Chemistry 2nd Paper · Classes 11-12 · Chapter 1 · 1.3.8
Graham's Law of Diffusion.
Loosen a gas tap and the smell takes a while to cross the room, but it does cross. Let two gases go together: which arrives first? The answer is one square root. The light one, then the heavy one.
1. Spreading out, and squeezing through a hole
Gas molecules are always rushing about in straight lines in every direction, so they keep colliding. Gases that do not react therefore blend into one homogeneous mixture. This spreading is called diffusion: the spontaneous and even spreading of the molecules or particles of a solid, liquid or gas from a place of higher density to a place of lower density.
Leave perfume in a closed room and after a while the scent is everywhere. Drop grains of KMnO₄ in water and the MnO₄⁻ ions diffuse outward, slowly turning the water purple.
Effusion is a different thing. Under high pressure from outside, a gas escapes forcefully and in one direction through a narrow hole in its container. The pressure is high inside, and low or a vacuum outside.
They differ in three ways. Basis: diffusion comes from the random, spontaneous movement of molecules, with the same air pressure inside and outside the container; effusion comes from a pressure difference between inside and outside. Examples: the smell of flowers or of H₂S spreading in air is diffusion; air leaking through a hole in a car tyre tube is effusion. Nature: by kinetic theory, diffusion is a slow spontaneous process at equal air pressure; effusion is the fast movement of gas molecules under high pressure. Now sort seven statements.
Diffusion and effusion
2. Graham's law
In 1846 the Scottish chemist Thomas Graham experimented on the diffusion of gases and gave a law. It holds for both diffusion and effusion.
The law: at constant pressure and temperature, the rate of diffusion r of a gas is inversely proportional to the square root of its molar mass M: r ∝ 1/√M. For two gases at the same temperature and pressure, r₁/r₂ = √(M₂/M₁), that is, r₁√M₁ = r₂√M₂.
The reason lies in molecular speed. Molecules of lower molar mass have a higher probable speed, so a light gas diffuses faster.
The law can also be written with density: at constant pressure and temperature r₁/r₂ = √(d₂/d₁) = √(M₂/M₁), that is, r ∝ 1/√d. Today the term molar mass is preferred to density.
Three things fix the rate: (1) the molar mass or density of the gas; (2) its temperature and pressure; (3) for effusion, the size of the hole and the size of the gas molecule.
3. NH₃ and HCl in a race
A race between two gases. Guess where the white ring will form, then let them go.
Where does the white ring form?
The two fronts are a model drawn from r ∝ 1/√M.
In the experiment of Figure 1.12, a glass tube open at both ends has cotton wet with conc. HCl at one end and cotton wet with conc. NH₃ solution at the other. The two gases rush towards each other, and where they meet, white smoke of NH₄Cl forms. The smoke appears 10 cm from the HCl end and 15 cm from the NH₃ end. So r_NH₃/r_HCl = 15/10 = 1.5, and the law gives √(36.5/17) = 1.465 ≈ 1.5. Drag the handle to change the molar mass of the second gas and watch which way the smoke moves.
Where the smoke forms
The smoke position is drawn from the law, a model and not a measurement of you.
4. What the law is used for
Graham's law has four uses: (1) finding the molar mass of a gas; (2) separating or enriching the parts of a gas mixture by their different diffusion rates, which is called atmolysis; (3) warning devices that detect methane in coal mines; (4) separating the isotopes of one element.
In the Second World War uranium was turned into UF₆ gas (b.p. 329 K) and ²³⁵U was separated from ²³⁸U by diffusion. Natural uranium is 0.7% ²³⁵U and 99.3% ²³⁸U. ²³⁵UF₆ (M = 349) diffuses faster than ²³⁸UF₆ (M = 352), and r(²³⁵UF₆)/r(²³⁸UF₆) = √(352/349) = 1.0043. The edge is tiny.
5. Working the numbers
When two gases are at different pressures (temperature fixed), the rate is proportional to pressure and inversely proportional to the square root of molar mass: r_A ∝ P_A/√M_A. So r_A/r_B = (P_A/P_B) × √(M_B/M_A). When P_A = P_B this falls back to r_A/r_B = √(M_B/M_A).
At the same pressure, for equal volumes of gas the effusion times go the opposite way to the rates: t₂/t₁ = r₁/r₂ = √(M₂/M₁). The gas that takes less time is the lighter one. Double the time means four times the molar mass: a gas taking twice as long as O₂ has M = 32 × 2² = 128.
