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
Explanation:
I assume that by velocity you mean root-mean-square speed,
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,
So, you know that both gases are kept under the same conditions, which can only mean that they have the same absolute temperature
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