According to Graham's law of effusion, how much faster does 79Br2 effuse compared to 81Br2?

• A. 79Br2 effuses about 1.013 times faster than 81Br2.
• B. 79Br2 effuses about 0.987 times as fast as 81Br2.
• C. 79Br2 effuses about 1.025 times faster than 81Br2.
• D. 79Br2 and 81Br2 effuse at the same rate.

Answer: (A)

Verified step by step guidance

1. Recall Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, this is expressed as: \[\text{Rate}_1 \propto \frac{1}{\sqrt{M_1}}\] and \[\text{Rate}_2 \propto \frac{1}{\sqrt{M_2}}\]
2. To find how much faster 79Br2 effuses compared to 81Br2, set up the ratio of their effusion rates using Graham's law: \[\frac{\text{Rate}_{79Br_2}}{\text{Rate}_{81Br_2}} = \sqrt{\frac{M_{81Br_2}}{M_{79Br_2}}}\]
3. Determine the molar masses of the two isotopic molecules. Since each molecule is diatomic bromine, multiply the atomic mass of each isotope by 2: \[M_{79Br_2} = 2 \times 79\] \[M_{81Br_2} = 2 \times 81\]
4. Substitute the molar masses into the ratio expression: \[\frac{\text{Rate}_{79Br_2}}{\text{Rate}_{81Br_2}} = \sqrt{\frac{2 \times 81}{2 \times 79}} = \sqrt{\frac{81}{79}}\]
5. Calculate the square root of the ratio to find how many times faster 79Br2 effuses compared to 81Br2. This will give you the numerical factor to compare their effusion rates.