The reaction betweeen mercury and oxygen to form mercuric oxide is written below.

2 Hg(l) + O2(g) ----> 2 HgO(s)

The DH for the reaction is -43,000 calories/mole and clearly exothermic. What is the sign for the entropy change? Is the reaction spontaneous? Justify your answer.

The DS for the reaction at 298 Kelvin is a small negative number because the reactants gas and liquid are forming an organized solid. Remember DS is negative for changes that go from
gas --> liquid;
liquid ---> solid

All of the calculations assume T = 298 Kelvin unless the number is stated differently.

Substitution into the definition for free energy is shown here. The TDS term is very small. It will not cancel the overall large negative value from the enthalpy.

The DG will be large and negative.

DG=-43,000 calories/mole - (298 K)(small negative DS )

Because the entropy change is small and negative the -47,000 calories dominates the free energy change.

The free energy change is a large negative number. The reaction is spontaneous. The entropy change is negative.