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Difficulty: MediumSulfur Allotropes, Hydrogen Sulfide, and Sulfur(IV) Oxide

Under the same conditions of temperature and pressure, 50 cm350\text{ cm}^3 of sulfur(IV) oxide diffuses through a porous plug in 20 seconds20\text{ seconds}. How long will it take for an equal volume of methane gas (CH4CH_4) to diffuse through the same plug under identical conditions? [H=1,C=12,O=16,S=32][H = 1, C = 12, O = 16, S = 32]

  1. 10 seconds10\text{ seconds}Answer
  2. B
    5 seconds5\text{ seconds}
  3. C
    40 seconds40\text{ seconds}
  4. D
    80 seconds80\text{ seconds}

Answer

The time required for an equal volume of methane gas to diffuse is 10 seconds10\text{ seconds}.
According to Graham's Law of diffusion, the time required for a fixed volume of gas to diffuse is directly proportional to the square root of its molar mass (t∝Mt \propto \sqrt{M}). Since the molar mass of sulfur(IV) oxide (SO2SO_2) is 64 g mol−164\text{ g mol}^{-1} and that of methane (CH4CH_4) is 16 g mol−116\text{ g mol}^{-1}, the ratio of their diffusion times is 1664=12\sqrt{\frac{16}{64}} = \frac{1}{2}. Therefore, methane takes half as long as sulfur(IV) oxide (20×0.5=10 seconds20 \times 0.5 = 10\text{ seconds}).

Step-by-Step Solution

1
Calculate the molar masses of sulfur(IV) oxide (SO2SO_2) and methane (CH4CH_4).
M(SO2)=32+(2×16)=64 g mol−1M(SO_2) = 32 + (2 \times 16) = 64\text{ g mol}^{-1} and M(CH4)=12+(4×1)=16 g mol−1M(CH_4) = 12 + (4 \times 1) = 16\text{ g mol}^{-1}.
Molar masses are required to apply Graham's Law of diffusion.
2
Set up Graham's Law equation relating diffusion time (tt) to molar mass (MM) for equal volumes.
t(CH4)t(SO2)=M(CH4)M(SO2)\frac{t(CH_4)}{t(SO_2)} = \sqrt{\frac{M(CH_4)}{M(SO_2)}}.
The rate of diffusion is inversely proportional to the square root of density or molar mass (R=Vt∝1MR = \frac{V}{t} \propto \frac{1}{\sqrt{M}}).
3
Substitute the known values into the equation and solve for t(CH4)t(CH_4).
t(CH4)20=1664=14=12\frac{t(CH_4)}{20} = \sqrt{\frac{16}{64}} = \sqrt{\frac{1}{4}} = \frac{1}{2}, so t(CH4)=20×12=10 secondst(CH_4) = 20 \times \frac{1}{2} = 10\text{ seconds}.
Because methane has a smaller molar mass, it diffuses faster and takes less time to diffuse.

Key Concept

Graham's Law of Diffusion
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Topics

Sulfur Allotropes, Hydrogen Sulfide, and Sulfur(IV) Oxide

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