Sure, let's solve this problem step-by-step using concepts from chemistry, specifically Graham's law of effusion.
1. Identify the molar masses of the gases:
- The molar mass of hydrogen sulfide [tex]\(\left( H_2S \right)\)[/tex] is 34.08 grams per mole.
- The molar mass of ammonia [tex]\(\left( NH_3 \right)\)[/tex] is 17.03 grams per mole.
2. Apply Graham's law of effusion:
- According to Graham's law, the rate of effusion of a gas is inversely proportional to the square root of its molar mass. The formula is given by:
[tex]\[
\frac{\text{rate}_1}{\text{rate}_2} = \sqrt{\frac{M_2}{M_1}}
\][/tex]
where [tex]\(\text{rate}_1\)[/tex] and [tex]\(\text{rate}_2\)[/tex] are the effusion rates of the two gases, and [tex]\(M_1\)[/tex] and [tex]\(M_2\)[/tex] are their molar masses respectively.
3. Set up the comparison of effusion rates:
- To determine which gas effuses faster, we compare the effusion rates of H[tex]\(_2\)[/tex]S and NH[tex]\(_3\)[/tex].
Taking [tex]\(H_2S\)[/tex] as gas 1 and [tex]\(NH_3\)[/tex] as gas 2:
[tex]\[
\frac{\text{rate}_{NH_3}}{\text{rate}_{H_2S}} = \sqrt{\frac{M_{H_2S}}{M_{NH_3}}}
\][/tex]
4. Calculate the ratio:
- Plugging in the molar masses:
[tex]\[
\frac{\text{rate}_{NH_3}}{\text{rate}_{H_2S}} = \sqrt{\frac{34.08}{17.03}}
\][/tex]
This ratio evaluates to approximately 1.4146.
5. Interpret the ratio:
- The ratio [tex]\(\sqrt{\frac{34.08}{17.03}} \approx 1.4146\)[/tex] tells us that [tex]\(\text{rate}_{NH_3}\)[/tex] is about 1.4146 times the [tex]\(\text{rate}_{H_2S}\)[/tex].
- Since this value is greater than 1, it indicates that ammonia ([tex]\( NH_3 \)[/tex]) effuses faster than hydrogen sulfide ([tex]\( H_2 S \)[/tex]).
6. Conclusion:
- Hence, ammonia ([tex]\(NH_3\)[/tex]) has the higher effusion rate.
Based on these calculations, the correct answer is:
[tex]\[ \boxed{NH_3} \][/tex]
