Overview

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. In the reaction A plus B gives products, one A meets one B, okay? So double the concentration of A, and the rate should double, right? But in one problem of this chapter, doubling A makes the rate four times bigger! How is that possible? Hold on to that question. Five chapters are done, so it is time to test yourself. This chapter gives you three sets of mixed problems in the JEE Main style. Let us see how to use them, and solve this puzzle together. So, let's begin.

0:37How to use this chapter

So, what do JEE Main questions on kinetics look like? Short numericals, of four types. First, rates from stoichiometry: the balanced equation links the rates of all the species. Okay? Then orders from experimental data, integrated rate laws and half-lives, and Arrhenius calculations, where the temperature changes the rate constant. Four types in all. Right? Quick question. Our puzzle from the start, doubling A and comparing the rates: which of these four types is it? Orders from data, yes, exactly! But see, the question never names the type. Spotting it is your job. Set one is easy, with nine solved problems, Set two is medium, with eight, and Set three is hard, with six. Clear? Then each set ends with two practice problems for you. Got it? And the real JEE questions, from twenty twenty one to twenty twenty six? They are at the end of Chapters one to five; the problems here are original, in the same style. With me?

1:46Name the type first: a quick one from Set 1

Now, name the type before you calculate. Here is a quick one from Set one. Two N two O five gives four N O two plus O two. The small numbers below count the atoms, and the arrow means gives. N two O five is used up at zero point zero two molar per second; the capital M on screen means molar, moles per litre. Find how fast N O two and O two form. This is rates from stoichiometry. So, N O two also forms at zero point zero two, the same speed, right? No, no, careful here! The coefficients matter. The rate of reaction, small r, equals minus one half, times d N two O five over d t. That equals one quarter times d N O two over d t, and d O two over d t. Square brackets mean concentration, d of it over d t is its rate of change, and the minus sign shows a reactant used up. Right? So N O two forms at zero point zero two times four over two: zero point zero four. And O two forms at zero point zero two times one half: zero point zero one molar per second. One line each, once you name the type. Got it?

3:05Take R=8.314: an Arrhenius one from Set 1

Next, one value for the whole chapter. R, the gas constant, equals eight point three one four joules per mole per kelvin. On screen, J is joules, mol to the power minus one means per mole, and K to the power minus one means per kelvin. Okay? You need R in Arrhenius calculations. 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. Here l n is the natural log, k one and k two are the rate constants at the two temperatures T one and T two, and capital E with a small a below it is the activation energy. The round brackets hold the difference, one over T one minus one over T two. Right? Quick check. Those two temperatures: in degrees Celsius, or in kelvin? Kelvin, always. Correct! Now a quick one from Set one: a rate constant quadruples, becomes four times, when the temperature goes from three hundred to three hundred and twenty kelvin; the arrow here just means to. One over three hundred, minus one over three hundred and twenty, is one over four thousand eight hundred. So the natural log of four equals E a over eight point three one four, times that. Clear? So, E a equals four thousand eight hundred, times eight point three one four, times the natural log of four. The wavy equals sign means approximately: about fifty five thousand three hundred and twenty joules per mole, or fifty five point three kilojoules per mole. With me?

4:47Example 1: time for 75% of a first order reaction

Now, Example one, from Set one. For a first-order reaction, k equals zero point zero six nine three per minute. On screen, min to the power minus one means per minute. How long does seventy five percent completion take? Pause the video, and try it first. Done? Okay, let's check. Integrated laws and half-lives. First the half-life, t half, written t with a small one over two below it. For first order, t half equals zero point six nine three over k. Zero point six nine three over zero point zero six nine three: ten minutes. Okay? Now, fifty percent takes ten minutes. Seventy five percent is one and a half times fifty percent, so fifteen minutes, right? No, no, careful here! That works only if the same amount reacts every minute. Seventy five percent complete means twenty five percent remains, one quarter. Right? And for first order, the half-life does not depend on the starting concentration. So every half-life takes the same ten minutes. The arrows show them: one hundred percent to fifty in ten minutes, then fifty to twenty five in another ten. Clear? So, two half-lives: twenty minutes, not fifteen.

6:11Example 2: rate law and k from initial rate data

Got it? Remember our puzzle? Here it is, from Set two. For A plus B gives products, there are three runs. Run one: A and B both at zero point one, rate two times ten to the power minus three. Run two: A at zero point two, B at zero point one, rate eight times ten to the power minus three. Run three: A at zero point one, B at zero point three, rate six times ten to the power minus three molar per second. Find the rate law and k. Pause the video, and try it first. Done? Okay, let's check. One A and one B in the equation, so the rate is just k times A times B, first order in each, right? No, no, careful here! That is the trap. Order is found by experiment, not from the balanced equation. So write r equals k times A to the power x, times B to the power y, and let the data decide x and y. Okay? Compare runs one and two: only the concentration of A changes. It doubles, and the rate goes from two to eight, four times. Four is two to the power two, and the double arrow means so: x equals two. That was our puzzle: double A, and the rate becomes four times! Right? Now runs one and three: only the concentration of B changes. It triples, and the rate goes from two to six, three times. Three to the power one, so y equals one. Clear? So the rate law: r equals k times A squared, times B, okay? Second order in A, first order in B, just as the data said. With me? Now k, from run one. k equals two times ten to the power minus three, over zero point one squared, times zero point one. The bottom is one thousandth, so k equals two. The units, L squared mol to the power minus two s to the power minus one, mean litre squared, per mole squared, per second. Got it?

8:25Quick revision: four types, R = 8.314, T in kelvin

So, quick revision. Four types, three sets from easy to hard: name the type first. Take R as eight point three one four joules per mole per kelvin, with temperatures in kelvin. And remember: order comes from experiment, and the first-order half-life does not depend on A naught, the starting concentration. Now open Set one, and begin. All the best!