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Avogadro’s Law Revisited

Avogadro’s Law Revisited

Sometimes we can take advantage of a simplifying feature of the stoichiometry of gases that solids and solutions do not exhibit: All gases that show ideal behavior contain the same number of molecules in the same volume (at the same temperature and pressure). Thus, the ratios of volumes of gases involved in a chemical reaction are given by the coefficients in the equation for the reaction, provided that the gas volumes are measured at the same temperature and pressure.

We can extend Avogadro’s law (that the volume of a gas is directly proportional to the number of moles of the gas) to chemical reactions with gases: Gases combine, or react, in definite and simple proportions by volume, provided that all gas volumes are measured at the same temperature and pressure.

For example, since nitrogen and hydrogen gases react to produce ammonia gas according to \({\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\phantom{\rule{0.4em}{0ex}}⟶\phantom{\rule{0.4em}{0ex}}2{\text{NH}}_{3}\left(g\right),\) a given volume of nitrogen gas reacts with three times that volume of hydrogen gas to produce two times that volume of ammonia gas, if pressure and temperature remain constant.

The explanation for this is illustrated in the figure below. According to Avogadro’s law, equal volumes of gaseous N2, H2, and NH3, at the same temperature and pressure, contain the same number of molecules. Because one molecule of N2 reacts with three molecules of H2 to produce two molecules of NH3, the volume of H2 required is three times the volume of N2, and the volume of NH3 produced is two times the volume of N2.

This diagram provided models the chemical reaction written with formulas across the bottom of the figure. The reaction is written; N subscript 2 plus 3 H subscript 2 followed by an arrow pointing right to N H subscript 3. Just above the formulas, space-filling models are provided. Above N H subscript 2, two blue spheres are bonded. Above 3 H subscript 2, three pairs of two slightly smaller white spheres are bonded. Above N H subscript 3, two molecules are shown composed each of a central blue sphere to which three slightly smaller white spheres are bonded. Across the top of the diagram, the reaction is illustrated with balloons. To the left is a light blue balloon which is labeled “N subscript 2”. This balloon contains a single space-filling model composed of two bonded blue spheres. This balloon is followed by a plus sign, then three grey balloons which are each labeled “H subscript 2.” Each of these balloons similarly contain a single space-filling model composed of two bonded white spheres. These white spheres are slightly smaller than the blue spheres. An arrow follows which points right to two light green balloons which are each labeled “N H subscript 3.” Each light green balloon contains a space-filling model composed of a single central blue sphere to which three slightly smaller white spheres are bonded.

One volume of N2 combines with three volumes of H2 to form two volumes of NH3.

Example

Reaction of Gases

Propane, C3H8(g), is used in gas grills to provide the heat for cooking. What volume of O2(g) measured at 25 °C and 760 torr is required to react with 2.7 L of propane measured under the same conditions of temperature and pressure? Assume that the propane undergoes complete combustion.

Solution

The ratio of the volumes of C3H8 and O2 will be equal to the ratio of their coefficients in the balanced equation for the reaction:

\(\begin{array}{l}{\text{C}}_{3}{\text{H}}_{8}\left(g\right)+5{\text{O}}_{2}\left(g\right)\text{ }\text{ }⟶\text{ }\text{ }3{\text{CO}}_{2}\left(g\right)+4{\text{H}}_{2}\text{O}\left(l\right)\\ \text{1 volume}+\text{5 volumes}\phantom{\rule{3.5em}{0ex}}\text{3 volumes}+\text{4 volumes}\end{array}\)

From the equation, we see that one volume of C3H8 will react with five volumes of O2:

\(2.7\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{L}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{3}{\text{H}}_{8}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{\text{5 L}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{2}}{1\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{L}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{3}{\text{H}}_{8}}}\phantom{\rule{0.2em}{0ex}}=\text{13.5 L}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{2}\)

A volume of 13.5 L of O2 will be required to react with 2.7 L of C3H8.

Example

Volumes of Reacting Gases

Ammonia is an important fertilizer and industrial chemical. Suppose that a volume of 683 billion cubic feet of gaseous ammonia, measured at 25 °C and 1 atm, was manufactured. What volume of H2(g), measured under the same conditions, was required to prepare this amount of ammonia by reaction with N2?

