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dalton200166
28.12.2020 •
Chemistry
Consider the following data:
Three entries. 1: Delta H f 1 of C O 2 equals negative 393.5 kilojoules per mole. 2: Delta H f 1 H 2 O equals negative 241.82 kilojoules per mole. 3: Delta H f 1 C 2 H 2 equals negative 84.68 kilojoules per mole.
Using these data, calculate for this chemical reaction. Round to the nearest whole number.
2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(g)
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Ответ:
∆H ° rxn =-2855.56 kJ
Further explanationGiven
ΔHf CO₂ = -393.5 kJ/mol
ΔHf H₂O = -241.82 kJ/mol
ΔHf C₂H₆ = - 84.68 kJ/mol
Reaction
2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(g)
Required
ΔHrxn=
Solution
∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)
∆H ° rxn = (4.-393.5+6.-241.82)-(2.-84.68)
∆H ° rxn = (-1574-1450.92)-(-169.36)
∆H ° rxn =-3024.92+169.36
∆H ° rxn =-2855.56 kJ
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