Well they've given you the enthalpies of combustion for each of "CH3OH(l), C(s) and H2(g)"

So you have the following 3 reactions:


Which you'll have to find out on your own.

Now, you WANT the standard enthalpy of formation of CH3OH...ie. for the reaction

? C + ? H₂ + ? O₂ --> CH₃OH

Now, you can rearrange the first three, and add them together in some multiples, etc...and use enthalpy rules and stuff (like if you reverse a reaction, what happens to the enthalpy?)..

Hopefully this helps.  This should definitely get you started - and if you need anything else, let us know.

P.S. If, in the future, you can answer others people's questions and give back to the forum we will love you more   than if you just pass by and use us  

I am also unsure where to begin with this question.  I have tried to use some sort of q=mcdt and  pv=nrt, but I am just not sure.  

If the standard enthalpy of combustion of C2H2(g), is -1300kJ/mol, how many litres of C2H2 measured at 25°C and 2.62 atm must be burned to liberate 10.0x10^6 kJ of heat?

I am also unsure where to begin with this question.  I have tried to use some sort of q=mcdt and  pv=nrt, but I am just not sure.  

If the standard enthalpy of combustion of C2H2(g), is -1300kJ/mol, how many litres of C2H2 measured at 25°C and 2.62 atm must be burned to liberate 10.0x10^6 kJ of heat?

Ok so if you want 10 x10^6 kJ of heat, and the heat released is 1300 kJ/mol, how many moles of C2H2 would you need?

Knowing the moles you need, determine the volume.  

I'm not sure why you subtracted?

You want to find how many moles you need for 10 x10^6 kJ of energy.

1 mol = 1300 kJ

? mol = 10 x10^6 kJ

*Lil' Moddy of my Own* Great Job!

Alrighty then, we'll work through it until someone who actually knows what they're doing comes along .

1 mol = 1300 kJ

? mol = 10 x10^6 kJ

Just a little ratio here.

(10 x10^6 kJ) / (1300 kJ.mol) = 7692.3 mol

So you need 7692.3 mol of C2H2 to release 10 x10^6 kJ of energy.

Now you know that:

n = 7692.3 mol
P = 2.62 atm or 265.4 kPa
R = 8.31 kPaL/molK <-- (just remembered the units )
T = 25°C or 298K
V = ?

back to the FIRST question, is where I'm going to go.  I'm going to use {H} for delta H

Reaction 1: 2 CH₃OH + 3 O₂ --> 2 CO₂ + 4 H₂O -- {H} = -726 kJ/mol
Reaction 2: C +  O₂ -->  CO₂ + 0 H₂O -- {H} = -394 kJ/mol
Reaction 3: 2 H₂ + O₂ --> 0 CO₂ + 2 H₂O -- {H} = -286 kJ/mol

Throwing the H2O in as a product was a red herring.  Sorry bout that, I just saw it myself.

Anyways, check this out:

We want the enthalpy for C + 2 H₂ + 1/2 O₂ --> CH₃OH -- THIS is what makes it enthalpy of formation - it is the the reaction where all the reactants are the elements in their 'natural' form (as they are found in nature) and the only product is the compound you want.

Anyways, if you take Reaction 1, and reverse in, AND halve it...you get

Reaction 4: 1 CO₂ + 2 H₂O --> CH₃OH + 3/2 O₂ -- {H} = -(-726) kJ/mol = 726 kJ/mol

Now we took the negative of the {H} because we reversed the reaction.  We DIDN'T divide by two...because the units on {H} are already PER MOLE...

Now, having done this a couple times, I realize that the water is going to cancel out in the final equation (if you don't get this the first time, no worries.  Do a couple more, see if it comes to you, and if it doesn't, let us know, we'll explain)...Anyways it's like an 'instinct' thing.

So yeah.  Let's take Reaction 3:

Reaction 5: 2 H₂ + O₂ --> 0 CO₂ + 2 H₂O -- {H} = -286 kJ/mol

And in the same way we won't do anything to reaction 2:

Reaction 2: C +  O₂ -->  CO₂ + 0 H₂O -- {H} = -394 kJ/mol

Now, let's recap

Reaction 2: C +  O₂ -->  CO₂ + 0 H₂O -- {H} = -394 kJ/mol
Reaction 5: 2 H₂ + O₂ --> 0 CO₂ + 2 H₂O -- {H} = -286 kJ/mol
Reaction 4: 1 CO₂ + 2 H₂O --> CH₃OH + 3/2 O₂ -- {H} = -(-726) kJ/mol = 726 kJ/mol

We can add all these together and get:

1 CO₂ + 2 H₂O + 2 H₂ + O₂ + C + O₂ --> CO₂ + 0 H₂O + 0 CO₂ + 2 H₂O + CH₃OH + 3/2 O₂

Which cancels out to:

C + 2 H₂ + 1/2 O₂ --> CH₃OH

Which is the formation reaction for CH₃OH !

So we added reactions 2, 5, and 4 together to get the formation reaction and we'll use Hess' Law to get the Enthalpy of Formation (sum the reactions --> sum the enthalpies)

So {H}_f = (726 kJ/mol) + (-394 kJ/mol) + (-286 kJ/mol) = 46 kJ/mol

And there you have it.

P.S. Normally, we don't go around solving entire questions for people like this.  We like to guide people along.  But you seemed to understand the other guy's thing when he explained it all out, hopefully this does the same.

P.P.S. It was murder to type that out.  Now I know why very few of the 'experienced' people don't use subscripts.  It almost made me cry   ... LOL

wait a tick, I got 46kJ/mol when I first tried it, but aparently that's not the answer.
My sample midterm exam gives -240 as the answer.