delta H formation / Hess Law

[Contach]

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Hi, can som1 set up this question for me please? I don't understand why deltaH products - deltaH reactants doesn't work for this...

Given the following information, what is the heat of formation (in kcal/mol) of methane?

C + O2 --> CO2 deltaH = -94 kcal/mol
H2 + 1/2 O2 --> H2O deltaH = -68 kcal/mol
CO2 + 2 H2O --> 2 O2 + CH4 deltaH = 213 kcal/mol

thanks.

 

[supraman]

Heat formation of methane: CH4

So the reactants will only have C and H2

1. ==>C + O2 --> CO2 deltaH = -94 kcal/mol
2 x[H2 + 1/2 O2 --> H2O deltaH = -68 kcal/mol] = -136
2. ==> 2H2 + O2 --> 2H2O = -136
3. ==>CO2 + 2 H2O --> 2 O2 + CH4 deltaH = 213 kcal/mol

Since 2 H20 is found on both sides (lines 2 and 3), they can be crossed out. The O2 in the first and second line equations can add together to give 2 O2 to cross out with the 2 O2 in the products of the third equation. CO2 is also the same on both sides in line 1 and 2. So therefore you just add the Delta H's and you should get -17 kcal/mol, corect me if I'm wrong anyone.

Your 27 in gchem is nuts!

 

[chessxwizard]

Please don't tell me that you're retaking the DAT. It worries me.

 

[Northside]

Please don't tell me that you're retaking the DAT. It worries me.

lol, I think he's prepping for the Canadian DAT.

 

[Danny289]

Please don't tell me that you're retaking the DAT. It worries me.

lol, I think he's prepping for the Canadian DAT.

I think he must retake, his TS is a little in low side

 

[Contach]

lol, I think he's prepping for the Canadian DAT.

yup, I'm taking the Canadian DAT.. I really don't want to though haha.

 

[Contach]

Heat formation of methane: CH4

So the reactants will only have C and H2

1. ==>C + O2 --> CO2 deltaH = -94 kcal/mol
2 x[H2 + 1/2 O2 --> H2O deltaH = -68 kcal/mol] = -136
2. ==> 2H2 + O2 --> 2H2O = -136
3. ==>CO2 + 2 H2O --> 2 O2 + CH4 deltaH = 213 kcal/mol

Since 2 H20 is found on both sides (lines 2 and 3), they can be crossed out. The O2 in the first and second line equations can add together to give 2 O2 to cross out with the 2 O2 in the products of the third equation. CO2 is also the same on both sides in line 1 and 2. So therefore you just add the Delta H's and you should get -17 kcal/mol, corect me if I'm wrong anyone.

Your 27 in gchem is nuts!

yup. -17 is correct.. and it makes sense to do it in this way...

but why won't doing deltaH products - deltaH reactants work?

To answer my own question: We do not know the deltaH formation of methane, but we do know the delta H formation of CO2 and H2O, unfortunately without the methane, we cannot do it in this way.

Awesome thanks.