21 Apr 2018

Henry’s law

  Henry’s law


Henry’s law states that at a constant temperature, the solubility of a gas in a liquid is directly proportional to the pressure of the gas.
The most commonly used form of Henry’s law states that
“the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution”.
This is expressed as:
 p = KH x Here KH is the Henry’s law constant.
Characteristics of KH
KH is a function of the nature of the gas.
Higher the value of KH at a given pressure, the lower is the solubility of the gas in the liquid. 
KH values increase with increase of temperature indicating that the solubility of gases increases with decrease of temp.



Applications of Henry’s law

1. In the production of carbonated beverages.

2. In the deep sea diving.

3. For climbers or people at high altitudes. 



Raoult’s Law as a special case of Henry’s Law 

According to Raoult’s law,

 pi = xi   pi0  

In the solution of a gas in a liquid, one of the components is so volatile that it exists as a gas.
 Its solubility according to Henry’s law,

 p = KH x. 

Thus, Raoult’s law becomes a special case of Henry’s law in which KH becomes equal to pi0 .



Henry’s law
  • Henry’s Law establishes aquantitative relation betweenpressure and solubility of a gasin a solvent.
  • This law is for gas-liquid solution.
  • According to the law at a constant temperature, thesolubility of a gas in a liquid is directly proportional to thepressure of the gas.
  • It can also be stated as - The partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution.
  • Mathematically,
KH x where K= Henry’s law constant.
  • At same temperature different gasses have different KH
  • It is a function of the nature of the gas. At a given pressure increasing value of KH implies lower solubility of the gas in the liquid.
  • Value of KH increases with the increase in temperature therefore solubility of gases increases with decreasing temperature. Due to this reason cold water is more sustainable for aquatic life than warm water.
 Problem: H2S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H2S in water at STP is 0.195 m, calculate Henry's law constant.
Solution:  
It is given that the solubility of H2S in water at STP is 0.195 m, i.e., 0.195 mol of H2S is dissolved in 1000 g of water.
Moles of water = 1000g / 18g mol-1 = 55.56 mol 
∴Mole fraction of H2S, x =  Moles of H2S / Moles of H2S+Moles of water 0.195 / (0.195+55.56)= 0.0035 
At STP, pressure (p) = 0.987 bar
According to Henry's law:
KH x  ⇒ KH = p / x
= 0.0987 / 0.0035 bar

= 282 bar