The heat of combustion of ethane C₂H₆: -1559.9 kJ / mol
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:
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
a combustion reaction
https://brainly.com/question/12079874
the type of chemical reaction
https://brainly.com/question/2846778
The combustion of alkanes, alkenes, and alkynes
https://brainly.com/question/6818114
Complete combustion of a sample of a hydrocarbon
https://brainly.com/question/5636208
The heat of combustion of ethane is -1559.7 J/mol.
The heat of combustion is the heat evolved when 1 mole of a substance is burnt in oxygen under standard conditions.
We have the following information;
ΔH ° f H₂O: -285.8 kJ
ΔH ° f CO₂: -393.5 kJ
ΔH ° f C₂H₆: -84.7 kJ
The equation of the reaction is;
2C₂H₆ (g) + 7O₂ (g) ⟶ 4CO₂ (g) + 6H₂O (l)
Hence;
ΔH°reaction = [4(-393.5) + 6(-285.8)] - [2(-84.7) + 7(0)]
ΔH°reaction =[(-1574) + (-1714.8)] + 169.4
ΔH°reaction = -3119.4 J/mol
If 2 moles of C₂H₆ produces -3119.4 J/mol
1 mole of C₂H₆ produces 1 mole * -3119.4 J/mol/2 moles
= -1559.7 J/mol
Learn more about heat of combustion: https://brainly.com/question/14143095
