Chapters 16 and 17
Thermodynamics – the study of energy and energy transfer.
Thermochemistry – the study of energy involved in chemical reactions.

Law of Conservation Of Energy
– the total energy in the universe is constant.
- energy cannot be created or destroyed.
∆ Energy
Universe
= 0
Energy CAN:
- be transferred from one substance to another
- or be converted into various forms.

In order to study energy changes – we need to define what part of the universe we are dealing with.
System - the part of the universe that is being studied or observed.
Surroundings – everything else in the universe.
Ex. In this chemical rxn – the system is the reactants and products in the flask
- the surroundings is the flask, the student, the air, and the hand holding the flask.

Therefore: Universe = System + Surroundings
∆ E universe
= ∆ E system
+ ∆ E surroundings
= 0
This relationship is known as the 1 st Law of Thermodynamics:
“Any change in the energy of a system is accompanied by an equal and opposite change in the energy of the surroundings”
∆ E system
= - ∆ E surroundings the energy the system gains = the energy the surroundings loses
OR
- ∆ E system
= ∆ E surroundings the energy the system loses = the energy the surroundings gains

Depending on how the system is separated from its surroundings, systems are defined in three different ways:
Open System – system is open to the surroundings – both matter and energy can be exchanged between the two.
Ex. A reaction in an open beaker.
Closed System – a system where energy can move between the system and the surroundings – however, matter cannot move.
Ex. A reaction in a stoppered Erlenmeyer flask.
Isolated System – a system is completely isolated from its surroundings – neither matter nor energy is exchanged between the two.
Ex. A reaction in a calorimeter.

1) Kinetic Energy - energy in motion
(always accompanied by a temperature change)
2) Potential Energy – energy that is stored
(changes in state and chemical rxns)
•
•
The SI unit for both types of energy is the joule (J).
One joule is equal to 1 kg∙m 2 /s 2
• Kinetic Energy is measured in J.
• Potential Energy is measured in kJ.

• Temperature, T, is a measure of the average kinetic energy of the particles that make up a system.
• Temperature is measured in either Celsius degrees ( o C) or kelvins (K).
• Temperature change in a substance is an indication of a change in kinetic energy.
- Symbolized by ∆T
∆T = T f
– T i change in temp = temp. final – temp. initial

• Heat, q, refers to the transfer of kinetic energy between objects with different temperatures.
• It is measured in joules.
• When a substance absorbs heat, the average kinetic energy of the particles increases – therefore the temperature increases.
Reconsidering the 1 st Law of Thermodynamics: loses.
q system
= -q surroundings
The heat the system gains is equal to the heat the surroundings absorbs.
- q
OR system
= q surroundings
The heat the system loses is equal to the heat the surroundings

When substances with different temperatures come in contact, kinetic energy is transferred from the particles of the warmer substance to the particles of the cooler substance.
Heat Transfers:
- from hot chocolate to your mouth
- from hot chocolate to the mug
- from the mug to your fingers
- from the hot chocolate to the air
You blow on the hot chocolate
- moves cool air over the hot chocolate and the hot chocolate transfers heat to the air.

Temperature and Energy Transfer
• Heat moves from greater to lower temperatures
• The greater the heat difference the greater the energy transfer
Mass and Energy Transfer
• Mass is directly related to heat transfer.
• Is used in calculations – designated symbol m.
Types of Substances and Energy Transfer
• In calculating heat transfer, we do not use the type of substance as a variable, but rather we use a variable that reflects the individual nature of different substances – the specific heat capacity .

Specific Heat Capacity (c) - the amount of energy required to change one gram of a substance by one degree Celsius.
• This reflects how well a substance stores energy.
• Units are J/g∙ o C
• A substance with a large specific heat capacity can absorb and release more energy that a substance with a smaller specific heat capacity.
• Water has a very large specific heat capacity
= 4.184 J/g∙ o C
See Table on Page 632

Three variables determine heat transfer:
1) temperature
2) mass
3) specific heat capacity
Mathematical Relationship:

Complete Practice Problems #1-4 Page 634

Calorimetry
• the technological process of measuring the changes in kinetic energy.
• uses a calorimeter
Calorimeter – a basic calorimeter contains water, a thermometer, and an isolated system.
We will use a coffee-cup calorimeter or a Poor Man’s Calorimeter

