Phase Changes
In order for a substance to move between the states of matter; for example, to turn from a
solid into a liquid, which is called fusion, or from a gas to a liquid (vaporization), energy
must be gained or lost. The heat of fusion (symbolized Hfus) of a substance is the amount of
energy that must be put into the substance for it to melt. For example, the heat of fusion of
water is 6.01 kJ/mol, or in other terms, 80 cal/g. The heat of vaporization, not
surprisingly, is the amount of energy needed to cause the transition from liquid to gas, and it
is symbolized Hvap. You will not be required to memorize heat of fusion or vaporization
values for the exam.
Changes in the states of matter are often shown on phase diagrams, and you will probably see
at least one of two different types of phase diagrams on the SAT II Chemistry exam. Let’s
start with the phase diagram for water. The phase diagram for water is a graph of pressure
versus temperature. Each of the lines on the graph represents an equilibrium position, at
which the substance is present in two states at once. For example, anywhere along the line
that separates ice and water, melting and freezing are occurring simultaneously.
The intersection of all three lines is known as the triple point (represented by a dot and a T
on the figure). At this point, all three phases of matter are in equilibrium with each other.
Point X represents the critical point, and at the critical point and beyond, the substance is
forever in the vapor phase.
This diagram allows us to explain strange phenomena, such as why water boils at a lower
temperature at higher altitudes, for example. At higher altitudes, the air pressure is lower,
and this means that water can reach the boiling point at a lower temperature. Interestingly
enough, water would boil at room temperature if the pressure was low enough!
One final note: If we put a liquid into a closed container, the evaporation of the liquid will
cause an initial increase in the total pressure of the system, and then the pressure of the
system will become a constant. The value of this final pressure is unique to each liquid and is
known as the liquid’s vapor pressure. Water has a relatively low vapor pressure because it
takes a lot of energy to break the hydrogen bonds so that molecules enter the gas phase.
Water and other liquids that have low vapor pressures are said to be nonvolatile.
Substances like rubbing alcohol and gasoline, which have relatively high vapor pressures, are
said to be volatile.
Example
What happens to water when the pressure remains constant at 1 atm but the temperature
changes from -10ºC to 75ºC?
Explanation
Looking at the phase change diagram for water and following the dashed line at 1 atm, you
can see that water would begin as a solid (ice) and melt at 0ºC. All of the water would be in
liquid form by the time the temperature reached 75ºC.
The second type of phase change graph you might see on the SAT II Chemistry exam is called
a heating curve. This is a graph of the change in temperature of a substance as energy is
added in the form of heat. The pressure of the system is assumed to be held constant, at
normal pressure (1 atm). As you can see from the graph below, at normal pressure water
freezes at 0ºC and boils at 100ºC.
The plateaus on this diagram represent the points where water is being converted from one
phase to another; at these stages the temperature remains constant since all the heat energy
added is being used to break the attractions between the water molecules.
Specific Heat
On the SAT II Chemistry test, you might see a diagram that looks something like this one,
and you might come across a question that asks you to calculate the amount of energy needed
to take a particular substance through a phase change. This would be one of the most difficult
questions on the exam, but you might see something like it, or at least part of it. If you were
asked to do this, you would need to use the following equation:
energy (in calories) = mCp DT
where m = the mass of the substance (in grams)
Cp = the specific heat of the substance (in cal/g ºC)
DT = the change in temperature of the substance (in either Kelvins or ºC, but make sure all
your units are compatible!)
As you can see, this requires that you know the specific heat of the substance. A substance’s
specific heat refers to the heat required to raise the temperature of 1 g of a substance by
1ºC. You will not be required to remember any specific heat values for the exam.
Work through the example below to get a feel for how to use this equation.
Example
If you had a 10.0 g piece of ice at -10ºC, under constant pressure of 1 atm, how much energy
would be needed to melt this ice and raise the temperature to 25.0ºC?
Explanation
First, the temperature of the ice would need to be raised from -10ºC to 0ºC. This would
require the following calculation. The specific heat for ice is 0.485 cal/g ºC. Substituting in
the formula
energy = mCp DT; energy = (10.0 g) (0.485 cal/g ºC) (10.0ºC) = 48.5 cal
So 48.5 calories are needed to raise temperature.
Next, we must calculate the heat of fusion of this ice: we must determine how much energy is
needed to completely melt the 10 g of it.
energy = mHfus
energy = (10.0 g) (80 cal/g) = 800 cal
So 800 cal of energy are needed to completely melt this sample of ice.
Next, we need to see how much energy would be needed to raise the temperature of water
from 0ºC to 25ºC. The specific heat for liquid water is 1.00 cal/g ºC. So again use
energy = mCp DT to get energy = (10.0 g) (1.00 cal/g ºC) (25.0ºC) = 250 cal
Finally, add together all of the energies to get the total: 48.5 + 800 + 250 = about 1100
calories are needed to convert the ice to water at these given temperatures.
