CHAPTER 6 THERMOCHEMISTRY
6.1 THE NATURE OF ENERGY
Know definitions of the followings terms:
Energy – capacity to do ____________ or produce _____________.
Law of Conservation of Energy – energy can be converted from one form to another but cannot be
_________________ or ____________________.
Potential Energy – energy due to _________________________ or __________________________.
Kinetic Energy – energy due to the _________________ of a object and depends on the ______________ of the object and its ____________________. KE = _____________
Heat – transfer of energy between 2 objects due to a ___________________ difference.
Work – force acting over a __________________________.
State Function - a property of a system that depends only on its ______________________ state.
Chemical Energy
System – the part of the universe on which we wish to focus attention.
Surroundings – include everything else in the universe.
Exothermic – process that results in the ___________________ of energy (heat)
Endothermic – process that _______________ energy (heat) from its surroundings.
Example: Write a balanced equation for the combustion of methane:
The above reaction is considered to be _________________ since energy is flowing __________ of the system. Energy lost by a system is ______________ by the surroundings.
Reactions that absorb energy from the surroundings are said to be ____________________.
Thermodynamics – the study of ______________________ and its interconversions.
First Law of Thermodynamics – The energy of the universe is ____________________.
Internal Energy – (symbol E) – of a system can be defined as the sum of the ___________________ and
_______________________ energies of all the “particles” in the system
Equation: w is _____________ if the system does work q is _______________ for endothermic processes q is _______________ for exothermic processes w is _____________if work is done on the system

6.2 ENTHALPY AND CALORIMETRY/6.4 STANDARD ENTHALPIES OF FORMATION
Enthalpy, H is the transfer of energy in a reaction. It is a state function
At constant pressure, the change in enthalpy ____ of the system is equal to the energy flow as heat.
The terms heat of reaction and change in enthalpy are used interchangeably for reactions studied at constant pressure.
For a chemical reaction, the enthalpy change is given by the following equation:
If ∆H is ____the reaction is endothermic. If ∆H is _____the reaction is exothermic.
The standard enthalpy of formation__________ of a compound is defined as a change in _______________ that accompanies the formation of _______ mole of a compound from its
_______________ with all substances in their ____________________ ________________.
A degree sign on a thermodynamic function indicates the corresponding process has been carried out under standard conditions.
Standard State Conditions:
For an element: the standard state is the form which the element exists at 25 o C and 1 atmosphere
For a compound:
-The standard state of a gaseous substance is a pressure of exactly 1 atm
-For a substance present in solution, the standard state is a concentration of exactly 1M
-For a pure substance in the condensed state, the standard state is the pure liquid or solid
The ∆H f o for an element in its standard state equals __________.
*Refer to Appendix Four, pp A21-A24 in textbook for ∆H f o values
Example: Write the equation that show the standard formation of NO
2
(g) from its elements.
The change in enthalpy for a given reaction can be calculated from the enthalpies of formation of the reactants and products:
Δ H o reaction
= Σn p
Δ H f o (Products) – Σn r
Δ H f o (Reactants)
Example: Using the standard enthalpies of formation from the table, calculate the enthalpy change for the following reaction:
4NH
3
(g) + 7O
2
(g)  4NO
2
(g) + 6H
2
O (l)
STANDARD HEATS OF FORMATION (APPENDIX 4) (in kJ/mol)
NH
3
(g)
O
2
(g)
-46
0
H
2
O (l)
NO
2
(g)
-286
34

6.2 CALORIMETRY
There are four methods by which we can determine the ∆H for a reaction.
#1 Using Bond Energies: ∆H o rxn
= ∑BE(bonds broken) - ∑BE(bonds formed) (Chapter 8)
#2 Using Enthalpies of Formation: ∆H o rxn
= ∑n∆H o f products - ∑n∆H o f reactants (from last section)
#3 Using Hess’s Law given the pathway/mechanism for a reaction. (Section 6.3)
The fourth method by which an enthalpy change can be determined – through the use of
CALORIMETRY – a method of measuring __________ flow between system and surroundings.
CALORIMETER – a device used to measure __________ flow between system and surroundings.
HEAT CAPACITY – (C p
) the amount of heat necessary to raise the temperature of an object by one degree.
SPECIFIC HEAT CAPACITY – (c)
If the heat capacity is given per gram of substance, it is called specific heat capacity. Unit:
For H
2
O ( ł
):
COFFE CUP CALORIMETER
The measurement of heat using a simple calorimeter such as that shown in the figure below is an example of
_______________ ________________ calorimetry, since the pressure – atmospheric pressure – remains constant during the process.
At constant pressure q (heat) = ∆H
EQUATION USED TO DETERMINE HEAT (q)
EXAMPLE #1
A 5.037 g piece of iron heated to 100 0 C is placed in a coffee cup calorimeter that initially contains 27.3 g of water at 21.2
o C. If the specific heat capacity of iron is 0.45 J/ o C g, what is the final temperature of the mixture?