• The system is isolated – no heat is exchanged with the surroundings outside the calorimeter.
• The amount of heat exchanged with the calorimeter itself is small enough to be ignored.
• If something dissolves in or reacts with the calorimeter water, the solution still retains the properties of water.
 specific heat capacity and density

Therefore: q system
= -q surroundings
The heat the system absorbs is equal to the heat that the surroundings released.
- q
OR system
= q surroundings
The heat the system releases is equal to the heat the surroundings absorbs.
Sample Problem on Page 663
Page 664 - #1-4

Now we will consider changes in the potential energy of a system....
Enthalpy (∆H)
• refers to the potential energy change of a system during a process such as a chemical or physical change.
• measured at constant pressure
• units are kJ/mol

• The enthalpy change of a chemical reaction represents the difference between the potential energy of the products and the potential energy of the reactants.
• In chemical rxns, potential energy changes result from the breaking or the forming of bonds.
• A chemical bond is caused by the attraction between electrons and the nuclei of the atoms. Energy is stored in bonds.
▫
▫

Net Absorption of Energy = Endothermic Rxn
Net Release of Energy = Exothermic Rxn

Chemists define the total energy of a substance at a constant pressure as its enthalpy, H.
They use relative enthalpy of the reactants and the products to determine the change in enthalpy that accompanies the reaction.
Consider:
N
2
+ O
2
→ 2NO triple bonds double bonds single bonds
This rxn absorbs energy less energy is released when NO bonds are formed than the energy required to break the double and the triple bonds

∆H
Rxn
- enthalpy of reaction
- enthalpy change of a rxn
- dependent on conditions of temperature and pressure
- it is the energy change associated with the reaction of
1 mole of a compound with another compound.
∆H o
Rxn will represent the enthalpy of rxn at SATP
(SATP = 1 bar, 25 o C)
** people will say “heat of reaction”
** O means nought – means standard conditions or standard state

a balanced chemical equation that indicates the amount of heat absorbed or released in a chemical rxn.
ENDOTHERMIC
- because heat is absorbed in an endothermic rxn, the heat term is included on the reactant side of the equation.
117.3 kJ + MgCO
EXOTHERMIC
3(s)
→ MgO
(s)
+ CO
2(g)
- because heat is released in an exothermic rxn, the heat term is included on the product side of the equation.
H
2 (g)
+ ½ O
2 (g)
→ H
2
O
( l )
+ 285.8 kJ

- for this method, the heat term is included on the right side of the equation.
- the sign for the heat term must be included to show when the equation is endothermic or exothermic.
ENDOTHERMIC
MgCO
3(s)
→ MgO
(s)
+ CO
2(g)
EXOTHERMIC
H
2 (g)
+ ½ O
2 (g)
→ H
2
O
( l )
∆H o = +117.3 kJ
∆H o = - 285.8 kJ

Enthalpy Diagram for an Exothermic Rxn
You should always include:
• Title
• Correct graph
• reactants
• products
• y-axis
• x-axis
• ∆H

• In thermochemistry :
- formation
- combustion
- melting
- freezing
- condensation
- vaporization
- solution

• In a formation reaction, a substance is formed from its elements in their standard states.
∆H o form
= standard molar enthalpy of formation
- the quantity of energy that is absorbed or released when 1 mol of a substance is formed directly from its elements in their *standard states.
- it is a value – a number.
* The standard state of an element is its most stable form at SATP.

Formation Equation for water (liquid)
Thermochemical Equation:
1 H
2 (g)
+ ½ O
2 (g)
→ 1 H
2
O
( l )
+ 285.8 kJ
Separate Equation Method:
1 H
2 (g)
+ ½ O
2 (g)
→ 1 H
2
O
( l )
∆H o form
= -285.8 kJ

• In a combustion equation, a substance burns in oxygen to form carbon dioxide and water.
∆H o comb
= the molar enthalpy of combustion
- the quantity of energy that is absorbed or released when 1 mol of a compound burns in oxygen to produce carbon dioxide and water.
- reactants and products should be in their standard states.

