Henry Law, Its Limitation and Active Mass

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In this video you will study about Henry Law, Its limitations, What is Active Maas, how it is calculated with the help of example

1. HENERY’S LAW Vinod Dahiya TUTORIAL : Advanced PHYSICAL Chemistry

2. WHAT IS HENERY’S LAW CONTENT - I Given by William Henry in 1803. This law states that: The mass of a gas that dissolve in given mass of a solvent at a constant temperature is directly proportional to the pressure of the gas at equilibrium with the solution provided. M ∝ P M = KP = where K is constant G(P) Temperature Constant

3. CONTENT - II Applicable to ideal gases only. However the law can be applied to real gases at low pressure, where real gas approaches the behavior of ideal gases. Not applicable to those gases which undergo a chemical change in the situation. NH 3 + H 2 O = NH 4 OH LIMITATIONS OF HENERY’S LAW

4. CONTENT - II This law cannot be applied to those gases which dissolve into the ions in the solution: HCl gas (H + , Cl - ) A soda water bottle contains CO2 gas dissolved in water at higher pressure and following equilibrium is created: CO 2 (gas) ⇌ CO 2 (in solution) LIMITATIONS OF HENERY’S LAW

5. CONTENT - II Since the bottle is sealed at higher pressure, there is plenty of gas dissolved in water and the gas pressure above the solution is very high. When the bottle is opened, the pressure gets lowered and becomes equal to atmospheric pressure. Therefore, the dissolved CO2 gas freeze out rapidly to reach a new equilibrium corresponding to atmospheric pressure. LIMITATIONS OF HENERY’S LAW

6. CONTENT - II Active mass is defined as the part of concentration which is involved in the reaction and active mass of the solid is constant. [ ] = n/V [ A ] = nA / V and [ B ] = nB / V [ ] = (weight/molecular-weight)/(volume/1) = (weight / V) X 1 / molecular-weight = density / molecular-weight at constant temperature, d is constant ACTIVE MASS

7. HENRY’S LAW QUESTION & ANSWERS CONTENT - II If 20g hydrogen is there in 40lt container, then find out the active mass of H 2 ? [ H 2 ] = n / V = weight / (molecular-weight X volume) [ H 2 ] = 20 / 2 X 40 [ H 2 ] = 0.25 molL -1