Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! Let us start with a quick guess. Here is a reaction with an activation energy of twenty kilojoules per mole, running at three hundred kelvin. Now we double the temperature, to six hundred kelvin. Does the rate constant also double, or does it grow even more? Keep your guess; we will check it in the last example. Today is Part A of our chemical kinetics problem bank, at JEE Main level. Five short problems, and most of them hide a small trap. We sharpen three tools, then solve two problems with them. Let's begin.

0:38Tool 1: the order and the units of k

First tool: the rate law. Rate equals k times A to the power m, times B to the power n. The square brackets mean concentration, k is the rate constant, and the small raised letters m and n are the orders. The overall order is m plus n. And remember, the orders are found by experiment, not read from the balanced equation, except for an elementary step. Okay? Problem one. Rate equals k times A to the power one half, times B to the power three halves. Fractional powers are allowed; orders can be zero, fractional, even negative. So tell me, what is the overall order? Two, correct! One half plus three halves is four halves, which is two. So this reaction is second order overall, even with fractional powers. Right? Now the units of k. For order n, k is in mole per litre, raised to the power one minus n, times per second. On screen, mol is mole, capital L is litre, small s is second, and a power of minus one means per. With me? So for our order two, what are the units of k? One minus two is minus one. Mole per litre, to the power minus one, simply flips over: litre per mole. So k is in litre per mole per second. Clear?

2:04Tool 2: first order and how complete it is

Second tool: the first-order law. k t equals the natural log of A naught over A; l n on screen means natural log. Here A naught, the A with a small zero below it, is the starting concentration, and A is the concentration left at time t. Got it? Problem two. For a first-order reaction, find t ninety nine percent over t ninety percent; the small label below each t says how much has reacted. Now, ninety nine is only a little more than ninety, and the wavy equals sign on screen means approximately. So the ratio is about ninety nine over ninety, one point one, right? No, no, careful here! The law uses what remains, not what has reacted. x percent complete means one hundred minus x percent remains. Okay? So at ninety percent, ten percent is left, and A naught over A is ten. At ninety nine percent, only one percent is left, and A naught over A is one hundred. With me? Now k is the same, so the times are in the ratio of the natural logs. The natural log of one hundred is two times the natural log of ten. So the ratio is exactly two! See, the last nine percent takes as long as the first ninety. Right?

3:21Tool 3: the half-life, zero order versus first order

Third tool: the half-life. For zero order, the concentration falls in a straight line: A equals A naught, minus k t. Clear? Now put A equal to half of A naught. Then t half, written t with a small one over two below it, equals A naught over two k. So for zero order, a bigger start means a longer half-life. Okay? Compare first order. t half equals zero point six nine three over k, and A naught does not appear at all. Start with any amount; the half-life stays the same. Keep this in mind for Example one. Got it? Problem three. For a zero-order reaction, t half is thirty seconds when A naught is zero point six molar; capital M means molar. So what is k? Zero point six over thirty: zero point zero two, right? Oh no, careful here! The formula has a two below: k equals A naught over two times t half. Zero point six over two times thirty is zero point six over sixty: zero point zero one. The units are molar per second, because a zero-order k has the units of a rate. With me?

4:32Example 1: A0 and half-life from two first-order readings

Now, Example one: problem four. In a first-order reaction, A is zero point five molar at one hour, and zero point one two five molar at three hours; small h on screen means hour. Find A naught and the half-life. Pause the video, and try it first. Done? Okay, let's check. Zero point five over zero point one two five is four. So the concentration became one quarter in three minus one, that is two hours. Right? Now think. For first order, each half-life halves whatever is left. How many half-lives turn the amount into one quarter? Two, exactly! One half times one half is one quarter. So two half-lives fit in two hours, and the half-life is one hour. And since a first-order half-life does not depend on the starting amount, every single hour halves it. Clear? Now A naught. The first reading given is zero point five molar. So A naught is zero point five, right? No, no, wait! That reading was taken at one hour, not at time zero. Time zero is one half-life earlier. Okay? So go back one half-life, and double it: two times zero point five, which is one point zero molar. Starting amount one molar, half-life one hour. Got it?

5:55Example 2: Arrhenius factor when T doubles, 300 K to 600 K

Now the harder one, Example two: problem five, and it is our puzzle. Here, a reaction has capital E with a small a below it, the activation energy, equal to twenty kilojoules per mole. By what factor does k rise from three hundred to six hundred kelvin? Capital K on screen is kelvin. Pause the video, and try it first. Done? Okay, let's check. Arrhenius at two temperatures: the natural log of k two over k one equals E a over R, times one over T one, minus one over T two. R is the gas constant, k one goes with T one, and k two with T two. And the temperatures must be in kelvin. Okay? Now, our guess from the start. The temperature doubles, from three hundred to six hundred. So k also doubles, right? No, no, careful here! Let us calculate and see. R is eight point three one four joules per mole per kelvin; J is joules on screen. So capital E a must be in joules too: twenty thousand joules per mole, not twenty. With me? Now the bracket. One over three hundred, minus one over six hundred. Over six hundred, that is two minus one: just one over six hundred. Right? So the natural log of k two over k one is twenty thousand over eight point three one four, times one over six hundred. That works out to approximately four point zero one. Clear? So k two over k one is e to the power four point zero one: about fifty five. Puzzle solved! For this reaction, doubling the temperature makes k about fifty five times bigger, not two times. The temperature works through that exponential, so its effect is huge.

7:40Quick revision: units of k, half-life, Arrhenius traps

Got it? So, quick revision. The overall order is the sum of the powers, and k is in mole per litre, to the power one minus n, times per second. x percent complete means one hundred minus x percent left, and a first-order half-life does not depend on A naught. The zero-order half-life is A naught over two k; and in Arrhenius, temperature in kelvin, E a in joules. Now go and try the two practice problems on this page. All the best!