AP Chemistry
Mr. Markic
Page 1 of 9
Chapter 18 - Entropy, Free Energy, and Equilibrium
Spontaneous Physical and Chemical Processes
• A waterfall runs downhill
• A lump of sugar dissolves in a cup of coffee
• At 1 atm, water freezes below 00C and ice melts above 00C
• Heat flows from a hotter object to a colder object
• A gas expands in an evacuated bulb
• Iron exposed to oxygen and water forms rust
Does a decrease in enthalpy mean a reaction proceeds spontaneously?
Spontaneous reactions
CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (l) ΔH0 = -890.4 kJ/mol

H+ (aq) + OH- (aq)
H2O (s)

NH4NO3 (s)
H2O (l) ΔH0 = -56.2 kJ/mol
H2O (l) ΔH0 = 6.01 kJ/mol

NH4+(aq) + NO3- (aq) ΔH0 = 25 kJ/mol
Entropy (S) is a measure of the randomness or disorder of a system.
Order  S 
Disorder  S 
ΔS = Sf - Si
If the change from initial to final results in an increase in randomness
ΔS > 0
Sf > S i
For any substance, the solid state is more ordered than the liquid state and the liquid state is more
ordered than gas state
Processes that lead to an increase in entropy (ΔS > 0)
Ssolid < Sliquid << Sgas
H2O (s)

H2O (l)
ΔS > 0
AP Chemistry
Mr. Markic
Page 2 of 9
Entropy
State functions are properties that are determined by the state
of the system, regardless of how that condition was achieved.
energy, enthalpy, pressure, volume, temperature, entropy
Sample Exercise
Predict whether the entropy change is greater or less than zero for each of the following processes:
(a) Freezing ethanol
(c) Dissolving glucose in water
(b) Evaporating a beaker of liquid bromine at
room temperature
(d) Cooling nitrogen gas from 80°C to 20°C
How does the entropy of a system change for each of the following processes?
(a) Condensing water vapor
(c) Heating hydrogen gas from 600C to 800C
(b) Forming sucrose crystals from a
supersaturated solution
(d) Subliming dry ice
First Law of Thermodynamics
Energy can be converted from one form to another but energy cannot be created or destroyed.
Second Law of Thermodynamics
The entropy of the universe increases in a spontaneous process and remains unchanged in an
equilibrium process.
Spontaneous process:
Equilibrium process:
ΔSuniv = ΔSsys + ΔSsurr > 0
ΔSuniv = ΔSsys + ΔSsurr = 0
Entropy Changes in the System (ΔSsys)
The standard entropy of reaction (ΔS0) is the entropy change for a reaction carried out at 1 atm
and 250C.
aA + bB 
cC + dD
ΔS0rxn = [cS0(C) + dS0(D)] – [aS0(A) + bS0(B)]
ΔS0rxn = ΣnS0(products) - ΣmS0(reactants)
Sample Exercise
From the following standard entropy values in Appendix 3, calculate the standard entropy changes for
the following reactions 25°C
(a) CaCO3(s)  CaO(s) + CO2(g)
[CaCO3(s) = 92.9 J/Kmol, CaO(s) = 39.8 J/Kmol, CO2(g) = 213.6 J/Kmol]
(b) N2(g) + 3H2(g)  2NH3(g)
[N2(g) = 192 J/Kmol, H2(g) = 131 J/Kmol, NH3(g)= 193 J/Kmol]
AP Chemistry
Mr. Markic
Page 3 of 9
(c) H2(g) + Cl2(g)  2HCl(g)
[H2(g) = 131 J/Kmol, Cl2(g) = 223 J/Kmol, HCl(g) = 187 J/Kmol]
(d) 2CO (g) + O2 (g)  2CO2 (g)
[2CO (g) =197.9 J/Kmol, O2 (g) = 205.0 J/Kmol, CO2 (g) = 213.6 J/Kmol]
Entropy Changes in the System (ΔSsys)
When gases are produced (or consumed)
• If a reaction produces more gas molecules than it consumes, ΔS0 > 0.
