyst. See the difference? Now the check. Delta G, the little triangle delta before capital G, is the Gibbs energy change of the reaction. The catalyst changes the path, and makes the reaction much faster. So does it change delta G too? No, it does not change delta G at all. Look at the two ends. Both curves start at the same reactants, and end at the same products. The catalyst changes only the path and its activation energy, not the energies of the reactants and products. With me? So delta G does not change. Neither does delta H, the gap you see on the right, nor capital K, the equilibrium constant. The catalyst only helps the reaction reach equilibrium sooner. Got it? So here is the trap in statement form. No catalyst can make a non-spontaneous reaction spontaneous. Non-spontaneous stays non-spontaneous. Clear?

4:59Example 1: why the rate falls while k stays the same

Now, two examples from this part of the problem bank. Example one. Why does the rate of most reactions fall as time passes, while k stays the same? Pause the video, and try it first. Done? Okay, let's check. Start from the rate law again: r equals k, times the concentration of A to the power m, times the concentration of B to the power n. Clear? Now, the rate is falling as time passes. And k sits right there in the rate law. So k must be falling too, right? No, no, careful here! Ask what really changes. As the reaction runs, the reactants are used up, so the concentrations of A and B fall. Smaller concentrations give a smaller rate. With me? But k is the rate constant. It depends only on the temperature, and on the catalyst, not on concentration. So at a fixed temperature, k stays the same, even while the rate falls. Got it? And why does the question say most reactions? For a zero order reaction, the rate simply equals k, with no concentration term at all. So there, the rate does not fall as time passes. See, every word in the question matters. Clear?

6:18Example 2: why an elementary step is never zero order

And now our statement from the start, as Example two. Can an elementary reaction be of zero order? Pause the video and try it first. Here is a hint: think back to check one. Done? Okay, let's check. In check one, we saw that orders can be zero. So an elementary reaction can be zero order too, right? Oh no, careful here! That line was about orders in general, found by experiment. For an elementary step, there is a special rule: the order equals the molecularity. Okay? So the question becomes this. What values can a molecularity take? One, two, or rarely three. Correct! It counts the species colliding in one step, and zero species cannot collide. So a molecularity is never zero. Right? So the order of an elementary step is greater than or equal to one; that is the sign you see on screen. It can never be zero. Our opening statement is false! With me? Then where does zero order come from? It appears only for complex reactions, never for a single elementary step. Got it? For example, surface reactions at saturation. Remember N H three, ammonia, with the small three counting the hydrogen atoms, breaking down on hot platinum, P t? The metal surface is saturated with adsorbed ammonia, so the rate does not depend on the gas pressure: zero order. And that is the whole answer to Example two.

7:53Quick revision: molecularity, pseudo first order, catalyst

So, quick revision. Molecularity is defined only for an elementary step; for a complex reaction, we speak of its slowest step. The rate falls as the concentrations fall, but k depends only on temperature and the catalyst. In ester hydrolysis, water is in large excess, so it is pseudo first order; and a catalyst never changes delta G, the Gibbs energy change. Finally, for an elementary step, the order equals the molecularity, at least one, so never zero. Now go and try the two practice problems on this page. Read every word of each statement. All the best!