# 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# - themolar massof the gas

Now, the root-mean-square speed is usually expressed is *meters per second*, **relative** to that of nitrogen gas,

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 fornitrogen gas

#v_("rms H"_ 2"S") = sqrt((3RT)/(M_("M H"_2"S"))) -># theroot-mean-square speed forhydrogen 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