• If the total number of gas molecules diminishes, ΔS0 < 0.
• If there is no net change in the total number of gas molecules, then ΔS0 may be positive or
negative BUT ΔS0 will be a small number.
Sample Exercise
Predict whether the entropy change of the system in each of the following reactions is positive or
negative.
(a) 2H2(g) + O2(g)  2H2O(l)
(b) NH4Cl(s)  NH3(g) + HCl(g)
(c) H2(g) + Br2(g)  2HBr(g)
Practice Exercise
Discuss qualitatively the sign of the entropy change expected for each of the following processes:
(a)
I2(s)  2I(g)
(b) 2Zn(s) + O2(g)  2ZnO(s)
(c) N2(g) + O2(g)  2NO(g)
Review of Concepts
Consider the gas-phase reaction of A2 (blue) and B2 (orange) to form AB3.
(a) Write a balanced equation for the reaction.
(b) What is the sign of S for the reaction?
AP Chemistry
Mr. Markic
Page 4 of 9
Entropy Changes in the Surroundings (ΔSsurr)
Exothermic Process
ΔSsurr > 0
Endothermic Process
ΔSsurr < 0
Third Law of Thermodynamics
The entropy of a perfect crystalline substance is zero at the
absolute zero of temperature.
S = k ln W
W=1
S=0
Gibbs Free Energy
Spontaneous process:
ΔSuniv = ΔSsys + ΔSsurr > 0
Equilibrium process:
ΔSuniv = ΔSsys + ΔSsurr = 0
For a constant-temperature process:
Gibbs free energy (G)
ΔG = ΔHsys -TΔSsys
ΔG < 0
The reaction is spontaneous in the forward direction.
ΔG > 0
The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse
direction.
ΔG = 0
The reaction is at equilibrium.
The standard free-energy of reaction (ΔG0 ) is the free-energy change for a reaction when it occurs
under standard-state conditions.
aA + bB
cC + dD
0
Grxn
= [ cG0f (C) + dG0f (D) ] - [ aG0f (A) + bG0f (B) ]
0
Grxn
=  nG0f (products) -  mG0f (reactants)
Standard free energy of formation (ΔG0) is the free-energy change that occurs
when 1 mole of the compound is formed from its elements in their standard states.
ΔG0 of any element in its stable form is zero.
AP Chemistry
Mr. Markic
Page 5 of 9
Sample Exercise
Calculate the standard free-energy changes for the following reactions at 25°C
(a) CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
[CH4(g) = -50.8 kJ/mol, O2(g) = 0 kJ/mol, CO2(g) = -394.4 kJ/mol H2O(l) = -237.2 kJ/mol]
(b) 2MgO(s)  2Mg(s) + O2(g)
[MgO(s) = -569.6 kJ/mol, Mg(s) = 0 kJ/mol, O2(g) = 0 kJ/mol]
Practice Exercise
What is the standard free-energy change for the following reaction at 25 0C? Is the reaction
spontaneous at 25 0C? 2C6H6 (l) + 15O2 (g)  12CO2 (g) + 6H2O (l)
[C6H6 (l) = 124.5 kJ/mol, O2 (g) = 0 kJ/mol, CO2 (g) = -394.4 kJ/mol, H2O (l) = -237.2 kJ/mol]
ΔG = ΔH - TΔS
Review of Concepts
(a) Under what circumstances will an endothermic reaction proceed spontaneously?
(b) Explain why, in many reactions in which both the reactant and product species are in the solution
phase, ΔH often gives a good hint about the spontaneity of a reaction at 298 K.
Temperature and Spontaneity of Chemical Reactions
CaCO3 (s) 
CaO (s) + CO2 (g)
Equilibrium Pressure of CO2
ΔH0 = 177.8 kJ/mol
ΔS0 = 160.5 J/K·mol
ΔG0 = ΔH0 – TΔS0
At 25 oC, ΔG0 = 130.0 kJ/mol
ΔG0 = 0 at 835 oC
AP Chemistry
Mr. Markic
Page 6 of 9
Gibbs Free Energy and Phase Transitions
ΔG0 = 0 = ΔH0 – TΔS0
H2O (l)
ΔS =
𝛥𝐻
𝑇

