Chapter 18, Problem 18.106QP

Cobalt(II) chloride hexahydrate, CoCl₂·6H₂O, is a bright pink compound, but in the presence of very dry air it loses water vapor to the air to produce the light blue anhydrous salt CoCl₂. Calculate the standard free-energy change for the reaction at 25°C:

${\text{CoCl}}_2\cdot 6{\text{H}}_2\text{O}(s)\stackrel{}{\rightleftharpoons }{\text{CoCl}}_2(s)+6{\text{H}}_2\text{O}(g)$

Here are some thermodynamic data at 25°C:

What is the partial pressure of water vapor in equilibrium with the anhydrous salt and the hexahydrate at 25°C? (Give the value in mmHg.) What is the relative humidity of air that has this partial pressure of water? The relative humidity of a sample of air is

$\text{Relative humidity=}\frac{{\text{partial pressure of H}}_2\text{O}(g)\text{in air}}{\text{vapor pressure of water}}\times 100\%$

What do you expect to happen to the equilibrium partial pressure over the hexahydrate as the temperature is raised? Explain.