(2*-393.5+4*-242)-(2*-293+0)
Delta H+ -1169 kJ/mol
Methanol is an organic solvent and is also used as fuel in some automobile engines. From the following data, calculate the standard enthalpy of formation of methanol.
2CH3OH + 3O2 --> 2CO2 + 4H2O, (delta H = -1452.8 kJ/mol
3 answers
I don't agree with the answer provided by ash.
dHxn = (n*dHf products) - 2x = -1452.
Solve for x for dHf CH3OH.
dHxn = (n*dHf products) - 2x = -1452.
Solve for x for dHf CH3OH.
just had this question on my webassign-
products
h2o is liquid: dH= -286 kJ (x4)
CO2 (g): dH= -393.509 kJ (x2)
reactants:
CH3OH (l): dH= 2dH (unknown)
O2 (g): dH= 0 (x3) (0 bc its elemental)
dH(rxn) = dH(prod) - dH(react)
-1452.8= -1931.018 - 2dH(CH3OH)
-2dH(CH3OH) = 478.218
dH(CH3OH) = -239.109
products
h2o is liquid: dH= -286 kJ (x4)
CO2 (g): dH= -393.509 kJ (x2)
reactants:
CH3OH (l): dH= 2dH (unknown)
O2 (g): dH= 0 (x3) (0 bc its elemental)
dH(rxn) = dH(prod) - dH(react)
-1452.8= -1931.018 - 2dH(CH3OH)
-2dH(CH3OH) = 478.218
dH(CH3OH) = -239.109