Second-Order Reactions and Comparisons

Transcript

0:00Introduction

Hello everyone! One small puzzle to start. Three reactions: zero order, first order and second order. They start with the same concentration and have exactly the same half-life. We wait for two half-lives. Is one quarter of the reactant left in each? Hold that thought. Today we learn second order reactions, and compare all three orders side by side. Comparisons like this come up often in JEE kinetics questions. So, let's begin.

0:32Second order: the integrated rate law

So, here is the law for a second order reaction. The square brackets around A mean the concentration of A, in moles per litre, at time t. The little zero below marks the starting concentration, and we read it as A naught. So, one over the concentration of A, minus one over A naught, equals k times t, where k is the rate constant. Okay? But notice when it holds. For a single reactant, or for two reactants that start with equal concentrations. If the starting concentrations differ, this simple form does not apply. Right? Now, the half-life. It is written t with a small one half below it, and we call it t half. At t half, the concentration has fallen to half of A naught. Put that into the law. What do you get? One over half of A naught is two over A naught. So, two over A naught, minus one over A naught, equals k times t half. With me? Two minus one is one, so t half equals one over k times A naught. Look carefully: the starting concentration sits in the denominator. So tell me. If we double the starting concentration, what happens to t half? It is halved, yes! Oh no, careful here! Many students carry over the first order habit, where t half never changes. That is not true for second order. Got it?

2:00Half-life and the starting concentration

Now let us put all orders together. For a reaction of order n, t half is proportional to A naught, to the power one minus n. The curly sign in the middle means proportional to. Okay? Zero order: n is zero, so the power is one. So t half is proportional to A naught itself: double the start, and the half-life doubles too. Now, first order, n equals one. What is the power? One minus one is zero, and anything to the power zero is one! So the first order half-life does not depend on the starting concentration at all. Very good! And second order: one minus two is minus one. So t half is proportional to one over A naught, exactly what we just found. Clear?

2:48Time for 75% over time for 50%

Next, another way to compare the orders. Here, t seventy five percent, with seventy five percent written small below the t, is the time for seventy five percent of the reactant to react, and t fifty percent, written the same way, is the time for half. For any order n other than one, their ratio is four to the power n minus one, minus one, over two to the power n minus one, minus one. At n equals one this gives zero over zero, so first order we work out separately. Okay? Zero order first. The reactant falls at a steady rate, so time is proportional to the amount reacted: zero point seven five over zero point five, one point five. Now, first order. Seventy five percent reacted means one quarter left. How many half-lives is that? Two half-lives, correct! Half, and then half of that half. So the ratio is exactly two. With me? And second order: put n equals two. Four minus one, over two minus one: three. So one point five, two and three, for zero, first and second order. Right?

4:00Pseudo-first-order reactions

One last idea. Sometimes one reactant is present in a large excess, often the solvent itself, like water. As the reaction runs, does the concentration of water change much? Hardly at all, exactly! Take the hydrolysis of an ester. The rate equals k, times the concentration of the ester, times the concentration of H two O; the small two counts the hydrogen atoms. Since water hardly changes, that equals k prime, k with the little dash, times the concentration of the ester. Okay? So k prime equals k times the concentration of water. The reaction is second order, but it behaves as first order. We call it pseudo first order, and pseudo means false. Got it? Two standard examples: the acid hydrolysis of ethyl acetate, and the inversion of cane sugar.

4:55Example 1: second order half-life, one third left at 20 min

Right? Let's solve a problem. Here, a second order reaction has k equal to zero point five litre per mole per minute: capital L is litre, m o l is mole, and the minus one powers mean per. The starting concentration is zero point two molar, where capital M means moles per litre. Find the half-life, and the concentration of A after twenty minutes. Pause the video and try it first. Done? Okay, let's check. t half equals one over k times A naught. Zero point five times zero point two is zero point one, and one over zero point one is ten. So the half-life is ten minutes. Got it? Now, twenty minutes is exactly two half-lives. So one quarter of zero point two is left, which is zero point zero five molar, right? No, no, careful here! That quarter rule works only for first order. For second order, use the law itself. One over the concentration of A equals one over zero point two, plus k times t, which is zero point five times twenty. Five plus ten: fifteen. Okay? So the concentration of A is one over fifteen, about zero point zero six seven molar, not zero point zero five. With me? And look: one fifteenth is exactly one third of zero point two. One third left, not one quarter. Keep that in mind for our puzzle. Clear?

6:31Example 2: amount left after two half-lives, orders 0, 1, 2

Now, our puzzle as a question. Zero, first and second order reactions start at the same A naught and have the same half-life. What fraction remains in each at t equals two t half? The graph shows the concentration of A over A naught against time, and all three curves cross the dashed half line together. Pause the video and try it first. Done? Okay, let's check. Zero order is the straight line. It loses the same amount, half of A naught, in every half-life. So after two half-lives: one minus two times one half, which is zero. Nothing is left! Now, first order: what fraction is left after two half-lives? One half, squared: one quarter. Correct! That is the one quarter you probably expected for all three. Okay? Now second order, the tricky one. Since t half equals one over k times A naught, k times A naught times t half equals one. Clear? Now put t equals two t half in the second order law, and multiply through by A naught. So, A naught over the concentration of A equals one, plus k times A naught times two t half. That is one plus two, which is three. With me? One third at second order. Zero, one quarter, one third. So the answer to our puzzle is no! Only first order has one quarter left. Zero order has just run out, and second order has the most left, just as the graph shows. Got it?

8:07Quick revision: second order, t75 by t50, pseudo first order

So, quick revision. Second order: one over the concentration of A, minus one over A naught, equals k t, and t half is one over k times A naught. The half-life goes as A naught to the power one minus n, and the seventy five to fifty percent time ratio is one point five, two and three for zero, first and second order. With one reactant in large excess, the reaction becomes pseudo first order. Now go and try the practice problems on this page. All the best!