Question #495b7

1 Answer
Feb 22, 2017

#DeltaH=+2034 kJ mol^-1# or #DeltaH=+726 kJ mol^-1#


If the reaction in question is exactly as you have written the answer should be +2034 kj per mol. But when hydrogenating ethene generally the reaction is: #H_2 + C_2H_4=C_2H_6# and then the answer would be + 726 kJ so take the answer that fits best with your question.

To calculate this we must use Bond energies (or Bond enthalpies). My values were taken from my course's data booklet and I advise to check your data booklet too to see that it corresponds.

The answer is found by comparing energies from "Breaking and making bonds".

Bonds broken= 4 H-H bonds, 4 C-H bonds, , 1 C-C double bond.
#DeltaH= 4(436) +4(414) + 614= +4014 kJ mol^1# (alt 1 H-H bond , then the answer becomes #436+4(414)+614=+2706 kJ mol^-1#
(as breaking is Endothermic, Delta H values are pos)
Bonds formed= 6 C-H bonds, 1 C-C bond
#6(-414) + (-346)=-1980 kJ mol^-1#
(As making is Exothermic, Delta H values are neg.)
Then the total enthalpy(energy) is:
#DeltaH=+4014-1980=+2034 kJ mol^-1#
#DeltaH=+2706-1980=+726 kJ mol^-1#

This picture also shows the Lewis structure of all molecules involved.