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Difficulty: MediumHydrogen: Preparation, Properties, Isotopes, and Water Hardness

Under the same conditions of temperature and pressure, how many times faster does protium gas (H2\text{H}_2) diffuse through a porous plug compared to tritium gas (T2\text{T}_2)? [Relative atomic masses: protium, 1H=1.0^{1}\text{H} = 1.0; tritium, 3H=3.0^{3}\text{H} = 3.0]

  1. 1.731.73Answer
  2. B
    3.003.00
  3. C
    0.580.58
  4. D
    9.009.00

Answer

Protium gas diffuses approximately 1.731.73 times faster than tritium gas.
The relative rate of diffusion of protium gas (H2\text{H}_2) relative to tritium gas (T2\text{T}_2) is governed by Graham's Law, which states that R1/R2=M2/M1R_1 / R_2 = \sqrt{M_2 / M_1}. Given M(H2)=2 g mol−1M(\text{H}_2) = 2\text{ g mol}^{-1} and M(T2)=6 g mol−1M(\text{T}_2) = 6\text{ g mol}^{-1}, the ratio is 6/2=3≈1.73\sqrt{6 / 2} = \sqrt{3} \approx 1.73. Thus, protium gas diffuses 1.731.73 times faster.

Step-by-Step Solution

1
Calculate the molar mass of diatomic protium gas (H2\text{H}_2) and tritium gas (T2\text{T}_2).
M(H2)=2×1.0=2.0 g mol−1M(\text{H}_2) = 2 \times 1.0 = 2.0\text{ g mol}^{-1} and M(T2)=2×3.0=6.0 g mol−1M(\text{T}_2) = 2 \times 3.0 = 6.0\text{ g mol}^{-1}.
Graham's law of diffusion requires the molar masses of the gaseous species.
2
Apply Graham's Law of diffusion: RH2RT2=MT2MH2\frac{R_{\text{H}_2}}{R_{\text{T}_2}} = \sqrt{\frac{M_{\text{T}_2}}{M_{\text{H}_2}}}.
RH2RT2=6.02.0=3.0\frac{R_{\text{H}_2}}{R_{\text{T}_2}} = \sqrt{\frac{6.0}{2.0}} = \sqrt{3.0}.
The rate of effusion or diffusion of a gas is inversely proportional to the square root of its molar mass.
3
Evaluate the square root to determine the ratio.
3.0≈1.73\sqrt{3.0} \approx 1.73.
This yields the relative diffusion rate of protium gas compared to tritium gas.

Key Concept

Graham's Law of Diffusion applied to Hydrogen Isotopes
Estimated Time:1m 15s
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Topics

Hydrogen: Preparation, Properties, Isotopes, and Water Hardness

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