Why is the reaction between ethylene, and diodine ENDOTHERMIC?

1 Answer
Aug 27, 2017

I presume it is a consequence of the bond strengths of #C=C# bonds, versus #C-H#, #C-I#, and #I-I# bonds........

Explanation:

This site gives me handle on #"element"-"element"# bond strengths.

#C=C#, #"bond strength"#, #614*kJ*mol^-1#.

#C-C#, #"bond strength"#, #348*kJ*mol^-1#.

#C-H#, #"bond strength"#, #413*kJ*mol^-1#.

#C-I#, #"bond strength"#, #216*kJ*mol^-1#.

#I-I#, #"bond strength"#, #251*kJ*mol^-1#.

For the hypothetical reaction......

#H_2C=CH_2 + I_2 rarr IH_2C-CH_2I#

We BREAK #1xxC=C# bond, and #1xxI-I# bond, i.e. #(614+251)*kJ*mol^-1=865*kJ*mol^-1#. We MAKE #1xxC-C# bond, and #2xxC-I# bonds, i.e. #(348+2xx216)*kJ*mol^-1=780*kJ*mol^-1#.

And since #DeltaH_"rxn"^@=Sigma_("bonds broken")-Sigma_("bonds made")#...we gots....

#DeltaH_"rxn"^@=(865-780)*kJ*mol^-1=85*kJ*mol^-1#

And thus the reaction as written is ENDOTHERMIC. I have not done the calculation for the reaction of bromine or chlorine with the olefin, however, I suspect that these reactions would be EXOTHERMIC, and thus enthalpy-favoured.