In the previous section, we identified two variables that affect the volume of a gas.

We learned that volume varies inversely with pressure. Boyle's Law states that under conditions of constant temperature, when the pressure increases by a certain factor, the volume decreases by the same factor. For example, if the pressure doubles, the volume decreases to ½ the original amount. The result is that the product of the pressure and volume is always a constant.

PV = k

We also learned that volume varies directly with temperature. For example, if the Kelvin temperature doubles from 300 K to 600 K, the volume of the gas will double. Charles' Law states that under conditions of constant pressure,  the ratio of volume to Kelvin temperature is a constant.

V/T = k'

We can combine these expressions to end up with the combined gas law,

PV/T = k"

While temperature and pressure can affect the volume of a gas, so can the amount or number of moles of gas. When we take moles of gas into account and combine it with Boyles and Charles Law, we end up with the Ideal Gas Law,

PV = nRT

P = Pressure (kPa or atm)
V = volume (liters)
n = moles of gas
T = Kelvin temp (K)
R = Ideal Gas Constant (8.314 l*kPa/mol*K)

or (0.08205  l*atm/mol*K)

Given any 3 of the four variables, you should be able to solve for the fourth. When using the Ideal Gas equation be sure that the units for R match the units for pressure. Also be sure that temperature is in Kelvin and volume is in liters.

Calculating Molar Mass and Density Using the Ideal Gas Equation

We can use the ideal gas equation to calculate the molar mass of a gas. Although there is no term for molar mass in the ideal gas equation, we know that n, the number of moles is equal to the mass (m) of a substance divided by its molar mass (M).

n = m/M

The ideal gas equation becomes

PV = (m/M)RT

We can solve the equation for the molar mass.

M = mRT/PV

We can also determine the density of a gas if we know the temperature, pressure and molar mass of the gas. There is no term for density in the ideal gas equation. However we know that the density is the ratio of the mass divided by the volume.

d = m/V

We can rearrange the above equation to solve for m/V or the density.

 d = MP/RT

In a mixture of gases, the total pressure is equal to the sum of the partial pressures of each gas in the mixture

P_t = pp₁ + pp₂ + ...

P_t = total pressure        pp = partial pressure

The partial pressure of a gas is equal to the product of the mole fraction (X) of that gas and the total pressure.

pp_a = X_aP_t

The mole fraction of a component in a mixture is similar to a percentage, except we don't multiply by 100. The mole fraction of a component in a mixture is determined by the following equation:

Many reactions have gases as either products or reactants. While it is difficult to measure the mass of a gas, it is fairly easy to measure the volume. If we know the volume of gas produced or consumed in a reaction, we can use the ideal gas equation to determine the number of moles of gas produced or consumed.

Stoichiometry problems with reactions involving gases are easily solved by combining the stoichiometry we learned in the unit on moles and the ideal gas equation.

Consider the following: Ca + 2 HCl --  CaCl₂ + H_2(g)

If we start with a 20.0 g of Ca, we can predict the volume of H₂ gas generated at STP, by the following steps:

• Step 1 Balanced equation Ca + 2 HCl --  CaCl₂ + H_2(g)

• Step 2: Convert mass of starting material to moles

• Step 3 Set up a Before: Change: After (BCA) table

                      Ca     +     2 HCl   -->      CaCl₂    +    H₂

Before               0.5 moles      XS                  0                  0

Change             - 0.5 moles     ?                   ?                     ?

After

• Step 4: Determine how much of the products are formed by using coefficients as guides. Then fill in BCA table.

moles HCl = 2x moles Ca

moles H2 = moles Ca

                               Ca     +     2 HCl   -->      CaCl₂    +    H₂

Before                 0.5 moles      XS                  0                  0

Change             - 0.5 moles     -1 mole       +0.5 moles    + 0.5 moles                 ?

After                       0 moles      XS-1mole      0.5 moles      0.5 moles

• Step 5: Convert moles of H2 to volume H2 using the ideal gas equation.

V = nRT/P = 0.5 moles | 8.314 lkPa | 273 K = 11.2 l H₂ gas
                                        | molK          | 101.3 kPa

One of the assumptions we make when we apply the equations and laws is that gases behave ideally. An ideal gas is a gas is a gas in which:

1) the gas particles do not interact with each other (no intermolecular forces)

2) the gas particles act as "point masses". They have no volume and can theoretically compress or contract to zero volume.

