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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
thanks for any help, I appreciate it
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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!
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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(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!
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!
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