Question

Ca(OH)2 dissolves exothermically in water and the molar solubility of Ca(OH)2 decreased as the temperature increases. What are the signs of Delta H and Delta S for dissolving Ca(OH)2?

Answer 1

First of all ,

The equilibrium constant,Ksp, for Ca(OH)2 is:

Ksp= [Ca2+][OH-]2

Remember that the calcium ion concentration is half the hydroxide ion concentration determined by the

titration.

The Gibbs Free Energy ΔG ̊, is related to the equilibrium constant (Ksp) by

ΔG ̊ = -RT ln Ksp

where R is the ideal gas constant(R = 8.3145 J/mol-K) andT is the absolute temperaturein Kelvin.

To find the enthalpy(ΔH ̊) andentropy(ΔS ̊) changes, the following equation is used

ΔG ̊ =ΔH ̊ - TΔS

where T is the absolute temperature. We assume that ΔH ̊ and ΔS ̊ do not change significantly with temperaturewhich is usually true for this small of a temperature change.

But if we consider above room temperature i.e at 100°C,

The average Ksp values give ∆G° values of +32.3 kJ /mol at room temperature and +43.60 kJ/mol at 100°C. The enthalpy of the reaction is ∆H°= -12.60 kJ/ mol and the entropy of the reaction is∆S°150.7 J mol^-1K^-1

So based on above data the signs of Delta H and Delta S for dissolving Ca(OH)2 will be postive up to rrom temp and then for exothermic reaction both signs are negative