Mixed Problems: Set 3

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. At three hundred kelvin, a first-order reaction is too slow for us. We want it ninety percent complete in ten minutes. So how much should we heat it? By fifty kelvin? By one hundred? The real answer is surprisingly small. This is Set three, the hard set, where each problem joins two ideas, and problems like these come up often in JEE Main. So we sharpen three tools, then solve two problems with them. Let's begin.

0:32Tool 1: the coefficient sits inside the rate

First tool. Say small a molecules of A give products; the arrow on screen means gives. Then the rate of reaction, small r, equals minus one over a, times d A over d t. The square brackets mean concentration, d of it over d t is how fast it changes, and the minus sign shows a reactant used up. The one over a divides by the coefficient, so the rate is the same for every species. Okay? Now, second order. One over A, minus one over A naught, equals k t. Here A naught, written A with a small zero below it, is the starting concentration. This holds for one reactant with coefficient one, or for two reactants that react one to one, with equal starting concentrations. Right? Quick question. For first order, the half-life does not depend on the starting concentration. What about second order? Does it depend on A naught? It does, yes! Put A equal to half of A naught, and t half equals one over k times A naught. We write t half as t with a small one over two below it. Double A naught, and the half-life halves. Keep this law and the coefficient ready; they meet in Example one.

1:49Tool 2: first order, and what complete means

Clear? Second tool, first order. k t equals the natural log of A naught over A; l n on screen means natural log. Or, k equals two point three zero three over t, times log to the base ten of a over a minus x. Here small a is the starting amount and x the amount reacted, so a minus x is what is left. With me? Now, problems rarely say how much is left; they say ninety percent complete. So tell me: when a reaction is ninety percent complete, what fraction of A is left? Ten percent, one tenth. Yes, exactly! x percent complete means one hundred minus x percent remains. So, A naught over A is ten. Got it? So k t equals the natural log of ten, about two point three zero three; the wavy equals sign means approximately. See, that famous two point three zero three is simply the natural log of ten. Okay? One more trap. If k is per second, time goes in seconds; if k is per minute, time goes in minutes. Keep k and t in the same time unit. Right?

3:05Tool 3: Arrhenius at two temperatures

Third tool: temperature. The Arrhenius equation: k equals capital A, times e to the power minus capital E a over R T, okay? Capital A is the frequency factor, capital E with a small a below it is the activation energy, R is the gas constant, and T is the temperature. Got it? For two temperatures, we use this form. The natural log of k two over k one equals E a over R, times one over T one, minus one over T two. k one goes with T one, and k two with T two; the small numbers below are just labels. Clear? Now think. If T two is higher than T one, is k two bigger or smaller than k one? Bigger, correct! One over T one is then larger than one over T two, so the bracket is positive, and k two is bigger. Now two rules. Temperatures always in kelvin. And R is eight point three one four joules per mole per kelvin; on screen, J is joules, mol is mole, K is kelvin, and the power minus one means per. So R is in joules, not kilojoules. Remember that for Example two. With me?

4:20Example 1: half-life for 2A to products, the 2k factor

Now, Example one, from Set three. For two A gives products, the rate equals minus one half times d A over d t, which equals k times A squared. k is zero point zero five litre per mole per second; L is litre and s is second. The starting concentration is zero point one molar, where capital M means molar. Find the half-life. Pause the video, and try it first. Done? Okay, let's check. Second order, so t half equals one over k times A naught. That is one over zero point zero zero five: two hundred seconds. Right? No, no, careful here! Read the rate definition again. It has a one half in front, because two molecules of A are used up each time. So minus d A over d t, how fast the reactant itself disappears, is two k times A squared. That is twice as fast as the rate. Okay? So the second-order law now carries two k, not k. One over A, minus one over A naught, equals two k t. Missing this factor of two is a common error. Got it? Now put A equal to half of A naught. t half equals one over two k times A naught. Two times zero point zero five times zero point one is zero point zero one, and one over that is one hundred. So the half-life is one hundred seconds, not two hundred. Clear?

5:52Example 2: temperature for 90% in 10 min, Ea 80 kJ/mol

Now, remember our puzzle from the start? Here it is, Example two, also from Set three. For a first-order reaction, k is two times ten to the power minus three per second at three hundred kelvin. Capital E a is eighty kilojoules per mole. At what temperature is it ninety percent complete in ten minutes? Pause the video, and try it first. Done? Okay, let's check. First, the k we need. k is per second, so ten minutes, min on screen, becomes six hundred seconds. Ninety percent complete means A naught over A is ten, so k two times six hundred equals the natural log of ten. So k two is three point eight four times ten to the power minus three per second. With me? Now Arrhenius, with k one at three hundred kelvin. The natural log of k two over k one is the natural log of three point eight four over two: zero point six five two. So the rate constant must nearly double. Clear? Next, E a is eighty kilojoules per mole. So we put eighty over eight point three one four into the equation, right? Oh no, careful here, that is the trap! R is in joules, so E a must be in joules too: eighty thousand joules per mole. So zero point six five two equals eighty thousand over eight point three one four, times one over three hundred, minus one over T, okay? Now solve for the bracket. Zero point six five two, times eight point three one four, over eighty thousand: six point seven eight times ten to the power minus five. Clear? So one over T is one over three hundred, minus that: three point two six six times ten to the power minus three. The double arrow means so: T is about three hundred and six kelvin. Our puzzle is solved: just six kelvin warmer, and the rate constant nearly doubles! Got it?

8:01Quick revision: 2k factor, percent complete, Ea in joules

So, quick revision. For two A gives products, with the one half in the rate, the law carries two k. x percent complete means one hundred minus x percent remains; keep k and t in the same time unit. And in Arrhenius, temperatures in kelvin, and E a in joules to match the gas constant. Now go and try the practice problems on this page. All the best!