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

answer: option (b) is the correct answer.

explanation:

a covalent bond is formed by sharing of electrons. whereas polarity arises due to difference in electronegativity of two atoms.

therefore, more is the electronegativity of an atom which combines with the titanium, more readily it will induce a partial positive and negative charge.

thus, out of the given options fluorine is most electronegative and has an incomplete octet. therefore, it will readily combine with titanium to share the electrons and complete its octet.