Overview

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Take a first order reaction. After some time, it is ninety percent complete. To reach ninety nine percent, only nine percent more has to react. So the extra time should be small, right? Maybe a tenth of the time we have already spent? Well, in fact, it is not small at all, and the true answer will surprise you. This is the unit problem bank for Chemical Kinetics. Today: how to use the bank, a few of its problems, and this puzzle, solved. So, let's begin.

0:35How to use this chapter

So, this chapter has three parts. Part A has five solved problems at JEE Main level, then two more for practice. Okay? The instruction: attempt Part A as a thirty minute test, after Chapters one to six. So sit with a clock, and open the solutions only at the end. Right? Then Part B: four longer problems, at JEE Advanced level, and one for practice. With me? And Part C: five concept checks, and two for practice. These are the ideas that are often tested in statement form, where one word can change the answer. Clear?

1:14Part A: name the idea before you calculate

Now, the real skill in a mixed test: before you calculate, name the idea. Here is a Part A problem. Rate equals k times A to the power half, times B to the power three over two. The square brackets mean concentration, and k is the rate constant. What is the overall order, and what are the units of k? The overall order is the sum of the powers: half plus three over two, that is two. Second order, yes! The double arrow means this leads to. And for second order, k is in litres per mole per second; the little minus ones on screen mean per. Got it? Next one. Zero order reaction: t half, the half-life, is thirty seconds; that is the t with one half written small below it. This is at A naught equal to zero point six M: the small zero below means the starting concentration, and capital M means moles per litre. Find k. Which formula links k and t half here? For zero order, t half equals A naught over two k. So k equals A naught over two t half. Zero point six over two times thirty, that is zero point six over sixty: zero point zero one moles per litre per second. Name the idea first, and Part A becomes fast. Right?

2:34Part C: read every word

Now, Part C, and these look easy, but they test whether your ideas are exact. Can an elementary reaction be of zero order? Elementary means it happens in one single step. So, yes or no? No, it cannot! For an elementary step, the order equals the molecularity: the number of reacting species taking part in that step. That is one, two, or rarely three. At least one, never zero. Okay? So zero order appears only for complex reactions, for example, surface reactions at saturation, where the surface is already full. With me? One more. Does a catalyst change delta G of a reaction? The triangle is the capital Greek letter delta, meaning change, so delta G is the change in Gibbs energy. What do you say? No, it does not. The catalyst changes only the path, and its activation energy, capital E a, with the small a written below. The energies of the reactants and products stay the same, so delta G stays the same. One word in the statement decides the answer, so in Part C, read every word. Clear?

3:41Example 1: t99 by t90 for a first order reaction

Now, Example one, from Part A, and this is our puzzle! For a first order reaction, find t ninety nine percent over t ninety percent. That is, the time to reach ninety nine percent completion, over the time to reach ninety percent. Pause the video, and try it first. Done? Okay, a quick first guess. Ninety nine over ninety is about one point one. So the two times are almost equal, right? No, no, careful here! Time does not grow in step with the percentage. For first order, the natural log of A naught over A equals k t. Here A, without the zero, is the concentration left at time t. So what matters is the ratio, A naught over A, okay? Ninety nine percent done means one percent left, so A naught over A is one hundred. So t ninety nine percent equals natural log of one hundred, over k. Right? Ninety percent done means ten percent left. The ratio is ten, so t ninety percent equals natural log of ten, over k. With me? Now divide, and k cancels. Natural log of one hundred is two times natural log of ten, so the ratio is exactly two! Puzzle solved: ninety nine percent takes twice the time of ninety percent. So the last nine percent takes just as long as the first ninety! Got it?

5:07Example 2: second order half-life with Arrhenius equation

Now a harder one, from Part B, JEE Advanced level. Read it with me. Second order reaction, activation energy capital E a equal to fifty kilojoules per mole. Its t half is forty minutes, at A naught equal to zero point one M and three hundred kelvin. Find t half at A naught equal to zero point two M, and three hundred and ten kelvin. Two things change at once. Pause the video and try it. Done? Okay, let's check. First, the starting concentration has doubled. Many students say: half-life does not depend on the starting concentration, so ignore it. Is that right here? Oh no, careful, that rule is for first order only! This reaction is second order, and for second order, t half equals one over k times A naught. Clear? So, double A naught, and t half is halved. That is one factor of two. Okay? Now the temperature. Higher temperature, bigger k. By how much? Arrhenius: the natural log of k two over k one equals capital E a over R, times one over T one, minus one over T two, in round brackets. R is the gas constant, and T one and T two must be in kelvin. Right? Put in the numbers. Fifty kilojoules becomes fifty thousand joules, and R is eight point three one four joules per kelvin per mole. Fifty thousand over eight point three one four, times one over three hundred minus one over three hundred and ten: zero point six four seven. So k two over k one is one point nine one. With me? Bigger k means shorter t half, so divide again. Forty over two times one point nine one: ten point five minutes. If you ignored the concentration, you got double this, and that was the trap. Two ideas in one problem: that is Part B for you. Got it?

7:05Quick revision: half-life vs order, molecularity, test plan

So, quick revision. Part A: a thirty minute test after Chapters one to six, and name the idea before you calculate. Part B: often two ideas at once; and t half depends on A naught in second order, not in first order. Part C: read every word; order equals molecularity only for an elementary step. Now open Part A, start your clock, and begin. All the best!