Lecture 1/27 method 2- using bond enthalpies to calculate deltaH of reaction

[Eden K 1B]

In the example for method 2 from lecture, each of the bonds that exist in the following equation were written out as:

CH2=CH2 (g) + H-Br (g) --> CH3-CH2Br (g)
C=C + H-Br --> C-C + C-H + C-Br

Why wasn't H2=H2 included on the reactants side?

[Wesley J the BioChem Kid]

Hey! I think it's because those four C-H bonds appear on both sides of the reaction. Therefore, when written out, the energy they produce/require is cancelled out or does not change going from reactants to products

[Tucker Gruber 2G]

Keep in mind that there is no H2=H2 bond on either side of the reaction (and this type of bond doesn't make sense or really exist in the first place). Like Wesley mentioned, there are 4 C-H bonds on the left side of the reaction and 5 C-H bonds on the right side of reaction, so those original 4 bonds can remain, and then we have to add the enthalpy of one more C-H bond on the right side of the reaction (as you showed in your example).
Here is what CH2=CH2 (usually written C2H4) looks like, so you can see the 4 C-H bonds:

And here is what CH3-CH2Br (usually written CH3CH2Br) looks like, so you can see the 5 C-H bonds:

Hope this helps!