[bleachandammonia]

Hi, I’ve been having issues to understand this question about energy change, so here it goes.

Calculate the energy change in the following reaction:
a) burning 1 mole of ethanol - C2H5OH + 3O2 —> 2CO2 + 3 H2O

So first you need to calculate the energy required to break the bonds between the reactant atoms:

1 C-C bond = 348
5 C-H bonds = (5 x 413) = 2065
1 O-H bond = 463
1 C-O bond = 360
3 O=O bonds = (3 x 498) = 1494
   Total = 4730 kJ

Then we need to do the same for the bonds that are being made (in the products). But here’s my problem, my text book says those are the bonds in the products:

4 C=O bonds = (4 x 743) = 2972
6 O-H bonds = (6 x 463) = 2778
  Total = 5750 kJ

However, looking at the equation above I see 2CO2, wouldn’t that give us, along with the 4 C=O bonds, 1 C-C bond? Even though carbon makes only 4 bounds, I don’t know if it makes sense for them to be separated. Or are the two molecules separated?

Thank you for your attention, and please help me!!

[mjc123]

2CO₂ means two molecules of CO₂. They are separate - not bonded together! (What were you going to do about the 3O₂ and 3H₂O?)

[bleachandammonia]

Oh I get it! So there isn’t supposed to be a bond there. Ok, I just have one more question... the next one asks me to calculate the energy change in a hydrogenation of 1 mole of ethene, so the steps should be the same... C₂H₄ + H₂ —> C₂H6

but in the answer, my text book only counts the bounds of H-H and C=C when counting the reactants section. I don’t understand why the H₄ part of ethene isn’t included in the sum. Does that have to do with the word hydrogenation in the question? Because in the other questions all bonds were included...

Anyways, thanks for your answer, it has already helped.

[mjc123]

Well, you could break and form all the bonds. But that would mean breaking 4 C-H bonds and forming 6. Why not just say you form 2, and leave the other 4 in place?

(Note: the C-H bonds of ethene are not exactly the same as the C-H bonds of ethane. But at this level you treat them as the same, to a first approximation.)