Enter your answer in the provided box. 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. 2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(l) ΔH o rxn = −1452.8 kJ/mol

-477.4 kJ/mol

Explanation:

Let's consider the following thermochemical equation.

2 CH₃OH(l) + 3 O₂(g) → 2 CO₂(g) + 4 H₂O(l)  ΔH°rxn = −1452.8 kJ/mol

We can find the standard enthalpy of formation (ΔH°f) using the following expression.

ΔH°rxn = 2 mol × ΔH°f(CO₂(g)) + 4 mol × ΔH°f(H₂O(l)) - 2 mol × ΔH°f(CH₃OH(l)) - 3 mol × ΔH°f(O₂(g))

2 mol × ΔH°f(CH₃OH(l)) = 2 mol × ΔH°f(CO₂(g)) + 4 mol × ΔH°f(H₂O(l)) - ΔH°rxn - 3 mol × ΔH°f(O₂(g))

2 mol × ΔH°f(CH₃OH(l)) = 2 mol × (-393.5 kJ/mol) + 4 mol × (-285.8 kJ/mol) - (-1452.8 kJ) - 3 mol × 0 kJ/mol

ΔH°f(CH₃OH(l)) = -477.4 kJ/mol

1.43 x 10^-11 j the guy commented the awnser just making sure u see it .