Question

Calculate the standard change in Gibbs free energy of the following reactions at Standard Ambient Temperature and Pressure (SATP where T = 25°C and P = 1 atm) and label them as spontaneous or nonspontaneous.

(a) 2 SO2(g) + O2(g) 2 SO3(g): ΔΗ.-197.8 k , dS.-188.0 J/K kJ, and the reaction is Selectv (b) 2 C6H6(/) + 15 O2(g) 12 CO2(g) + 6 H2O(/); ΔΗ.-6535.0 kJ, S.-439.2 J/K kJ, and the reaction is Select (c) C(diamond); C(graphite); ΔΗ--19 ki, as 3.31/K kJ, and the reaction is Selectv (d) SiO2(s) si(s) + O2(g); ΔΗ. +910.7 ki, AS. +182.53/K kJ, and the reaction is Selectv

Answer 1

Using the formula delta G^0 = delta H^0 - T delta S^0 IF delta G^0 < 0 rightarrow Spontaneous delta G^0 > 0 non - spontaneous 1KJ = 1000J (a) delta H^0 = -197.8kJ delta S^0 = -1.88 J/k T = 273 + 25 = 298 k delta G^0 = -197.8 - 298 times (- 188)/1000 KJ = -197.8 + 56.02 = -141.78 KJ spontaneous b. delta H^0 = -6535 KJ, delta S^0 = -439.2 delta G^0 = -6535 - 298(- 439.2)/100 = -6535 + 130.88 = -6404.1 kJ < 0 spontaneous c. delta H^0 = -1.9kJ delta S^0 = 3.3J/k delta G^0 = -1.9 - 298 times 3.3/1000 = -1.9 - 0.98 = -2.88 kJ = -2.9kJ Spontaneous d. delta H^0 = 910.7kJ delta S^0 = 182.5 delta G^0 = -910.7 - 298 times 182.2/1000 = 910.7 - 64.4 = 856.3 KJ > 0 So non spontaneous