Chapter 20 Entropy and Free Energy - Spontaneity of Reaction
Thermodynamics: Entropy,
Free Energy
Direction of Chemical Reactions
1. The Second Law of Thermodynamics:
Predicting Spontaneous Change
First Law and enthalpy, Second Law and entropy, Standard Entropy
and Third Law
2. Calculating the Change in Entropy of a Reaction
The standard entropy of reaction, entropy change and equilibrium
3. Entropy, Free Energy, and Work
Free energy change and spontaneity, Standard free energy changes
4. Free Energy, Equilibrium, and Reaction Direction
Important criterions
Spontaneous reaction, in chemical reaction terms, is one which occurs with the
system releasing free energy in some form (often, but not always, heat) and moving
to a lower energy, hence more thermodynamically stable, state.
Enthalpy, H, is the sum of the internal energy and the product of the pressure and
volume of a thermodynamic system.
Entropy, S, is a state function which measures the degree of disorder (randomness)
of a system. The larger the disorder, the greater the value of S in units of J/K.
Free Energy is the measure of the spontaneity of a process and of the useful energy
available from it.
The First Law of Thermodynamics indicates that to conserve energy, the change in
energy must equal the heat transferred plus the work performed, E = q + w.
The Second Law of Thermodynamics states that for a spontaneous process,
∆Suniverse = ∆Ssystem + ∆Ssurroundings > 0 (∆Suniverse positive)
Spontaneous reaction and exothermic /endothermic
Common mistake – exothermic is spontaneous.
Figure 1
A spontaneous endothermic chemical reaction.
water
Ba(OH)2.8H2O(s) + 2NH4NO3(s)
Ba2+(aq) + 2NO3-(aq) + 2NH3(aq) + 10H2O(l)
H0rxn = + 62.3 kJ
The Concept of Entropy (S)
Entropy refers to the state of order, measures the
degree of disorder of a system. .
A change in order is a change in the number of ways of arranging the particles,
and it is a key factor in determining the direction of a spontaneous process.
more order
solid
more order
crystal + liquid
more order
crystal + crystal
less order
liquid
gas
less order
ions in solution
less order
gases + ions in solution
Figure 2
Spontaneous expansion of a gas
stopcock
closed
evacuated
1 atm
In thermodynamic terms, a
change in order is the
number of ways of
arranging the particles.
The change in order is likely
the key factor to determine
the direction of a
spontaneous reaction.
The number of possible
arrangement is increased.
stopcock
opened
0.5 atm
0.5 atm
The number is of ways of
arranging a system is
related to the entropy.
1877 Ludwig Boltzman
S = k ln W
where S is entropy, W is the number of ways of arranging the
components of a system, and k is a constant (the Boltzman constant),
R/NA (R = universal gas constant, NA = Avogadro’s number.
A system with relatively few equivalent ways to arrange its
components (smaller W) has relatively less disorder and low entropy.
A system with many equivalent ways to arrange its components
(larger W) has relatively more disorder and high entropy.
Suniverse = Ssystem + Ssurroundings > 0
This is the second law of thermodynamics.
Figure 4
Random motion in a crystal
The third law of
thermodynamics.
A perfect crystal has
zero entropy at a
temperature of
absolute zero.
Ssystem = 0 at 0 K
Predicting Relative S0 Values of a System
1. Temperature changes
S0 increases as the temperature rises.
2. Physical states and phase changes
S0 increases as a more ordered phase changes to a less ordered phase.
3. Dissolution of a solid or liquid
S0 of a dissolved solid or liquid is usually greater than the S0 of the
pure solute. However, the extent depends upon the nature of the
solute and solvent.
4. Dissolution of a gas
A gas becomes more ordered when it dissolves in a liquid or solid.
5. Atomic size or molecular complexity
In similar substances, increases in mass relate directly to entropy.
In allotropic substances, increases in complexity (e.g. bond flexibility)
relate directly to entropy.
Thermodynamics:
Entropy, Free Energy, and the Direction of Chemical Reactions
1 Three Laws of Thermodynamics:
First Law : ∆Universe = 0, Second Law: ∆S>0, G<0, Third law :
Ssolid=0
2 Calculating the Change in Entropy of a Reaction
∆S0 = mS0products -  nS0reactants
Combine standard molar entropies to find the standard entropy
for a reaction
3 Entropy, Free Energy, and Work
G = H-TS, ∆ G= ∆H- ∆TS, ∆G= ∆H- T∆S, ∆G0= ∆H0 - T∆S0,
4 Free Energy, Equilibrium, and Reaction Direction
Lecture 2
Summary
Spontaneous reaction happens at a specific condition without continuous
input of energy.
