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

1 Answer
Jun 17, 2016

Answer:

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

Explanation:

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#.