Mixed Problems: Set 2

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take a gas, call it A, that breaks up into three gas molecules. We start with A alone at a pressure of one hundred, and after ten minutes the total pressure is two hundred. It has doubled! So, is all of A used up? And ten minutes later, what will the pressure be? Three hundred? Four hundred? Hold on to these questions. This is Set two of the mixed problems: eight medium questions in the JEE Main style, and many of them use first-order kinetics. So first we sharpen the first-order tools, then we see how the half-life tells you the order, and then we solve two problems from the set, our puzzle included. So, let's begin.

0:46The first-order toolkit for Set 2

So, the first-order law. Square brackets mean concentration, and the small zero below means the start, so we say A naught for the starting concentration. The natural log of A naught over A equals k times t, where k is the rate constant and t is the time. Okay? Now a quick question. The half-life is the time for half of the reactant to go. If you start with twice as much A, does the first-order half-life change? No, it does not! This is t half, written t with a small one over two below it. t half equals natural log of two over k, that is zero point six nine three over k. There is no A naught in it, so the first-order half-life is independent of the starting concentration. For zero and second order, the half-life does depend on it. Remember that. Right? And two small habits. When a question says x percent complete, it means one hundred minus x percent remains. And keep k and t in the same time unit, minutes with per minute, seconds with per second. Clear?

1:54Half-life tells the order: a quick one from Set 2

Now, the half-life can also tell you the order. t half is proportional to A naught to the power one minus n, where n is the order. This sign, like an open alpha, means proportional to. Okay? So, zero order: t half is proportional to A naught. First order: the power is zero, so t half stays constant. And second order: t half is proportional to one over A naught. With me? Here is a quick one from Set two. The half-life is fifty seconds when A naught is zero point four molar, and one hundred seconds at zero point one molar. Capital M means molar, moles per litre. The concentration falls to a quarter, and the half-life doubles. It goes up, so it must be second order, right? No, no, careful here! For second order, a quarter of the concentration would make the half-life four times. Here it only doubles. So we use the formula. n equals one, plus log of t one over t two, over log of a two over a one. Here t one and t two are the two half-lives, a one and a two are the two starting concentrations, and log is to the base ten. Clear? Fifty over one hundred is one half, and zero point one over zero point four is one quarter. One quarter is one half squared, so the log of a quarter is twice the log of a half. n equals one plus one half: three over two. An order of three halves! Fractional orders are fine, because order comes from experiment, not from the equation. Got it?

3:31Example 1: half-life when 30% complete in 20 min

Now, Example one, from Set two. Read it with me. The reaction is first order, and it is thirty percent complete in twenty minutes. Find its half-life. Pause the video, and try it first. Done? Okay, let's check. Thirty percent in twenty minutes. So fifty percent takes twenty times fifty over thirty, about thirty three point three minutes, right? Oh no, careful here, that works only if the same amount reacts every minute. In first order, the rate is k times the concentration of A, so it slows down as A is used up. So we use the integrated law. Thirty percent complete means seventy percent remains, so A naught over A is one hundred over seventy. Okay? Now k equals one over twenty, times the natural log of one hundred over seventy. Time is in minutes, so k is in per minute: min to the power minus one on screen. Right? Then t half equals natural log of two over k. Put k in, and t half equals twenty times natural log of two, over natural log of one over zero point seven. One over zero point seven is the same as one hundred over seventy. With me? Twenty times natural log of two is thirteen point eight six, and the natural log of one over zero point seven is zero point three five six seven. The wavy equals sign means approximately: about thirty eight point nine minutes. See, more than thirty three point three, as it should be, because the reaction slows down. Clear?

5:10Example 2: half-life and total pressure for A to 2B + C

Now remember our puzzle? Here it is, as Example two from Set two. Gas A gives two B plus C, and it is first order. The small g in brackets means gas, and the arrow means gives. We start with A alone at one hundred m m, that is millimetres of mercury, and the total pressure is two hundred m m after ten minutes. Find the half-life, and the total pressure after twenty minutes. Pause the video, and try it first. Done? Okay, let's check. The pressure went from one hundred to two hundred. It doubled. So all of A is used up, right? No, no, careful here! Each A gives three gas molecules: two of B and one of C, okay? So let the pressure of A fall by x. Then A has one hundred minus x left, B has two x, and C has x. Add them up: p with a small t below, the total pressure at time t, equals one hundred plus two x. Got it? Now set one hundred plus two x equal to two hundred. The double arrow means so: x equals fifty. Then p with a small A below, the pressure of A itself, is one hundred minus fifty, just fifty. Only half of A is gone, not all of it! Right? And half of A gone in ten minutes is exactly the half-life. So t half equals ten minutes. Clear? Now twenty minutes. In the first ten minutes, fifty of A reacted. So in the next ten minutes, another fifty reacts, and A is completely finished, right? No, no, that is the trap again! The first-order half-life does not depend on the starting amount, so every ten minutes, half of what is left goes. Twenty minutes is two half-lives: one hundred, then fifty, then twenty five. So the pressure of A is one quarter of one hundred, twenty five, and x equals seventy five. With me? Finally, p t equals one hundred plus two times seventy five: two hundred and fifty m m. Not three hundred, not four hundred: two hundred and fifty. Puzzle solved! The pressure doubled with only half of A gone, because each A gives three molecules. Got it?

7:32Quick revision: half-life, order and gas pressure

So, quick revision. First order: t half is zero point six nine three over k, whatever the starting amount. How t half changes with A naught gives you the order. x percent complete means one hundred minus x percent remains, and in first order, never scale the time linearly. And for gases, write the total pressure in terms of x, then find the pressure of the reactant itself. Now go and try the practice problems on this page. You can do them. All the best!