Neither first law of thermodynamic or the sign of ∆H predicts the direction
of reaction.
The spontaneous reaction involves a change in the universe from more to
less order.
Entropy is the measure of disorder and directly related to the number of
ways of arrangement for components in the system.
The second law of thermodynamic states that the entropy of universe
increases in a spontaneous process.
Entropy value are absolute since the pure single crystal has a zero entropy at
absolute zero degree.
Standard molar entropy is dependant on the temperature, phase change,
dissolution, atomic size, and molecular complexity.
Sample Problem
Predicting Relative Entropy Values
PROBLEM: Choose the member with the higher entropy in each of the following pairs,
and justify your choice [assume constant temperature, except in part (e)]:
(a) 1mol of SO2(g) or 1mol of SO3(g)
(b) 1mol of CO2(s) or 1mol of CO2(g)
(c) 3mol of oxygen gas (O2) or 2mol of ozone gas (O3)
(d) 1mol of KBr(s) or 1mol of KBr(aq)
(e) Seawater in midwinter at 20C or in midsummer at 230C
(f) 1mol of CF4(g) or 1mol of CCl4(g)
PLAN: In general less ordered systems have higher entropy than ordered systems
and entropy increases with an increase in temperature.
SOLUTION:
(a) 1mol of SO3(g) - more atoms
(d) 1mol of KBr(aq) - solution > solid
(b) 1mol of CO2(g) - gas > solid
(e) 230C - higher temperature
(c) 3mol of O2(g) - larger #mols
(f) CCl4 - larger mass
Important criterions
Spontaneous reaction, in chemical reaction terms, is one
which occurs with the system releasing free energy in some
form (often, but not always, heat) and moving to a lower
energy, hence more thermodynamically stable, state.
Entropy, S, measures the degree of disorder of a system.
The larger the disorder, the greater the value of S in units of
J/K.
Free Energy, G, is the measure of the spontaneity of a
process and of the useful energy available from it.
When a reaction occurs, chemists expect to learn how to predict and
calculate the change in entropy using the following equation:
∆S0 = mS0products -  nS0reactants
Sample Problem
PROBLEM:
Calculating the Standard Entropy of Reaction, S0rxn
Calculate S0rxn for the combustion of 1mol of propane at 25 0C.
C3H8(g) + 5O2(g)
3CO2(g) + 4H2O(l)
PLAN: Use summation equations. It is obvious that entropy is being lost
because the reaction goes from 6 mols of gas to 3 mols of gas.
SOLUTION:
Find standard entropy values in the Appendix or other table.
S = [(3 mol)(S0 CO2) + (4 mol)(S0 H2O)] - [(1 mol)(S0 C3H8) + (5 mol)(S0 O2)]
S = [(3 mol)(213.7J/mol*K) + (4 mol)(69.9J/mol*K)] - [(1 mol)(269.9J/mol*K) +
(5 mol)(205.0J/mol*K)]
Lecture 2
S = - 374 J/K
Determining Reaction Spontaneity
In many spontaneous reaction, the system become more ordered at a
given condition, ∆S0rxn<0.
The Second Law of Thermodynamics indicated that the decrease in
entropy of a system only can occur when the increase the entropy in
surrounding overweigh them.
The entropy change of the surrounding is related to an opposite change
in the heat of the system. ∆S0surr  - qsys
The entropy change of the surrounding is also reversely proportional to
the temperature of the heat transferred. ∆S0surr  1/T
Combining these changes, we obtain the equation:
∆S0surr = ∆S0
Lecture 2
surr =
-
qsys
T
∆ Hsys
T
When a pressure is
constant, the heat is ∆H.
( ∆H = q + ∆ PV)
Sample Problem
PROBLEM:
Determining Reaction Spontaneity
At 298K, the formation of ammonia has a negative S0sys;
N2(g) + 3H2(g)
2NH3(g)
S0sys = -197 J/K
Calculate S0rxn, and state whether the reaction occurs
spontaneously at this temperature.
