Bond Enthalpy and Atomisation

Transcript

0:00Introduction

Hello everyone! Let's start with a small puzzle. Methane has four carbon hydrogen bonds, and all four are exactly alike. Now, pull the hydrogen atoms off one at a time. Will every bond need the same energy to break? Most students say yes. Same bonds, same price, right? Hold that thought. By the end of this video you will see the real numbers, and you will know which one to use. Bond enthalpies come up often in JEE thermochemistry, so let's begin.

0:33Bond dissociation enthalpy

So, first, what is a bond dissociation enthalpy? It is the enthalpy needed to break one mole of a particular bond, in gaseous molecules, into gaseous fragments. The arrow on screen means gives. And notice the condition: gas in, gas out, okay? Here is the simplest one. H two gas, where the small two means two hydrogen atoms joined, gives two hydrogen atoms. The small g in brackets means gas. It needs four hundred and thirty six kilojoules per mole; the little minus one after mol means per. Clear? Now think. When you break a bond, does the molecule take energy in, or give energy out? Taken in, exactly! You have to pull the atoms apart, so you must supply energy. That is why bond enthalpies are always positive. Right? And the other way round? When atoms join to form a bond, energy comes out. Forming bonds releases energy. Keep this sign in mind; we need it in Example two. Got it?

1:36Mean bond enthalpy

Now, back to our puzzle. Methane, C H four; the small four means four hydrogen atoms. It has four C H bonds, and the little dash between C and H stands for one bond. We break them one after another. So, does each break need the same energy? No, no, it does not! In a polyatomic molecule, a molecule with many atoms, the successive bonds of the same kind need different energies. Four identical bonds, four different numbers. Surprising, right? So which number do we use in calculations? The average. We call it the mean bond enthalpy, okay? For the C H bond in methane, the mean is four hundred and fifteen kilojoules per mole. In Example one, we will work out this average ourselves, from real numbers. With me?

2:28Enthalpy of atomisation

Now, the enthalpy of atomisation. We write delta H, where delta, the little triangle, means change in, with a small subscript a for atomisation. It is the enthalpy change to break a substance completely into gaseous atoms. Every single bond, gone. Clear? For methane gas, that means one carbon atom plus four hydrogen atoms, all as gas. Every bond in the molecule is broken. Okay? So, delta a H of methane is simply the mean C H bond enthalpy, isn't it? Oh no, careful! Count the bonds: four C H bonds, and all four break. So delta a H of methane equals four times the mean C H bond enthalpy; the cross sign means times. Got it? Now, what about a metal, like sodium? Atomising it means turning the solid metal straight into gaseous atoms. What do we call that process? Sublimation, yes! So, for a metal, the enthalpy of atomisation equals its enthalpy of sublimation. Clear?

3:35Topper's trick: enthalpy of formation of a gaseous atom

Now, a trick toppers use. Delta f H, with a small subscript f for formation, of gaseous hydrogen atoms. Breaking H two needs four hundred and thirty six, so delta f H of hydrogen atoms is four hundred and thirty six, right? No, careful! One H two gives two hydrogen atoms. So for one mole of atoms, we break only half a mole of H two: that is the one half you see on screen. So delta f H equals half times four hundred and thirty six, which is two hundred and eighteen kilojoules per mole. Clear? And that is the trick. For an element that is a diatomic gas, delta f H of its gaseous atom equals half its bond enthalpy. Diatomic means two atoms in each molecule, like hydrogen, oxygen and chlorine. Careful: bromine is a liquid and iodine is a solid, so for them this half rule alone is not enough. Okay? So, oxygen: the double line between the two oxygens means a double bond, and it needs four hundred and ninety eight. Half of it is two hundred and forty nine. Chlorine: the chlorine chlorine bond needs two hundred and forty two, and half is one hundred and twenty one kilojoules per mole. Keep these handy; we use two hundred and forty nine in Example two. With me?

4:56Example 1: mean C-H bond enthalpy of methane

Okay, Example one, and it answers our puzzle. The four successive C H dissociation enthalpies of methane are four hundred and thirty nine, four hundred and sixty two, four hundred and twenty four, and three hundred and thirty eight kilojoules per mole. Find the mean C H bond enthalpy, and the enthalpy of atomisation of methane. Pause the video, and try it first. Done? Okay, let's check. First, look at these numbers: four identical bonds, four different energies. That is our puzzle, solved! Now add all four: one thousand six hundred and sixty three. Right? The mean is one thousand six hundred and sixty three over four: four hundred and fifteen point seven five. The wavy equals sign means approximately, so about four hundred and sixteen kilojoules per mole. Very close to the four hundred and fifteen we saw; different data tables differ by a kilojoule or so. Perfect! Now atomisation. We just learnt that delta a H is four times the mean. So, four times four hundred and sixteen, one thousand six hundred and sixty four, right? Close, but careful! Four hundred and sixteen is a rounded number. Atomisation breaks all four bonds, so delta a H is simply their sum: one thousand six hundred and sixty three kilojoules per mole. Four times the unrounded mean, four hundred and fifteen point seven five, gives exactly the same. Clear?

6:38Example 2: enthalpy of formation of water vapour

Now Example two, a harder one. The mean O H bond enthalpy is four hundred and sixty three kilojoules per mole. Using the bond enthalpy of H two, four hundred and thirty six, and delta f H of oxygen atoms, two hundred and forty nine, estimate delta f H of water vapour, H two O gas. Pause and try. Hint: break everything into atoms first. Done? Okay, let's check. Formation means water vapour made from its elements: H two gas, plus half a mole of O two gas, gives H two O gas. Okay? Step one, break into atoms. H two into two hydrogen atoms: four hundred and thirty six. Half O two into one oxygen atom: that is just delta f H of oxygen atoms, two hundred and forty nine, our topper's trick! Total, six hundred and eighty five, taken in. With me? Step two, form the water molecule. The O H bond enthalpy is four hundred and sixty three, so four hundred and sixty three comes out, right? No, no, wait! In water, the oxygen holds two hydrogens, so there are two O H bonds. Two times four hundred and sixty three: nine hundred and twenty six. And forming bonds releases energy, so this much comes out. Got it? So delta f H is what went in minus what came out: six hundred and eighty five minus nine hundred and twenty six, which is minus two hundred and forty one kilojoules per mole. Negative: making water vapour gives out heat. And notice, the question said estimate: mean bond enthalpies are only averages, and they work for gases only. That's it!

8:30Quick revision: bond, mean and atomisation enthalpies

So, quick revision. One: a bond enthalpy breaks one mole of bonds, gas to gas, and it is always positive. Two: the mean bond enthalpy is an average, and the enthalpy of atomisation breaks every bond, so for methane it is four times the mean C H bond enthalpy. Three: for a diatomic gas element, delta f H of its gaseous atom is half its bond enthalpy. Now go and try the practice problems on this page. All the best!