∆H ° rxn =-2855.56 kJ
<h3>Further explanation</h3>
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
<em>∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants) </em>
∆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
C, im pretty sure if im wrong my bad
