Given the following heats of combustion. ch3oh(l) + 3/2 o2(g) co2(g) + 2 h2o(l) δh°rxn = -726.4 kj c(graphite) + o2(g) co2(g) δh°rxn = -393.5 kj h2(g) + 1/2 o2(g) h2o(l) δh°rxn = -285.8 kj calculate the enthalpy of formation of methanol (ch3oh) from its elements.

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

Explanation:

The formation reaction of CH_3OH will be,

The intermediate balanced chemical reaction will be,

..[1]

..[2]

..[3]

Now we will reverse the reaction 3, multiply reaction 2 by 2  then adding all the equations, Using Hess's law:

We get :

..[1]

..[2]

[3]

The expression for enthalpy of formation of will be,

The standard enthalpy of formation of methanol is, -238.7 kJ/mole

the awnser is group 12