Answer: D: -216 KCal. Solution is as follows:
DHdiss° = [energy of bonds broken] - [energy of bonds formed]
In the equation, you have the following bonds:
4C-H (CH4) + 2 O=O --> O=C=O + 2 H-O-H ***OR***
4C-H + 2O=O --> 2 C=O + 4 O-H
Using the heats of formation provided, the equation becomes
DHdiss° = [(4*87) + (2*118)] - [(2*178) +(4*111)] = -216KCal
Key Takeaway: Tabulated values are always reported as positive by convention. The formula presented the avoids the issue of +Q heat, which is a proxy for internal energy, which is essentially enthalpy H. The +Q heat absorbed into the system and -Q heat output from the system is the reason the final product is a negative value.
