The heat of combustion of ethane C₂H₆: -1559.9 kJ / mol
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)
ΔH∘rxn = ΔH∘f of the product (s) if ∆Hf ° (reactants) = 0
The elements in standard conditions are not included in the enthalpy calculations because the enthalpy of those elements under the standard conditions is zero.
The reaction of the combustion of hydrocarbons with oxygen produces CO₂ and H₂O (water vapor
)
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
Learn more
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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