Chemistry
Energy Unit
Chemistry
Learning Objectives Energy
Essential knowledge and skills:
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Graph and interpret a heating curve (temperature vs. time).
Interpret a phase diagram of water.
Calculate energy changes, using molar heat of fusion and molar heat of vaporization.
Calculate energy changes, using specific heat capacity.
Distinguish between an endothermic and exothermic process.
Essential understandings:
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Vapor pressure is the pressure of the vapor found directly above a liquid in a closed container. When the vapor pressure equals
the atmospheric pressure, a liquid boils. Volatile liquids have high vapor pressures, weak intermolecular forces, and low boiling
points. Nonvolatile liquids have low vapor pressures, strong intermolecular forces, and high boiling points.
Solid, liquid, and gas phases of a substance have different energy content. Pressure, temperature, and volume changes can
cause a change in physical state. Specific amounts of energy are absorbed or released during phase changes.
A fourth phase of matter is plasma. Plasma is formed when a gas is heated to a temperature at which its electrons dissociate
from the nuclei.
A heating curve graphically describes the relationship between temperature and energy (heat). It can be used to identify a
substance’s phase of matter at a given temperature as well as the temperature(s) at which it changes phase. It also shows the
strength of the intermolecular forces present in a substance.
Molar heat of fusion is a property that describes the amount of energy needed to convert one mole of a substance between its
solid and liquid states. Molar heat of vaporization is a property that describes the amount of energy needed to convert one mole
of a substance between its liquid and gas states. Specific heat capacity is a property of a substance that tells the amount of
energy needed to raise one gram of a substance by one degree Celsius. The values of these properties are related to the strength
of their intermolecular forces.
Chemical reactions are exothermic reactions (heat producing) and endothermic reactions (heat absorbing).
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Exothermic reactions– Reactions that release/lose energy (examples: burning, a beaker getting warm)
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the products have less energy than the initial reactants
The ∆ H rxn is ALWAYS negative
Ex: 4Fe + 3O2  2 Fe2O3 ∆H rxn = -1625 kJ
Endothermic reactions – Reactions that gain/absorb energy (ex: cooking, decomposition of water)
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The products have more energy than the initial reactants.
The ∆ H rxn is ALWAYS positive for endothermic reactions!
Ex: NH4NO3 (s)  NH4+ (aq) + NO3-(aq) ∆H rxn = +27 kJ
Entropy (S) – a measure of the disorder or randomness of the particles that make up a system
Law of Disorder – Spontaneous processes ALWAYS proceed in a way that the entropy of the universe increases
(increase disorder)
Entropy changes associated with state of matter:
Solids – molecules are tightly packed and cannot move (more order)
Liquids – molecules have some freedom to move (some order)
Gas – unrestricted movements (very little order)
Entropy increases as a substance changes from solidliquidgas
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Ex: As water (l) vaporizes (g) – entropy increases
Ex: As a metal (s) melts to liquid (l) – entropy increases
WORKSHEET
REACTION ENERGY
1. Fill in the blanks on the reaction coordinate
diagram with the appropriate letters.
Not all letters will be used
A.
B.
C.
D.
reactants
products
energy released
energy absorbed
Specify whether each reaction is exothermic (EXO) or endothermic (ENDO).
2. ________ The reaction shown in the diagram to the right.
3. ________ The burning of wood to produce a hot flame.
4. ________ 4Fe(s) + 3O2(g)  2Fe2O3(s) + energy
5. ________ A test tube that feels cold to the touch after two substances have been mixed.
Worksheet 2
1.
Is the overall reaction as shown exothermic or endothermic?
2.
What is the activation energy for the forward reaction?
3.
What is the activation energy for the reverse reaction?
4.
What is the enthalpy change of reaction (∆H) for the forward reaction?
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5.
What is the ∆H for the reverse reaction?
6.
Is the reverse reaction exothermic or endothermic?
7.
Which species or set of species has the highest potential energy?
8. What is ∆H for the reaction:
X2Y2  X2 + Y2 ?
9. Which do you think would be faster, the forward reaction or the reverse reaction?
10. Show the ∆H, the Activation Energy for the forward reaction and the Activation Energy
for the reverse reaction on the graph above.
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1. The graph below is a potential energy diagram
for the hypothetical reaction:
Is the forward reaction endothermic
or exothermic? Calculate the value
of ΔH for this reaction.
A + B  C + D
Is the reverse reaction endothermic
or exothermic? Calculate the value
of ΔH for this reaction.
What is the value of the potential
energy of the activated complex?
Calculate the activation energy for
the forward reaction.
Draw a dashed line to show the
effect of adding a catalyst to the
system
2.
On the graph below, draw a potential
energy diagram for the following
reaction:
Q + R  S + T
given the following information:
the potential energy of Q + R is 150 kJ
the potential energy of S + T is 250 kJ
the potential energy of the activated complex is
375 kJ
Is the forward reaction endothermic
or exothermic? Calculate the value
of ΔH for this reaction.
Calculate the activation energy for
the forward reaction.
Draw a dashed line to show the
effect of adding a catalyst to the
system.
