Use the Born–Haber cycle and data from Appendix IIB and
Table 9....

Question

Use the Born–Haber cycle and data from Appendix IIB and 
Table 9.3 to calculate the lattice energy of CaO. (ΔHsub for calcium 
is 178 kJ>mol; IE1 and IE2 for calcium are 590 kJ>mol and 
1145 kJ>mol, respectively; EA1 and EA2 for O are -141 kJ>mol 
and 744 kJ>mol, respectively.)

Accepted Answer

hey everyone in this example we're told that the Born haber cycle is an approach to studying energy changes during the formation of a crystalline cycle. So we need to use the born haber cycle to determine the lattice energy for magnesium chloride. Using the below values. So what we should recall is that our formula for the entropy of formation delta H degree F. Is going to equal our some of our heat of sublimation added to our first ionization energy which is then added to our second ionization energy and then added to our first bond energy for our cl to bond. Then added to our 2nd or rather two times our activation energy for our given chlorine adam. And then we're going to add to that our entropy of our lattice. And so again we want to find our lattice energy for magnesium chloride. So we're going to use this process that we've outlined above. So we're going to plug in all the values we know we know our entropy of formation which is given to us as negative 6 41.3 kg jewels Permal. And this is set equal to our sublimation energy. Or sorry, a sublimation entropy which is given to us as a value of 48 kg joules per mole. We're then adding this to our first ionization energy given as 7 37 kg jewels Permal which is then added to our second ionization energy given as 1450 kg joules per mole. And then added to that figure. We have our bond energy of our chlorine cl two bonds Which is going to be equal to a value given as to 39 kg joules per mole. And then added to this quantity, we're going to take two multiplied by our given activation energy for chlorine of negative 3 48.5 kg joules per mole. And then lastly we will go ahead and add our entropy of our lattice, which is what we are solving for. So we can go ahead and simplify this. So that on our left hand side we would still have our entropy of formation which we have as negative 6 41.3 kg jewels Permal. Whereas on the right hand side we're going to add up all of these illegal Permal quantities here which represent our different variables. And this should give us a total of 1877 kg joules per mole added to our entropy of our lattice. And so to isolate for entropy of our lattice. We're going to subtract 1008 or sorry about that. We're going to subtract 1877 kg joules per mole from both sides of our equation. And so this is going to result in our entropy of our lattice equaling the difference between our left hand side here And that will be a total of negative 2518.3 kayla jewels per mole. And so this will complete this example as our final answer as our lattice energy for our magnesium chloride. So this will correspond to choice. Be in our multiple choice. I hope that everything I reviewed was clear. But if you have any questions, please leave them down below and I'll see everyone in the next practice video.

Use the Born–Haber cycle and data from Appendix IIB and Table 9.3 to calculate the lattice energy of CaO. (ΔHsub for calcium is 178 kJ>mol; IE1 and IE2 for calcium are 590 kJ>mol and 1145 kJ>mol, respectively; EA1 and EA2 for O are -141 kJ>mol and 744 kJ>mol, respectively.)

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