Vapour Density and the Degree of Dissociation

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. N two O four gas has a molar mass of ninety two grams per mole, so its vapour density should be half of that: forty six. Now take a flask of N two O four at equilibrium with N O two, and measure the vapour density. You get only thirty eight point three three! The gas looks lighter, but not a single gram has left the flask. So how is that possible? By the end of this class, that one number will tell you exactly how much of the gas has broken up. You will meet this vapour density method in JEE numericals too. So, let's begin.

0:42Vapour density follows the moles

So, let's start with the basic idea. Vapour density equals one half times the molar mass; that small fraction on screen is one half, and the cross means times. For a pure gas, just halve its molar mass. Okay? Now think. In a closed flask, some molecules split into smaller ones. The mass does not change at all, but the number of moles goes up. So what happens to the vapour density? It falls, yes, exactly! Since mass is conserved, vapour density is proportional to one over the total moles. The curly sign on screen means proportional to. Right? So, more moles, smaller vapour density. Fewer moles, larger vapour density. Hold on to this one idea; the whole lesson is built on it.

1:32Dissociation: A to nB, alpha = (D - d)/((n - 1)d)

Clear? Now, dissociation. One molecule of A breaks into n molecules of B; the two half arrows mean the reaction is reversible. Just like in the last lesson, we start with one mole of A, okay? Let alpha, this Greek letter, be the degree of dissociation: the fraction of A that breaks up. So tell me, how many moles are there in total at equilibrium? One minus alpha of A is left over. And each mole that breaks gives n moles of B, so n alpha moles of B form. Add them up: one plus, n minus one in brackets, times alpha. With me? Now, two labels. Capital D is the vapour density of undissociated A, the gas before any of it breaks. Small d is the observed value at equilibrium. Got it? Now use our big idea: vapour density goes as one over the moles. Capital D goes with one mole, and small d goes with the total. So capital D over small d equals one plus, n minus one, times alpha. Right? Rearrange it, and here is the formula. Alpha equals capital D minus small d, over, n minus one, times small d. Write this down. Clear? And one careful point: n is the moles of products that one mole of A gives. For example, P C l five gives P C l three plus C l two, two moles of products, so n is two. Okay?

3:06Association: nA to An, alpha = n(d - D)/((n - 1)d)

Now the opposite case: association. Here n molecules of A join up into one bigger molecule, A with a small subscript n. For example, in the vapour, acetic acid, C H three COOH, pairs up: two molecules join into one dimer, written in brackets with a small two. With me? So now tell me. Molecules join, so the number of moles goes down. Will the observed small d be bigger or smaller than capital D? Bigger, yes, exactly! Fewer moles, larger vapour density. So small d is greater than capital D; that is the greater than sign on screen. Right? And the formula changes. Alpha equals n times, small d minus capital D, over, n minus one, times small d. See, it is small d minus capital D on top now, so alpha comes out positive. Got it? Here, capital D is the vapour density of the monomer, the single molecule on its own. Oh no, careful: not of the dimer! Mixing these up is a common slip. Okay? And one bonus: the same formulas hold with molar masses in place of vapour densities. Why? Because vapour density is just half the molar mass, so the ratio does not change.

4:28Example 1: alpha of N2O4 from vapour density 38.33

Clear? Now, remember our puzzle from the start? Let's finish it. The vapour density of an equilibrium mixture of N two O four and N O two is thirty eight point three three. Find alpha for N two O four. Pause the video, and try it first. Done? Okay, let's check. N two O four gives two N O two, so one mole gives two moles of products: n equals two. Right? Capital D is for undissociated N two O four. Its molar mass is ninety two, so ninety two over two: forty six. That is the forty six from our puzzle. Okay? Now the formula. Forty six minus thirty eight point three three, over, two minus one, times thirty eight point three three. That is seven point six seven over thirty eight point three three: zero point two zero. So twenty percent of the N two O four has split. Clear? And a quick check. Capital D over small d: forty six over thirty eight point three three is one point two zero, which is one plus alpha. So one mole of gas became one point two moles. Mystery solved: same mass, more moles, so the gas looks lighter!

5:50Example 2: NH3 dissociation and the per-mole n trap

Now a slightly harder one. Ammonia decomposes: two N H three gives N two plus three H two. The equilibrium mixture has vapour density five. Find the degree of dissociation. Pause the video and try it. Hint: be careful with n. Alright. Two moles of ammonia give one plus three, four moles of products. So n is four, right? No, no, wait! This is the trap. In our formula, n is the moles of products that one mole of A gives. So write the equation per mole of ammonia: N H three gives one half N two plus three halves H two. With me? So n equals one half plus three halves: two, not four! Got it? Capital D: ammonia has molar mass seventeen, so seventeen over two, eight point five. Okay? Now, alpha equals eight point five minus five, over, two minus one, times five. Three point five over five: zero point seven zero. So seventy percent of the ammonia has decomposed. See how one small n decides the whole answer?

7:04Quick revision: dissociation and association formulas

So, quick revision. Vapour density is proportional to one over the total moles, because mass is conserved. Dissociation: alpha equals capital D minus small d, over, n minus one, times small d, with n counted per mole of A, okay? Association: alpha equals n times, small d minus capital D, over, n minus one, times small d, with capital D for the monomer. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!