What is the total potential energy
content of the activated complex?
What is the activation energy of the
reverse reaction?
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Phase Diagrams
A phase diagram is a graphical way to depict the effects of pressure and temperature on the phase of a
substance: whether it’s a solid, liquid, or gas
The CURVES indicate that BOTH phases exist on these lines:
Melting/Freezing: Any point on this
line (pressure & temperature) the
substance is both solid and liquid
Sublimation/Deposition: Any point
on this line (pressure & temperature)
the substance is both solid and gas
Vaporization/Condensation: Any
point on this line (pressure &
temperature) the
substance is both liquid and gas.
NOTE: the vapor pressure curve
ends at the critical point, the
temperature above which the gas cannot be liquefied. This critical point is called a supercritical fluid –
indistinguishable between gas or liquid (neither one)
The TRIPLE POINT is the condition of temperature and pressure where ALL THREE phases exist in
equilibrium (solid, liquid, gas)
The phase diagram for water is different!
Phase diagram Worksheet 1
Use the following phase diagram for problems (1) – (7)
Critical point
225 atm
PRESSURE
1 atm
Water (liquid)
Ice
(solid)
Water Vapor (gas)
0.006 atm
Triple
point
0.01 °C
100°C 374 °C
TEMPERATURE
(1)
List the phase changes a sample of ice would go through if heated to its critical
temperature at one atmosphere pressure.
(2)
At what range of pressure will water be a liquid at temperatures above its normal
boiling point?
(3)
In what phase does water exist at its triple point?
(4)
How does the melting point of water change as the pressure increases from 1 atm?
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(5)
What does the line separating the solid phase from the gas phase represent?
(6)
What does the line separating the liquid phase from the gas phase represent?
(7)
What is the vapor pressure of liquid water at 100 oC?
Phase Diagram Worksheet 2
Refer to the phase diagram below when answering the questions on this worksheet:
NOTE: “Normal” refers to STP – Standard Temperature and Pressure.
A phase diagram is a graph of physical state versus temperature and pressure. The lines dividing the
physical state represent conditions where a change of physical state occurs. At a point on the line, both
physical states (the ones on either side of the line) exist.
FIRST: Label the following on the diagram below: solid, liquid, vapor (gas), critical point, triple point,
solid-gas line, liquid-gas line, liquid-solid line, using arrows, indicate the direction of melting, freezing,
boiling, condensing, sublimation, deposition.
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1) What are the values for temperature and pressure at STP? T= ________, P= ________
2) What is the normal freezing point of this substance? ________
3) What is the normal boiling point of this substance? ________
4) What is the normal melting point of this substance? ________
5) What is the phase (s, l, g) of a substance at 2.0 atm and 100 °C? _________
6) What is the phase (s, l, g) of a substance at 0.75 atm and 100 °C? _________
7) What is the phase (s, l, g) of a substance at 0.5 atm and 100 °C? _________
8) What is the phase (s, l, g) of a substance at 1.5 atm and 50 °C? _________
9) What is the phase (s, l, g) of a substance at 1.5 atm and 200 °C? _________
10) What is the phase (s, l, g) of a substance at 1.5 atm and 800 °C? _________
11) What is the temperature and pressure at the triple point? T= ________, P= _______
12) If a quantity of this substance was at an initial pressure of 1.25 atm and a temperature of 300ºC was
lowered to a pressure of 0.25 atm, what phase transition(s) would occur? ____________
13) If a quantity of this substance was at an initial pressure of 1.25 atm and a temperature of 0ºC was
lowered to a pressure of 0.25 atm, what phase transition(s) would occur? ____________
14) If a quantity of this substance was at an initial pressure of 1.0 atm and a temperature of 200ºC was
lowered to a temperature of -200ºC, what phase transition(s) would occur? ____________
15) If a quantity of this substance was at an initial pressure of 0.5 atm and a temperature of 200ºC was
lowered to a temperature of -200ºC, what phase transition(s) would occur? ____________
16) If this substance was at a pressure of 2.0 atm, at what temperature would it melt? ____________
17) If this substance was at a pressure of 2.0 atm, at what temperature would it boil? ___________
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18) If this substance was at a pressure of 0.75 atm, at what temperature would it melt? ___________
19) If this substance was at a pressure of 0.75 atm, at what temperature would it boil? ___________
20) At what temperature do the gas and liquid phases become indistinguishable from each other? ______
21) At what pressure would it be possible to find this substance in the gas, liquid, and solid phase? _____
22) If I had a quantity of this substance at a pressure of 1.00 atm and a temperature of -100ºC, what phase
change(s) would occur if I increased the temperature to 600ºC and the pressure stayed constant? At
what temperature(s) would they occur?
22) If I had a quantity of this substance at a pressure of 2.00 atm and a temperature of -150ºC, what phase
change(s) would occur if I decreased the pressure to 0.25 atm? At what pressure(s) would they
occur?
23) Describe how the melting point of the substance changes with the external pressure. Based on the
slope of the melting-point curve in the phase diagram, would you characterize the solid phase of the
substance as more dense or less dense than the liquid phase?
