https://youtu.be/rxmovBZQF6g 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)
Temperatur
e Constant

3. CONTENT - II
1) 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.
2) Not applicable to those gases which undergo a chemical
change in the situation. NH3 + H2O = NH4OH
LIMITATIONS OF HENERY’S LAW

4. CONTENT - II
3) This law cannot be applied to those gases which dissolve
into the ions in the solution:
HCl gas (H+, Cl-)
4) A soda water bottle contains CO2 gas dissolved in water at
higher pressure and following equilibrium is created:
CO2 (gas) ⇌ CO2 (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 H2?
[ H2 ] = n / V = weight / (molecular-weight X volume)
[ H2 ] = 20 / 2 X 40
[ H2 ] = 0.25 molL-1