\({\text{N}}_{2}\left(g\right)+3{\text{H}}_{2}\left(g\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{NH}}_{3}\left(g\right)\)

Solution

Because equal volumes of H2 and NH3 contain equal numbers of molecules and each three molecules of H2 that react produce two molecules of NH3, the ratio of the volumes of H2 and NH3 will be equal to 3:2. Two volumes of NH3, in this case in units of billion ft3, will be formed from three volumes of H2:

\(683\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{billion}\phantom{\rule{0.2em}{0ex}}{\text{ft}}^{3}\phantom{\rule{0.2em}{0ex}}{\text{NH}}_{3}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{\text{3 billion}\phantom{\rule{0.2em}{0ex}}{\text{ft}}^{3}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}}{2\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{billion}\phantom{\rule{0.2em}{0ex}}{\text{ft}}^{3}\phantom{\rule{0.2em}{0ex}}{\text{NH}}_{3}}}\phantom{\rule{0.2em}{0ex}}=1.02\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{billion}\phantom{\rule{0.2em}{0ex}}{\text{ft}}^{3}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\)

The manufacture of 683 billion ft3 of NH3 required 1020 billion ft3 of H2. (At 25 °C and 1 atm, this is the volume of a cube with an edge length of approximately 1.9 miles.)

Example

Volume of Gaseous Product

What volume of hydrogen at 27 °C and 723 torr may be prepared by the reaction of 8.88 g of gallium with an excess of hydrochloric acid?

\(2\text{Ga}\left(s\right)+6\text{HCl}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}2{\text{GaCl}}_{3}\left(aq\right)+3{\text{H}}_{2}\left(g\right)\)

Solution

To convert from the mass of gallium to the volume of H2(g), we need to do something like this:This figure shows four rectangles. The first is shaded yellow and is labeled “Mass of G a.” This rectangle is followed by an arrow pointing right to a second rectangle which is shaded pink and is labeled “Moles of G a.” This rectangle is followed by an arrow pointing right to a third rectangle which is shaded pink and is labeled “Moles of H subscript 2 ( g ).” This rectangle is followed by an arrow pointing right to a fourth rectangle which is shaded lavender and is labeled “Volume of H subscript 2 ( g ).”

The first two conversions are:

\(8.88\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{g Ga}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{1\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{mol Ga}}}{69.723\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{g Ga}}}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{\text{3 mol}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}}{2\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{mol Ga}}}\phantom{\rule{0.2em}{0ex}}=0.191\phantom{\rule{0.2em}{0ex}}{\text{mol H}}_{2}\)

Finally, we can use the ideal gas law:

\({V}_{{\text{H}}_{2}}={\left(\cfrac{nRT}{P}\right)}_{{\text{H}}_{2}}=\phantom{\rule{0.2em}{0ex}}\cfrac{0.191\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{mol}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{0.08206 L}\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{atm}\phantom{\rule{0.2em}{0ex}}{\text{mol}}^{\text{−1}}\phantom{\rule{0.2em}{0ex}}{\text{K}}^{\text{−1}}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{300 K}}{0.951\phantom{\rule{0.2em}{0ex}}\require{cancel}\cancel{\text{atm}}}\phantom{\rule{0.2em}{0ex}}=\text{4.94 L}\)

Susan Solomon

Atmospheric and climate scientist Susan Solomon (see the picture below) is the author of one of The New York Times books of the year (The Coldest March, 2001), one of Time magazine’s 100 most influential people in the world (2008), and a working group leader of the Intergovernmental Panel on Climate Change (IPCC), which was the recipient of the 2007 Nobel Peace Prize. She helped determine and explain the cause of the formation of the ozone hole over Antarctica, and has authored many important papers on climate change. She has been awarded the top scientific honors in the US and France (the National Medal of Science and the Grande Medaille, respectively), and is a member of the National Academy of Sciences, the Royal Society, the French Academy of Sciences, and the European Academy of Sciences. Formerly a professor at the University of Colorado, she is now at MIT, and continues to work at NOAA.

A photograph is shown of Susan Solomon sitting next to a globe.

Susan Solomon’s research focuses on climate change and has been instrumental in determining the cause of the ozone hole over Antarctica. (credit: National Oceanic and Atmospheric Administration)

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