24) Explain the significance of the triple point?
Specific Heat Capacity (Cp)
Definition: The amount of heat required to raise the temperature of one gram of a substance 1°C.
Formula for heat capacity:
Mass of a substance (g or kg)
Q = m c ∆T
Specific heat capacity
(J/g°C) This value will be
given to you!
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“Delta T” – the change in
temperature (final – initial)
Example: How much energy is required to raise the temperature of 15.0 g of water 14.0°C?
Q = mc∆T
m = 15.0 g
∆T = 14.0°C
c = 4.18 J/g°C
* the mass of the substance and the change in temperature is provided right in the question. The
specific heat capacity (c) is provided in the table above.
Therefore:
Q = (15.0g) (4.18 J/g°C) ( 14.0°C)
Q = 8780 J (3 sf)
Specific Heat Worksheet 1
Q = mc∆T, where Q = heat energy, m = mass, and T = temperature Remember, ∆T = (Tfinal – Tinitial).
Show all work and proper units. Answers are provided at the end of the worksheet without units.
1. A 15.75-g piece of iron absorbs 1086.75 joules of heat energy, and its temperature changes from
25.0°C to 175.0°C. Calculate the specific heat capacity of iron.
2. How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22.0°C to
55.0°C, if the specific heat of aluminum is 0.90 J/g°C?
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3. To what temperature will a 50.0 g piece of glass raise if it absorbs 5275 joules of heat and its specific
heat capacity is 0.500 J/g°C? The initial temperature of the glass is 20.0°C.
4. Calculate the heat capacity of a piece of wood if 1500.0 g of the wood absorbs 6.750×104 joules of
heat, and its temperature changes from 32.0°C to 57.0°C.
5. 100.0 mL of 4.0°C water is heated until its temperature is 37.0°C. If the specific heat of water is 4.18
J/g°C, calculate the amount of heat energy needed to cause this rise in temperature.
6. 25.0 g of mercury is heated from 25.0°C to 155.0°C, and absorbs 466.0 joules of heat in the process.
Calculate the specific heat capacity of mercury.
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7. What is the specific heat capacity of silver metal if 55.0 g of the metal absorbs 47.3 calories of heat
and the temperature rises 15.0°C? (1 calorie = 4.18 J)
8. If a sample of chloroform is initially at 25°C, what is its final temperature if 150.0 g of chloroform
absorbs 1.0 kilojoules of heat, and the specific heat of chloroform is 0.96 J/g°C?
Answers in random order without units:
0.14; 1.8; 32; 3.0102; 0.4600; 1.38104; 0.240; 231
Specific Heat Worksheet 2
1. What is the specific heat of a substance that absorbs 2.50 x 103 joules of heat when a sample of 1.0
x 104 g of the substance increases in temperature from 10.0°C to 70.0°C?
2. How many grams of water would require 2.20 x 104 joules of heat to raise its temperature from
34.0°C to 100.0°C? The specific heat of water is 4.18 J/g∙°C
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3. If 200. grams of water is to be heated from 24.0°C to 100.0°C to make a cup of tea, how much
heat must be added? The specific heat of water is 4.18 J/g∙°C
4. A block of aluminum weighing 140. g is cooled from 98.4°C to 62.2°C with the release of 1080
joules of heat. From this data, calculate the specific heat of aluminium.
5. A cube of gold weighing 192.4g is heated from 30.0°C to some higher temperature, with the
absorption of 226 joules of heat. The specific heat of gold is 0.030 J/g∙°C. What was the final
temperature of the gold?
6. A total of 54.0 joules of heat are absorbed as 58.3 g of lead is heated from 12.0°C to 42.0°C.
From these data, what is the specific heat of lead?
7. The specific heat of wood is 2.03 J/g∙°C. How much heat is needed to convert 550. g of wood at 15.0°C to 10.0°C?
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8. What is the total amount of heat needed to change 2.25 kg of silver at 0.0°C to 200.0°C? The
specific heat of silver is 0.129 J/g∙°C
9. Granite has a specific heat of 0.790 J/g∙°C. What mass of granite is needed to store 1.50 x 106 J
of heat if the temperature of the granite is to be increased by 15.5°C?
10. A 55.0 kg block of granite has an original temperature of 15.0°C. What will be the final
temperature of this granite if 4.5 x 104 kJ of heat energy are added to the granite?
Phase Changes
Condensed states: solid and liquid phases where particles are in contact with one another
– In solids and liquids, molecules are attracted to one another.
→These attractions are called intermolecular forces.
Energetics of Melting and Evaporation
– To undergo a change in physical state (e.g. solid →liquid or liquid →gas), particles in the
solid or liquid must have enough thermal energy (associated with motion) to overcome the
attractive forces it experiences with surrounding molecules.
– Depending on the liquid or solid, the attractive forces may be intermolecular forces, or
they may be chemical bonds (e.g. ionic bonds, covalent bonds, or metallic bonds).
