Calculate the standard molar enthalpy for the complete combustion of liquid ethanol (c2h5oh) using the standard enthalpies of formation of the reactants and products.

The standard molar enthalpy for the complete combustion of liquid ethanol (C₂H₅OH) : -1366.7 kJ  mol⁻¹

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

Delta H solution

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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

the product of the pressure and volume is also constant

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.