EXAMPLE #2
Suppose we mix 50.0 mL of 1.0 M HCl at 25 o C with 50.0 mL of 1.0 M NaOH also at 25 o C in a calorimeter. After the reactants are mixed by stirring, the temperature is observed to increase to 31.9
o C. Calculate the ∆ H rxn
in kJ / mol of water produced. The density of the resulting solution is 1.02 g/mL.
This problem is a two part problem. The first part is working with stoichiometry and the second is the thermochemistry part.
STOICHIOMETRY PART
Write the net ionic equation for the reaction:
THERMOCHEMISTRY PART
We have now found the mole part of the answer but we also need to know the heat (kJ) required to raise the temperature of a given amount of water by 1 o C.
Now use the appropriate data and calculate the heat released by the reaction in kJ/mol
How much energy would have been released if twice these amounts of solutions had been mixed?
The heat of a reaction is an __________________ property meaning it depends directly on the amount of substance used – in this case the amount of reactants. In contrast, an __________________ property is not related to the amount of substance. _________________ is an intensive property.
NOTE:
If two reactants at the same temperature are mixed and the resulting solution gets warmer, this means the reaction is
__________________. An endothermic reaction __________ the solution.
EXAMPLE #3
When 1.00 L of 1.00M Ba(NO
3
)
2
solution at 25 o C is mixed with 1.00L of 1.50 M Na
2
SO
4
solution at 25 o C in a calorimeter, the white solid BaSO
4
forms and the temperature increases to 28.1
o C. Assuming that the calorimeter absorbs only a negligible quantity of heat, that the specific heat capacity of the solution is 4.18 J/ o C g, and that the density of the final solution is 1.00 g/mL, calculate the enthalpy change per mole (kJ/mol) of BaSO
4
formed.

6.3 HESS’S LAW
Another method to determine ∆H to determine is to use what is known as Hess’s Law.
Since enthalpy is a state function (meaning it is pathway independent) – in going from a particular set of reactants to a particular set of products, the change in enthalpy is independent of the pathway taken by the reactants. This principle is known as Hess’s Law.
In other words, Hess’s Law states that if a reaction is carried out in a series of steps, ∆H for the overall reaction will equal the sum of the enthalpy changes for the individual steps
Example:
68 kJ + N
2
(g) + 2 O
2
(g)  2NO
2
(g)
One possible pathway (reaction mechanism) is the following:
N
2
(g) + O
2
(g)  2 NO (g)
___ _2NO(g) + O
2
(g)  2 NO
2
(g)
Net reaction: N
2
(g) + 2O
2
(g)  2 NO
2
(g)
∆ H
∆ H
∆ H
1
2
= 180 kJ/mol
= -112 kJ/mol rxn
= 68 kJ/mol
Intermediate – a species that is neither a product or reactant in the net reaction – is formed and consumed in the reaction process.
In the above sequence, ___________ is an intermediate.

Note: The above 2 steps represent one possible pathway by which the reactants become products. There can be multiple other pathways that give the same result. This is why enthalpy is a state function – independent of pathway taken.
Practice Problem:
Given the following data, calculate the ∆ H rxn
for the reaction: S (s) + O
2
(g)

SO
2
(g)
S (s) + 3/2 O
2
(g)

SO
3
(g) ∆ H = -395.2 kJ/mol
2SO
2
(g) + O
2
(g)  2SO
3
(g) ∆ H = -198.2 kJ/mol
2 characteristics of ∆ H for a reaction:
1) If a reaction is reversed, the sign of ∆ H is also reversed.
2) The magnitude of ∆ H is directly proportional to the quantities of reactants and products of the
reaction. If the coefficients in a balanced equation are multiplied by an integer, the value of
∆ H is multiplied by the same integer.

