Bond Enthalpy

[Jua Ellie Ha 1K]

While calculating for the net enthalpy change of C2H5Br formation during lecture, Dr. Lavelle added the enthalpy of each bond involved in the reaction. Can someone explain why he didn't include the bond enthalpy of C-H of the reactant?

[kamrin schultz 2J]

On the reactant side, there are 4 C-H bonds and on the product side, there are 5 C-H bonds. Since 4 bonds are broken and 5 are formed, this is essentially equal to 1 bond formed.

[Jennifer Liu 2A]

If you rewatch the posted lecture, there's a section that might help in which he talks about which bonds are broken and which bonds are formed. The C--H bond of the reactant wasn't broken, so he didn't need to include the bond enthalpy (the bonds broken were C=C and H--Br and the bonds formed were C--C, C--H, and C--Br).

[Katherine Phan 1J]

The only bonds broken on the reactant side are the C-C and H-Br bonds. Since there are 4 C-H bonds in the reactant and 5 C-H bonds in the product, you can basically consider those 4 bonds as "cancelling" each other out for a net formation of 1 C-H bond in the product (-412 kJ/mol).

[Alexa Le 3F]

When using the method of bond enthalpies to calculate the change in enthalpy, the easiest thing to do is to see which bonds are being broken, which requires energy and is thus a positive value. To see this more clearly, I would recommend drawing out the structures for all of the molecules, so you can see which bonds are being broken. For example, with the bonds on the reactant side, the bonds C-H are not being broken, because they are still present on the product side. However, the bond C-C is broken, and we know this because on the product side they only form a single bond, while on the reactant side they formed a double bond. This would not make sense, so you would essentially have to count that as a broken bond as well. You would do the same for H-Br, as this bond is broken to form the new bonds with carbon. Hope that helps!