PLAN: S0universe must be > 0 in order for this reaction to be spontaneous, so
S0surroundings must be > 197 J/K. To find S0surr, first find Hsys; Hsys =
Hrxn which can be calculated using H0f values from tables.
S0universe = S0surr + S0sys.
SOLUTION:
H0rnx = [(2 mol)(H0fNH3)] - [(1 mol)(H0fN2) + (3 mol)(H0fH2)]
H0rnx = -91.8 kJ
S0surr = -H0sys/T = -(-91.8x103J/298K) = 308 J/K
S0universe = S0surr + S0sys = 308 J/K + (-197 J/K) = 111 J/K
S0universe > 0 so the reaction is spontaneous.
Lecture 2
Gibbs Free Energy (G)
G, the change in the free energy of a system, is a
measure of the spontaneity of the process and of the
useful energy available from it.
G0system = H0system - TS0system
G < 0 for a spontaneous process
G > 0 for a nonspontaneous process
G = 0 for a process at equilibrium
G0rxn = S mG0products - S nG0reactants
Lecture 2
Sample Problem
PROBLEM:
Calculating G0 from Enthalpy and Entropy Values
Potassium chlorate, a common oxidizing agent in fireworks and
matchheads, undergoes a solid-state disproportionation reaction when
heated.
Note that the oxidation number of Cl in the reactant is higher in one of the products and
lower in the other (disproportionation).
+5

4KClO3(s)
+7
-1
3KClO4(s) + KCl(s)
Use H0f and S0 values to calculate G0sys (G0rxn) at 250C for this reaction.
PLAN:
Use Appendix B values for thermodynamic entities; place them into
the Gibbs Free Energy equation and solve.
SOLUTION:
H0rxn = S mH0products - S nH0reactants
H0rxn = (3 mol)(-432.8 kJ/mol) + (1 mol)(-436.7 kJ/mol) (4 mol)(-397.7 kJ/mol)
H0rxn = -144 kJ
Lecture 2
Sample Problem
Calculating G0 from Enthalpy and Entropy Values
continued
S0rxn = S mS0products - S nS0reactants
S0rxn = (3 mol)(151 J/mol*K) + (1 mol)(82.6 J/mol*K) (4 mol)(143.1 J/mol*K)
S0rxn = -36.8 J/K
G0rxn = H0rxn - T S0rxn
G0rxn = -144 kJ - (298K)(-36.8 J/K)(kJ/103 J)
G0rxn = -133 kJ
Lecture 2
Entropy, Free Energy, Work
Free Gibbs energy is the measure of the spontaneity of a process.
• ∆G0sys = ∆H0sys - T∆S0sys
• T∆S0univ = - (∆H0sys - T∆S0sys)
• The sign for ∆G0sys and ∆S0univ can tell us the spontaneity of a
reaction.
• ∆G0sys < 0 for a spontaneous reaction;
• ∆G0sys = 0 for a reaction at equilibrium;
• ∆G0sys > 0 for a nonspontaneous reaction;
• ∆S0univ > 0 for a spontaneous reaction;
• ∆S0univ = 0 for a reaction at equilibrium;
• ∆S0univ < 0 for a nonspontaneous reaction.
Lecture 2
The spontaneity of reaction and the sign of ∆H, ∆G, ∆S.
∆G0 = ∆H0 - T∆S0
∆H0
∆S0
-T∆S0
∆G0
Description
-
+
-
-
Spontaneous at all T
+
-
+
+
Non-spontaneous at all T
+
+
-
+ or -
Spontaneous at high T
Non-spontaneous at low T
-
-
+
+ or -
Spontaneous at low T
Non-spontaneous at high T
Determining the effect of temperature on G
PROBLEM: An important reaction in production of sulfuric acid is the oxidation of
SO2 (g) to SO3(g),
2SO2(g) + O2(g)
2SO3(g)
At 298K, ∆G0 = -141.6 kJ/mol, ∆H0 = -198.4 kJ/mol, ∆S0 = -187.9kJ/mol,
(a). Use the data to predict if this reaction is spontaneous reaction at 25 C.
(b). Predict how ∆G0 changes when temperature increases.
(c). Assume ∆H0 and ∆S0 are constant with increasing T, is the
reaction spontaneous at 900 C?
