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

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dsdrajlin dsdrajlin · Answer 1 · 2020-04-16T02:55:58+02:00

Answer:

-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

adioabiola adioabiola · Answer 2 · 2021-12-05T04:09:56+01:00

From the reaction given; the standard enthalpy of formation methanol is; 726.4 kJ/mol

According to the question;

The chemical reaction given involves the combustion of methanol to produce CO2 and H2O as follows;

However, when the reaction occurs in the opposite direction as follows;

By definition; the standard enthalpy of formation is the amount of energy required for the formation of one mole of that substance.

Dividing the whole chemical equation by 2; we have;

In which case; the standard enthalpy of formation of methanol becomes; 1452.8/2 kJ/mol = 726.4 kJ/mol