Answer to Question #334015 in Physical Chemistry for tma

Question

The enthalpy of combusbon for methane at 298.15 K is -726.1 kJ mol-1 .What is its enthalpy of

formation? (given: ∆H_f (C0₂ ) =-393.5 kJmol^-1

and ∆H_f (H₂O) = -285.8 kJmol^-1·.

Answer 1

CH₄ + 2O₂ = CO₂ + 2H₂O ∆_rH = -726.1 kJ mol^-1

∆_rH = -726.1 kJ mol^-1 = ∆_fH(CO₂) + 2∆_fH(2H₂O) - 2∆_fH(O₂) - ∆_fH(CH₄) = -393.5 - 2*285.8 - 2*0 - ∆_fH(CH₄) = -965.1 - ∆_fH(CH₄)

∆_fH(CH₄) = 726.1 - 965.1 = -239 kJ mol^-1