# 14.15: Diffusion and Effusion and Graham's Law

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- 53835

We usually cannot see gases, so we need ways to detect their movements indirectly. The relative rates of diffusion of ammonia to hydrogen chloride can be observed in a simple experiment. Cotton balls are soaked with solutions of ammonia and hydrogen chloride (hydrochloric acid) and attached to two different rubber stoppers. These are simultaneously plugged into either end of a long glass tube. The vapors of each travel down the tube at different rates. Where the vapors meet, they react to form ammonium chloride \(\left( \ce{NH_4Cl} \right)\), a white solid that appears in the glass tube as a ring.

## Graham's Law

When a person opens a bottle of perfume in one corner of a large room, it doesn't take very long for the scent to spread throughout the entire room. Molecules of the perfume evaporate and the vapor spreads out to fill the entire space. **Diffusion** is the tendency of molecules to move from an area of high concentration to an area of low concentration until the concentration is uniform. While gases diffuse rather quickly, liquids diffuse much more slowly. Solids essentially do not diffuse.

A related process to diffusion is effusion. **Effusion** is the process of a confined gas escaping through a tiny hole in its container. Effusion can be observed by the fact that a helium-filled balloon will stop floating and sink to the floor after a day or so. This is because the helium gas effuses through tiny pores in the balloon. Both diffusion and effusion are related to the speed at which various gas molecules move. Gases that have a lower molar mass effuse and diffuse at a faster rate than gases that have a higher molar mass.

Scottish chemist Thomas Graham (1805 - 1869) studied the rates of effusion and diffusion of gases. **Graham's law** states that the rate of effusion or diffusion of a gas is inversely proportional to the square root of the molar mass of the gas. Graham's law can be understood by comparing two gases (\(\ce{A}\) and \(\ce{B}\)) at the same temperature, meaning the gases have the same kinetic energy. The kinetic energy of a moving object is given by the equation \(KE = \frac{1}{2} mv^2\) where \(m\) is mass and \(v\) is velocity. Setting the kinetic energies of the two gases equal to one another gives:

\[\frac{1}{2} m_A v_A^2 = \frac{1}{2} m_B v_B^2\]

The equation can be rearranged to solve for the ratio of the velocity of gas \(\ce{A}\) to the velocity of gas \(\ce{B}\) \(\left( \frac{v_A}{v_B} \right)\).

\[\frac{v_A^2}{v_B^2} = \frac{m_B}{m_A} \: \: \: \text{which becomes} \: \: \: \frac{v_A}{v_B} = \sqrt{\frac{m_B}{m_A}}\]

For the purposes of comparing the rates of effusion or diffusion of two gases at the same temperature, the molar masses of each gas can be used in the equation for \(m\).

Example 14.15.1

Calculate the ratio of diffusion rates of ammonia gas \(\left( \ce{NH_3} \right)\) to hydrogen chloride \(\left( \ce{HCl} \right)\) at the same temperature and pressure.

**Solution****:**

*Step 1: List the known quantities and plan the problem.*

__Known__

- Molar mass \(\ce{NH_3} = 17.04 \: \text{g/mol}\)
- Molar mass \(\ce{HCl} = 36.46 \: \text{g/mol}\)

__Unknown__

- Velocity ratio \(\frac{v_{NH_3}}{v_{HCl}}\)

Substitute the molar masses of the gases into Graham's law and solve for the ratio.

*Step 2: Solve.*

\[\frac{v_{NH_3}}{v_{HCl}} = \sqrt{\frac{36.46 \: \text{g/mol}}{17.04 \: \text{g/mol}}} = 1.46\]

The rate of diffusion of ammonia is 1.46 times faster than the rate of diffusion of hydrogen chloride.

*Step 3: Think about your result.*

Since ammonia has a smaller molar mass than hydrogen chloride, the velocity of its molecules is greater and the velocity ratio is larger than 1.

## Summary

- The processes of gas diffusion and effusion are described.
- Graham's law relates the molecular mass of a gas to its rate of diffusion or effusion.

## Contributors

CK-12 Foundation by Sharon Bewick, Richard Parsons, Therese Forsythe, Shonna Robinson, and Jean Dupon.