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EVAPORATION AND CONDENSATION
vaporization: liquid → gas
– From a molecular viewpoint, a molecule “escaping” from the liquid to gaseous state
As the liquid evaporates, more molecules go into the gas phase.
vapor: The gas above a liquid when the liquid and gaseous states are both present
vaporization: liquid + heat → vapor
condensation: vapor → liquid + heat
Liquid-Gas Equilibrium:
liquid + heat
vapor
When the molecules in the liquid have enough energy, they escape to the gas phase.
– In a closed system, when enough vapor exists above the liquid, some
gaseous molecules condense back to the liquid.
– Ultimately, the rate of vaporization = the rate of condensation.
→The system has reached a state of dynamic equilibrium in which the forward
process occurs at the same rate as the reverse process.
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In an open system, molecules in the liquid have enough energy to escape to the gas
phase and continue to escape in a process called evaporation.
– The vaporized molecules continue to escape → little or no condensation occurs.
→Ultimately, all of the liquid is converted into a gas.
– Since vaporization requires energy, the liquid molecules take energy from
the surroundings, so the temperature of the surroundings may decrease.
→Why evaporation is a “cooling process” and is used to reduce body temperature.
Evaporation and Condensation
Vapor Pressure (v.p.): pressure exerted by gas molecules above a liquid
– For a molecule to go from liquid to gas, it has to break the intermolecular forces with
other liquid molecules around it
→Weaker the intermolecular forces are easier to break
→more gas molecules →higher vapor pressure
→Stronger the intermolecular forces are harder to break
→fewer molecules go from liquid to gas →lower vapor pressure.
Example: Consider the two closed systems above. In which container, (a) or (b), does
the liquid have stronger intermolecular forces? Explain why.
Boiling Point:
temperature where vapor pressure of liquid is equal to the external
pressure (usually atmospheric pressure)
For a liquid to boil, its vapor pressure must equal the atmospheric pressure.
– Note that atmospheric pressure is the pressure exerted by air molecules (i.e., the
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gas molecules in the atmosphere).
→Since intermolecular forces influence vapor pressure, they also influence boiling point.
→Weaker the intermolecular forces →more gas molecules →higher vapor pressure
→Less energy is needed to get vapor P = atmospheric P →lower boiling point
→Stronger the intermolecular forces →fewer gas molecules →lower vapor pressure
→More energy is needed to get vapor pressure = atmospheric pressure
→higher boiling point
– normal boiling point (b.p.) is the b.p. when atmospheric pressure is 1 atm (760
mmHg)
– e.g. Water boils at 100°C at sea level (where atmospheric pressure is ~1 atm) but at
~95°C in Denver where atmospheric pressure is much lower (~0.85 atm).
Volatile and Nonvolatile Liquids
volatile: liquids that evaporate easily at a given temperature
– Rubbing alcohol is more volatile than water at room temperature.
nonvolatile:
–
liquids that do not evaporate easily
Motor oil is virtually nonvolatile at room temperature.
Ex. 1: Acetone is the active ingredient in nail polish remover. Acetone is a polar molecule
that experiences dipole-dipole forces.
a. Would acetone’s vapor pressure be higher or lower than water’s at
room temperature? Explain why.
b. Would acetone be more volatile or less volatile than water? Explain why.
c. The intermolecular forces in acetone are stronger/weaker than water’s.
d. Acetone’s boiling point would be higher/lower than water’s.
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Relationship between boiling point, enthalpy of vaporisation, intermolecular forces and
vapour pressure
(a) What do you think enthalpy of vaporisation is?
(b) What would be the sign of enthalpy of vaporisation?
(c) What do you think the relationship is between enthalpy of vaporisation and vapour pressure?
Explain that relationship.
(d) Use this link to read about how vapour pressure and boiling are linked. When does a liquid
boil? Include vapour pressure in your answer.
http://www.chem.purdue.edu/gchelp/liquids/boil2.html
(e) Use the table below to investigate the relationship between vapour pressure and boiling
point? Explain that relationship.
compound
CH4
C2H4
C3H8
C4H10
C6H14
C8H18
C10H22
F2
Cl2
Br2
Hvap (kJ/mol)
9.2
14
18.1
22.3
28.6
33.9
35.8
6.52
20.4
30.7
Boiling point (°C)
-161
-89
-30
0
68
125
160
-188
-34.6
59
compound
HF
HCl
HBr
HI
H2O
H2S
NH3
PH3
SiH3
Hvap (kJ/mol)
30.2
15.1
16.3
18.2
40.6
18.8
23.6
14.6
12.3
Boiling point (°C)
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-84
-70
-37
100
-61
-33
-88
-112
(f) Study the graph below which shows the relationship between vapour pressure and
temperature for three different liquids. Use the information in the graph only to explain why
pentane has a lower boiling point.
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(g) In one sentence, describe the relationship between boiling point, vapour pressure, enthalpy of
vaporisation and intermolecular forces.
Latent heat
If you have a glass of a cool drink, well supplied with ice, you can expect its temperature to drop
until it is close to 0 ºC. You also can expect (and can easily check with a thermometer) that it will
remain cold, regardless of the outside temperature, as long as there remains some unmelted ice in
the drink. Only after all the ice has melted will the temperature of the drink begin to rise. Why is
this?