Combustion of Propane
Thermochemical Equation:
1 C
3
H
8 (g)
+ 5 O
2 (g)
→ 3 CO
2 (g)
+ 4 H
2
O
(l)
+ 2323.7 kJ
Separate Equation Format:
1C
3
H
8 (g)
+ 5 O
2 (g)
→ 3 CO
2 (g)
+ 4 H
2
O
(l)
∆H o comb
= -2323.7 kJ
Combustion Of Butane
Thermochemical Equation :
1 C
4
H
10 (g)
+ O
2 (g)
2
→ 4 CO
Separate Equation Format :
2 (g)
+ 5 H
2
O
(l)
+ 3003.0 kJ
1C
4
H
10 (g)
+ O
2 (g)
2
→ 4 CO
2 (g)
+ 5 H
2
O
(l)
Page 643 #15-18
∆H o comb
= -3003.0 kJ

•
How much heat is released when 50.00 g of methane forms from its elements?
50.00 g CH
4 x
1 molCH
4
16.05
gCH
4 x

74.6
kJ
1 molCH
4
 
232.4
kJ change from use the molar enthalpy grams to moles (obtained from the chart)
232.4 kJ was released.

How much heat is released when 50.00 g of methane combusts completely?
50.00 g CH
4 x
1 molCH
4
16.05
gCH
4 x

965.1
kJ
1 molCH
4
 
3007
Page 645 #19-23
Remember that: at SATP: 1 mol of a gas = 22.4 L of the gas
1L = 1000 mL

• Changes in state involve changes in the potential energy of a system.
• Changes that you know: melting, freezing, etc.
• The temperature of the system undergoing the change remains constant but because the energy is being absorbed or released as heat, the temperature of the surroundings often changes.

• In general, the energy change associated with changes in state are much smaller than those of chemical changes.
▫ Why? Intermolecular forces (L.D., D.D., H.B.) are easier to break than intramolecular forces (ionic, covalent, metallic bonds).
▫ State changes for water animation
▫ Solid changing state

• ∆H o melt
= the molar enthalpy of melting (fusion)
- the quantity of energy that is absorbed when 1 mol of a compound changes state from a solid to a liquid.
Ex. ∆H o melt
= 27.2 kJ
27.2 kJ + NaCl
(s)
→ NaCl
( l )
• ∆H o fre
= the molar enthalpy of freezing
- the quantity of energy that is released when 1 mol of a compound changes state from a liquid to a solid.
Ex. ∆H o fre
= - 27.2 kJ
NaCl
( l )
→ NaCl
(s)
+ 27.2 kJ

• ∆H o vap
= the molar enthalpy of vaporization
- the quantity of energy that is absorbed when 1 mol of a compound changes state from a liquid to a gas.
207 kJ + NaCl
( l )
→ NaCl
(g)
• ∆H o cond
= the molar enthalpy of condensation
- the quantity of energy that is released when 1 mol of a compound changes state from a gas to a liquid.
NaCl
(g)
→ NaCl
( l )
+ 207 kJ
Note: ∆H o melt
∆H o vap
= - ∆H o fre
= - ∆H o cond
See Table 16-4 on Page 647

• ∆H o soln
= the molar enthalpy of solution
- the quantity of energy that is absorbed or released when 1 mol of a compound dissolves in a solvent.
• Dissolving can be exothermic or endothermic.
• Manufacturers use this information to make hot packs and/or cold packs.
• Hot Packs – process of dissolving is exothermic
• Cold Packs – process of dissolving is endothermic
Ex. 25.7 kJ + NH
4
NO
3 (s)
→ NH
4
NO
3 (aq)
Ex. CaCl
2 (s)
→ CaCl
2 (aq)
+ 82.8 kJ

• We can re-look at the theory for calorimetry when considering state changes and chemical rxns.
When a measured amount of hot water is placed in a calorimeter and ice is added to it...
• heat travels from the water into the ice.
• the energy the ice absorbs is used to melt the ice.
q water
= + q ice
The heat the water releases is equal to the heat the ice absorbs
Better:
The heat the water releases is equal to the enthalpy the ice absorbs

• Water undergoes a kinetic energy change (there is a temperature change) we can use q = mc∆T
Then, - q water
= + q ice
• Ice undergoes a potential energy change (no temperature change) we can use stoichiometry.
• Note: We do not mass the ice in an analytical balance.
Take the mass of water when the lab is finished and subtract the initial mass of water and you will obtain the mass of ice that melted.

