The heat of combustion of ethane C₂H₆: <u>-1559.9 kJ / mol
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<h3>Further explanation
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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:
<h3>∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)
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ΔH∘rxn = ΔH∘f of the product (s) if ∆Hf ° (reactants) = 0
<em>The elements in standard conditions are not included in the enthalpy calculations because the enthalpy of those elements under the standard conditions is </em><u><em>zero.
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The reaction of the combustion of hydrocarbons with oxygen produces CO₂ and H₂O (water vapor
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Hydrocarbon combustion reactions (specifically alkanes)
[tex] \large {\box {\bold {C_nH _ (_2_n _ + _ 2_) + \frac {3n + 1} {2} O_2 ----> nCO_2 + (n + 1) H_2O}}} [ /tex]
From data on the internet:
∆H ° f H₂O: -285.8 kJ
°H ° f CO₂: -393.5 kJ
°H ° f C₂H₆: -84.7 kJ
Combustion of ethane, C₂H₆
2C₂H₆ (g) + 7O₂ (g) ⟶ 4CO₂ (g) + 6H₂O (l)
= {[4 (∆H ° f CO₂ (g))] + [6 (∆H ° f H₂O (l))]} - {2 [∆H ° f C₂H₆ (g)] + [(7) (∆ H ° f O₂ (g))]}
= {[4 (-393.5 kJ)] + [6 (-285.8 kJ]} - {2 [-84.7 kJ] + [(0 kJ)]}
= -1574 kJ -1714.8 kJ +169.4 kJ
= 3119.8 kJ (for 2 mol C₂H₆)
for 1 mole C₂H₆ ---> 3119.8 kJ: 2 = = -1559.9 kJ / mol
<h3>Learn more
</h3>
a combustion reaction
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the type of chemical reaction
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The combustion of alkanes, alkenes, and alkynes
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Complete combustion of a sample of a hydrocarbon
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