The velocity of nitrogen is 55.6 cm/s. Determine the the rate at which hydrogen sulfide would travel under these same experimental conditions?

Question

Wyatt Atkins · Accepted Answer

#"51 cm s"^(-1)#

I assume that by velocity you mean root-mean-square speed, #v_"rms"#, which for an ideal gas kept under a temperature #T# is equal to

#color(blue)(|bar(ul(color(white)(a/a)v_"rms" = sqrt((3RT)/M_M)color(white)(a/a)|)))#

Here

#R# - the universal gas constant #M_M# - the molar mass of the gas

Now, the root-mean-square speed is usually expressed is meters per second, #"m s"^(-1)#, but you don't have to convert it because you're going to express the root-mean-square speed of hydrogen sulfide, #"H"_2"S"#, relative to that of nitrogen gas, #"N"_2#.

So, you know that both gases are kept under the same conditions, which can only mean that they have the same absolute temperature #T#.

You can thus say that

#v_("rms N"_ 2) = sqrt( (3RT)/M_("M N"_2)) -># the root-mean-squares speed for nitrogen gas

#v_("rms H"_ 2"S") = sqrt((3RT)/(M_("M H"_2"S"))) -># the root-mean-square speed for hydrogen sulfide

Divide these two equations to get

#v_("rms H"_ 2"S")/v_("rms N"_ 2) = sqrt( color(red)(cancel(color(black)(3RT)))/M_("M H"_2"S") * M_("M N"_2)/color(red)(cancel(color(black)(3RT))))#

#v_("rms H"_ 2"S")/v_("rms N"_ 2) = sqrt(M_("M N"_ 2)/M_("M H"_2"S"))#

This is equivalent to

#v_("rms H"_ 2"S") = v_("rms N"_2) * sqrt(M_("M N"_ 2)/M_("M H"_2"S"))#

The molar masses of the two gases are

#M_("M N"_2) = "28.0134 g mol"^(-1)#

#M_("M H"_2"S") = "34.089 g mol"^(-1)#

You will thus have

#v_("rms H"_2"S") = "56 cm s"^(-1) * sqrt((28.0134 color(red)(cancel(color(black)("g mol"^(-1)))))/(34.089color(red)(cancel(color(black)("g mol"^(-1))))))#

#v_("rms H"_2"S") = color(green)(|bar(ul(color(white)(a/a)color(black)("51 cm s"^(-1))color(white)(a/a)|)))#

The answer is rounded to two sig figs.

Now, does the result make sense?

The root-mean-square speed is inversely proportional to the square root of the molar mass of the gas, which means that the heavier the molar mass, the lower the root-mean-square speed for gases kept under the same absolute temperature #T#.

Olivia Anton · Answer

undefined The rate of travel for different gases under the same experimental conditions is determined by their respective molar masses. Using Graham's law of effusion, the ratio of the rates is inversely proportional to the square root of the molar masses.

$$ \text{Rate}_{\text{H2S}} = \sqrt{\frac{\text{Molar Mass}_{\text{N2}}}{\text{Molar Mass}_{\text{H2S}}} \times \text{Rate}_{\text{N2}}} $$

Given the molar mass of nitrogen (N2) is approximately 28 g/mol and the molar mass of hydrogen sulfide (H2S) is approximately 34 g/mol, the rate of hydrogen sulfide can be calculated using the provided velocity of nitrogen.