Question

Methanol (CH3OH) is an organic solvent and is also used as a fuel in some automobile engines. From the following data, calculate the standard enthalpy of formation of methanol.
2 CH3OH(l) + 3 O2(g) → 2 CO2(g) + 4 H2O(l) ΔH°rxn = -1452.8 kJ

I've evaluated standard heats of formation for all of the compounds and subtracted and such.

2(-393.5)+4(-286)-3(0)-2X=-1452.8
I got X=-239.1, but WebAssign doesn't want to accept that. It says that 253.7 is within 10% of the answer whereas 285.8 is outside of the 10% boundary.

Answer 1

Oh, and in case it wasn't obvious already, the answer is meant to be in kJ/mol.

Answer 2

it's 239.1

the problem is that you didn't use a second bracket

2(-393.5)+4(-286)-[3(0)-2X]=-1452.8

2(-393.5)+4(-286)+2X=-1452.8

Answer 3

No, since it should be 2(-393.5)+4(-286)-[3(0)+2X]=-1452.8 if I do it that way. WebAssign does say that the answer is negative.

Answer 4

I'm going by what you wrote.

\(H_{rxn}=\Sigma H^{products}_f-\Sigma H^{products}_f\)

you wrote: 2(-393.5)+4(-286)-3(0)-2X=-1452.8

but it should be 2(-393.5)+4(-286)-[3(0)-2X]=-1452.8

which then simplifies to 2(-393.5)+4(-286)+2X=-1452.8

Answer 5

that second H should be reactants not products

Answer 6

No, it should be [3(0)+2X] since it's the SUM of the standard heats of formation of the reactants.

Answer 7

oh you already expanded the bracket. why didn't you say so

Answer 8

Trust me, the sign is not the issue here. It's the number that's apparently off. But whatever. I guess even you guys here on this forum can't pinpoint the problem...I'll just bring it up to my teacher's attention tomorrow I guess. Thanks anyway.

Answer 9

The medal's for trying. :) I help a lot on the math forums and I know how much those mean. :)

Answer 10

lol i don't care 2 bits about medals but thanks !

i'm not sure whats wrong though, everything seems to be in order