Real gases consist of particles that do have volume and can interact with each other under the right conditions. A real gas can act like an ideal gas if:

• the temperature is well above the boiling point for that gas.
• the pressure is low enough to allow for enough space between gas particles

Much work in the lab involves solutions. If we are interested in how many molecules, atoms or ions that are in a given amount of solution, we need to find a better way of expressing concentration instead of percent or grams per liter.

We define concentration as the amount of solute per volume of solution. Molarity is a unit of concentration defined as moles solute / liter solution.

Example:

29.25 g NaCl is dissolved in 500 ml of solution. Calculate the molarity.

29.25 g NaCl  | 1 mole NaCl | 1000 ml  = 1 mol NaCl  =   1 M NaCl
  500 ml           |  58.5 g NaCl | 1 liter              liter

We can think of molarity as a conversion factor that we can use to convert volume of solution to moles of solute

Example:

You have 200 ml of a 4 M NaCl solution. Determine the number of moles and the mass of NaCl.
 

200 ml | 1 liter      | 4.0 mole NaCl  = 0.8 moles NaCl
             |  1000 ml | liter
 

           0.8 moles NaCl | 58.5 g NaCl = 46.8 g NaCl
                                       | 1 mole NaCl

Chemical Reactions and Energy

In any chemical reaction, energy is either absorbed or released.

In an exothermic reaction, energy from the chemical potential account is transferred to the thermal account. Eventually energy moves into the surroundings, which become somewhat warmer than what they were. 

In an endothermic reaction, energy is absorbed and the surroundings become somewhat colder than what they were. In this case,  energy is transferred from the thermal to the chemical account. 

The demonstration we did with the Ba(OH)₂ and (NH₄)₂SCN where the beaker froze to the surface was endothermic.

If we look at any reaction we usually need to break the "old" bonds and then form "new" bonds. Bond breaking is always endothermic. Energy is required. Bond formation on the other hand is exothermic.

If we get more energy back from bond formation than what was required for bond breaking, the reaction is exothermic.

To determine whether a reaction is exothermic or endothermic, we need to calculate the DH or change in enthalpy of the reaction. In the reaction, we need to determine the difference between the DHf or enthalpy of formation of the products and reactants.

DH_rxn = SDH_{f products}- SDH_{f reactants}

Consider the reaction H₂SO₄ + 2 NaOH -- Na₂SO₄ + 2 H₂O

The DH of the reaction is calculated by  DH_rxn = SDH_{f products}- SDH_{f reactants}
  

DH_rxn = (DH_fNa2SO4 + 2DH_fH2O) -( DH_fH2SO4 +2DH_fNaOH)
    ( [-1384] + 2[-286] ) - ( [-908] + 2[-427] )

DH_rxn = -190 kJ/2mol NaOH = -95 kJ/mol NaOH

The enthalpy of formation of a substance, DH_f is how much energy is released or absorbed when 1 mole of that substance is made from its constituent elements.
 

• Substances with a negative DH_f have less enthalpy than substances with a positive DH_f. High energy molecules like glucose and gasoline have positive enthalpies of formation
• Also substances with a negative DH_f are formed by an exothermic reaction of their elements whereas substances with a positive DH_f are made by an endothermic reaction of their elements. Low energy molecules like CO₂ and water have negative enthalpies of formation.

We can combine our knowledge of moles and thermodynamics to calculate the heat released or absorbed when a specific amount of substance reacts or is formed. There are basically 4 steps to solving these problems.

1) Determine the balanced equation. H₂SO₄ + 2 NaOH -- Na₂SO₄ + 2 H₂O

2) Determine the DH of the reaction as written. Include the DH term in the equation. 

DH_rxn = -190 kJ/2mol NaOH = -95 kJ/mol NaOH

3) Determine the number of moles of limiting reactant or substance formed.

6.2 g NaOH |  1 mole NaOH   =   0.155 mole NaOH
                     |  40 g NaOH

4) Use the DH as a conversion factor to determine the amount of energy released in the reaction.

0.155 mole NaOH | -190 kJ             = - 14.7 kJ
                                | 2 mole NaOH

Resources