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Graham's Laws of Effusion and Diffusion

Key Concepts

Diffusion is the passage of a substance throughout another medium, eg, gases from an open bottle of perfume diffuse through the air.
Effusion of a gas is its passage through a pinhole or orifice.
Graham's Law of Diffusion: the rate of diffusion of one gas through another is inversely proportional to the square root of the density of the gas.
Graham's Law of Effusion: the rate of effusion of a gas is inversely proportional to the square root of the density of the gas.
Graham's Laws of Diffusion and Effusion:
rate = constant x (1 ÷ √d)
where d = density
for gases A and B under the same conditions,
Rate_A
= √d_B
Rate_B √d_A

Since density is proportional to molecular mass (formula weight):
Rate_A
= √MM_B
Rate_B √MM_A

where MM = molecular mass (formula weight)

Examples

Quantitatively compare the rates diffusion for equal moles of hydrogen gas and oxygen gas at the same temperature and pressure.
Organize the data:
MM_H₂ = 2
MM_O₂= 32
Write the equation:
Rate_H₂
= √MM_O₂
Rate_O₂ √MM_H₂

Substitute the known values:
Rate_H₂
= √32
Rate_O₂ √2

Rearrange the equation and solve:
Rate_H₂ = Rate_O₂ √ 16 = 4 x Rate_O₂
Hydrogen gas will diffuse 4 times faster than oxygen gas.
Gas X effuses through a pinhole at a rate of 4.73 x 10^-4mol s^-1.
Methane gas, CH_4(g), effuses through the same pinhole at a rate of 1.43 x 10^-3 mol s^-1 under the same conditions of temperature and pressure.
What is the molecular mass (formula weight) of gas X?
Organize the data:
MM_X = ? MM_CH₄= 16

rate_X = 4.73 x 10^-4 mol/s rate_CH₄= 1.43 x 10^-3 mol/s

Write the equation:
Rate_X
= √MM_CH₄
Rate_CH₄ √MM_X

Rre-arrange this equation to solve for molecular mass of gas X:
√MM_X
= rate_CH₄ √MM_CH₄
Rate_X

Substitute in the known values:
√MM_X
= 1.43 x 10^-3 √16
4.73 x 10^-4

√MM_X
= 1.43 x 10^-3 x 4
4.73 x 10^-4

Solve for the square root of MM_X:
√MM_X = 12.093
Solve for the molecular mass of X by squaring both sides of the equation:
MM_X = 12.093² = 146.24 g/mol

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