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AP Chemistry
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5  Thermochemistry & Thermodynamics
NOTES
5.1 The Nature of Energy
 Thermodynamics is the study of energy and its transformations
o Thermochemistry is the relationship between chemical reactions and energy changes involving
heat
 Energy is the capacity, or ability, to do work or transfer energy. There are 2 types:
o Kinetic energy, Ek, is the energy of motion and depends on the mass, m, and speed, v, of the
object
Ek = ½ mv2
o Potential energy is the energy of an object based on its position relative to another object. In
short, it is the potential to perform work
o Work is the force applied to move an object with mass for a distance
w=Fxd
 Force is any kind of push or pull on an object
o Heat is energy that causes an object’s temperature to increase
 Joule is the SI unit for energy (1 J = 1 kg·m2/s2). It is small so we usually use kilojoules (kJ)
o calorie (cal) is a non-SI unit but is often used ( 1 cal = 4.184 J )
 A system refers to a specific area or reaction that we single out to observe; everything not in a
system is the surroundings
o An open system allows for exchange of energy and matter between the system and the
surroundings
o A closed system allows energy (work and heat) to be exchanged with the surroundings but not
matter. Most thermochemistry deals with closed systems
o An isolated system doesn’t allow any exchange between the two
5.2 The First Law of Thermodynamics
 The First Law of Thermodynamics basically means energy is conserved. If energy is lost by a
system, it goes to its surroundings and vice versa but it has not gone away.
 Internal energy, E, is the sum of all energies, both kinetic and potential, in a system
o The change in internal energy, ΔE, and is the only part that matters although it comes from
ΔE = Efinal - Eintial
o In a chemical reaction, Eintial is the reactants and Efinal is the products
*Fig 5.5 shows an energy diagram for a reaction.
o Positive ΔE means the system has gained energy from the surroundings
o Negative ΔE means the system has lost energy to the surroundings
o When heat, q, is added or work is done on a system, internal energy increases
ΔE = q + w
*Table 5.1 tells how sign changes affect heat, work, and internal energy. Any positive shows that
energy is entering the system
 Endothermic processes involve a system absorbing heat


Exothermic processes involve a system losing heat
State function is the property of a system that depends on its state, such as the temperature and
pressure
o Its value depends only on the difference between the system’s final state and initial state, or its
present state and not how it got there
*Fig 5.8 shows how internal energy is a state function
5.3 Enthalpy
 Pressure-volume work is the amount of energy involved in the compression and expansion of a gas
w = -P ΔV ; where ΔV = Vfinal - Vintial
 Enthalpy, H, is the heat content of a chemical; it is a state function
ΔH = Hfinal – Hinitial
 The relationship between energy and enthalpy is that
ΔE = ΔH – PΔV
o ΔE is almost equal to ΔH because PΔV is typically very small
5.4 Enthalpies of Reaction
 The enthalpy of a reaction, or heat of a reaction, can be found by
ΔH = Hproducts – Hreactants ; ΔH can also be written ΔHrxn
o ΔH means the reaction endothermic and – ΔH means the reaction exothermic
 Enthalpy is an extensive property, meaning that it depends on the amount of substance involved. For
example, if we burned 2 moles of CH4 instead of just 1 in the below reaction, then the enthalpy
would double as well and ΔH = -1780 kJ
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)
ΔH = -890 kJ
 Enthalpy of a reaction is equal and opposite for the reverse of a reaction. For example, if we flipped
the above reaction so that the products were the reactants and vice versa, then ΔH = +890 kJ
 Enthalpy change depends on the state of matter. For example, the production, or formation, of liquid
H2O has a lower ΔH than the formation of gaseous H2O
*Appendix C lists the enthalpies of formation, ΔHf, for many substances
 You can use the relationship between the molar ratio of a reaction and the ΔH of the reaction to find
the heat, q, of a specified amount
*Sample Exercise 5.4 shows to how to solve for q in a reaction given the balanced reaction and it’s ΔH
5.5 Calorimetry
 Calorimetry is the measurement of heat flow.
o It is measured using a calorimeter.