=
H2O (g)
40.79 𝑘𝐽/𝑚𝑜𝑙
373 𝐾
= 1.09 x 10-1 kJ/K·mol = 109 J/K·mol
Sample Exercise
The molar heats of fusion and vaporization of benzene are 10.9 kJ/mol and 31.0 kJ/mol, respectively.
Calculate the entropy changes for the solid  liquid and liquid  vapor transitions for benzene. At 1
atm pressure, benzene melts at 5.5°C and boils at 80.1°C.
Practice Exerice
The molar heats of fusion and vaporization of argon are1.3 kJ/mol and 6.3 kJ/mol, and argon’s
melting and boiling points are -190°C and -186°C respectively. Calculate the entropy changes for
fusion and vaporization.
Review of Concepts
Consider the sublimation of iodine (I2) at 45°C in a close flask shown here. If the enthalpy of
sublimation is 62.4 kJ/mol, what is the ΔS for sublimation?
Gibbs Free Energy and Chemical Equilibrium
ΔG = ΔG0 + RT lnQ
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature (K)
Q is the reaction quotient
At Equilibrium
ΔG = 0
Q=K
0 = ΔG0 + RT lnK
ΔG0 = - RT lnK
Free Energy Versus Extent of Reaction
ΔG0 < 0
ΔG0 > 0
AP Chemistry
Mr. Markic
Page 7 of 9
ΔG0 = - RT lnK
Sample Exercise
Using data listed in Apx 3, calculate the equilibrium constant (KP) for the following reaction at 25°C
2H2O(l)  2H2(g) + O2(g)
ΔG°rxn = 474.4 kJ/mol ,
Practice Exercise
Calculate the equilibrium constant (KP) for the reaction at 25°C
2O3(g)  3O2(g)
ΔG° = -326.8 kJ/mol
In chapter 16 we discussed the solubility product of slightly soluble substances. Using the solubility
product of silver chloride at 25°C (1.6 x 10-10), calculate G° for the process:
AgCl(s)  Ag+(aq) + Cl-(aq)
Calculate G° for the following process at 25°C, the Ksp of BaF2 is 1.7 x 10-6:
BaF2(s)  Ba2+(aq) + 2F-(aq)
AP Chemistry
Mr. Markic
Page 8 of 9
Sample Exercise
The equilibrium constant (KP) for the reaction
N2O4(g)  2NO2(g)
is 0.113 at 298 K, which corresponds to a standard free energy change of 5.40 kJ/mol. In a certain
experiment, the initial pressures are PNO2 = 0.122 atm and PN2O4 = 0.453 atm. Calculate G for the
reaction at these pressures and predict the direction of the net reaction.
Practice Exercise
The ΔG° for the reaction: H2(g) + I2(g) 2HI(g)
Is 2.60 kJ/mol at 25C. In one experiment, the initial pressures are PH2 = 4.26 atm, PI2 = 0.024 atm,
and PHI = 0.23 atm. Calculate ΔG for the reaction and predict the direction of the net reaction
AP Chemistry
Mr. Markic
Page 9 of 9
Big Idea 5: The laws of thermodynamics describe the essential role of energy and explain and predict the direction
of changes in matter
Duration: Early April
Textbook Chapter: 17 – Entropy, Free Energy and Equilibrium
Enduring Understanding
Essential Knowledge
5.E: Chemical or physical processes are driven by a
decrease in enthalpy or an increase in entropy, or both.
6.D: The equilibrium constant is related to temperature
and the difference in Gibbs free energy between reactants
and products.
5.E.1: Entropy is a measure of the dispersal of matter and
energy.
5.E.2: Some physical or chemical processes involve both a
decrease in the internal energy of the components (ΔH° < 0)
under consideration and an increase in the entropy of those
components (ΔS°> 0). These processes are necessarily
“thermodynamically favored” (ΔG°< 0).
5.E.3: If a chemical or physical process is not driven by both
entropy and enthalpy changes, then the Gibbs free energy
change can be used to determine whether the process is
thermodynamically favored.
5.E.4: External sources of energy can be used to drive change in
cases where the Gibbs free energy change is positive.
5.E.5: A thermodynamically favored process may not occur due
to kinetic constraints (kinetic vs. thermodynamic control).
6.D.1: When the difference in Gibbs free energy between
reactants and products (ΔG°) is much larger than the thermal
energy (RT), the equilibrium constant is either very small (for
ΔG° > 0) or very large (for ΔG° < 0). When ΔG° is comparable to
the thermal energy (RT), the equilibrium constant is near 1.
Learning Objectives
3.11 The student is able to interpret observations regarding macroscopic energy changes associated with a reaction or process
to generate a relevant symbolic and/or graphical representation of the energy changes.
5.3 The student can generate explanations or make predictions about the transfer of thermal energy between systems based
on this transfer being due to a kinetic energy transfer between systems arising from molecular collisions.
5.4 The student is able to use conservation of energy to relate the magnitudes of the energy changes occurring in two or more
interacting systems, including identification of the systems, the type (heat versus work), or the direction of energy flow.
5.12 The student is able to use representations and models to predict the sign and relative magnitude of the entropy change
associated with chemical or physical processes.
5.13 The student is able to predict whether or not a physical or chemical process is thermodynamically favored by
determination of (either quantitatively or qualitatively) the signs of both ΔH° and ΔS°, and calculation or estimation of ΔG°
when needed
5.14 The student is able to determine whether a chemical or physical process is thermodynamically favorable by calculating
the change in standard Gibbs free energy.
5.15 The student is able to explain how the application of external energy sources or the coupling of favorable with
unfavorable reactions can be used to cause processes that are not thermodynamically favorable to become favorable.
5.16 The student can use LeChatelier’s principle to make qualitative predictions for systems in which coupled reactions that
share a common intermediate drive formation of a product.
5.18 The student can explain why a thermodynamically favored chemical reaction may not produce large amounts of product
(based on consideration of both initial conditions and kinetic effects), or why a thermodynamically unfavored chemical
reaction can produce large amounts of product for certain sets of initial conditions.
6.2 The student can, given a manipulation of a chemical reaction or set of reactions (e.g., reversal of reaction or addition of
two reactions), determine the effects of that manipulation on Q or K.
6.24 The student can analyze the enthalpic and entropic changes associated with the dissolution of a salt, using particulate
level interactions and representations.
6.25 The student is able to express the equilibrium constant in terms of ΔG° and RT and use this relationship to estimate the
magnitude of K and, consequently, the thermodynamic favorability of the process.