Chapter 11.4 : Effusion and Diffusion

1. Effusion and Diffusion Chapter 11.4 Objective: State Graham’s Law of effusion Determine the relative rates of effusion of two gases of known molar masses. State the relationship between the molecular velocities of two gases and their molar masses.

2. Definitions Effusion Process whereby the molecules of a gas confined in a container randomly pass through a tiny opening in the container. Diffusion The gradual mixing of two gases due to their sp0ontaneous, random motion Effusion

3. Graham’s Law of Effusion Rate of effusion: Depends on: Velocity of gas molecules Mass of molecules 1 2 M A v A 2 = 1 2 M B v B 2 M A v A 2 = M B v B 2 v A 2 v B 2 = M A M B v A v B = M A M B SO: Rate of effusion of A Rate of effusion of B = M A M B Defined as: The rates of effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses .

4. Application of Graham’s Law Lighter gases (lower Molar mass or densities) diffuse faster than heavier gases. Also provides a method for determining molar masses. Rates of effusion of known and unknown gases can be compared to one another Rates of effusion of different gases

5. Problem 1 Compare the rates of effusion of hydrogen and oxygen at the same temperature and pressure. Rate of effusion of H 2 Rate of effusion of O 2 = M H 2 M O 2 = 32.00 g/mol 2.02 g/mol = 3.98 ***Remember that the molar masses are inversely related ***Find the molar masses of each ***Expressed like this Hydrogen effuses 3.98 times faster than oxygen

6. Problem 2 A sample of hydrogen effuses through a porous container about 9 times faster than an unknown gas. Estimate the molar mass of the unknown gas. Rate of effusion of H 2 Rate of effusion of unknown = M H 2 M unknown 2.02 g/mol M unknown = 9 2 2 81 = 2.02 g/mol M unknown x 2.o2 g/mol 2.o2 g/mol x M unknown = 160 g/mol