The standard molar enthalpy for the complete combustion of liquid ethanol (C₂H₅OH) : <u>-1366.7 kJ mol⁻¹</u>
<h3>Further explanation</h3>
The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation
The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)
Determination of the enthalpy of formation of a reaction can be through a calorimetric experiment, based on the principle of Hess's Law, enthalpy of formation table, or from bond energy data
Delta H reaction (ΔH) is the amount of heat / heat change between the system and its environment
(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)
The value of ° H ° can be calculated from the change in enthalpy of standard formation:
∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)
We complete the data values for ∆Hf° of compounds
∆H ° f C₂H₅OH (ℓ): -277.7 kJ
/mol
∆H ° f 2CO₂ (g): -393.5 kJ
/mol
∆H ° f H₂O (g): -285.8 kJ
/mol
°H ° f 3O₂ (g): 0 (ΔHf ° (O₂ (g)) is zero for free elements)
Combustion reaction that occurs :
C₂H₅OH (ℓ) + 3O₂ (g) -> 2CO₂ (g) + 3H₂O (g)
then :
ΔHrxn ° = (2ΔHf ° (CO₂ (g) + 3ΔHf ° (H₂O (l)) - (ΔHf ° (C₂H₅OH (l)) + 3ΔHf ° (O₂ (g))
ΔHrxn ° = [(2 × -393.5 + 3 × -285.8) - (-277.7 + 0)] kJ mol⁻¹
ΔHrxn ° = -1366.7 kJ mol⁻¹
<h3>Learn more
</h3>
Delta H solution
brainly.com/question/10600048
an exothermic reaction
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as endothermic or exothermic
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an exothermic dissolving process
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Keywords: Delta H reaction (ΔH), enthalpy of standard formation
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