2 CH3OH(l) + 3 O2(g) → CO2(g) + 2 H2O(g) Calculate ∆G at 400 °C using the Gibbs Free Energy equation. State whether or not the reaction will be spontaneous. 2 CH3OH(l) + 3 O2(g) → CO2(g) + 2 H2O(g) ∆H (-238.7) (0) (-393.51) (-241.8) ∆S (0.1268) (0.205) (0.2136) (0.1887) (2(-241.8) + (-393.51)) – (2(-238.7) + 3(0)) ΔH° = -399.71 kJ (2(0.1887) + (0.2136)) – (2(0.1268) + 3(0.205)) ΔS° = -0.2776 kJ/K ∆G = ∆H° - T∆S° ∆G = (-399.71) – (673)(-0.2776) ΔG = -212.9 kJ spontaneous
1) ∆G = ∆H° - T∆S° a) ∆G = (-638.4) – (298)(0.1569) ΔG = -685.2 kJ spontaneous b) ∆G = (-57.2) – (298)(-0.1759) ΔG = -4.8 kJ spontaneous
C3H8 (l) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) 2) C3H8 (l) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) ∆H (-118.9) (0) (-393.5) (-285.8) ∆S (0.171) (0.2052) (0.21379) (0.06995) (4(-285.8) + 3(-393.5)) – ((-118.9) + 5(0)) ΔH° = -2204.8 kJ (4(0.06995) + 3(0.21379)) – ((0.171) + 5(0.2052)) ΔS° = -0.2758 kJ/K ∆G = ∆H° - T∆S° ∆G = (-2204.8) – (298)(-0.2758) ΔG = -2122 kJ spontaneous
Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) 3) Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g) ∆H (-822.0) (-110.53) (0) (-393.5) ∆S (0.090) (0.1976) (0.0273) (0.21379) (2(0) + 3(-393.5)) – ((-822.0) + 3(-110.53)) ΔH° = -26.91 kJ (2(0.0273) + 3(0.213679)) – ((0.090) + 3(0.1976)) ΔS° = 0.01497 kJ/K ∆G = ∆H° - T∆S° -31.3 = (-26.91) – (T)(0.01497) T = 293 K 20°C
N2(g) + 3 F2(g) → 2 NF3(g) 4) ∆G = ∆H° - T∆S° (2(-131.2)) – ((0) + 3(0)) ΔH° = -262.4 kJ (2(0.2607) ) – ((0.1915) + 3(0.2027)) ΔS° = -0.2782 kJ/K ∆G = ∆H° - T∆S° 0 = (-262.4) – (T)(-0.2782) T = 943 K 670°C