# Electron affinty from born haber cycle, is my cycle correct?

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#1
Heres the question

calculate electron affinity of chlorine.

 Process Enthalpy change/kJmol-1 Ca(s) à Ca(g) +190 Ca(g) à Ca2+(g) + 2e +1730 1/2Cl2(g) à Cl(g) +121 Ca2+(g) + 2Cl-(g) à CaCl2(s) -2184 Ca(s) + Cl2(g) à CaCl2(s) -795

Here is my cycle, is it correct?
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#2
thanks
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6 years ago
#3
I don't like your layout, but there you have it.

Don't forget that your unknown refers to 2x electron gain enthalpy of Cl, hence whatever value you get will need halving.
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#4
I haven't been taught this yet pigster, I am being taught this next week but I will miss the class due to work commitments. So I just used a youtube video for reference.

Any chance you could show me a better way to set it out?

many thanks
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6 years ago
#5
(Original post by hurricaneee)
I haven't been taught this yet pigster, I am being taught this next week but I will miss the class due to work commitments. So I just used a youtube video for reference.

Any chance you could show me a better way to set it out?

many thanks
Here's an interactive on Born Haber showing how they are set out...

... and here's a Born Haber cycle of magnesium chloride (very similar to calcium chloride)
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#6
Thankyou for the links, very greatly appreciated!
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6 years ago
#7
There are a few of things you HAVE to ensure you do:
• atomise before ionise.
• ionise before electron gain.
• atomise before electron gain.
• have both ions before lattice enthalpy.

Apart from that, they can go in any order.
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#8
(Original post by Pigster)
There are a few of things you HAVE to ensure you do:
• atomise before ionise.
• ionise before electron gain.
• atomise before electron gain.
• have both ions before lattice enthalpy.

Apart from that, they can go in any order.

thanks for the adviuce pigster, I just quickly constructed a new cycle.

Am I on the right track?
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#9
I know the answer is-386.5kj mol -1, but i'm no getting.

I was under the impression thats your first calculation would be equal to the formation energy -795kj mol-1? Im not getting close to that.....
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6 years ago
#10
(Original post by hurricaneee)
I know the answer is-386.5kj mol -1, but i'm no getting.

I was under the impression thats your first calculation would be equal to the formation energy -795kj mol-1? Im not getting close to that.....
You would have to show us what you are doing - it's impossible to troubleshoot without the trouble ...

Actually, looking at your diagram, you have not included the electron affinity of the chlorine atoms, which should be negative.

You have included it with the double ionisation of calcium ... which is wrong.

Look at the Born-Haber cycle of the magnesium chloride that I posted earlier ...
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#11
(Original post by charco)
You would have to show us what you are doing - it's impossible to troubleshoot without the trouble ...

Actually, looking at your diagram, you have not included the electron affinity of the chlorine atoms, which should be negative.

You have included it with the double ionisation of calcium ... which is wrong.

Look at the Born-Haber cycle of the magnesium chloride that I posted earlier ...
I mistakenly put in Ca(g) instead of Ca(s).

As for the electron affinity of chlorine

Im a bit confused now, I havent put the elctron affinitys of chlorine in as is that not what im trying to find out from the cycle?

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6 years ago
#12
Charco: in the OP, +1730 is given as the value for IE1 & IE2 combined, hence why they're shown together, which is (given the data) the correct thing to do.

hurricaneee: the top line of your last picture should say Ca2+(g) + 2e- + 2Cl(g), then there should be a line down to what you've got on your top line. The difference between your top line and the missing line represents 2x electron gain enthalpy of Cl. There should then be a line down, i.e. -2184.
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#13
Ok thanks mate, I wil get this drawn out, redo the calculations and hope everything comes good.

cheers!
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#14
thanks
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6 years ago
#15
(Original post by Pigster)
Charco: in the OP, +1730 is given as the value for IE1 & IE2 combined, hence why they're shown together, which is (given the data) the correct thing to do.

hurricaneee: the top line of your last picture should say Ca2+(g) + 2e- + 2Cl(g), then there should be a line down to what you've got on your top line. The difference between your top line and the missing line represents 2x electron gain enthalpy of Cl. There should then be a line down, i.e. -2184.
Pigster, if you look at the diagram that he's drawn he puts:

Ca(g) + Cl2 --> Ca2+ + 2Cl-

He has included the ionisation energies with the electron affinities of chlorine in the +1730 kJ
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#16
For whatever reason my calculations do not add up. I know I should be getting -795kj mol-1 then dividing it by two for my answer but my numbers just arnt adding up.

Am I doing something wrong or have I just been given the wrong answer?

cheers
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6 years ago
#17
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#18
im getting -336.5kj mol-1 for my answer, my mark scheme says the answer is -386.5kJmol-1.....
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6 years ago
#19
Given that lattice enthalpy = delta H formation - (everything else)
then
total electron affinity = delta H formation - (everything else)

So total electron affinity = -795 -190-1730-(2 x 121) - (-2184) = -773
but this is for 2Cl --> 2Cl-
so divide by 2 = -773 /2 = -386.5 kJ/mol
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6 years ago
#20
(Original post by hurricaneee)
im getting -336.5kj mol-1 for my answer, my mark scheme says the answer is -386.5kJmol-1.....

Here is the complete Born-Haber cycle of calcium chloride

The maths is just substituting the values given ...
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