Is anyone out there smart enough to solve this problem?

Question

The heat of formation of CH3OH(l) = -238.6kJ / mol, of CO2(g) = -393.5 kJ / mol, and of H2O(g) = -241.8 kJ / mol. What is delta H for the heat of combustion of methanol?

I don't know how to setup and solve this problem. Is there any smart people out there?

Answer 1

Combustion of MeOH is:

2 CH3OH + 3 O2 --> 2 CO2 + 4 H2O

delta H = n delta H products - n delta H reactants

= [ 2(-393.5 kJ/mol) + 4(-241.8 kJ/mol) ]
- [ 2(-238.6 kJ/mol) + 3(0 kJ/mol) ]

= -1277 kJ for 2 moles

=> -638.5 kJ for 1 mole MeOH

Answer 2

The reaction for the combustion of 1 mole of methanol is:

CH3OH (l) + 3/2 O2 (g) ----> CO2 (g) + 2 H2O (g)

delta H = sum Hf products - sum Hf reactants
delta H = (-393.5) + 2*(-241.8) - (-238.6) = -638.5 kJ/mol

O2 has no heat of formation by definition (there is no energy of formation for an element in its natural physical state at room temperature).

Answer 3

im guessing its the sq. root of waffles? for all i be attentive to it might desire to be waffles... I heavily have not got any theory. there are symbols that make no senese. and what's a modulo? im clever and that i've got not got any theory what the hell you're speaking approximately.