Mixed Problems: Set 2

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Pulling one electron off a gaseous potassium atom costs four hundred and nineteen kilojoules per mole. Giving that electron to a bromine atom gives back only three hundred and twenty five. So making the two ions costs energy overall, right? And yet, when potassium bromide forms from its elements, energy is released, three hundred and ninety four kilojoules per mole! So where does that energy come from? This is Set two of the mixed problems: eight medium questions in the JEE Main style. First we sharpen two tools, then we solve three problems from the set, our puzzle included. So, let's begin.

0:44Ionisation enthalpy: one electron at a time

First, the ionisation enthalpy. Take a gaseous atom, call it capital X, where the small g in brackets means gas. The arrow shows one electron pulled out, the e minus on screen, leaving the ion X plus, with a small raised plus sign. The energy needed is delta i H one: capital Greek delta, a subscript i for ionisation, capital H for enthalpy, and a subscript one for the first electron. It is always positive: pulling an electron out costs energy. Okay? Pulling out a second electron, then a third, and so on, gives the successive ionisation enthalpies. Delta i H one is less than delta i H two, which is less than delta i H three; the less than signs show the order, and the three dots mean it goes on. Always, because each electron is removed from a more positive ion. Right? And watch for the huge jump. Once the valence electrons are gone, the next electron must come from the noble-gas core, and the value shoots up. Clear? So count the electrons removed before the jump. That is the number of valence electrons. The double arrow means this gives the group: one valence electron, group one; two, group two; three, group thirteen, and so on. With me?

2:00A quick one from Set 2: which group?

Here is a quick one from Set two. An element has successive ionisation enthalpies of eight hundred and one, two thousand four hundred and twenty seven, three thousand six hundred and sixty, and twenty five thousand and twenty six, in kilojoules per mole, the k J mol to the power minus one on screen. Identify its group. Look, the second value is roughly three times the first! So the jump is after the first electron, and it is group one, right? No, no, careful here, because some rise is always there. So compare the ratios. The wavy equals sign means approximately. The second is about three times the first, and the third about one and a half times the second. But the fourth is about six point eight times the third. That is the real jump, and it comes after the third electron. Got it? So three valence electrons means group thirteen, not group three. The element is boron. Clear?

3:00Born–Haber cycle: finding the lattice enthalpy

Alright, now the Born Haber cycle. Take a solid metal M, the small s in brackets, reacting with one half of X two, a gas, where the small two below means two atoms in the molecule. It gives the solid, M X, okay? Delta f H, with a small f below, is the enthalpy of formation. It equals delta sub H of M, the sublimation of the metal, plus one half of delta diss H of X two, the dissociation, plus delta i H of M, the ionisation enthalpy. Only half the dissociation, because one M X needs just one X atom. Right? Then plus delta e g H of X, with e g below for electron gain enthalpy. For most elements it is negative, because energy is released. And finally, plus delta latt H, the lattice enthalpy. Got it? The lattice enthalpy is the enthalpy of forming the solid from the gaseous ions. Energy is released here, so it is negative. Clear? So the lattice enthalpy is delta f H minus the sum of all the other steps. And watch the signs: every step goes in with its own sign. With me?

4:10Example 1: energy to turn 0.1 mol of Mg(g) into Mg2+(g)

Now, Example one, from Set two. Find the energy needed to convert zero point one mole of magnesium, M g, with the small g in brackets for gas, into magnesium two plus ions, with the raised two plus, also as gas. Delta i H one is seven hundred and thirty eight, and delta i H two is one thousand four hundred and fifty one kilojoules per mole. Pause the video, and try it first. Done? Okay, let's check. We want magnesium two plus, so we use delta i H two. So zero point one times one thousand four hundred and fifty one gives one hundred and forty five point one kilojoules, right? Oh no, careful here, because delta i H two starts from magnesium plus, not from the atom. The two arrows on screen show it: the atom first becomes magnesium plus, then magnesium two plus, and we pay for both steps. Okay? So per mole, add them. Seven hundred and thirty eight plus one thousand four hundred and fifty one gives two thousand one hundred and eighty nine kilojoules per mole. Right? But we have only zero point one mole, m o l on screen. The cross means times: zero point one times two thousand one hundred and eighty nine gives two hundred and eighteen point nine kilojoules. So we avoided two traps, forgetting the first step and forgetting the zero point one. Got it?

5:30Example 2: lattice enthalpy of KBr by Born-Haber cycle

Now, Example two from Set two, and this is our puzzle! For potassium bromide, K B r on screen, delta f H is minus three hundred and ninety four. The sublimation of potassium is eighty nine, and delta i H of potassium is four hundred and nineteen. The enthalpy to form gaseous bromine atoms from one half of liquid bromine, B r two with the small l for liquid, is one hundred and twelve. And delta e g H of bromine is minus three hundred and twenty five, all in kilojoules per mole. Find the lattice enthalpy. Pause the video, and try it first. Done? Okay, let's check. Our cycle had one half of the dissociation enthalpy. So we take half of one hundred and twelve, that is fifty six, right? No, no, wait, read the question again! The one hundred and twelve already takes one half of liquid bromine to one mole of gaseous bromine atoms. The half is already inside it, so do not halve it again. Okay? Now add the other steps. Eighty nine, plus four hundred and nineteen, plus one hundred and twelve, and then minus three hundred and twenty five for the electron gain enthalpy, with its own sign. The total is two hundred and ninety five. With me? Put it in the cycle. Minus three hundred and ninety four equals two hundred and ninety five, plus delta latt H, okay? So delta latt H equals minus three hundred and ninety four, minus two hundred and ninety five, which gives minus six hundred and eighty nine kilojoules per mole. And that solves our puzzle! Making the gaseous ions costs two hundred and ninety five, but bringing them together into the solid releases six hundred and eighty nine. That pays for everything, with three hundred and ninety four to spare. Got it?

7:22Quick revision: IE jump, Mg2+ energy, lattice enthalpy

So, quick revision. Successive ionisation enthalpies always rise, and the huge jump gives the group. To make magnesium two plus, add both ionisation steps, then multiply by the moles. And in a Born Haber cycle, the lattice enthalpy is delta f H minus the sum of the other steps, each with its own sign. Now go and try the practice problem on this page. You can do it. All the best!