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When a solid substance changes from the solid phase to the liquid phase, energy must be supplied in order
to overcome the molecular attractions between the constituent particles of the solid. This energy must be
supplied externally, normally as heat, and does not bring about a change in temperature. We call this
energy latent heat (the word "latent" means "invisible"). The latent heat is the energy released or
absorbed during a change of state. With this in mind, we define the specific latent heat of fusion:
"The latent heat of fusion of a substance is the amount of heat required to convert unit mass of the solid
into the liquid without a change in temperature."
Latent heat of vaporisation:
A change of state from liquid to vapour at constant temperature also requires the input of energy, called
the latent heat of vaporisation. This implies that while a liquid undergoes a change to the vapour state at
the normal boiling point, the temperature of the liquid will not rise beyond the temperature of the boiling
point.
The latent heat of evaporation is the energy required to overcome the molecular forces of attraction
between the particles of a liquid, and bring them to the vapour state, where such attractions are minimal.
The definition of the specific latent heat of vaporisation is
'The latent heat of vaporisation is the amount of heat required to convert unit mass of a liquid into the
vapour without a change in temperature."
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The values for water for both the heat of fusion and the heat of vaporization are high compared with those
for other substances. This indicates strong attractive forces between the molecules. The strong forces of
attraction – hydrogen bonding – between water molecules causes water to have this
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high latent heat
high specific heat
high heat of vaporization, and
large difference between the melting point and the boiling point.
Boiling point of substance is temperature where vapor pressure of liquid equals atmospheric pressure.
Several properties of a liquid depend on intermolecular forces
VAPOR PRESSURE. The partial pressure exerted by a vapor in equilibrium with its liquid phase is the
equilibrium vapor pressure. The equilibrium vapor pressure depends on the intermolecular forces and
temperature. The stronger the intermolecular forces, the lower the vapor pressure. The equilibrium vapor
pressure is higher at higher temperature.
MOLAR HEAT OF VAPORIZATION. Vaporization is endothermic. The energy required to vaporize
one gram of substance is called heat of vaporization (the unit is J/g) the energy required to vaporized one
mole is called molar heat of vaporization (the unit is kJ/mol). The molar heat of vaporization of a liquid
with strong intermolecular attraction is higher than the molar heat of vaporization of a liquid with weak
intermolecular attraction.
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THE BOILING POINT. The boiling point is that temperature at which the vapor pressure of the liquid is
equal to the pressure above its surface. The boiling point of a substance depends on the surrounding
pressure. The temperature at which the vapor pressure is equal to one atmosphere is called the normal
boiling point. The boiling point depends on the intermolecular forces. The stronger the intermolecular
forces, the higher the boiling point.
VISCOSITY. Viscosity is an internal resistance of a liquid to flow. Liquids with strong intermolecular
attractions are more viscous than liquids with weak intermolecular attractions.
SURFACE TENSION. A molecule at the surface of a liquid experiences a net attraction into the body of
the liquid. Molecules at the surface of a liquid possess a higher potential energy than those within the body
of the liquid. To minimize this energy, a liquid tends to have minimum surface. The resistance of a liquid
to an increase in its surface area is called the surface tension. Liquids with strong intermolecular attractions
have higher surface tension than liquids with weak intermolecular attractions.
Questions: Heat of Fusion and Heat of Vaporization
Water (H2O)
Ethanol (C2H5OH)
Mercury (Hg)
Specific Heat (Cp)
Heat of Fusion (ΔHfus)
4.18 J/g °C
2.44 J/g °C
0.14 J/g °C
6.02 kJ/mol
5.02 kJ/mol
2.29 kJ/mol
Heat of Vaporization
(ΔHvap)
40.7 kJ/mol
38.6 kJ/mol
59.1 kJ /mol
1) How much heat is required to increase the temperature of 20.0 grams of water by 26.0 °C?
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2) How much heat is required to melt 2.50 moles of ice?
3) How much heat is transferred when 57.0 grams of mercury cools from 76.0°C to 18.0°C?
4) If you have 27.0 grams of 100°C liquid water, how much heat is required to turn it into water vapor?
5) How much energy is transferred when 70.0 grams of ethanol is heated from 21.0 °C to 68.0°C?
6) How much heat is transferred when 400.0 grams of mercury (Hg) is vaporized?
7) How much heat is transferred when 400.0 grams of mercury is cooled from 280.0°C to 30.0°C?
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8) How much heat is transferred when 230.0 grams of ethanol is frozen?
9) How much heat is transferred when 100.0 grams of ethanol condenses from a gas to a liquid?
10) How much heat is transferred when 1.85 mole of mercury is heated from 80.0°C to 125.0°C?
11) How much heat is required to melt 40.0 grams of ice and then increase its temperature from 0.0°C to
25.5°C?
12) How much heat is required to heat 2.0 moles of water from 25.0°C to its boiling point and then turn it
into water vapor?