Sample Problem:
A student takes 100.0 mL of water at 51.8
o C and places it in a calorimeter. Ice is added to the water until it reaches 0.0
o C.
At that point, all the ice is removed and the water is remeasured in a graduated cylinder. The final volume of water is 165.7 mL. Calculate the molar enthalpy of melting.

q w q w
= m w c w
∆T w
= (100.0 g)(4.184 J/g∙ o C)(-51.8
o C)
= -21673.12 J = -21670 J
- q water
= + q ice
-21670 J = +21.67 kJ
∆H melt
=
 21.67
3.646

5.94
/
You need the moles of ice:
165.7 mL – 100.0 mL = 65.7 mL
65.7 mL x 1 molH O
2
18.02
gH O
2

3.646
molH O
2

Problems
1. A student adds 7 ice cubes to 101.6 mL of water 38.4
o C.
When the ice melted, the temperature of the water is
0.0
o C. The final volume of water is 150.3 mL.
Calculate the molar enthalpy of melting of water.
= 6.04 kJ/mol
2. A student takes 225.4 mL of water at 52.6
o C and places it in a calorimeter. Ice is added to the water until it reaches 0.0
o C. At that point, all the ice is removed and the water is re-measured in a graduated cylinder. The final volume of water is 371.2 mL. Calculate the molar enthalpy of melting.
= 6.13kJ/mol

• When a measured amount of water is placed in a calorimeter and a solid substance is dissolved in it...
• Energy will travel from the water to the solid or from the solid to the water.
Exothermic Dissolving
• the heat the water gains is equal to the enthalpy the process of dissolving releases.
Endothermic Dissolving
• the heat the water loses is equal to the enthalpy the process of dissolving absorbed.

Exothermic Dissolving
+ q water
= - q soln
Endothermic Dissolving
- q water
= + q soln
Use stoichiometry – using the mass of the substance dissolved – to determine the molar enthalpy of solution for the substance.

When a 4.25 g sample of ammonium nitrate dissolves in
60.0 g of water in a calorimeter, the temperature drops from 22.0
o C to 16.9
o C. Calculate the ∆H soln for ammonium nitrate.
Water q=mc∆T
= (60.0g)(4.184J/g o C)(-5.1
o C)
= -1280 J
- q w
=q soln
-1280 J = 1.280 kJ
Ammonia Nitrate
4.25 g x 1 mol NH
4
NO
80.06 g NH
4
3
NO
3
= 0.05309 mol
∆H soln
= kJ = 1.280 kJ = 24 kJ/mol mol 0.05309 mol

1. When a 1.80 g sample of magnesium hydroxide dissolves in 108.9 g of water in a calorimeter, the temperature rises from 22.0
o C to
83.9
o C. Calculate the ∆H soln for magnesium hydroxide.
= -914 kJ/mol
2. A 9.96 g sample of lithium chloride was dissolved in 101.2 mL of water at 22.2
o C. When the salt is dissolved, the temp. of the soln was 41.3
o C. Calculate the molar enthalpy of solution for lithium chloride.
= -34.4 kJ/mol
3.
A student dissolves 1.96 g of NaOH in 99.7 mL of water at 23.4
o C.
After the NaOH dissolves, the temp. of the water rises to 28.7
o C.
What is the molar enthalpy of solution?
= -45 kJ/mol

• Coffee-cup calorimetry can be used to study changes in dilute aqueous solutions.
• The water in the calorimeter absorbs (or provides) the energy that is released (or absorbed) by the chemical rxn.
+ q soln
rxn
soln
rxn
Same calculations as melting and solution... We just need to determine the limiting reagent when calculating the molar enthalpy.

Sample Problem Page 665
• Copper sulfate, CuSO
4
, reacts with sodium hydroxide, NaOH, in a double displacement reaction. A precipitate, Cu(OH)
2
, is formed.

4
→
2
2
4
50.0 mL of 0.300 mol/L CuSO
4 reacts with 51.8 mL of 0.600 mol/L
NaOH. The initial temperature of both solutions is 21.4
o C. After mixing the solutions in a coffee-cup calorimeter, the highest temperature reached is 24.6
o C. Determine the enthalpy change for the reaction and then write a thermochemical equation.

