General Chemistry

Question

Calculate the enthalpy of formation of C₁₂H₂₆ (g), ΔH_{f, given the following information:}

2C₁₂H₂₆ (g) + 37O₂ (g) → 24CO₂ (g) + 26H₂O (l) , ΔH_f =-16,170.0 kJ/mol

H₂O (l) , ΔH_f = -285.8 kJ/mol

CO₂ (g) , ΔH_f = -393.5 kJ/mol

Answer 1

Reverse reaction and divide by 2 to get...

12 CO₂ + 13 H₂O ==> C₁₂H₂₆ + 18 1/2 O₂ ∆H_f = + 8085 kJ/mol

∆Hº_f = ∑∆Hº_f products - ∑∆Hº_f reactants = (x + 0) - [(12x-393.5)+(13x-285.8)] = 8085

x - (-4722 + -3715) = 8085

x + 8437 = 8085

x = -352 kJ/mole