Okay Lisa, you owe me BIG on this
But here we go. I made a mistake
Go to my huge-*ss post above, with all the reactions, and you'll notice,
Reaction 5: 2 H2 + O2 --> 0 CO2 + 2 H2O -- {H} = -286 kJ
/molThis isn't really true. It should be -286 kJ/mol of H
2...so...
Reaction 5b: H2 + 1/2 O2 --> 0 CO2 + 1 H2O -- {H} = -286 kJ/mol H
2But since we need 2 mols of H2, then
Reaction 5c: 2 H2 + O2 --> 0 CO2 + 2 H2O -- {H} = -572 kJ/2 mol H
2 = -572 kJ/mol O
2Note how we just slid in a single O
2 for 2 H
2. We can do that, because it's in the equation (like a titration)
Reaction 2: C + O2 --> CO2 + 0 H2O -- {H} = -394 kJ/mol
is fine. One mole of O2. We're cool.
Now comes a tricky part again
Reaction 4: 1 CO2 + 2 H2O --> CH3OH + 3/2 O2 -- {H} = -(-726) kJ/mol = 726 kJ
/mol CH3OHSo {H} = 726 kJ/(1 mol CH3OH) = 726 kJ/(3/2 mol O2) = 484 kJ/mol O2
That each of the equations has O2 in it. So we'll use O2 as the main compound thing. Units are important, keep that in mind.
Anyways, we now have
Reaction 2: C + O2 --> CO2 + 0 H2O -- {H} = -394 kJ/mol
O2Reaction 5: 2 H2 + O2 --> 0 CO2 + 2 H2O -- {H} = -572 kJ/mol
O2Reaction 4: 1 CO2 + 2 H2O --> CH3OH + 3/2 O2 -- {H} = 484 kJ/mol
O2TOTALLY notice that each is in terms of kJ/mol O
2 - before we had the one reaction in terms of kJ/mol H
2 and the other one...and that's not the same unit. So you can't add them together.
NOW when you add up the numbers you get
(484 kJ/mol O
2) + (-572 kJ/mol O
2) + (-394 kJ/mol O
2) = -482 kJ/mol O
2Which is for C + 2 H2 + 1/2 O2 --> CH3OH
And so, again with the conversion:
-482 kJ/(mol O
2) = -482 kJ/(2 mol CH3OH) = -241 kJ/mol CH3OH
Which is your answer.
P.S. It is less important to note that: (726) + (-394) + (-286) + (-286) = -240
But this is NOT the correct way to do it.
P.P.S. Did you know that here at Waterloo, we have T-Shirts that say, "Friends don't let friends go to Western"?
- They might be all over the place. Maybe even at Western