Page 667
1. Consider the following rxn:
HCl
(aq)
+ NaOH
(aq)
→ NaCl
(aq)
+ H
2
O
(l)
The chemist uses a coffee-cup calorimeter to neutralize completely 61.1 mL of 0.543 mol/L HCl with 62.5 mL of 0.543 mol/L of NaOH. The initial temperature of both solutions is 17.8
o C. After neutralization, the highest recorded temperature is
21.6
o C.
2. Consider the following rxn:
Mg
(s)
+ 2 HCl
(aq)
→ MgCl
2 (aq)
+ H
2 (g)
The chemist uses a coffee-cup calorimeter to react completely 0.500 g of magnesium ribbon with 100.0 mL of 1.00 mol/L of HCl. The initial temperature of the HCl is
20.4
o C. After the reaction is complete, the highest recorded temperature is 40.7
o C.
3. Consider the following rxn:
HNO
3 (aq)
+ KOH
(aq)
→ KNO
3 (aq)
+ H
2
O
(l)
∆H rxn
= -53.4 kJ/mol
The chemist uses a coffee-cup calorimeter to neutralize completely 55.0 mL of 1.30 mol/L of HNO3 with 58.9 mL of 1.50 mol/L of KOH. The initial temperature of both solutions is 21.4
o C. What is the final temperature of the mixture?

• Chemists use heating and cooling curves to represent temperature and phase changes in substances.
• Temperature goes on the y-axis
• Time goes on the x-axis
• Key points – boiling points, melting points, freezing points, etc.
• Flat lines - shows a state change
- potential energy changes
- no temperature change
• Slanted lines – shows temperature change
- changes in kinetic energy
• Curves do not need to show all state changes – you may select a temp range of your choice.

Heating Curve for Water
Cooling Curve for Water
Areas 1, 3, 5
- temp. changes
- Kinetic energy changes
Areas 2 and 4
- no temp. change
- potential energy changes
Heating Curve for Water
Temp. Range -15 o C to 120 o C

• When calculating energy changes that occur through both potential and kinetic energy – you need to calculate the changes for each area, convert the different units to one unit and then add them all together.
• For areas with kinetic energy changes –
Use q = mc∆T
• For areas with potential energy changes –
Use stoichiometry

Page 654
Strategy:
1. Draw a quick curve to determine how many types of energy are involved.
2. Use q=mc∆T for changes in Kinetic Energy.
3. Use stoichiometry for changes in Potential Energy.
4. Convert the numbers to the same units and add together.

Sample Problem #2
A sample of solid mercury having a mass of 3.45 g is heated from -55.7
o C to 68.4
o C. The melting point of mercury is -38.8
o C. The specific heat capacity of liquid mercury is 0.140 J/g∙ o C and the specific heat capacity of solid mercury is 0.145 J/g∙ o C. What is the total heat gained by the metal?
= 463 J
Problems #30-34 Page 655

• For some chemical reactions we cannot use coffee-cup calorimetry to determine the enthalpy change.
• These reactions may be:
Too explosive
Too slow
Not aqueous in water
• In this case – there are three possible methods to use :
▫ Hess’ Law
▫ Using Enthalpies of Formation
▫ Using Bond Energies

• Hess’ Law – the enthalpy change of a physical or chemical process depends only on the beginning conditions (reactants) and the end conditions (products). The enthalpy change is independent of the pathway of the process and the number of intermediate steps in the process. It is the sum of all the enthalpy changes of all the individual steps that make up the process.
• Hess’ Law is valid because enthalpy is a state function.
A state function is a property that is determined only by the current conditions of the system – it is not dependent on the path taken by the system to reach those conditions.

Ex. Carbon dioxide can be formed by two different pathways:
Pathway #1
C then, CO
(s)
(g)
+ ½ O
+ ½ O
2 (g)
2 (g)
→
→
CO
CO
Pathway #2
C
(s)
+ O
2 (g)
→ CO
2 (g)
(g)
2 (g)
∆H o = -110.5 kJ
∆H o = -283.0 kJ
∆H o = -393.5 kJ
Adding the two equations from Pathway #1 will result in the same equation as Pathway #2

•
•

• Use the following formula:
∆H o rxn
= ∑(∆H o form PRODUCTS
) - ∑ (∆H o form REACTANTS
)
Consider:
CH
4(g)
+ 2 O
2 (g)
→ CO
2 (g)
+ 2 H
2
O
(g)
Practice Problems #20-22 Page 688

• Use the following formula:
∆H o rxn
= ∑( Bond Energies
REACTANTS
) - ∑ ( Bond Energies
PRODUCTS
)
Consider:
CH
4(g)
+ 2 O
2 (g)
→ CO
2 (g)
+ 2 H
2
O
(g)
Practice Problems #23-26 Page 690