# born Haber cycle problem

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#1
why is the enthalpy of atomisation for Cl2 x2? The enthalpy of atomisation for Cl2 should produce 2 Cl atoms regardless? And this is for the splitting of only 1 molecule of Cl2, so I have no idea why it is measured as 2x the enthalpy of atomisation.

thanks for any help, I appreciate it
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1 month ago
#2
I think you are confusing enthalpy of atomisation with bond enthalpy. The atomisation of chlorine is given formula: 1/2 Cl2(g)—> Cl(g). It’s not Cl2(g) —> Cl(g).

In the Born-Haber cycle above, you have 2 x 1/2 Cl2(g) —> 2Cl(g) so you must do 2 x enthalpy of atomisation. Hope this helps!
Last edited by KaziMahathir; 1 month ago
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#3
(Original post by KaziMahathir)
I think you are confusing enthalpy of atomisation with bond enthalpy. The atomisation of chlorine is given formula: 1/2 Cl2(g)—> Cl(g). It’s not Cl2(g) —> Cl(g).

In the Born-Haber cycle above, you have 2 x 1/2 Cl2(g) —> 2Cl(g) so you must do 2 x enthalpy of atomisation. Hope this helps!
thank you man appreciate it
1
1 month ago
#4
(Original post by pondering-soul)
thank you man appreciate it
No problem
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