Lattice Enthalpy and the Born–Haber Cycle

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Pulling the outer electron off a sodium atom in the gas phase costs four hundred and ninety six kilojoules per mole. Chlorine takes it and gives back only three hundred and forty nine. So making the ions costs energy, right? Yet sodium and chlorine form sodium chloride and release four hundred and eleven kilojoules per mole! So who pays the bill? Today's answer: lattice enthalpy, and the Born Haber cycle that finds it. It comes up often in JEE, so let's begin.

0:37Lattice enthalpy: gaseous ions into a solid

So, look at this reaction. The g in brackets means gas, and s means solid. The little plus on sodium means a positive ion, and the minus on chlorine, a negative chloride ion. The arrow: the gaseous ions come together into solid sodium chloride. Okay? Now tell me. Is heat taken in here, or given out? Given out, yes! That is the lattice enthalpy. The triangle is the capital Greek letter delta, meaning change in, H is enthalpy, and the small word lattice says which change. One mole of ionic solid from its gaseous ions: negative. Right? The reverse, breaking the solid back into gaseous ions, is the lattice dissociation enthalpy: positive. Read which one the question means. Clear? But here is the catch. Gaseous ions cannot be brought together in a calorimeter, the instrument that measures heat, so we cannot measure it directly. Every other step can be measured, so Hess's law gives us this one. With me?

1:43The Born–Haber cycle for sodium chloride

Now, the Born Haber cycle for sodium chloride. Each flat line is an energy level, and each arrow a step, with its value in kilojoules per mole. We start here: solid sodium, plus half a mole of chlorine gas, C l two. Half, because one sodium chloride needs only one chlorine atom, okay? Now climb the green arrows. Sublimation: solid sodium to sodium gas. Half bond dissociation: chlorine molecules split into atoms. Ionisation: sodium loses an electron, the e with a minus sign. Upward steps take energy in, so they are positive. Got it? Then down the orange arrows. Electron gain: chlorine takes the electron and gives energy out. Then the long orange arrow: the gaseous ions lock into the solid. That is the lattice enthalpy, minus seven hundred and eighty seven, the biggest drop! We will prove that number in Example one. Downward steps are negative, right? Now the dark arrow on the left: delta f H, the small f for formation, straight from the elements to sodium chloride. Two routes, same start, same end. So what about their totals?

2:59Hess's law around the cycle

Yes, exactly! The totals are equal: Hess's law. Delta f H equals delta sub H, plus half delta diss H, plus delta i H, plus delta e g H, plus delta lattice H, okay? Sub is sublimation, diss is bond dissociation, i is ionisation, and e g is electron gain. Every term on the right is measured, except delta lattice H, the only unknown. One equation, one unknown. Clear? Now a common slip. Bond dissociation breaks a whole C l two molecule into two atoms. For one sodium chloride, how many chlorine atoms do we need? Just one! Correct! So take half the chlorine chlorine bond enthalpy; the dash between the two chlorines is the bond. Using the whole of it is a very common mistake. Count the atoms. Got it? And calcium chloride, C a C l two? Two chlorine atoms: the whole bond enthalpy goes in once, and the electron gain of chlorine twice. Same rule, okay?

4:06What makes lattice enthalpy large?

Now, what makes a lattice enthalpy large? First, charge: larger charges, larger lattice enthalpy. Larger means the magnitude, ignoring the minus sign. Right? Second, size: smaller ions, larger magnitude. Okay? Quick test. Sodium fluoride, chloride, bromide and iodide: the same sodium ion each time. Which has the largest lattice enthalpy? Sodium fluoride, yes! Fluoride is the smallest of the four negative ions. So the order goes fluoride, chloride, bromide, iodide, with the greater than sign between them. Clear? Now, magnesium oxide against sodium fluoride. Which is larger, and by a little or by a lot? Magnesium oxide, by a lot! Magnesium is M g two plus, and oxide is O two minus: double charges on both. This sign, like two greater than signs together, means much greater than. With me? And a large lattice enthalpy means a high melting point, and hardness. Magnesium oxide melts at about two thousand eight hundred degrees Celsius, the little circle and C, and sodium fluoride at about nine hundred and ninety. Okay?

5:22Example 1: lattice enthalpy of NaCl from Born-Haber

Now, Example one: let's prove that number from the diagram, and solve our puzzle from the start. Find the lattice enthalpy of sodium chloride. Delta f H, minus four hundred and eleven. Sublimation of sodium, one hundred and eight. Half bond dissociation of chlorine, one hundred and twenty one. Ionisation of sodium, four hundred and ninety six. Electron gain of chlorine, minus three hundred and forty nine. All in kilojoules per mole, the k J mol to the power minus one. Pause the video, and try it. Done? Okay, let's check. Let U be the lattice enthalpy. Hess's law: minus four hundred and eleven equals one hundred and eight plus one hundred and twenty one plus four hundred and ninety six, minus three hundred and forty nine, plus U, okay? Electron gain goes in with its minus sign. The first three add to seven hundred and twenty five. Minus three hundred and forty nine: three hundred and seventy six. Right? So U is minus four hundred and eleven, minus three hundred and seventy six. Minus seven hundred and eighty seven kilojoules per mole. Perfect! Now, our puzzle. Making the ions costs a net three hundred and seventy six. The lattice gives back seven hundred and eighty seven. The difference, four hundred and eleven, is exactly the heat released! The lattice pays the bill. Mystery solved!

6:56Example 2: lattice enthalpy of MgO, two ionisations

Now a harder one: magnesium oxide. Delta f H, minus six hundred and two. Sublimation of magnesium, one hundred and forty eight. First and second ionisation, seven hundred and thirty eight and one thousand four hundred and fifty one. Half delta diss H of O two, two hundred and forty nine. First and second electron gain of oxygen, minus one hundred and forty one and plus seven hundred and ninety eight. Pause and try. Done? Let's check. Ionisation of magnesium: just seven hundred and thirty eight, right? No, no, wait! Magnesium becomes M g two plus: it loses two electrons. So both ionisations go in, and they add to two thousand one hundred and eighty nine. Okay? Same for oxygen: O two minus means two electrons gained. And a second electron gain always costs energy, so it is plus seven hundred and ninety eight. Minus one hundred and forty one plus seven hundred and ninety eight: plus six hundred and fifty seven. Got it? Now Hess's law: minus six hundred and two equals one hundred and forty eight, plus two thousand one hundred and eighty nine, plus two hundred and forty nine, plus six hundred and fifty seven, plus U, right? The known steps total three thousand two hundred and forty three. So U is minus six hundred and two, minus three thousand two hundred and forty three: minus three thousand eight hundred and forty five kilojoules per mole! Clear? That is nearly five times sodium chloride's seven hundred and eighty seven. The double charges more than pay for both extra costs. Very good!

8:45Quick revision: lattice enthalpy, Born-Haber and trends

So, quick revision. One: lattice enthalpy, gaseous ions to one mole of ionic solid, is negative. Two: in the Born Haber cycle, delta f H is the sum of all the steps; count the atoms and keep the signs. Three: larger charges and smaller ions mean a larger lattice enthalpy and a higher melting point. Now go and try the practice problems on this page. All the best!