Question

Calculate ∆H, ∆S, and ∆G for the following reactions at 298K and at 350K.

CH₄(g) + 2O₂(g) -> CO₂(g) + 2H₂O(g)

HCl(g) -> H⁺(aq) + Cl^-(aq)

HSO₄^-(aq) -> H⁺(aq) + SO₄^2-(aq)

Answer 1

CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g)

DHrxn = (DH0fCO2 + 2*DH0fH2O) - (DH0fCH4+2*DH0fO2)

      = (-393.51+2*-241.8)-(-74.8+2*0)

      = -802.31 kj

DS0rxn = (S0CO2+2*S0H2O)-(S0CH4+2*S0O2)

        = (213.6+2*188.7)-(186.2+2*205.0)

        = -5.2 j/mol.k

DG0 = DH0-TDS0

    = (-802.31)-(298*-5.2*10^-3)

    = -800.76 kj

DG0 = DH0-TDS0

    = (-802.31)-(350*-5.2*10^-3)

    = -800.49 kj

HCl(g) -> H+(aq) + Cl-(aq)

DH0rxn = (0+(-167.08))-(-92.31)

       = -74.77 Kj

DS0rxn = (0+56.5)-(186.8)

        = -130.3 j/mol.k

DG0 = DH0-TDS0

     = (-74.77)-(298*-130.3*10^-3)

     = -35.94 kj

DG0 = DH0-TDS0

     = (-74.77)-(350*-130.3*10^-3)

     = -29.165 kj

HSO4-(aq) -> H+(aq) + SO42-(aq)

DH0rxn = (0+-909.34)-(-886.9)

       = -22.44 kj

DS0rxn = (0+20.1)-(131.8)

        = -111.7 j/mol.k

DG0 = DH0-TDS0

    = (-22.44)-(298*-111.7*10^-3)

    = 10.85 kj

DG0 = (-22.44)-(350*-111.7*10^-3)

    = 16.655 kj