# B H Cycle for Calcium Iodide

Okay so we’re going to look at the lattice enthalpy of dissociation for calcium iodide so calcium is in group two and it will have a plus two charge iodine is in group seven and it will have a minus charge so calcium iodide is see a i 2 and if we do the equation for this then calcium is a solid and it will react with iodine and idean in its standard state is a solid

And it will give calcium iodide as a solid okay so if we draw the enthalpy equation for this so this would be calcium in its solid state plus iodine in its solid state and it would give us calcium iodide in its solid state so we know this so if i then do my transformations just on the metal i’m going to turn calcium into a gas and then going to take one electron

Off calcium and ionize it then going to take a second electron off calcium and ionizer okay now i’m going to do the next transformation on the iodine i’m just going to use a different color so it can actually see what i’m doing so the id n– stays as a solid stays as a solid and stays as a solid now the transformation i’m going to do on iodine here because i’m

Going to turn it into a gas so i to guess and then what i’m going to do is i’m going to break the iodine bond and i’m gonna make iodine atoms okay the calcium can stay as it is co2 plus as a gas and two electrons ca 2 plus as a gas and two electrons and i’m just going to rule off okay so now what’s going to happen is the iodine is going to pick up those electrons

And just look this to iodine’s two electrons so they’re gonna pick up that’s going to be quite exothermic yep yes exothermic so i’ve got two iodine ions now and i’ve got my calcium 2 plus and then they’re gonna come together and that’s my lattice enthalpy of dissociation that lattice enthalpy of formation right so let’s put some figures on here so the question

Is here we go number five calculate the first electron affinity of iodine so that’s this one here we’re going to have to do that given that the lattice enthalpy of dissociation okay so it’s giving me that is plus two oh five four so we know if we’re going to come this way it must be minus 2i five four so i’m gonna i’m gonna come this way 205 oh it won’t as long

As you treat it as long as this is the only one we look at five four okay and it’s enthalpy of formation this is minus five three five okay we can’t see that very well i’ll choose a different four okay so this is atomization of calcium plus one nine three this is ionization of calcium plus five nine oh this is the second ionization of calcium plus one one five

Okay so now we’re on to iodine so to ayat’ to take the solid to a gas we are looking at that’s not atomization so if we go okay so i have taken this solid to a gas so basically our sublimed it but actually because this is a non-metal i would turn it from its standard state to its atoms like this and the ionized at the rock bond breaking here our atomization is

To create one mole of atoms and as you can see i’ve created two moles of atoms so to go from here to here is plus 107 times two okay so they say this just goes to show that even teachers can make a mistake so going from solid to gas for iodine and we’re going to go from the solid to the atoms so if this is the atomization okay so we’ve got to find this first

Ionization first electron affinity so that’s the electron affinity that we’re going to find and everything else we have got so we can set up our equation we will be going this route so i work so on this route here is the same as going around this route here so i’m going to set up this equation though we’ve got – five three five equals and i’ve got plus 1 9 3 +

5 9 5 + 1 1 5 i plus 2 times 107 and we’ve not just got one electron affinity we’ve got two electron affinity so i’m going to put two because i’ve got two iodine iodine atoms going to two iodide ions and then i’m going with this ri so – 205 for so all i need to do is simplify these numbers and then i can rearrange okay so i’ve simplified all of these numbers and –

5 3 5 equals 93 + 2 electron affinities so if i just rearrange the equation so – 5 3 5 – 93 equals 2 electron affinities and i can do those maths so this is minus 6 2 8 equals 2 electron affinities so the electron of the first electron affinity for iodine is minus 6 2 8 divided by 2 which equals minus 3 1 4 kilojoules per mole

Transcribed from video
B H Cycle for Calcium Iodide By Dr KV Oates