CHAPTER 16 SPONTANEITY, ENTROPY AND FREE ENERGY
16.1 – Spontaneous Processes and Entropy
Entropy (denoted by symbol _____) – measure of the ____________________ or _________________
It is a state function.
The natural progression of things is from ___________ to _____________, from _______________ entropy to _______________entropy.
Spontaneous Processes – a process that occurs ____________ outside intervention.
A driving force for a spontaneous process is an______________ in the entropy.
A driving force for a spontaneous process is a ________________ in enthalpy.
If both of these driving forces are present, a process is always ____________________.
If both of these driving forces are absent, a process is always _____________________.
If one driving force is present, and the other is absent, temperature will be a factor as to whether the process will be spontaneous or not (will discuss this more later in the chapter.)
16.2 Entropy and the Second Law of Thermodynamics
Second Law of Thermodynamics – In any spontaneous process, there is always an ________________ in the entropy of the universe.
Predicting the sign of ∆S
+∆S means entropy is ____________________ -∆S means entropy is ____________________
Entropy _____________ with increases in temperature and volume.
Entropy _______________ with the freedom of motion of molecules. Therefore S (g) ____ S ( l
) ____ S (s)
Generally, when a solid is dissolved in a solvent, entropy _____________________.
Examples: For each of the following pairs, choose the substance with the greater entropy a) N
2
(g) at 273 K or N
2
(g) at 373 K (same volume) b) N
2
(g) at 1 atm or N
2
(g) at 1.0 x 10 -2 atm (same temp)
Examples : Predict the sign of ∆S for the following a) Dry ice sublimes b) Sodium chloride crystals dissolve in water c) Nitrogen monoxide (g) + Oxygen (g)  Nitrogen dioxide (g)

16.5 Entropy Changes in Chemical Reactions
∆S o = ∑nS o products
- ∑ nS o reactants
Third law of thermodynamics – the entropy of a perfect crystalline solid at 0 K is zero
(This is an unattainable ideal – taken as standard but never actually observed)
S o values are given in Appendix Four (pp A21- A24) along with ∆H f o values
Unlike standard enthalpies of formation, ∆H f o
, standard molar entropies, S o , are not zero.
Example: Predict the sign of ∆S o , then calculate ∆S o for the following reaction:
Decomposition of silver oxide (s)
16.4/16.6 Free Energy and the Spontaneity of Chemical Reactions
A spontaneous reaction is one that moves in the forward direction (to the right)
All processes are driven by 2 factors:



A _______________ in enthalpy (release of heat energy,
∆H < 0
). This means an _____________reaction is favorable for the reaction to be spontaneous.
An ________________ in entropy (increase in disorder,
∆S > 0
) This means that an __________ in entropy is favorable for a reaction to be spontaneous.
A reaction that is spontaneous will produce a significant quantity of ________________.
You can also therefore conclude that if a reaction is ______________ or if entropy is _____________ that these are not favorable for the process to be spontaneous.
Would the following reaction be spontaneous? Explain.
3 A(s) + 2 B (l)

5 C (g) + energy
Would the following reaction be spontaneous? Explain.
Energy + A (g) + 2 B (g)  3 C (s)
The spontaneity of a reaction involves two thermodynamic properties: ___________ and _____________.
Willard Gibbs came up with an equation, combining both enthalpy and entropy contributions, that provided a means to describe energy content and therefore a means to evaluate the spontaneity of a reaction when that energy content changes. The energy content of a substance was termed the Gibbs free energy and it was defined by the
Gibbs free energy equation: G = H – T x S
For processes that occur at a constant temperature, the change in free energy (∆G) is given by the equation:
∆ G = ∆H – T x ∆S
____ is free energy ____ is enthalpy ____ is entropy ____is the temperature in Kelvin
The change in Gibbs free energy associated with a chemical reaction is a useful indicator of whether the reaction will proceed spontaneously. Since the change in free energy is equal to the maximum useful work which can be accomplished by the reaction:

