Question

for the reaction 2SO2+O2=2SO3 what is the entropy change

Answer 1

For the reaction

2 SO3(g) → 2 SO2(g) + O2(g)

∆H◦r = +198 kJ·mol−1 at 298 K.


I think not sure.

Answer 2

ENTHALPY OF REACTION
[2ΔHf(SO3 (g))] - [2ΔHf(SO2 (g)) + 1ΔHf(O2 (g))]
[2(-395.72)] - [2(-296.83) + 1(0)] = -197.78 kJ
-197.78 kJ (exothermic)

ENTROPY CHANGE
[2ΔSf(SO3 (g))] - [2ΔSf(SO2 (g)) + 1ΔSf(O2 (g))]
[2(256.65)] - [2(248.11) + 1(205.03)] = -187.95 J/K
-187.95 J/K (decrease in entropy)

FREE ENERGY OF REACTION (AT 298.15 K)
From ΔGf° values:
[2ΔGf(SO3 (g))] - [2ΔGf(SO2 (g)) + 1ΔGf(O2 (g))]
[2(-371.08)] - [2(-300.19) + 1(0)] = -141.78 kJ
-141.78 kJ (spontaneous)

From ΔG = ΔH - TΔS:
-141.74 kJ (spontaneous)

EQUILIBRIUM CONSTANT, K (AT 298.15 K)
6.9209141967e+024
This process is favorable at 25°C.