Question ba287

Jun 24, 2015

The rate of effusion for water and hydrogen sulfide is 1.375.

Explanation:

To determine the rate of effusion for a gas, you can use Graham's Law

$\text{rate} \propto \frac{1}{\sqrt{{M}_{M}}}$, where

${M}_{M}$ - the molar mass of the gas.

So, the rate of effusion of a gas is proportional to $\frac{1}{\sqrt{{M}_{M}}}$. This tells you that the heavier the gas molecules are, i.e. the bigger their molar mass is, the slower the effusion rate.

In your case, you have to determine the ratio between the effusion rate of water and the ffusion rate of hydrogen sulfide. Using Graham's Law, you can write

"ratio" = r_(H_2O)/r_(H_2S) = 1/sqrt(M_("M"H_2O)) * sqrt(M_("M"H_2S)#

${r}_{{H}_{2} O} / {r}_{{H}_{2} S} = \sqrt{{M}_{\text{M"H_2S))/sqrt(M_("M} {H}_{2} O}}$

This means that you get

${r}_{{H}_{2} O} / {r}_{{H}_{2} S} = \sqrt{34.08 \cancel{\text{g/mol"))/sqrt(18.02cancel("g/mol}}} = \sqrt{\frac{34.08}{18.02}} = \textcolor{g r e e n}{1.375}$

In other words, water has a faster effusion rate than hydrogen sulfide.