Question

An experiment to determine compound partitioning among air, water and solid phases was conducted by adding 1 gram of the compound to a reactor that had 1 m3 of air, 10- 3 m3 of water, and 100 g of solid. The reactor was at atmospheric pressure and at a constant temperature of 20C. The compound has a molecular weight of 100 g/mole. Once equilibrium is reached, the air has a mass concentration of 0.500 g/m3 , and the water has a mass concentration of 200 g/m3 .

a) Determine the value of the Henry law constant Hc with units of Pa/(mole/m3 )

b) Given that the adsorption isotherm between water and the solid phase for this compound is linear, determine the isotherm constant

c) If the solid were completely removed from the air and water, and then resuspended in 10-3 m3 of initially clean water, what would be the contaminant concentration in water once equilibrium is reached?

Answer 1

(a): Given the volume of air, V = 1 m3

Pressure, P = 1 atm = 1.01x10⁵ Pa

Temperature, T = 20 DegC = 293 K

Now the total number of moles of gaseus air can be calculated from ideal gas equation

PV = n_tRT

=> n_t = PV / RT =  1.01x10⁵ Pax 1 m3 / 8.314 JK-1mol-1 x 293 K = 41.46 mol

Given the mass concentration of the compound =  0.500 g/m3

Hence 0.500 g of the compound is present in 1 m3 of air.

Given the molecular mass of the compound = 100 g/mole

Hence moles of the compound = mass / molecular mass = 0.500 g / 100 g/mole = 0.005 mol

Hence partial pressure in air due to the compound, P = mole fraction x Pt

=> P = (0.005 mol / 41.46 mol) x 1.01x10⁵ Pa = 12.18 Pa

Given the mass concentration of the compound in liquid = 200 g/m3 = 200 g / 100g/mol/m3 = 2 mol/m3

Hence mole concentration of the compound in liquid, Caq = 2 mol/m3

Now applying Henry's Law

K_H = P / Caq =  12.18 Pa / 2 mol/m3 = 6.09 Pa/(mole/m3 ) (answer)

(b): Given the mass of the adsorbent(solid), m = 100 g

equilibrium concentration of adsorbate(compound) in liquid, C = 200 g / m3

mass of adsorbent in solid, x = 1 g - (0.5 g + 200g/m3 x 10^-3m3) = 1g - (0.5g+0.2g) = 0.3 g

Given that the graph is linear. Hence

x/m = KxC

=> K = x / Cm = 0.3 g / 200 g / m3 x 100 g = 1.5 x 10^-5 m3 / g (answer)