General Chemistry

In the previous post, we talked about the Graham’s law which shows that the rate of effusion of a gas is inversely proportional to the square root of its molar mass:

 

\[{u_{{\rm{rms}}}}\;{\rm{ = }}\;\sqrt {\frac{{{\rm{3RT}}}}{{\rm{M}}}} \]

 

One application of Graham’s law is the formula obtained for two gases correlating their molar masses and effusion rates:

 

This allows for determining the molar mass of unknown gas by comparing its effusion rate to standard gas. 

Below are some practice examples of the Graham’s law of effusion:

 

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Practice

1.

What is the ratio of the effusion rates of hydrogen gas (H2) and carbon dioxide (CO2) at the same pressure and temperature?

answer

4.7

Solution

According to the Graham’s law of effusion:

\[\frac{{{\rm{rate}}\;{{\rm{H}}_{\rm{2}}}}}{{{\rm{rate}}\;{\rm{C}}{{\rm{O}}_{\rm{2}}}}}\;{\rm{ = }}\;\sqrt {\frac{{{{\rm{M}}_{{\rm{C}}{{\rm{O}}_{\rm{2}}}}}}}{{{{\rm{M}}_{{{\rm{H}}_{\rm{2}}}}}}}} \]

\[\frac{{{\rm{rate}}\;{{\rm{H}}_{\rm{2}}}}}{{{\rm{rate}}\;{\rm{C}}{{\rm{O}}_{\rm{2}}}}}\;{\rm{ = }}\;\sqrt {\frac{{{\rm{44}}{\rm{.0}}\;{\rm{g/mol}}}}{{{\rm{2}}{\rm{.0}}\;{\rm{ g/mol}}}}} \; = \;4.7\]

2.

The rate of effusion of an unknown gas is 9.20 mL/min. Under identical conditions, the rate of effusion of pure nitrogen (N2) gas is 14.65 mL/min. Identify the unknown gas using the Graham’s law.
a) O2 b) C3H8 c) C4H10  d) NO2  e) Cl2

answer

Cl2

Solution

\[\frac{{{\rm{rate}}\;{\rm{X}}}}{{{\rm{rate}}\;{{\rm{N}}_{\rm{2}}}}}\;{\rm{ = }}\;\sqrt {\frac{{{{\rm{M}}_{{{\rm{N}}_{\rm{2}}}}}}}{{{{\rm{M}}_{\rm{X}}}}}} \]

 

You can square both sides of the equation to get rid of the square root:

 

\[{\left( {\frac{{{\rm{rate}}\;{\rm{X}}}}{{{\rm{rate}}\;{{\rm{N}}_{\rm{2}}}}}} \right)^2}\;{\rm{ = }}\;\frac{{{{\rm{M}}_{{{\rm{N}}_{\rm{2}}}}}}}{{{{\rm{M}}_{\rm{X}}}}}\]

 

Next, rearrange the equation to get an expression for the molar of the unknown gas (Mx):

\[{{\rm{M}}_{\rm{X}}}\; = \,\frac{{{{\rm{M}}_{{{\rm{N}}_{\rm{2}}}}}}}{{{{\left( {\frac{{{\rm{rate}}\;{\rm{X}}}}{{{\rm{rate}}\;{{\rm{N}}_{\rm{2}}}}}} \right)}^2}}}\]

 

\[{{\rm{M}}_{\rm{X}}}\; = \,\frac{{{\rm{28}}{\rm{.0}}\;{\rm{g/mol}}}}{{{{\left( {\frac{{{\rm{9}}{\rm{.20}}\;{\rm{mL/min}}}}{{{\rm{14}}{\rm{.65}}\;{\rm{mL/min}}}}} \right)}^{\rm{2}}}}}\;{\rm{ = }}\;{\rm{71}}{\rm{.0}}\;{\rm{g/mol}}\]

 

The molar mass of the unknown gas is 71 g/mol, and therefore, it is chlorine.

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3.

A sample of krypton effuses from a container in 95 seconds. The same amount of an unknown gas requires 55 seconds. Identify the unknown gas.

answer

Mx = 28.1 g/mol

From common gases, this can be N2 or CO.