The calculator below holds four examples from the book's problems; pick one and the numbers fill in. Problem 1.49: an unknown gas and Cl₂ effuse in 60 s and 73 s, so M₁ = 71 × (60/73)² = 48. Problem 1.52: 100 mL of gas A takes 293 s and CO₂ takes 230 s, so M₁ = 44 × (293/230)² = 71.41. Problem 1.53: an unknown gas diffuses 2.92 times as fast as NH₃, so M₁ = 17/(2.92)² = 1.99 ≈ 2, and the gas is H₂. Do not mix rates with times: each field's name says which mode it belongs to. Numbers you type go nowhere. The notebook is yours.
Heavier, slower
The numbers come from the book's problems; the working is the law's model.
Problem 1.54: if 1.0 L of H₂ effuses in 5 minutes at STP, r₁ = 0.2 L/min. Then r₁/r₂ = √(32/2.016) = 3.98, so r₂ = 0.05 L/min, and O₂ effuses 0.25 L in 5 minutes.
Problem 1.47: H₂ (M = 2.016), HD (3.022), D₂ (4.028). r(HD)/r(D₂) = √(4.028/3.022) = 1.154 and r(H₂)/r(D₂) = √(4.028/2.016) = 1.414. Taking D₂ as 1, the relative rates are H₂ (1.414) > HD (1.154) > D₂ (1.000).
Problem 1.48: air is 80% N₂ and 20% O₂ by volume. Its density relative to H₂ = (28/2 × 0.8) + (32/2 × 0.2) = 11.2 + 3.2 = 14.4. Its effective molar mass = 28 × 0.8 + 32 × 0.2 = 28.8.
Problem 1.51: the effusion rates of the unknown gas and Cl₂ are in the ratio 6 : 5, and the density of Cl₂ is 36. From 6/5 = √(36/d₁), d₁ = 25, so the molar mass = 2 × 25 = 50.
Problem 1.55: at 27°C, gas A (0.5 g, 760 mm Hg, 300 mL) and gas B (0.6 g, 770 mm Hg, 400 mL). The working gives M_A = 41 g mol⁻¹ and M_B = 36.42 g mol⁻¹. B has the smaller molar mass, so B diffuses faster: r_B/r_A = √(41/36.42) = 1.061.
Problem 1.50: pure O₂ takes 80 s to effuse and O₂ with 20% ozone takes 85 s. M₂ = 32 × (85/80)² = 36. Then from 0.2x + 25.6 = 36 (0.8 × 32 = 25.6), x = 52, which is the value printed in the book. The true molar mass of ozone is 48, so this value is not used in exam questions. What you take from it is the method.
6. Who is fastest?
Picking the fastest from a list is easy: the gas of least molar mass is the fastest. Among HCl (36.5), NH₃ (17), CH₄ (16) and Cl₂ (71), CH₄ has the highest rate. Gases of equal molar mass, like N₂ and CO (both 28), diffuse at equal rates. Below, with the molar masses beside them, put the gases in order from fastest to slowest.
Fastest to slowest
Some molar masses for the working (g mol⁻¹): H₂ 2, He 4, Ne 20, N₂ 28, O₂ 32, F₂ 38, Ar 40, CO₂ 44, Cl₂ 71, CH₄ 16, C₂H₆ 30, C₃H₈ 44, C₄H₁₀ 58. The gas of smaller molar mass diffuses faster, so among Ne, F₂ and Cl₂, Ne is the fastest.
A few more: SO₂ 64, NO₂ 46, N₂O 44, NO 30, CO 28, C₂H₄ 28, NH₃ 17, HCl 36.5, PH₃ 34, BF₃ 67.8. Gases of equal molar mass diffuse at equal rates, as N₂, CO and C₂H₄ do at 28 each.
If the relative density of air (to H₂) is 14.4, H₂ diffuses √14.4 = 3.79 times as fast as air. So in the time that 2.0 m³ of air leaves through a hole, 2.0 × 3.795 = 7.59 m³ of H₂ leaves.
7. Check yourself
Three questions in the exam's style. Start with the first.
Question 1
The rate of diffusion is related to the-
The second may want a little working as well as thought.
Question 2
By which process does the smell of a perfume spread through a room?
The last comes at the note from another side.
Question 3
Which gas has the higher rate of diffusion?
The whole note on seven cards.
Seven cards
Source: Chemistry 2nd Paper, Classes 11-12 (NCTB), Chapter 1 “Environmental Chemistry”, section 1.3.8.