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Dalton's Law of Partial Pressures

Assume that y...

Question

Assume that you have a sample of a gas containing $H_{2}, H D$ and $D_{2}$ that you want to separate into pure components $(H =^{1} H and D =^{2} H)$ . What are the relative rates of diffusion of the three gases according to Graham’s law?

A

H_{2} > H D > D_{2}

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B

D_{2} > H D > H_{2}

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C

D_{2} > H_{2} > H D

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D

All have same rate

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Solution

The correct option is A $H_{2} > H D > D_{2}$
According to Graham's law of diffusion,
$Rate of diffusion = \frac{1}{\sqrt{Molar mass}}$

Since $D_{2}$ is the heaviest of the three molecules, it will be the slowest to diffuse and let us call its relative rate as $1.00$ . We can then compare the rate of diffusion of $H D$ and $H_{2}$ with $D_{2}$ by using Graham's law of diffusion.

Comparing $H D$ with $D_{2}$ we have,
$\frac{Rate of H D diffusion}{Rate of D_{2} diffusion} = \sqrt{\frac{Molar mass of D_{2}}{Molar mass of H D}} = \sqrt{\frac{4.0}{3.0}} = 1.15$

Comparing $H_{2}$ with $D_{2}$ we have,
$\frac{Rate of H_{2} diffusion}{Rate of D_{2} diffusion} = \sqrt{\frac{Molar mass of D_{2}}{Molar mass of H_{2}}} = \sqrt{\frac{4.0}{2.0}} = 1.41$

Thus, the relative rates of diffusion are $H_{2} (1.41) > H D (1.15) > D_{2} (1.00)$

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