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Molar Enthalpies Worksheet
1. If ΔHfus of sodium is 2.60 kJ/mol, how much energy is required to melt 145.50 g of sodium?
2. The ΔHfus for water is 6.01 kJ/mol. How much energy is needed to melt 5.28 x 1024 ice molecules?
3. The ΔHsolid for mercury is -2.30 kJ/mol. How much energy is lost when 9.26 g of mercury freezes.
4. The ΔHvap of NH3 is 23.3 kJ/mol. How much energy is needed to vaporize 100.0 g of NH3.
5. a. Jessica has 250.0 g of water that she needs to boil. If the ΔHvap is 40.66 kJ/mol, how much heat
energy will she need to boil her water?
6. Yash needs to burn 875.0 g of glucose (C6H12O6). If the molar heat of combustion is -2800.0 kJ/mol,
how much heat will be released when the glucose burns?
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7. Given that the ΔHcombustion for hydrogen is -286 kJ/mol, how much heat is released when Ben blows
up a balloon containing 4.50 L of hydrogen gas at STP?
8. The ΔHfus of ethanol (C2H5OH) is 4.94 kJ/mol. How much heat is needed to freeze 250.0 mL of
ethanol if the density of ethanol is 0.789 g/mL?
Heating/cooling curves:
The diagram shows the uptake of heat by 1 kg of water, as it passes from ice at -50 ºC to steam at
temperatures above 100 ºC, affects the temperature of the sample.
A: Rise in temperature as ice absorbs heat.
B: Absorption of latent heat of fusion. Intermolecular bonds are breaking.
C: Rise in temperature as liquid water absorbs heat.
D: Water boils and absorbs latent heat of vaporisation. Intermolecular bonds are breaking.
E: Steam absorbs heat and thus increases its temperature.
The above is an example of a heating curve. One could reverse the process, and obtain a cooling curve.
The flat portions of such curves indicate the phase changes.
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Thermochemistry: Energy changes involving phase changes
Sample Problem: How much energy is needed to convert 23.0 grams of ice at -10.0C into steam at
109.0C?
When solving problems involving phase changes, it is helpful
to draw a diagram to visualize the different steps involved.
Each of these steps (1-5) is associated with an energy change
that is reflected in a calculation.
Step 1: If we start by looking at melting a specific amount of ice, the amount of heat needed to bring the
ice up to 0C is calculated by the equation Q=mCT with:
Q = amount of energy (J)
m = mass (grams)
C = specific heat capacity, Cice = 2.10 J/gC
T = change in temperature (C)
So, for this sample problem: Q = 23.0 grams  2.10 J/gC  10.0C = 483 J
Step 2: The next step occurs while the ice is melting. There is no temperature change so we cannot use the
same equation as before. A particular amount of ice takes a constant amount of heat to melt it. This
energy is used to break those hydrogen bonds that hold the water in the crystal structure that makes it ice.
This amount of heat is called the heat (or enthalpy) of fusion. Heat of fusion, Hfus = 6.02 kJ/mol for
water.
23.0 grams/18.02 g/mol  6.01 kJ/mol = 7.68 kJ
Step 3: During this step there is a temperature change again, so we use the same equation as in Step 1,
H=mCT to calculate the energy changes. However, we are now working with a different state of
matter, liquid water, so the specific heat value is different. Cwater = 4.18 J/gC. The temperature change is
the difference between the melting point and the boiling point, so 0.0C to 100.0C.
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Solving this step, we get H = 23.0 grams  4.18 J/gC  100C = 9610 J
Step 4: While the water is boiling, the temperature does not change. The heat of vaporization is the
amount of energy it takes to vaporize a particular quantity of a substance. For water, Hvap = 40.7
kJ/mol.
23.0 grams/18.0 g/mol  40.7 kJ/mol = 52.0 kJ
Step 5: Once the water is in the gas phase, a temperature change occurs again to increase the steam from
100C to 109C, so the equation H=mCT is used once more. However, we’re using a different state of
matter now, water vapor, so the specific heat value is Csteam = 2.00 J/gC.
Solving this last step, H = 23.0 grams  2.00 J/gC  9.0C = 410.0 J
Now, add all 5 steps together. At the end you should probably convert to kJ.
483 J (step 1) + 7.682 kJ (step 2) + 9614 J (step 3) + 51.980 kJ (step 4) + 414.0 J (step 5) = 70173.9 J or
70.2 kJ
Not all problems will go through all 5 steps. Diagram it out and choose the appropriate ones
Practice Problems:
1. Calculate the heat needed to raise 27.0 g of water from 10.0 °C to 90.0 °C.
2. How many kJ are required to heat 45.0 g of H2O at 25.0°C and then boil it all away?
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3. Calculate the amount of energy used when 33.3 grams of ice at 0.00 °C is converted to steam at 150.0
°C.
4. How many joules of heat are needed to change 50.0 grams of ice at -15.0 °C to steam at 120.0 °C?
5. Calculate the number of joules given off when 32.0 grams of steam cools from 110.0 °C to ice at 40.0 °C.
6. What amount of ice must be added to 540.0 g of water at 25.0 °C to cool the water to 0.0 °C and have
no ice.
