For the equilibrium #CuSO_4. 5H_2O(s) rightleftharpoons CuSO_4. 3H_2O(s) + 2H_2O(g)#, #K_p = 2.25*10^(-4) atm^2# and vapour pressure of water is #22.8# Torr at 298K?

#CuSO_4. 5H_2O(s#) is efflorescent (i.e loses water) when relative humidity is?
A) less than 33%
B) less than 50%
C) less than 60%
D) less than 66.6%

1 Answer
Truong-Son N.
Jan 17, 2018

The pure vapor pressure of water at #"25"^@ "C"# is around #"23.8 torr"#. If right now it is #"22.8 torr"#, we're either not at #"298 K"#, or there's a typo...

The relative humidity is:

#phi = P_(H_2O)/(P_(H_2O)^"*") xx 100%#

where #"*"# simply indicates pure solvent. #P_(H_2O)# is the vapor pressure above the water.

The solids are simply #1# in the equilibrium constant, so...

#K_p = P_(H_2O,eq)^2 = (0 + 2x)^2 = 2.25 xx 10^(-4) "atm"^2#

#=> 2x = 1.50 xx 10^(-2) "atm"#

And so the equilibrium partial pressure seems to be

#2x = P_(H_2O) =# #"11.4 torr"#

And this relative humidity seems to be:

#color(blue)(phi) = "11.4 torr"/"23.76 torr" xx 100% = color(blue)(48.0%)#

And this is less than #50%#. If the relative humidity was any higher, then the atmosphere would be saturated (i.e. #Q_p > K_p#, so #Q_p darr#, making more pentahydrate). If it is lower, then the equilibrium would shift to release water vapor (i.e. #Q_p < K_p#, so #Q_p uarr#, making more trihydrate as well).

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