Homework 8.73 a

[hannah_flagg_3F]

In the solution manual, it says the only bonds broken are double bonds between two carbon atoms, however, there are also hydrogen molecules in the formula for the reactants, why is this?

[Mikaela Gareeb 3G]

I'm not sure why the solution manual doesn't break the C-H bonds that are on the reactants side (as well as the products side), but this is how I did it:
break 1 C-C triple bond: (3 mol)(837 kJ/mol)
break 2 H-C single bonds: (3 mol)(412 kJ/mol)
form 6 C-C resonance bonds: -(6 mol)(518 kJ/mol)
form 6 C-H single bonds: -(6 mol)(412 kJ/mol)
Delta H of formation= -597 kJ/mol
At first I thought that the solution manual didn't include the C-H bonds because there were the same number of them on both sides of the equation, but with further inspection that did not turn out to be the case; however, with the method above you still get the correct answer.

[Anthony_Imperial_1G]

The total number for C-H bonds for both sides is six. (3mol C2H2= 6 CH).

The only things breaking and forming are the C-C triple bond and C-2 double bond. If you compare the products and reactants you can see what you need to calculate to get your answer. In this case, you can ignore the carbon-hydrogen bonds.

for reactants --> 3mol x 837 kJ = 2511
for products --> 6 mol x 518 kJ x -1 = -3108

together they add up to -597 kJ.

[Absolute Zero 1O]

I suggest drawing out the lewis structure for each molecule to see what bonds are being broken and formed. Do not simply assume that everything in the reactants is being broken and everything in the products is being formed. Dr. Lavelle does a great job explaining this. If you head over to bruin cast website, click on chem 14b lecture 1 and go to the lecture from Wed. 1/11/2017. And seek to 24 minutes, there he goes over an example that is really comprehensive on this topic.