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Chemistry A-Level Born Haber Cycles

My understanding of enthalpy change of atomisation is that it is the enthalpy change when one mole of gaseous atoms are formed from an element (e.g 1/2 Cl2 (g) = Cl (g) )
So when you create a born haber cycle for MgCl2, you double the enthalpy change of atomisation (as you are forming 2 moles of Cl)

But why is it that when you create a born haber cycle for MgO, you halve the enthalpy change of atomisation?
I have this step as 1/2O2 (g) = O(g) which I thought was the enthalpy change of atomisation of oxygen? Is this a mistake in the mark scheme or have I made a mistake?
Reply 1
Avatar for gingur
gingur
13
To my best understanding, you are correct. The Chemguide revision sheet for thermodynamics has 1.2O2 (g) -> O (g) as the example for enthalpy of atomisation (https://chemrevise.files.wordpress.com/2022/01/1.8-revision-guide-thermodynamics-aqa.pdf)

Excuse my formatting lmao
Original post by N70
My understanding of enthalpy change of atomisation is that it is the enthalpy change when one mole of gaseous atoms are formed from an element (e.g 1/2 Cl2 (g) = Cl (g) )
So when you create a born haber cycle for MgCl2, you double the enthalpy change of atomisation (as you are forming 2 moles of Cl)

But why is it that when you create a born haber cycle for MgO, you halve the enthalpy change of atomisation?
I have this step as 1/2O2 (g) = O(g) which I thought was the enthalpy change of atomisation of oxygen? Is this a mistake in the mark scheme or have I made a mistake?

You have to be careful about which energy changes you are given, as the definitions are all-important.

As you say, the enthalpy of atomisation refers to the formation of 1 mol of gaseous atoms from an element in its standard state.

However, the bond enthalpy term refers to the energy needed to break 1 mol of covalent bonds in a gaseous molecule.

So:
The bond dissociation enthalpy
Cl2(g) ==> 2Cl(g) - the energy change = 242 kJ mol-1

Whereas the atomisation enthalpy
1/2Cl2(g) ==> Cl(g) - the energy change = 121 kJ mol-1

I expect that you have been given the bond enthalpy of oxygen, not the enthalpy of atomisation.

Here's a run through of the MgO Born-Haber cycle:




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(edited 2 years ago)
Reply 3
Avatar for N70
N70
OP
4
Original post by gingur
To my best understanding, you are correct. The Chemguide revision sheet for thermodynamics has 1.2O2 (g) -> O (g) as the example for enthalpy of atomisation (https://chemrevise.files.wordpress.com/2022/01/1.8-revision-guide-thermodynamics-aqa.pdf)

Excuse my formatting lmao


Original post by charco
You have to be careful about which energy changes you are given, as the definitions are all-important.

As you say, the enthalpy of atomisation refers to the formation of 1 mol of gaseous atoms from an element in its standard state.

However, the bond enthalpy term refers to the energy needed to break 1 mol of covalent bonds in a gaseous molecule.

So:
The bond dissociation enthalpy
Cl2(g) ==> 2Cl(g) - the energy change = 242 kJ mol-1

Whereas the atomisation enthalpy
1/2Cl2(g) ==> Cl(g) - the energy change = 121 kJ mol-1

I expect that you have been given the bond enthalpy of oxygen, not the enthalpy of atomisation.

Here's a run through of the MgO Born-Haber cycle:




---------------------------------------------------------------
Signature
Colourful Solutions Chemistry Videos on YouTube

Thank you!! This makes a lot more sense!

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