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)

∆H ° rxn =-2855.56 kJ

Given

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

increased stability in the completion of fluorine's outer shell.

a polar bond is formed when electrons are shared unevenly between atoms