Question

What is the heat of combustion of ethane, c2h6, in kilojoules per mole of ethane? Enthalpy of formation values can be found in this list of thermodynamic properties?

Answer 1

The balanced combustion reaction for one mole of ethane will be

C2H6 + 3.5 O2  ---> 2CO2(g) + 3H2O (l)

Enthalpy of reaction of combustion = ∑ΔHf (products)- ∑ΔHf (reactant)

ΔHf = enthalpy of formation

The enthalpy of formation of oxygen = 0 as it is an element in its native state

ΔHf C2H6 = -83.820 kJ / mole

ΔHf H2O = -285.830 kJ / mole

ΔHf CO2 = -393.509 kJ / mole

ΔHrxn = [2 X ΔHf CO2 + 3 X ΔHf H2O] - [ΔHf C2H6]

ΔHrxn = [2 X (-393.509) + 3 (-285.830)] - [-83.820]

ΔHrxn = -1560.688 kJ / mole

Answer 2

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

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