Born–Haber Cycles

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. To pull an electron off a sodium atom, you must put in four hundred and ninety six kilojoules per mole. When chlorine takes that electron, it gives back only three hundred and forty nine. So turning the gaseous atoms into ions is uphill: it costs one hundred and forty seven kilojoules per mole. And yet, when sodium and chlorine form solid sodium chloride, four hundred and eleven kilojoules per mole are given out. So who pays the bill? This cycle does appear in JEE questions, so let's find out.

0:39Two routes to the same ionic solid

So, here is the reaction. M is a metal like sodium; the small s in brackets means solid. X two, with a small two below, is a gas like chlorine; g means gas. The half in front? One M X needs one X atom, which is half an X two molecule. And the arrow leads to the solid, M X, okay? The direct route is this one reaction. Its enthalpy change is delta f H: the Greek capital delta for change, H for enthalpy, and a small f for formation. Right? Now let me ask you this. From the same elements, is there another way to reach the same solid? That's right, through ions: first turn everything into gaseous ions, then let them pack into the solid. That is the long route. With me? Now the key point. Both routes start at the same elements and end at the same solid, so the total enthalpy change must be the same. This loop is called a Born Haber cycle. Got it?

1:40The long route, step by step

Alright, the long route. Step one, sublimation: the solid metal turns into a gas. That is delta sub H of M, with sub written small, and it takes energy in. Okay? Step two: the X two molecules break into atoms. That is the dissociation enthalpy, delta diss H, and we take only half of it, for half a molecule. Right? Step three, from the previous chapter: the ionisation enthalpy, delta i H, with a small i. The metal atom loses an electron, the e with a minus sign, and becomes M plus; the raised plus is its charge. Clear? Step four, from this chapter: electron gain enthalpy, delta e g H, for the X atom. It takes that electron and becomes X minus, and for most elements, energy comes out here. With me? Now the last step. The plus and minus ions attract strongly, so when they come together into a solid, is energy given out, or taken in? It is given out, very good! This is the lattice enthalpy, delta latt H, the enthalpy of forming the solid from its gaseous ions. So in this course, delta latt H is negative. Some books quote the reverse step, breaking the solid into gaseous ions, and that value is positive. Got it?

3:01Add the steps: the Born–Haber equation

So, the two routes give the same total. Delta f H, the direct route, equals the five steps added up: delta sub H of M, plus half delta diss H of X two, plus delta i H of M, plus delta e g H of X, plus delta latt H, the lattice step. This is the Born Haber equation; write it down. Right? Now, a topper's trick. Often, the one missing value is the lattice enthalpy. So add the other four steps, and subtract that sum from delta f H, the formation enthalpy. Clear? Two habits save marks here: keep every sign as given, and take only half the dissociation enthalpy. With me?

3:42Example 1: lattice enthalpy of NaCl, half-dissociation trap

Now, example one. For sodium chloride, delta f H is minus four hundred and eleven. Sublimation of sodium is one hundred and eight, and delta i H of sodium is four hundred and ninety six. Dissociation of C l two is two hundred and forty two, and delta e g H of chlorine is minus three hundred and forty nine, all in kilojoules per mole; the minus one on mol means per mole. Find the lattice enthalpy. Pause the video and try it first. Done? Okay, let's check. The dissociation enthalpy of chlorine is two hundred and forty two, so that goes into the equation, right? Oh no, careful, that is the trap! One sodium chloride needs one chlorine atom, which is half a C l two molecule. So half times two hundred and forty two gives one hundred and twenty one. Clear? Now write the equation, with every sign. 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 delta latt H, the only unknown. Okay? Now the topper's trick: add the known steps. One hundred and eight, plus one hundred and twenty one, plus four hundred and ninety six, gives seven hundred and twenty five. Then subtract three hundred and forty nine, which leaves three hundred and seventy six. Right? So delta latt H equals minus four hundred and eleven, minus three hundred and seventy six, which gives minus seven hundred and eighty seven kilojoules per mole. Negative, as it should be, for ions forming a solid. Got it?

5:21The NaCl cycle as an energy ladder

Here is the same cycle as a picture from the course: each flat line an energy level, each arrow a step, in kilojoules per mole. The dark arrow on the left is delta f H, minus four hundred and eleven, straight down to solid sodium chloride. The equation on the right writes bond for the dissociation, and lattice in full. Okay? Climb the green arrows: sublimation, one hundred and eight; half bond dissociation, one hundred and twenty one; and ionisation, four hundred and ninety six. Together, seven hundred and twenty five go in. Right? Then come down the orange arrows: electron gain, three hundred and forty nine, and the long lattice arrow, seven hundred and eighty seven, our answer. Together, one thousand one hundred and thirty six come out. Clear? So the net change is seven hundred and twenty five, minus one thousand one hundred and thirty six, which gives minus four hundred and eleven, exactly the formation enthalpy. And that solves our puzzle! Making the ions was uphill, but the lattice enthalpy pays the whole bill. With me?

6:26Example 2: why MgO has a much larger lattice enthalpy

Now example two, a reasoning question. The lattice enthalpy of magnesium oxide is about minus three thousand eight hundred kilojoules per mole. Why is it so much larger in magnitude, that is, in size without the sign, than that of sodium chloride? Pause the video and try it first. Done? Okay, let's think. In magnesium oxide the charges are two plus and two minus, double those in sodium chloride. So the attraction simply doubles, right? No, no, wait, that is the trap! The attraction scales as the product of the charges, so we must multiply them. Okay? For sodium chloride, the plus or minus one on screen means charges of plus one and minus one, and one times one is one. For magnesium oxide, two times two gives four. So the charges alone make it four times, not two. Clear? And a second reason: M g two plus and O two minus, with their charges written raised, are also smaller ions, so they sit closer and attract even more strongly. Got it? So, about three thousand eight hundred, against seven hundred and eighty seven, is nearly five times larger: the charges give four, and the smaller size pushes it further. With me?

7:42Quick revision: Born-Haber equation and lattice enthalpy

So, quick revision. First, delta f H equals sublimation, plus half the dissociation, plus delta i H, plus delta e g H, plus delta latt H, all with their signs. Second, delta latt H is for gaseous ions forming the solid, and it is negative. And third, larger charges and smaller ions give a larger lattice enthalpy, in magnitude. Now go and try the practice problem on this page. All the best!