Question

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

Answer 1

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

Further explanation

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

Learn more

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 2

Answer : To calculate the standard molar enthalpy of complete combustion of liquid ethanol can be done by using the formula given below;

$C_{2} H_{5}OH + 3O_{2} ----\ \textgreater \ 2CO_{2} + 3 H_{2}O$

Now, the enthalpies of the elements in their standard state is zero always. 

So the $H_{reaction} = H_{products} - H_{reactants}$

Here we can avoid the heat of enthalpy of oxygen,carbon dioxide and water as they exists in their standard states.

So the $H_{reaction}$ = $H_{ C_{2}H_{5}OH}$