C2H6 + 3 1/2 02 forms 2CO2 + 3H20 . bond enthalpies , C-H = 413 , C-C=347 , H-O=464 , C=O= 805 O=O= 498 ,how do I work out the bond enthalpy?
can anyone explain how to do this and how they calculated the bond enthalpy total on either side from the formula , I don't know whether I should be taking in account the subscripts in the formula and which bonds are actually present and how many
- hcbiochemLv 76 months agoFavorite Answer
You actually need to think about the structures of the molecules and how the atoms are bound to each other. So, yes, you do need to take into account the subscripts, but also the structures.
Delta H reaction = sum of bonds broken - sum of bonds formed
6 C-H = 6(413) = 2478 kJ
1 C-C = 1(347) = 347 kJ
3.5 O=O = 3.5 (498) = 1743 kJ
Total of bonds broken = 4568 kJ
4 C=O = 4(805) = 3220 kJ
6 H-O = 6(464) = 2784 kJ
Total of bonds formed = 6004 kJ
Delta Hrxn = 4568 - 6004 = -1436 kJ