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Heating Curves
Below is a diagram showing a typical heating/cooling curve for water. It reveals a wealth of information
about the structure and changes occurring in water as it is heated or cooled through all three phases of
matter at different temperatures. At the top of the diagram are pictures representing the typical particle
arrangement as substances change through their states.

Heat removed
1. There is something clearly wrong about the particle spacing in the pictures at the top. What is it?
(Hint… the pictures were drawn for most materials, not for water specifically).
2. Identify by letter (A-E) in which section the following are found:
a. _______ Freezing (if cooling)
b. _______ Particles farthest apart
c. _______ Boiling
d. _______ Particle motion is most restricted
e. _______ Heat of fusion
f. _______ All areas where energy change is potential only
g. _______ Heat of vaporization
h. _______ All areas where particles move past each other
i. _______ Least kinetic energy
j. _______ All areas where kinetic energy is changing
k _______ most potential energy
l. _______ All areas where phase changes occur
m. _______ All areas in which the heat is making the particles move faster
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n. _______ All areas in which the heat is breaking the attractions or bonds between particles
o. _______ All areas in which the particles are not changing their speed
Phase Change Worksheet
The graph was drawn from data collected as a substance was
heated at a constant rate. Use the graph to answer the following
questions.
At point A, the beginning of observations, the substance exists in
a solid state. With each passing minute, (HEAT/
TEMPERATURE) is added to the substance. This causes the
molecules of the substance to move (MORE/ LESS) rapidly
which we detect by a (RISE / FALL) in the temperature of the substance. At point B, the temperature of
the substance is (5°C / 70°C). The solid begins to (BOIL /MELT). At point C, the substance is completely
in a (SOLID / LIQUID) state. The energy put to the substance between minutes 5 and 9 was used to
convert the substance from a (SOLID / LIQUID) to a (SOLID / LIQUID). This heat energy is called the
latent heat of fusion.
Between 9 and 13 minutes, the added energy increases the (ENERGY / TEMPERATURE) of the
substance. During the time from point D to point E, the liquid is (MELTING / BOILING). By point E,
the substance is completely in the (LIQUID / GAS) phase. The energy put to the substance between
minutes 13 and 18 converted the substance from a (LIQUID / GAS) state. This heat energy is called the
latent heat of vaporization. Beyond point E, the substance is still in the (LIQUID / GAS) phase, but the
molecules are moving (FASTER / SLOWER) as indicated by the increasing temperature.
QUESTION: Based on the graph and the table below, what is the substance? Support your answer with
evidence.
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Substance Melting point Boiling point
Bolognium
20 °C
100 °C
Unobtainium
40 °C
140 °C
Foosium
70 °C
140 °C
Phase Changes of Water
1. Does this graph represent water absorbing or releasing heat (kinetic energy)?
2. What letter(s) represent an increase in kinetic energy (motion)?
3. What segments (ex: B – C) represent the 3 phases of water
a. liquid
b. solid
c. gas
4. What letters represent the point at which the water begins to:
a. melt
b. boil
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5. What process begins at
point B
point D
6. For each of the following, indicate which line segment (ex: A – B) represents it.
a.
b.
c.
d.
Latent heat of evaporation
Latent heat of freezing
Latent heat of condensation
Latent heat of melting
7. Which line segment represents the following
a. water vapor changing to a water?
b. ice melting into water?
c. water changing to a water vapor?
d. water freezing into ice?
8. What changes phases by either absorbing or releasing energy. For each line segment indicate
whether it is absorbing or releasing energy.
Segment
Absorbing
Releasing
A–B
F–E
D–C
B–A
E –F
C-D
Heat Capacity and Molar Heat Worksheet
Specific Heat Capacity (water),
Latent heat of Fusion (water),
Latent Heat of Vaporization (water),
c = 4.18 (J/g°C)
∆Hfus = 6.02 (kJ/mol)
∆Hvap = 40.7 (kJ/mol)
1. Find the amount of heat energy needed to warm 250.0 g of water from 18.0°C to 98.0°C for coffee.
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2. i.
ii.
iii.
3.
Find the energy removed from 1.00 kg of water to cool it from 25.0°C to 0.0°C.
The molar heat of fusion (given above) is the heat involved to change 1 mole of
ice at 0.0°C into water at 0.0°C. How much heat must be removed from the water in a) to freeze
it?
How much heat must a freezer remove from 50.0 g of water at 20.0°C to make an ice
cube at 0.0°C?
A 200.0 g piece of metal at 19.0°C is added to 80.0 g of water at 60.0°C in a styrofoam cup (for
insulation). The temperature of the water cooled to 54.7°C.
i. Find the change in temperature for the water and for the metal. Assume that the temperature of
the water and he metal are the same
ii. Find the specific heat of the metal.
HINT: Start with the assumption that heat lost by the water was gained by the metal.
iii. What is the metal (Cu, c = 0.387 J/gºC; Au, c = 0.129 J/gºC; Fe, c = 0.4498 J/gºC; Ag, c =
0.245 J/gºC)?
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4. i.
A swimming pool measures 6.50 m x 12.0 m x 2.50 m. Calculate the heat capacity of the whole
pool (in kJ/°C), using the following information:
1 m3 = 1000 L, 1 L (water) = 1 kg.