Taking a look at the equation, ∆G = ∆H – T∆S, if both enthalpy and entropy are favorable ____∆H and ___∆S for spontaneity, then ∆G < 0 -- Gibbs free energy is _________________.
A ∆G < 0 (negative Gibbs Free Energy) means the reaction is ___________________ and is proceeding to the right making a significant quantity of products.
Free Energy and Spontaneity:
Reactions proceed in the direction that lowers their free energy (-∆G)
∆G o =∆ H o – T∆S o
∆H = (-), ∆S
= (+). Favorable enthalpy change (exothermic) and favorable entropy change (disorder increases)
Result? The reaction is ____________________ at all temperatures, ∆G ____ 0
∆H = (+) ,∆ S = (-).
Unfavorable enthalpy change (endothermic) and unfavorable entropy change (disorder decreases) Result? The reaction is ___________________ at all temperatures, ∆G ____ 0
∆H = (-) ,∆S = (-).
Favorable enthalpy change (exothermic) and unfavorable entropy change (disorder decreases)
Result? Depends on magnitude of enthalpy or entropy and temperature as to whether the reaction is spontaneous.
Generally, spontaneous at _______________ temperatures.
∆H = (+), ∆ S = (+).
Unfavorable enthalpy change (endothermic) and favorable entropy change (disorder increases)
Result? Depends on magnitude of enthalpy or entropy and temperature as to whether the reaction is spontaneous.
Generally, spontaneous at _______________ temperatures.
Standard Gibbs Energy Changes
The Gibbs energy term, G, is a state function , thus values can be defined for substances at specific conditions of temperature and pressure known as the standard state . In this case, we will have ∆G values associated with the formation of compounds from their elemental constituents, known as the standard Gibbs energy of formation,
∆G f o
.
Calculating Free Energy Change
What information will the calculation of ∆G 0 provide?



∆G 0 < 0, the reaction is spontaneous as written (i.e. goes to the right )
∆G 0 > 0, the reaction is nonspontaneous as written, spontaneous in the reverse direction.
∆G 0 = 0, the reaction is at equilibrium
1. ∆G o = ∆H o – T∆S o
Example Problem:
Consider the reaction: 2 SO
2
(g) + O
2
(g)  2 SO
3
(g) carried out at 25 o C and 1 atm. Using the following table data, determine whether the reaction is spontaneous at standard conditions.
What is (are) the driving force(s) for this reaction?
Substance
∆H f o kJ/mol S o (J/K mol)
SO
2
(g)
SO
3
(g)
O
2
(g)
-297
-396
0
248
257
205

2. ∆G 0 =Σn∆G f
0 (products) - Σn∆G f
0 (reactants)
Standard Gibbs free energy values can be used to calculate the standard Gibbs free energy change associated with a
reaction: Table of Thermodynamic Data-Appendix 4- back of text
Example Problem:
Methanol is a high octane fuel used in high-performing racing engines. The reaction that occurs is as follows:
2 CH
3
OH (g) + 3 O
2
(g)  2 CO
2
(g) + 4H
2
O (g)
Would expect this reaction to be exothermic or endothermic?
What sign would you predict for the ∆S o ? What sign would you predict for the ∆G 0 ?
Would predict this reaction to be spontaneous?
Using table data for ∆G f o , calculate the ∆G 0 for this reaction.
3. ∆G o can be calculated by manipulating known equations, as in Hess’s Law problems for ∆H
Example Problem:
Using the following data (at 25 o C), calculate ∆G o for the reaction:
N
2
(g) + O
2
(g)  2NO(g)
N
2
(g) + 2O
2
(g)  2NO
2
(g) ∆G
1 o = 104 kJ
2NO+ O
2
(g)

2NO
2
(g) ∆G
2 o = -70 kJ
How can we interpret the answer we obtained for ∆
G 0 ?

