Question

question 22
4 pts
which gas will have the greatest rate of
effusion at a given temperature?
o hbr
o ar
o nh₃
o ch₄

Explanation:

Step1: Recall Graham's Law of Effusion

Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass, i.e., \( \text{Rate} \propto \frac{1}{\sqrt{M}} \), where \( M \) is the molar mass of the gas. So, the gas with the smallest molar mass will have the greatest rate of effusion.

Step2: Calculate molar masses

• For \( \text{CH}_4 \): Molar mass of \( \text{C} = 12.01 \, \text{g/mol} \), molar mass of \( \text{H} = 1.008 \, \text{g/mol} \). So, \( M_{\text{CH}_4} = 12.01 + 4\times1.008 = 16.04 \, \text{g/mol} \).
• For \( \text{NH}_3 \): Molar mass of \( \text{N} = 14.01 \, \text{g/mol} \), molar mass of \( \text{H} = 1.008 \, \text{g/mol} \). So, \( M_{\text{NH}_3} = 14.01 + 3\times1.008 = 17.03 \, \text{g/mol} \).
• For \( \text{Ar} \): Molar mass of \( \text{Ar} = 39.95 \, \text{g/mol} \) (from periodic table).
• For \( \text{HBr} \): Molar mass of \( \text{H} = 1.008 \, \text{g/mol} \), molar mass of \( \text{Br} = 79.90 \, \text{g/mol} \). So, \( M_{\text{HBr}} = 1.008 + 79.90 = 80.91 \, \text{g/mol} \).

Step3: Compare molar masses

Comparing the molar masses: \( 16.04 \, (\text{CH}_4) < 17.03 \, (\text{NH}_3) < 39.95 \, (\text{Ar}) < 80.91 \, (\text{HBr}) \). Since \( \text{CH}_4 \) has the smallest molar mass, it will have the greatest rate of effusion.

Answer:

\(\text{CH}_4\) (the option corresponding to \(\text{CH}_4\))