 Heat capacity, C, is the amount of heat energy it takes to raise an object’s temperature by 1 °C/1 K
 Molar heat capacity, Cm, is the heat capacity of 1 mole of a substance
 Specific heat [capacity], Cs, is the heat capacity of 1 gram of a substance; units are J/g·°C
*Table 5.2 shows specific heat of some substances. Appendix B shows Cs for H2O
 We can calculate the heat change of a substance by using its specific heat
q = Cs x m x ΔT


A very common calorimeter is a Styrofoam cup because it does not absorb any of the reaction’s heat
so the exchange in energy is strictly between the system and its surroundings.
o It is under constant pressure because it is under atmospheric pressure
o We can assume that qreaction = -qsolution and that they are always opposite
A bomb calorimeter measures heat while keeping a constant volume
o Moles are not needed because the calorimeter’s specific heat is used
5.6 Hess’s Law
 Hess’s Law basically says that if a reaction happens in steps, then the ΔH of the overall reaction is
equal to the sum of the ΔH for the individual steps. For example:
Step 1:
Step 2:
Overall:
CH4(g) + 2 O2(g)
2 H2O(g)
CH4(g) + 2 O2(g) + 2 H2O(g)
→
→
→
CO2(g) + 2 H2O(g)
2 H2O(l)
CO2(g) + 2H2O(l) + 2H2O(g)
ΔH = -802 kJ
ΔH = -88 kJ
ΔH = -890 kJ
From the overall step, you cancel out items that appear on both sides of the arrow to get the net equation
CH4(g) + 2 O2(g)
→
CO2(g) + 2 H2O (l)
ΔH = -890 kJ

o Sometimes you will need to flip a step around so that the correct product is achieved. Sample
Exercise 5.9 shows this method.
o Sometimes you flip steps to make sure that the product of one step is the reactant of another step.
Sample Exercise 5.8 shows this method.
Enthalpy, H, is a state function which means that it is the same whether a reaction takes 1 or multiple
steps. For example, if you combine oxygen and hydrogen to get water and the ΔH = -285.8 kJ, then
it will equal that amount whether you do it in 1 step or 5 steps.
5.7 Enthalpies of Formation
 There are many types of enthalpy for each substance:
o Enthalpy of vaporization is the energy it takes to change a liquid to a gas
o Enthalpy of fusion is the amount of energy it takes to melt a solid substance
o Enthalpy of combustion is the amount of energy it takes to burn a substance in oxygen
o Enthalpy of formation (ΔHf )is the amount of energy it takes to create a substance from its
component substances (things that it is made from)
 Standard state is a substance’s state at 1 atm of pressure and 298 K
 Standard enthalpy change (ΔH° ) is the enthalpy change while all reactants and products are in their
standard states (the little ° means that it is referring to the standard state)
 Standard enthalpy of formation (ΔH°f) is the difference in enthalpy when one mole of a substance is
made from elements while both the product and its reactants are in their standard state
o The standard enthalpy of formation of any stable element, even the diatomics, is 0
*See Equation 5.25 for an example of the formation of a compound while in standard states
*See Table 5.3 and Appendix C for the ΔHf ° of many common substances
 You can use the enthalpies of formation for substances in a reaction to determine the enthalpy of the
reaction
ΔH°rxn = Σn ΔHf °(products) - Σm ΔHf °(reactants), where n and m are the coeffecients in the reaction
o Write an example of creating the enthalpy of reaction below from Pg 190
*See Sample Exercises 5.11 and 5.12 show how to use the enthalpies of formation to find the enthalpy
of a reaction and how to use the enthalpy of reaction to find the enthalpies of formation
19.1 Spontaneous Processes
 A spontaneous process is a process that will occur on its own and does not need outside forces to get
it going
o It is in a definite direction – goes only one way
o Can occur at any speed; thermodynamics only tells us what direction the reaction will go in and
how far it will go
o If a process is spontaneous, then the reverse is nonspontaneous
 For example, at room temperature, an ice cube spontaneously melts. But in order for us to
freeze it again, we have to change the heat, which means that we had to apply an outside
force
 A reversible process is a process in which a system can go back the way it was by simply doing the
opposite
o For example, if you
o A reversible change produces the maximum amount of work that a system can perform on its
surroundings
 An irreversible process is a process in which more than just doing the opposite thing makes the
system reverse
o For example, if you cook chicken,
o A spontaneous process is irreversible
o An isothermal process is a process with a constant temperature; it is typically used when dealing
with the pressure and volume changes of gases
19.2 Entropy and the Second Law of Thermodynamics
 Entropy (S) can generally be described as the amount of randomness, or disorder, in a system
o Think: A gas has more entropy than a solid because its particles are more unorganized
o Entropy, just like internal energy (E) and enthalpy (H), is also a state function meaning that the
state of the system makes a difference.
