Question

The Henry’s law constant for O2 in water is 49,500 bar at 298 K and 63,400 bar at 313 K.
a) Calculate the number of grams of dissolved O2 per kg of water at 298 K if pO2 = 0.21 bar.

b) Oxygen (nonpolar) and water (highly polar) are not expected to form ideal solutions. It is an

excellent approximation, however, to assume activity coefficient of O2(HL) = 1. Why?

c) The vast majority of substances that dissolve in water (such as sugar and salt) become more soluble with increasing temperature. Use the Henry’s law constants for O2 to illustrate that gases are less soluble in water as the temperature is raised.

d) Give a qualitative thermodynamic explanation for the surprising result that solubility of gases in liquids decreases with temperature. Hint: dG/dT = So. Most dissolution processes produce significant increases in entropy due to the mixing process. But for gas dissolution, such as O2(g)  O2(aq), the entropy change is large and negative (Why?)

Answer 1

(a). We know C= k*P k is henrys solubility constant but here units are in bar so Henrys volatility constant is used,

H=p/x

At 298 K ,Here x = 0.21/49500=4.24 * 10^-6

Now Ca= x*density of Water/18 = 2.36*10^-7 moles/cc= 0.0075g in 1 kg water (since density of water us 1 kg/litre)

(b). Its possible due to low solute concentration

(C). As clearly seen from the data given as temperature increases from 298K to 313 K henrys volatility law constant is increasing so solubility is decreasing.

(d). Its because of infinite dilution extent.