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

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :