To determine which gas has the higher effusion rate, we'll use Graham's law of effusion. Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. This means that a gas with a lower molar mass will effuse faster than a gas with a higher molar mass.
First, let's find the molar masses of each gas:
1. Calculating the molar mass of [tex]\( \text{H}_2\text{S} \)[/tex]:
- Hydrogen (H) has a molar mass of approximately 1 g/mol.
- Sulfur (S) has a molar mass of approximately 32 g/mol.
- Since [tex]\( \text{H}_2\text{S} \)[/tex] has 2 hydrogen atoms and 1 sulfur atom, its molar mass is:
[tex]\[
\text{Molar mass of } \text{H}_2\text{S} = 2 \times 1 \text{ g/mol} + 32 \text{ g/mol} = 34 \text{ g/mol}
\][/tex]
2. Calculating the molar mass of [tex]\( \text{NH}_3 \)[/tex]:
- Nitrogen (N) has a molar mass of approximately 14 g/mol.
- Hydrogen (H) has a molar mass of approximately 1 g/mol.
- Since [tex]\( \text{NH}_3 \)[/tex] has 1 nitrogen atom and 3 hydrogen atoms, its molar mass is:
[tex]\[
\text{Molar mass of } \text{NH}_3 = 14 \text{ g/mol} + 3 \times 1 \text{ g/mol} = 17 \text{ g/mol}
\][/tex]
Now, we compare the molar masses:
- Molar mass of [tex]\( \text{H}_2\text{S} = 34 \text{ g/mol} \)[/tex]
- Molar mass of [tex]\( \text{NH}_3 = 17 \text{ g/mol} \)[/tex]
Since the molar mass of [tex]\( \text{NH}_3 \)[/tex] (17 g/mol) is less than the molar mass of [tex]\( \text{H}_2\text{S} \)[/tex] (34 g/mol), according to Graham's law, [tex]\( \text{NH}_3 \)[/tex] will have a higher rate of effusion.
Therefore, the gas with the higher effusion rate is:
[tex]\[
\text{NH}_3
\][/tex]
