# Chemistry Question

Hi, i am a bit confused about a question:
Calculate the standard lattice enthalpy of CuBr2?
Can anyone help???
It all depends on whether it's the lattice enthalp of formation or breaking.

Of formation, it would be:

Cu{2+}(g) + 2Br{1-}(g) --> CuBr2(s)

Breaking would be with the arrow pointing in the other direction. Formation is exothermic and breaking is endothermic.
Sorry i forgot to mention that it is through a Born-Haber cycle for copper bromide,

the question also stated data:
atomisation energy of bromine: +111.9kJ mol -1
atomisation energy of copper: +338.3kJ mol -1
First ionisation energy of copper: +746.0kJ mol -1
Second ionisation energy of copper: +1958.0kJ mol -1
electron affinity of bromine: -324.6kJ mol -1
enthalpy change of formation of CuBr2: -141.8kJ mol -1

Calculate the lattice enthalpy of CuBr2 using the Born-Haber cycle?
Originally posted by Unregistered
Sorry i forgot to mention that it is through a Born-Haber cycle for copper bromide,

the question also stated data:
atomisation energy of bromine: +111.9kJ mol -1
atomisation energy of copper: +338.3kJ mol -1
First ionisation energy of copper: +746.0kJ mol -1
Second ionisation energy of copper: +1958.0kJ mol -1
electron affinity of bromine: -324.6kJ mol -1
enthalpy change of formation of CuBr2: -141.8kJ mol -1

Calculate the lattice enthalpy of CuBr2 using the Born-Haber cycle?

If they give you a diagram in the exam it's easier, but here goes.
The lattice enthalpy of formation starts with gaseous ions and finishes with copper bromide compound. Working backwards around the loop, you'll also get the same figure.

So you need to go:
1) against the electron affinity of bromine TWICE (as there are two bromine atoms)
2) against the second, then the first ionisation energies of copper
3) against the atomisation of copper
4) against the atomisation of bromine (only once, as bromine is naturally diatomic)
5) WITH the direction of formation, as that arrow points from the natural atoms to the compound, in the direction you are moving around the cycle.

Written as an equation, you get:

^H(lattice)
= -2(-342.6) - 1958 - 746 - 111.9 - 338.3 - 141.8
= -2160 kJmol -1

bear in mind this is the enthalpy of formation, lattice breaking would be +2160 kJmol-1
Thanks for your help, but this is what i did and our answers vary a bit, i am not sure,

(formation)-141.8= +(+338.3) + (+746.0) + (+1958.0) + (-324.6) + (+111.9) + lattice enthalpy
=(-141.8) = +2829.6
lattice enthalpy of copper bromide = -141.8 - 2829.6
=-2971.4 kJ mol -1

Am i doing it wrong?
Hi, I got -2646.8kJmol-1.

I did
Lattice enthalpy = deltaH (f)-[deltaH(at of Cu)+deltaH(i1 of Cu) + deltaH(i2) + deltaH (at of Br2) + (2*deltaH (ea of Br))]

= -141.8-(338.3+746+1958+111.9+(2*-324.6)
= -141.8-2505
= -2646.8kJmol-1
I think that you forgot to multiply the delta H (ea of Br) by 2, because you need 2 molecules of Br- ions and 1 of Cu 2+ ions in order to get CuBr2, I think.
But then I got a different answer to the previous person, so I don't know.
Originally posted by Xenon
Hi, I got -2646.8kJmol-1.

I did
Lattice enthalpy = deltaH (f)-[deltaH(at of Cu)+deltaH(i1 of Cu) + deltaH(i2) + deltaH (at of Br2) + (2*deltaH (ea of Br))]

= -141.8-(338.3+746+1958+111.9+(2*-324.6)
= -141.8-2505
= -2646.8kJmol-1

Numbers everywhere! Trying to work it out in windows calculator on the numpad wasn't the best way to do it, so don't trust my answer

The best way to do it is to draw the cycle and write S and F on the arrow you want to calculate. Then use the alternative route to go from S to F, and then you can see what direction the other arrows are pointing - making it easier to calculate.
Yes i did forget to multiply Br by 2, thanks Guest Rob and Xenon for helping me
umm, sorry but can i just ask who got the correct answer in the end? its -2646.8 KJ mol isn't it? that's what i just got.
Guys, I'm real sorry, but checking again, I think that I have done it wrong. Firstly, it has to be 2 times delta H(at of Cl), as well as 2 times delta H(ea1). But I'm guessing that the values given for bromine are for Br2, and therefore, you don't have to multiply any values by 2. Therefore the answer really is
-2971.4kJmol-1
Sorry, unregistered was right; you don't have to timesa any values by 2, cos in the data, the values are given as diatomic molecules.
Sorry for any confusion I might have (and probably have) caused. And hope this sorts you out.
Just to repeat again, ignore all of my last parts apart from the previous one, because my first answer was wrong. DO NOT MULTIPLY ANY VALUES BY 2!!!!!!!!!!!!
Sorry!
Ta for the clarification! damn born haber cycles!