PLAN:
Note the sign of ∆G0, ∆H0, and T∆S0
Use the equations ∆G0 = ∆H0 - T∆S0
Determining the effect of temperature on G
SOLUTION:
(a) Noting G = -141.6 kJ, is negative, so the reaction at
room temperature is spontaneous.
(b) Noting G = H – TS
= -198.4 - (- 0.1879T) = -198.4 + 0.1879T
If T increases, the G is less negative. Therefore, the
reaction is less spontaneous.
(c) Calculating G = H – TS at 900 C
= -198.4 - (- 0.1879T)
= -198.4 + 0.1879(900+273.15)
= 22.0 kJ
At 900 C, G is positive. Therefore, the reaction is non- spontaneous.
Free Energy, Equilibrium and Reaction Direction
•If Q/K < 1, then ln Q/K < 0; the reaction proceeds to the right (G < 0)
•If Q/K > 1, then ln Q/K > 0; the reaction proceeds to the left (G > 0)
•If Q/K = 1, then ln Q/K = 0; the reaction is at equilibrium ( G = 0)
G = RT ln Q/K = RT lnQ - RT lnK
Under standard conditions (1M concentrations, 1atm for gases), Q = 1 and ln
Q = 0 so
G0 = - RT lnK
Table The Relationship Between G0 and K at 250C
G0(kJ)
K
100
3x10-18
50
2x10-9
10
2x10-2
1
7x10-1
0
1
-1
1.5
-10
5x101
-50
6x108
-100
3x1017
-200
1x1035
Essentially no forward reaction;
reverse reaction goes to completion
Forward and reverse reactions
proceed to same extent
Forward reaction goes to
completion; essentially no reverse
reaction
REVERSE REACTION
9x10-36
FORWARD REACTION
200
Significance
Calculating G at Nonstandard Conditions
PROBLEM:
The oxidation of SO2,
2SO2(g) + O2(g)
2SO3(g)
is too slow at 298K to be useful in the manufacture of sulfuric acid. To
overcome this low rate, the process is conducted at an elevated temperature.
(a) Calculate K at 298K and at 973K. (G0298 = -141.6kJ/mol of reaction as
written using H0 and S0 values at 973K. G0973 = -12.12kJ/mol of reaction
as written.)
(b) In experiments to determine the effect of temperature on reaction
spontaneity, two sealed containers are filled with 0.500atm of SO2,
0.0100atm of O2, and 0.100atm of SO3 and kept at 250C and at 700.0C. In
which direction, if any, will the reaction proceed to reach equilibrium at each
temperature?
(c) Calculate G for the system in part (b) at each temperature.
PLAN: Use the equations and conditions. G0 = -RTlnK
Calculating G at Nonstandard Conditions
SOLUTION: (a) Calculating K at the two temperatures:
G0
= -RTlnK so
(G / RT)
Ke
0
(-141.6kJ/mol)(103J/kJ)
At 298, the exponent is -G0/RT = -
= 57.2
(8.314J/mol*K)(298K)
(G / RT)
Ke
0
= e57.2 = 7x1024
(-12.12kJ/mol)(103J/kJ)
At 973, the exponent is -G0/RT
= 1.50
(8.314J/mol*K)(973K)
(G / RT)
Ke
0
= e1.50 = 4.5
Calculating G at Nonstandard Conditions
2SO2(g) + O2(g)
2SO3(g)
pSO32
(b) The value of Q =
(pSO2)2(pO2)
(0.100)2
=
= 4.00
(0.500)2(0.0100)
Since Q is < K at both temperatures the reaction will shift right; for 298K
there will be a dramatic shift while at 973K the shift will be slight.
(c) The nonstandard G is calculated using G = G0 + RTLnQ
G298 = -141.6kJ/mol + (8.314J/mol*K)(kJ/103J)(298K)(Ln4.00)
G298 = -138.2kJ/mol
G973 = -12.12kJ/mol + (8.314J/mol*K)(kJ/103J)(973K)(Ln4.00)
G298 = -0.9kJ/mol
Thermodynamic
Important equations
• ∆G0sys = ∆H0sys - T∆S0sys
• ∆S0univ = - qsys/T
• ∆G0 = ∆H0 - T∆S0
• ∆S0 = mS0products -  nS0reactants
• ∆G0 = mG0products -  nG0reactants
• ∆G0 = - RT lnK
• ∆G = ∆G0 - RT lnQ