To determine which gas has the higher effusion rate, we can apply 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. In simpler terms, the lighter the gas (lower molar mass), the faster it effuses.
1. Determine the molar masses of the gases:
- Molar mass of [tex]\( H_2S \)[/tex] (Hydrogen Sulfide):
The atomic masses are approximately:
- Hydrogen (H): 1.01 g/mol
- Sulfur (S): 32.07 g/mol
So, the molar mass of [tex]\( H_2S \)[/tex]:
[tex]\[
\text{Molar mass } H_2S = 2 \times 1.01 + 32.07 = 34.08 \text{ g/mol}
\][/tex]
- Molar mass of [tex]\( NH_3 \)[/tex] (Ammonia):
The atomic masses are approximately:
- Nitrogen (N): 14.01 g/mol
- Hydrogen (H): 1.01 g/mol
So, the molar mass of [tex]\( NH_3 \)[/tex]:
[tex]\[
\text{Molar mass } NH_3 = 1 \times 14.01 + 3 \times 1.01 = 17.03 \text{ g/mol}
\][/tex]
2. Apply Graham's Law:
Graham's Law formula is given by:
[tex]\[
\frac{\text{Rate of effusion of gas 1}}{\text{Rate of effusion of gas 2}} = \sqrt{\frac{\text{Molar mass of gas 2}}{\text{Molar mass of gas 1}}}
\][/tex]
3. Calculate the effusion rates:
- For [tex]\( H_2S \)[/tex]:
[tex]\[
\text{Rate of effusion of } H_2S = \frac{1}{\sqrt{\text{Molar mass of } H_2S}} = \frac{1}{\sqrt{34.08}} \approx 0.171
\][/tex]
- For [tex]\( NH_3 \)[/tex]:
[tex]\[
\text{Rate of effusion of } NH_3 = \frac{1}{\sqrt{\text{Molar mass of } NH_3}} = \frac{1}{\sqrt{17.03}} \approx 0.242
\][/tex]
4. Compare the effusion rates:
From the calculated rates:
[tex]\[
0.171 \text{ for } H_2S \quad \text{ and } \quad 0.242 \text{ for } NH_3
\][/tex]
5. Conclusion:
Since [tex]\( 0.242 > 0.171 \)[/tex], ammonia [tex]\( (NH_3) \)[/tex] has a higher effusion rate than hydrogen sulfide [tex]\( (H_2S) \)[/tex].
Therefore, the gas with the higher effusion rate is [tex]\( NH_3 \)[/tex].
