Question

What is the ratio of the rate of effusion of the most abundant gas, nitrogen, to the lightest gas, hydrogen?

Short answer

Answer: The ratio of the rate of effusion of nitrogen to the rate of effusion of hydrogen is approximately 1/3.74.

Step-by-step solution

Identify the relevant information for each gas

For this problem, we need to know the molar masses of nitrogen (N2) and hydrogen (H2). The molar mass of nitrogen is 28 g/mol, and the molar mass of hydrogen is 2 g/mol.

Apply Graham's law of effusion

Graham's law of effusion states that the rate of effusion (R) of a gas is inversely proportional to the square root of its molar mass (M): \[ R \propto \frac{1}{\sqrt{M}} \]We are asked to find the ratio of the rate of effusion of nitrogen (R_N2) to the rate of effusion of hydrogen (R_H2). Using Graham's law, we can write this ratio as: \[ \frac{R_{N2}}{R_{H2}} = \frac{\frac{1}{\sqrt{M_{N2}}}}{\frac{1}{\sqrt{M_{H2}}}} \]

Insert the values of molar masses and simplify

Now, we can plug in the molar masses of nitrogen and hydrogen into the equation and simplify. \[ \frac{R_{N2}}{R_{H2}} = \frac{\frac{1}{\sqrt{28}}}{\frac{1}{\sqrt{2}}} \] By multiplying both the numerator and the denominator by \(\sqrt{28}\sqrt{2}\), we get: \[ \frac{R_{N2}}{R_{H2}} = \frac{1 \cdot \sqrt{2}}{1 \cdot \sqrt{28}} = \frac{\sqrt{2}}{\sqrt{28}} \] Now, simplify the fraction by dividing both the numerator and the denominator by the square root of 2: \[ \frac{R_{N2}}{R_{H2}} = \frac{\sqrt{2}}{\sqrt{28}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{2}{\sqrt{56}} = \frac{1}{\sqrt{14}} \]

Interpret the results

The ratio of the rate of effusion of nitrogen to the rate of effusion of hydrogen is \(\frac{1}{\sqrt{14}}\), meaning that nitrogen effuses at a rate approximately 1/3.74 (since \(\sqrt{14} \approx 3.74\)) times the rate of hydrogen. In other words, hydrogen effuses approximately 3.74 times faster than nitrogen.