Enthalpy
Often chemical changes result in either the release or the absorption of heat, and this change
in heat in the system is measured in terms of the system’s enthalpy (H). A reaction in
which there is a net absorption of heat energy is called an endothermic reaction, and in
this type of reaction energy is a reactant, and the change in enthalpy of the system, DH, has a
positive value. A reaction in which there is a net production of heat by the system is called an
exothermic reaction. In this type of reaction, energy is a product, and the change in
enthalpy of the system, DH, has a negative value. The figure below shows an exothermic
reaction—you can see that the products have lower energy than the reactants and that the
DH of the reaction has a positive value—890 kJ.
There are several different forms of enthalpy you might encounter on the SAT II Chemistry
exam, so make sure to study the following list so nothing surprises you on test day.
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Enthalpy of reaction (DHrxn)—The amount of heat absorbed or released by the
chemical reaction
Enthalpy of combustion (DHcomb)—The amount of heat absorbed or released by
combustion (burning; usually in the presence of O2)
Enthalpy of formation (DHf)—The amount of heat absorbed or released when 1
mole of a compound is formed from elements in their standard states
Enthalpy of fusion (DHfus)—The amount of heat that must be absorbed to melt 1
mole of solid to liquid at the normal melting point
Enthalpy of vaporization (DHvap)—The amount of heat that must be absorbed to
change 1 mole of liquid to gas at the normal boiling point
Some final notes about enthalpy before we move on. First of all, enthalpy is a state
function, meaning that its value is fixed when temperature, pressure, composition, and
physical form are specified. Second, at a constant pressure, DH = q, meaning that at constant
pressure, the enthalpy of a system is equal to the heat, in joules, of a system. Finally, the
enthalpy changes of a reaction can be calculated in several ways, including by using
stoichiometry, calorimetry, tables of standard values, Hess’s law, and the bond energies of
the substances involved. Let’s now move on to determining DH values of systems using the
above methods.
Heat Capacity and Specific Heat
We can determine the value of DH for a reaction in the lab by using a calorimeter, which
measures the heat flowing into and out of a system as the reaction proceeds. We’ll talk more
about this in chapter 11, the Laboratory chapter. The heat capacity of a substance is the
amount of heat energy it must consume in order to raise its temperature by 1K or 1ºC; it can
be expressed using either the units joules or calories. A calorie is defined as the amount of
heat needed to raise the temperature of 1.00 gram of water by 1.00ºC, and joules are the SI
units for energy; 1 calorie = 4.184 joules.
Every pure substance involved in a chemical reaction has a unique heat capacity, and the
heat capacity of 1 mol of a pure substance is known as its molar heat capacity (J/mol-K or
J/mol-ºC). The heat capacity of 1 gram of a substance is known as its specific heat (J/g-K).
The following equation relates the specific heat of a substance, the temperature change, the
mass of the substance, and how much energy was put into the system:
q = mCpDT
where
q = quantity of heat (joules or calories)
m = mass in grams
DT = Tf - Ti (final – initial)
Cp = specific heat capacity (J/g ºC)
One thing about specific heats that you should probably remember for the SAT II Chemistry
test is that the specific heat of liquid water is 4.184 J/g ºC (or 1.00 cal/g ºC), which is
unusually high (this is due to hydrogen bonding).
Example
How much energy would be needed to heat 450 grams of copper metal from a temperature of
25.0ºC to a temperature of 75.0ºC? The specific heat of copper at 25.0ºC is 0.385 J/g ºC.
Explanation
Given mass, two temperatures, and a specific heat, you have enough values to plug into the
specific heat equation
q = mCpDT
and plugging in your values you get
Example
q = (450 g) (0.385 J/g ºC) (50.0ºC)
= 8700 J
100.0 mL of 1.0 M NaOH and 100.0 mL of 1.0 M HCl are mixed in a calorimeter. Both
solutions were originally at 24.6ºC. After the reaction, the final temperature is 31.3ºC.
Assuming that the solution has a density of 1.0 g/cm3 and a specific heat capacity of 4.184
J/g ºC, calculate the enthalpy change for the neutralization of HCl by NaOH. Assume that no
heat is lost to the surroundings or the calorimeter.
Explanation
We are looking for energy in this problem too. Remember that the solution has a density of
1.08/cm3, and we have a total of 200 ml of solution, so the total mass to be considered is 200
g. You have all of the information you need, so again plug it into the formula q = mCpDT:
q = (200 g) (4.184 J/g ºC) (6.7 ºC)
The answer is -5.6 kJ/mol.