<u>Given:</u>
Mass of methanol = 1.02 g
<u>To determine:</u>
Enthalpy for the reaction of 1.02 g of methanol with excess O2
<u>Explanation:</u>
Balanced equation-
2CH3OH(g) + 3O2(g) → 2CO2(g) + 4H2O(g)
The reaction enthalpy is given as:
ΔHrxn = ∑nH°f(products) - ∑nH°f(reactants)
where n = number of moles
H°f = standard enthalpy of formation.
ΔHrxn = [2H°f(CO2(g)) + 4H°f(H2O(g))] - [2H°f(CH3OH(g)) + 3H°f(O2(g))]
= [2(-393.5) + 4(-241.8)]-[2(-201.5) + 3(0)] = -1351.2 kJ
Now, 1 mole of CH3OH = 32 g
The calculated ΔHrxn corresponds to 2 moles of CH3OH. i.e.
The enthalpy change for 64 g of Ch3OH = -1351.2 kJ
Therefore, for 1.02g gaseous methanol we have:
ΔH = 1.02 * -1351.2/64 = -21.5 kJ
Ans: The enthalpy for the given reaction is -21.5 kJ