Solution

According to the Graham’s law of effusion:

\[\frac{{{\rm{rate}}\;{\rm{Kr}}}}{{{\rm{rate}}\;{\rm{X}}}}\;{\rm{ = }}\;\sqrt {\frac{{{{\rm{M}}_{\rm{X}}}}}{{{{\rm{M}}_{{\rm{Kr}}}}}}} \]

Pay attention, in this problem, it is not rates that are given, it is the time each gas takes to effuse. Because, rate ∼ 1/t, we can write that:

\[\frac{{{{\rm{t}}_{\rm{X}}}}}{{{{\rm{t}}_{{\rm{Kr}}}}}}\;{\rm{ = }}\;\sqrt {\frac{{{{\rm{M}}_{\rm{X}}}}}{{{{\rm{M}}_{{\rm{Kr}}}}}}} \]

\[\frac{{{\rm{55}}}}{{{\rm{95}}}}\;{\rm{ = }}\;\sqrt {\frac{{{{\rm{M}}_{\rm{X}}}}}{{{\rm{83}}{\rm{.8}}}}} \]

\[{\left( {\frac{{{\rm{55}}}}{{{\rm{95}}}}} \right)^2}\;{\rm{ = }}\;\frac{{{{\rm{M}}_{\rm{X}}}}}{{{\rm{83}}{\rm{.8}}}}\]

\[0.335\;{\rm{ = }}\;\frac{{{{\rm{M}}_{\rm{X}}}}}{{{\rm{83}}{\rm{.8}}}}\]

Mx = 28.1 g/mol

From common gases, this can be N2 or CO.

The numbers makes sense because the unknown gas is lighter and therefore, the rate of effusion is higher and it takes less time to effuse.

4.

If a sample of Br2 vapor can effuse from an opening in a heated vessel in 46 s, how long will it take the same amount of He to effuse under identical conditions?

answer

7.3 s

Solution

According to the Graham’s law of effusion:

\[\frac{{{\rm{rate}}\;{\rm{B}}{{\rm{r}}_{\rm{2}}}}}{{{\rm{rate}}\;{\rm{He}}}}\;{\rm{ = }}\;\sqrt {\frac{{{{\rm{M}}_{{\rm{He}}}}}}{{{{\rm{M}}_{{\rm{B}}{{\rm{r}}_{\rm{2}}}}}}}} \]

\[\frac{{{\rm{rate}}\;{\rm{B}}{{\rm{r}}_{\rm{2}}}}}{{{\rm{rate}}\;{\rm{He}}}}\;{\rm{ = }}\;\sqrt {\frac{{{\rm{4}}{\rm{.00}}}}{{{\rm{159}}{\rm{.8}}}}} \; = \;0.158\]

\[\frac{{\rm{1}}}{{0.158}}\;{\rm{ = }}\;6.329\;\]

This is how much faster He effuses compared to Br2. Therefore, it will take that times less to effuse:

tHe = 46 s ÷ 6.329 = 7.3 s

 

5.

It has been demonstrated that 3.56 mL of an unknown gas effuses through a hole in the same time that 8.64 mL of argon does under the same conditions. Determine the molecular mass of the unknown gas.

answer

236 g/mol

Solution

The volume of the gas is directly proportional to the rate of the effusion. The larger the volume, the higher the rate of the effusion. Given this information, we can determine the ratio of effusion rates of Ar and the unknown gas.

8.64 mL/3.56 mL = 2.43

So, Ar effuses 2.43 times faster than the unknown gas and we can use this to set up an equation for the Graham’s law:

\[\frac{{{\rm{rate}}\;{\rm{Ar}}}}{{{\rm{rate}}\;{\rm{X}}}}\;{\rm{ = }}\;\sqrt {\frac{{{{\rm{M}}_{\rm{X}}}}}{{{{\rm{M}}_{{\rm{Ar}}}}}}} \; = \;2.43\]

\[{\left( {2.43} \right)^2}\;{\rm{ = }}\;\frac{{{{\rm{M}}_{\rm{X}}}}}{{{{\rm{M}}_{{\rm{Ar}}}}}}\;\]

Mx = 5.9049 x 39.9 = 236 g/mol

 

 

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