16.7 The Dependence of Free Energy on Pressure
Until now, we have worked with standard conditions. Now we will calculate free energy at nonstandard pressures.
For an ideal gas, enthalpy is not pressure dependent. ________________ is however, affected by pressure.
S low pressure
________ S high pressure
We have the following relationship between free energy and pressure:
G o
G = G o + RTln(P)
- free energy of a gas at a pressure of ____ atm. G - free energy of a gas at P atm
R - universal gas constant ( 8.31 J/mol K) T - Kelvin Temperature
∆G o
∆G = ∆G o + RTln (Q)
- free energy change for a reaction with all reactants and products at a pressure of ____ atm
∆G - free energy change for the reaction for the specified pressure of reactants and products
R - universal gas constant (8.31 J/mol K) T - Kelvin Temperature
Q - reaction quotient (using initial pressures and law of mass action)
Example Problem Calculate ∆ G at 298 K for the following reaction:
C (s) + H
2
O (g)

> CO (g) + H
2
(g)
Initial pressures are P
H2O
= 0.85 atm, P co
= 1.0 x 10 -4 atm, P
H2
= 2.0 x 10 -4 atm
16.8 Free Energy and Equilibrium
Thermodynamically, equilibrium occurs when
G products
= G reactants or ∆G = G products
– G rectants
= 0
Also at equilibrium Q _____ K
Therefore :
Qualitative Relationship between ∆G o and K for a given reaction
-When ∆G o = 0, the system is at equilibrium when the pressures of all the reactants and products are 1atm, which
means K _____ 1.
-When ∆G o < 0, then G o products is less than G o reactants , the system will shift to the right from standard conditions to
attain equilibrium, and K _____ 1.
- When ∆G o < 0, then G o reactants is less than G o products
, the system will shift to the left from standard conditions to
attain equilibrium, and K ______ 1.

Example Problem : The standard free energy change for the reaction: N
2
(g) + 3 H
2
(g)  2 NH
3
(g) is -32.9 kJ.
Calculate the equilibrium constant for this reaction at 25 o C.
Relating ∆G to a Phase Change at Equilibrium
Example Problem: The enthalpy of vaporization (the amount of energy needed to change one mole of a substance from its liquid phase to its vapor phase) of benzene (C
6
H
6
) is 30.8 kJ/mol at its boiling point. If the entropy change for this phase change is +87.3 J/K mol, what is the boiling point of benzene?
Review Free Energy Problem: The overall process for the reducing of iron (III) oxide to iron metal is represented by the following reaction: 2 Fe
2
O
3
(s) + 3 C (graphite)

4 Fe (s) + 3 CO
2
(g)
∆H o rxn
= + 467.9 kJ/mol ∆S o rxn
= +560.3 J/K mol ∆G o rxn
= +300.9 kJ/mol
For the reaction above:
(a) Is the ∆H o rxn
favorable or unfavorable for the process to be spontaneous?
(b) Is the ∆S o rxn
favorable or unfavorable for the process to be spontaneous?
(c) According to the value of ∆G o rxn
, is the reaction spontaneous or nonspontaneous?
(d) For the value of ∆G o rxn
, the value of the equilibrium constant, K, would be equal to, less than, or
greater than 1.
(e) Calculate the minimum temperature at which the reaction would become spontaneous.
(f) What happens to the value of ∆G o rxn as the temperature increases? More positive? More negative?
Stays the same? Use thermodynamic principles in your answer.
(g) Write the K p
expression for the above reaction. What happens to the value of K p
as temperature
increases for the reaction? Use principles of equilibrium in your answer.

ALL HOMWORK TO BE COMPLETED ON SEPARATE SHEET OF PAPER!!!!!!!!
HOMEWORK – CHAPTER 6
Homework – 2/21 (∆H) p. 282 #30,32,33 (a,b) pp. 285-86 #62,66,68
Homework – 2/22 (Calorimetry)
1. In a coffee-cup calorimeter, 50.0 mL of 0.100 M AgNO
3
and 50.0 mL of 0.075 M HCl are mixed. The
two solutions were initially at 22.60
o C, and the final temperature is 23.40
o C. Calculate the heat that
accompanies this reaction in kJ/mol of AgCl formed. Assume a density of 1.00 g/mL.
2. If 10.0 g of solid NaOH is added to 1.00 L of water at 25.0
o C in a constant pressure calorimeter, what will be the final
temperature of the solution? NaOH(s)

NaOH(aq) + 43 kJ
HOMEWORK – CHAPTER 16
p. 285 #56,58
Complete lab and turn in lab report by end of period
(∆S) pp. 829-830 #23,30, 34, 35, 38, 40
(∆G) pp. 830-831 #42, 45c, 50, 54, 56
AP FRQ #1 and #2
Mon 3/4 ELA TAKS
Tues 3/5 Ch 6 and 16 Review
Wed. 3/6 Test: Ch 6 and 16