 The Second Law of Thermodynamics basically says that all spontaneous processes (irreversible
processes) move in the direction that has an increase in entropy, or disorder, in the universe
o For example, ice melting is a spontaneous process and as it moves from a solid to a liquid, the
entropy increases
o For a reversible process
∆Suniverse = ∆Ssystem + ∆Ssurroundings = 0
o For an irreversible process
∆Suniverse = ∆Ssystem + ∆Ssurroundings > 0
19.3 The Molecular Interpretation of Entropy
 Atoms and molecules are in constant motion and thusly posses some type of entropy
 Entropy increases (+S or S > 0) because of the following:
o from solid to liquid or liquid to gas or solid to gas
o fewer moles/molecules of a gas to more moles/molecules of a gas
o simpler molecules to more complex molecules
o smaller molecules to longer molecules
o ionic solids with strong attractions to ionic solids with weaker attractions
o mixing a solid solute & a solvent to form a solution
o gas dissolved in water becoming an escaped gas
o generally an increase in temperature
o ANYTHING OPPOSITE the above is a decrease in entropy
*See Sample Exercises 19.3 and 19.4 for examples of whether entropy is increasing or decreasing
 The Third Law of Dynamics basically describes Absolute 0 (0 K). It states that a perfect solid has
no entropy, or no movement, at Absolute 0
 Entropy generally increases as temperature increases (Ssolid<Sliquid<Sgas)
*Figure 19.14 shows the entropy as a function of temperature in a phase change diagram
19.4 Entropy Changes in Chemical Reactions
 Standard molar entropy (S°) is the entropy of a substance in its standard state, which is the pure
substance at 1 atm and 298 K; the ° symbol represents that it is in its standard state. It has the units
J/mol·K. Keep in mind that:
o Molar entropies of gases are higher than the molar entropies of liquids
o Standard molar entropy increases as molar mass increases
o Standard molar entropy increases as the number of atoms in the molecules increase
 Just like ΔH°, we can calculate ΔS° using the S° of the substances
ΔS°rxn = Σn S°(products) - Σm S°(reactants), where n and m are the coefficients in the reaction
*Sample Exercise 19.5 shows how to calculate ΔS°rxn , which can also be simply called ΔS°
*Appendix C lists the S° of many substances
 We can relate entropy and enthalpy through the equation
ΔSsurroundings = -ΔHsystem/ T ; where T is the temperature in K
19.5 Gibbs Free Energy
 Gibbs free energy (G) is the amount of energy available in a chemical reaction and is also a state
function. It is also in the units kJ/mol, just like H
o It is related to enthalpy and energy through the equation
ΔG° = ΔH° - TΔS°; where T is the absolute temperature (K)
*Sample Exercise 19.8 shows how to use this equation to find ΔG°
 ΔG can also be used to determine if a reaction is spontaneous
o If ΔG < 0 then the reaction is spontaneous in the forward direction
o If ΔG = 0 then the reaction is at equilibrium
o If ΔG > 0 then the reaction is nonspontaneous in the forward direction; however, the reverse
reaction is spontaneous
 Standard free energy of formation is the free energy of a substance at its standard state
 Just like ΔH° and ΔS°, we can calculate ΔG° using the G° of the substances
ΔG°rxn = Σn G°(products) - Σm G°(reactants), where n and m are the coefficients in the reaction
*Sample Exercise 19.7 shows how to find ΔG° using the G° of each substance
*Appendix C lists the G° of many substances
19.6 Free Energy and Temperature
 In the equation ΔG° = ΔH° - TΔS°, temperature is used therefore the value of ΔG° is affected by the
temperature
 We can summarize the relationships between G, H, and S and how spontaneity of a reaction is
affected by them through the following table
ΔH°
ΔS°
ΔG°
Reaction Type
+
+
-
+
Always spontaneous /spontaneous at all temperatures
-
-
+ or -
+
+
+ or -
Never spontaneous / nonspontaneous at all temperatures
Spontaneous at low temps / nonspontaneous at high temps;
can decrease temp to make it spontaneous
Spontaneous at high temps / nonspontaneous at low temps;
can increase temp to make it spontaneous
*Table 19.4 offers an alternative version of this table
19.7 Free Energy and the Equilibrium Constant can be simplified to the below at
We will not study Equilibrium until Unit 5 however I will introduce some of the terminology so you will
be familiar with it when we revisit this section later.

We can find ΔG, which is the free energy at nonstandard conditions by using the ΔG°, which is at
the standard conditions of 298 K and 1 atm by using the following equation
ΔG = ΔG° + RT ln Q
o R is the ideal gas constant of 8.314 J/mol·K
o T is the absolute temperature
o Q is the reaction quotient and refers to the equilibrium expression of a specific reaction and is
basically [Products]/[Reactants]
 At equilibrium, we can solve for ΔG° using
ΔG° = -RT ln K; where K is the equilibrium constant
*Table 19.5 shows the relationship between ΔG° and K
*Appendix D lists some equilibrium constants, K, for several substances
Additional Notes: