CHM 101 GENERAL CHEMISTRY

FALL QUARTER 2008

Section 2

 

Lecture Notes – 11/5/2008

(last revised: 11/5/08, 4:00 PM)

 

 

5.4          Gas Stoichiometry: Now that we have the ideal gas law in our toolkit, we are equipped to handle problems in stoichiometry that involve measurements of the temperatures, pressures, volumes, and numbers of moles of gaseous reactants and products.

  • Sample Exercise 5.11 (p. 191): A sample of nitrogen gas (N2) has a volume of 1.75 L at STP (0 °C & 1 atm). How many moles of N2 do we have?

o       We could use the ideal gas law to solve this problem, and that would be a perfectly correct approach. But we could also start from our knowledge that one mole of gas has a volume of 22.42 L at STP. Here we use the second approach.

 

o       (If you don’t recall the figure, 22.42 L, you can always use the ideal gas law directly for a problem like this.)

  • Sample Exercise 5.12 (pp. 191-2): Quicklime (CaO) is produced by heating calcium carbonate (CaCO3) to drive off carbon dioxide (CO2). The reaction is written:

CaCO3 (s) ——> CaO (s) + CO2 (g)

Calculate the volume of CO2 at STP produced by the reaction (decomposition) of 152 g of CaCO3.

o       The equation clearly indicates that one mole of CO2 is generated for each mole of CaCO3 that decomposes. We can thus calculate the number of moles of CO2 we produce from the mass and molar mass of CaCO3 that decomposes:

 

o       Now we can convert 1.52 moles of CO2 to volume at STP. This time we’ll use the ideal gas law, even though we could work from a molar volume of 22.42 L.

 

  • Sample Exercise 5.13 (pp. 192-3): A sample of methane gas (CH4) with a volume of 2.80 L at 25 °C and 1.65 atm is mixed with a sample of oxygen gas with a volume of 35.0 L at 31 °C and 1.25 atm. The mixture is ignited to produce carbon dioxide (CO2) and water (H2O). Calculate the volume of carbon dioxide (CO2) that forms at a pressure of 2.50 atm and a temperature of 125 °C.

o       This is a limiting reactant problem where all the data are values of P, V, and T, and where none of the conditions match STP. We must use the ideal gas law all the way. We start by writing (and balancing) the equation for the reaction:

CH4 (g) + 2O2 (g) ——> CO2 (g) + 2H2O (g)

o       The number of moles of CH4 is:

 

o       The number of moles of O2 is:

 

o       Since 0.378 moles of O2 will react with the entire 0.189 moles of CH4, we see that CH4 is the limiting reactant, and there will be 0.189 moles of CO2 produced. We finish by calculating the volume of CO2 when it is measured at 2.50 atm and 125 °C:

 

5.5          Dalton’s Law of Partial Pressures: You will recall that we ”derived” Dalton’s law last time from the postulates of the kinetic molecular theory (KMT) of gases. We argued that since individual gas particles do not interact with each other, the particles of different kinds of gases would behave as if the other gas were not there. Hence each kind of gas would exert its own pressure on the walls of the container, and the total pressure on the container would be the sum of the individual pressures.

Dalton had no knowledge of the KMT. He obtained Dalton’s law from his careful experimentation on mixtures of gases.

  • The Law: Dalton’s law can be stated: For a mixture of gases in a container, the total pressure exerted is the sum of the pressures each gas would exert if it were alone in the container. Mathematically, the law can be expressed:

Here P1 stands for the partial pressure of gas 1, P2 stands for the partial pressure of gas 2 and so on, and Ptotal is the total pressure of the mixture. We can calculate these partial pressures from the ideal gas law:

,   ,   ,   …

 

Then the total pressure of the mixture is:

 

Here, ntotal is the sum of the numbers of moles of the individual gases. This result indicates that the total pressure of a mixture of gases depends on the total numbers of moles of gas particles but not on the identities of the gases. This is illustrated in Figure 5.12:

 

  • Sample Exercise 5.15 (p. 195): A 5.0 L scuba tank is charged with 46 L of helium gas (He) and 12 L of oxygen gas (O2), both measured at 25 °C and 1.0 atm. Calculate the partial pressure of each gas and the total pressure in the tank at 25 °C.

First we calculate the number of moles of each gas:

 

 

Next, we calculate the partial pressures of each gas in the 5.0 L tank (at 25 °C):

 

 

We finish by adding the two partial pressures to obtain the total pressure:

Notice that we could have avoided a bunch of arithmetic. Since the numbers of moles of the two gases remain constant, and the temperature remains at 25 °C, we could have applied Boyle’s law:

 

The superscript, 0, refers to the starting conditions, the superscript, 1, refers to the conditions in the scuba tank. Here are the direct calculations of the partial pressures:

 

 

  • Collecting a Gas Over Water: This is an important procedure you will probably apply on numerous occasions in chemistry lab. If you run a reaction that generates a gas that you wish to collect, you can pipe it through a collection tube and into a collection vessel, an inverted container initially filled with water, as illustrated in Figure 5.13, where the mixture of potassium chlorate (KClO3) and manganese (IV) oxide (MnO2) in the test tube will generate oxygen gas (O2) when heated:

Before the reaction starts to run, the collection vessel remains full of water, held in place by the pressure of the atmosphere bearing down on the surface of the water in the large pan.

When heat is applied to the test tube, the KClO3 decomposes to produce O2, which flows through the glass pipe and into the collection vessel, where it displaces liquid water and creates a vapor space.

Eventually, when all the KClO3 has decomposed and no more O2 is being produced, the volume of the vapor space in the collection vessel can be measured. Before this measurement can be used in a stoichiometric calculation, one major correction needs to be applied. One must compensate for the water vapor present in the vapor space and in equilibrium with the liquid water in the system.

Where does the water vapor come from? It comes from evaporation of liquid water in the system. This water will also condense out of the vapor space at a rate proportional to the partial pressure of water vapor in the vapor space. When the rate of condensation matches the rate of evaporation, an equilibrium is established, and the vapor pressure of water in the vapor space remains at a steady value, called the equilibrium vapor pressure. Extensive tables of equilibrium vapor pressures versus temperature have been compiled, and one can look up the value that needs to be applied in order to correct the measured volume for the volume of water vapor it contains.

One also needs to measure the pressure in the vapor space, in order to use the volume for any gas law calculations. Notice, however, that the apparatus in effect is a manometer, and if one adjusts the position of the collection vessel so that the water level inside it matches the water level in the large pan, then the pressure in the vapor space is the same as the measured barometric pressure.

Another factor needs to be considered. some of the gas in the vapor space of the collection vessel is actually air that was present in the test tube and the collection pipe before the start of the reaction. However, this volume of air is equal to the volume of oxygen from the reaction that now remains behind in the test tube and collection pipe. Thus there is no need to correct for it.

  • Sample Exercise 5.18 (pp. 198-9): Now let us apply the foregoing discussion to a stoichiometric problem.

A sample of potassium chlorate (KClO3) is heated in an apparatus like Figure 5.13. It decomposes according to the following reaction:

2KClO3 (s) ——> 2KCl (s) + 3O2 (g)

(The MnO2 shown in the diagram acts as a catalyst, but does not take part in the net reaction.) The oxygen was displaced over water at 22 °C at a total pressure of 754 torr. The volume of gas collected was 0.650 L, and the vapor pressure of water at 22 °C is known to be 21 torr. Calculate the partial pressure of O2 in the collected gas, and calculate the mass of KClO3 that was decomposed.

o       First, we calculate the partial pressure of O2, using Dalton’s law:

 

 

o       Next, we find the number of moles of O2 by using the ideal gas law:

 

o       Next, we consider the stoichiometry of the reaction to calculate the number of moles of KClO3:

 

 

o       Finally, we convert the number of moles of KClO3 to mass:

  • The Mole Fraction: We define the mole fraction as the ratio of the number of moles of a given component in a mixture to the total number of moles of all components in the mixture. We use the Greek letter, chi (χ), as a symbol for mole fraction. Thus, the mole fraction of component 1 in a mixture is:

 

Mole fractions can be defined for all components of any homogeneous mixture, whether the state of the mixture is a solid, liquid or gas. In the special case of a mixture of ideal gases (where Dalton’s law is true) we can write:

 

 

 

  • Sample Exercise 5.16 (p. 197): The partial pressure of oxygen (O2) is found to be 156 torr in an atmosphere with a total pressure of 743 torr. What is the mole fraction of the oxygen in this atmosphere?

o       We can use the equation:

 

o       All we need to do is plug in our data:

 

o       Notice that the mole fraction does not have any units; it is dimensionless. Also notice that we can use any convenient pressure unit, so long as the partial and the total pressure are expressed in the same unit of pressure.

  • Sample Exercise 5.17 (p. 198): The mole fraction of nitrogen (N2) in air is 0.7808. What is the partial pressure of  in air with an atmospheric pressure of 760. torr?

o       We can use the equation:

 

o       Only this time we need to rearrange it to:

o       Now we can plug in our data:

5.10     Chemistry in the Atmosphere

  • Composition of the Atmosphere: The composition of the atmosphere (actually dry, clean air near sea level) is shown in Table 5.4 from the text.

Nitrogen is the most abundant component, followed by oxygen. But a gas does not need to be very abundant to be important. Consider carbon dioxide. Its mole fraction seems negligible, but it is an absolutely necessary reactant for photosynthesis. On the other hand, its concentration in the atmosphere has roughly doubled since the beginning of the Industrial Revolution, and most scientists consider this increase to be a major cause of global warming.

  • Photochemical Smog: Clean air doesn’t necessarily stay that way. Our industrial civilization dumps significant quantities of reactive chemicals into our atmosphere. For example, internal combustion engines convert fuel into mechanical energy, but the combustion process also generates oxides of nitrogen and leaves unburned residues of fuel, both of which end up released into the atmosphere where they undergo a series of chemical transformations to produce what is collectively known as smog. See Figure 5.31.

  • Electricity from Coal: The most abundant fossil fuel we have in the USA is coal, and much of our electric power is generated in coal-burning power plants. The primary component of coal is carbon, but coal always contains various amounts of contaminants. (I have heard it said that every element in the periodic table can be found in coal. I have also heard that a typical coal-fired power plant releases more radioactivity than a nuclear generating plant of the same power generating capacity.)

One coal contaminant of special concern is sulfur. The sulfur in coal burns to form sulfur dioxide:

S (s, in coal) + O2 (g) ——> SO2 (g)

And the sulfur dioxide is released to the atmosphere along with carbon dioxide.

  • Acid Rain: We have just seen that internal combustion engines tend to produce oxides of nitrogen (sometimes abbreviated NOx) and that sulfur dioxide is a by-product of power generation from coal. But where does acid rain come from? And how do NOx and SO2 fit in?

In fact, rainwater is naturally acidic, but its acidity is very small. The source is atmospheric CO2. When raindrops form inside clouds, CO2 from the air tends to dissolve in them and ionize:

H2O (l) + CO2 (g) ——> H+ (aq) + HCO3— (aq)

But, the acidity of acid rain is very small, since carbonic acid (aqueous carbon dioxide) is a very weak acid.

If, however, raindrops form in air that has been contaminated with oxides of sulfur and nitrogen, the acidity of the rain is much higher. Nitrogen dioxide (a principal component of NOx) reacts in water to form nitrous and nitric acids:

2NO2 (g) + H2O (l) ——> HNO2 (aq) + HNO3 (aq)

While nitrous acid is weak, albeit much stronger than carbonic acid, nitric acid is one the strong mineral acids that ionizes completely in aqueous solution.

Sulfur dioxide will dissolve in water to produce sulfurous acid, a weak acid, but like nitrous acid, much stronger than carbonic acid:

SO2 (g) + H2O (l) ——> H2SO3 (aq)

However, airborne sulfur dioxide can and will react with atmospheric oxygen to produce sulfur trioxide:

2SO2 (g) + O2 (g) ——> 2SO3 (g)

And when sulfur trioxide reacts with water, it generates sulfuric acid, another strong, mineral acid that totally ionizes in water:

SO3 (g) + H2O (l) ——> H2SO4 (aq) ——> H+ (aq) + HSO4— (aq)

  • Some Consequences of Acid Rain: When acid rain falls to earth, it can cause considerable damage.

o       It can acidify lakes to the point where some forms of aquatic life can no longer survive.

o       It can neutralize the lime content of soil so as to interfere with plant growth.

o       An especially visible consequence of acid rain is the destruction of outdoor marble statues and the exteriors of limestone buildings. The chemistry is simple: marble and limestone are forms of calcium carbonate, which can react with sulfuric acid to form calcium sulfate:

CaCO3 (s) + H2SO4 (aq) ——> Ca2+ (aq) + SO42— (aq) + H2O (l) + CO2 (g)

Calcium sulfate is a more-or-less water soluble salt that can be washed away by rainwater. If it is protected from excessive amounts of water, it will crystallize as gypsum:

Ca2+ (aq) + SO42— (aq) + 2H2O (l) ——> CaSO4 · 2 H2O (s)

The two waters in the formula for gypsum are called waters of hydration. Gypsum is much softer than marble or limestone, and it easily binds soot and dust. The consequence is that limestone and marble become eroded and discolored when attacked by acid rain.