Question

The log Henry's law constant (units of L-atm/mole and measured at 25 C) for trichloroethylene is 1.03; for tetrachloroethylene, 1.44; for 1, 2-dimethylbenzene, 0.71; and for parathion, -3.42. (a) What is the dimensionless Henry's law constant for each of these chemicals? (b) Rank the chemicals in order of ease of stripping from water to air. If needed use ideal gas Law (pV= nRT). R = 0.08205 (atm times L)/(mole times K). The dimensionless Henry's law constant is equal to K_H(L_H_2 O/L_Air) = K_H(L - atm/mole)/RT Given that the constants we are given are in logarithm, we have to cancel the log by raising it to exponential power of 10 since it is a natural log or 10^log^K H. This gives us K_H (trichloroethylene) = 10^1.03 = 10.7 L-atm/mol. After finding the regular Henry's law constant, just plug the value into the formula with R = 0.08206 (L-atm/mol K) and T = 298 K. This gives us: K_H (trichloroethylene) = 0.44, K_H (tetrachloroethylene) =1.1, K_H (1, 2-dimethylbenzene) = 0.21, K_H (parathion) = 1.6 times 10-5. The ease of stripping from water to air depends on the value of the Henry's law constant. As the Henry's law constant increase, the harder it is to convert from aqueous solution to air and vice versa.

Answer 1

Solution:

The dimenssionless henrys law constant is given as,

K_H(water/Air) = K_H/RT   

Where, KH(water/Air) - Dimenssionless henrys law constant

KH - Logarithmic henrys law constant L - atm/mole

R - ideal gas constant

T - temperature

As it is mentioned in the problem log henrys law constant is given, to convert it into constants we have to take 10 to the power so that it will get as,

logK_H = 10^LogK_H

Now find out values for tricloroethylene = 10^1.03 = 10.71 l-atm/mol

for tetrachloroethylene = 10^1.44 = 27.54

for 1,2 - dimethylbenzene - 10^0.71 = 5.12

for parathion = 10^-3.42 = 3.8 * 10^-4

Now put these values one by one in equation 1 to calculate KH dimenssionless value,

For trichloroethylene:

K_H(trichloroethylene) = 10.71/(0.08205*298)

R = 0.08206 L-atm/mole

T = 25+273 = 298 K

K_H(trichloroethylene) = 0.438

For Tetrachloroethylene

K_H(tetrachloroethylene) = 27.54/(0.08206 * 298) = 1.1

For 1,2 dimethylbenzene

K_H(1,2 - dimethylbenzene) = 5.12/(0.08206*298) = 0.21

For Parathion

K_H(parathion) = 3.8 * 10^-4/(0.08206 * 298) = 1.6*10^-5

Now to decide ease of stripping from water to airon the basis of henrys law constant.

Henrys law constant increases ease of stripping decrreases means it is harder to strip water into air.

therefore,

Parathion>1,2-dimethylbenzene>trichloroethylene>tetrachloroethylene