You will need to find the volume of the pool and then using dimensional analysis find the mass of
water (kg) required to fill the pool and then use the specific heat constant for water (4.18
kJ/kg°C).
ii.
How much energy must be added to the pool to raise its temperature by 15.0°C?
iii.
Propane, C3H8, has a molar heat of combustion of 2200 kJ/ mol. What mass in kg of propane is
required to heat the water in b)? You will need to work out the number of moles of propane and
then convert moles to mass.
ANSWERS:
1. 83.6kJ
3. silver
2. i. -105 kJ ii. 334 kJ iii. 20.9 kJ
4. i. 8.15 x 10 5 (kJ/°C) ii. 1.22 x 107 kJ
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iii. 0.245 t (246 kg)
Review Worksheet
PART A – INTERMOLECULAR FORCES
1. Fill in the diagram (with high or low) to show how intermolecular forces influence the volatility, vapor
pressure, and boiling point of a substance.
weak
strong
When IMF are…
volatility is ____________
volatility is __________
vapor pressure is __________
vapor pressure is _______
boiling point is __________
boiling point is __________
PART B – VAPOR PRESSURE GRAPHS Use the graph below to answer the
following questions.
2. What is the vapor pressure of CHCl3 at 50C?
3. What
is
the
boiling
external pressure is 30 kPa?
point
of
H 2O
when
the
4. What is the normal boiling point of CCl4?
5. Which substance has the weakest IMF?
PART C – HEATING CURVES. Use the heating curve below to answer the
following questions.
6.
What
substance?
7. What is the boiling point of the substance?
8. Which letter represents heating of the solid?
9. Which letter represents heating of the vapor?
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is
the
melting
point
of
the
10. Which letter represents melting of the solid?
11. Which letter represents boiling of the liquid?
PART D – PHASE DIAGRAMS. Use the phase diagram for water below to answer the following questions.
12. What is the state of water at 2 atm and 50
13. What phase change will occur if the temperature is lowered
from 80C to -5C at 1 atm?
14. You have ice at -10C and 1 atm. What could you do in
order
cause
the
ice
to
sublime?
At standard temperature and pressure, bromine (Br2) is a red liquid. Bromine sublimes when the
temperature is –25 0C and the pressure is 101.3 kPa. The phase diagram for bromine is shown below.
Use this diagram for questions (1) – (9)
(1)
Label each region of the graph as solid, liquid, or gas.
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(2)
Label the triple point, normal melting point, and normal boiling point on the
graph and estimate their values in the spaces below.
Normal Melting Point =
_________________________
Normal Boiling Point = _________________________
Triple Point =
_________________________
(3)
Use the letter ‘A’ to label the line that gives the conditions for equilibrium
between solid and liquid phases of bromine, the letter ‘B’ to label the curve
that gives the conditions for equilibrium between liquid and vapor phases
of bromine, and the letter ‘C’ to label the curve that gives the conditions for
equilibrium between solid and vapor phases of bromine.
(4)
Describe how the melting point of bromine changes with the external pressure. Based on the
slope of the melting-point curve in the phase diagram, would you characterize the solid phase of
bromine as more dense or less dense than the liquid phase of bromine?
________________________________________________________
________________________________________________________
________________________________________________________
(5)
What is the boiling point of bromine when the external pressure is 75 kPa?
________________________________________________________
(6)
Explain the significance of the triple point?
________________________________________________________
________________________________________________________
________________________________________________________
(7)
Using arrows labeled ‘S’, ‘V’, and ‘M’, label those portions of the phase
diagram where sublimation, vaporization, and melting occur, respectively.
Circle the correct word in parentheses in each of the following sentences.
(8)
Bromine vapor at 15 0C (condenses, sublimes) when the pressure is
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raised to 50 kPa.
(9)
Bromine liquid at 70 kPa (vaporizes, freezes) when the temperature is
decreased to –15 0C.
Phase Diagram Review Worksheet 1
1) At a pressure of 60 atm and a temperature of -100ºC what state is carbon dioxide?
2) what state of matter would Carbon dioxide be if you had a bottle of 20ºC and a pressure of 100
atm?
3) At what temperature and pressure will all three phases coexist, the triple point?
4) If I have a bottle of CO2 at a pressure of 1 atm and temperature of 100ºC what will happen if I
drop the temperature to -120ºC?
5) At what temperature and pressure would CO2 sublime?
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Advanced Questions for Research Classes
1. Calculate the quantity of energy required to heat 21.00 g of water from –7.00 oC to
102.0 oC.
Cp(g) = 2.10 J / g oC
∆H Fusion = 6.02 kJ /mole
Cp(l) = 4.20 J / g oC
∆H Vaporization = 40.7 kJ/mole
Cp(s)= 2.10 J / g oC
2. CCl2F2 has a boiling point of -30. C, and a heat of vaporization of 0.165 kJ/g. The vapor and the
liquid have specific heats of 0.61 J/gK and 0.97 J/gK respectively. How much heat must be
evolved when 10.0 g of CCl2F2 is cooled from +40. C to -40. C?
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