Unit 18 – Thermochemistry and
Equilibrium
Previously on General Chemistry…
Thermodynamics is the study of heat and other forms of
energy involved in chemical or physical processes
• Energy is exchanged between system and surroundings
• Chemical reactions are systems
• Signs of energy changes are system-centric
• Reactions can be exothermic (q<0) or endothermic
(q>0)
• At constant pressure, DH = qP
• Calorimeters measure q
18 | 2
Previously on General Chemistry…
State function: function only of initial and final state
of system, not on path from initial to final state
Energy is a state function
18 | 3
Previously on General Chemistry…
First Law of Thermodynamics:
Energy of the Universe is Constant
DEsystem + DEsurroundings = 0
DEsystem = q + w
Energy cannot be created or destroyed
Calculation of q: calorimetry:
q = m Cs DT
Calculation of w:
PV work: w = −PDV (constant P)
18 | 4
Previously on General Chemistry…
Enthalpy, H - An extensive property of a
substance that can be used to obtain the heat
absorbed or evolved in a chemical reaction.
The change in enthalpy is relatable to heat
q = nΔH
N2(g) + 3H2(g) → 2NH3(g); ΔH = -91.8 kJ
ΔH = -91.8 kJ/(1 mol N2)
ΔH = -91.8 kJ/(3 mol H2) = -30.6 kJ/(1 mol H2)
18 | 5
Previously on General Chemistry…
Manipulations for Thermochemical Equations
1. When a chemical equation is reversed, the value
of ΔH is reversed in sign.
2. When a thermochemical equation is multiplied
by any factor, the value of ΔH for the new
equation is obtained by multiplying the ΔH in
the original equation by that same factor.
3. The ΔH of a reaction that is the sum of two or
more reactions is equal to the sum of the ΔH
values of the constituent reactions.
18 | 6
Previously on General Chemistry…
The standard enthalpy of formation, ΔHf°, is the
enthalpy change for the formation of one mole of the
substance from its elements in their reference forms
and in their standard states (1 atm, 25 °C). ΔHf° for an
element in its reference and standard state is zero.
H2(g) + 1/2O2(g) → H2O(l)
Hf° = –285.8 kJ
We can calculate the standard enthalpy of the reaction
using the standard enthalpies of formation, ΔHf°:
ΔHrxn° = ΣnpΔHf°(products) − ΣnrΔHf°(reactants)
18 | 7
Spontaneous Processes
Spontaneous process:
Processes that will occur under
given conditions
Nonspontaneous process:
Processes that require energy to
occur under given conditions
Spontaneity depends on dispersion of energy that
occurs during a process. 18 | 8
Spontaneous Processes
At room temperature…
Spontaneous Process
Non-spontaneous Process
18 | 9
The Energy Tax
More than 60% of energy used for electricity generation
is lost in conversion
Every energy transition results in a “loss” of energy
– an “Energy Tax” demanded by nature
– and conversion of energy to heat which is “lost” by heating
up the surroundings
Why?
18 | 10
Spontaneous Processes
First law of Thermodynamics: Energy is not created or
destroyed (Conservation of Energy)
We talk at exothermic processes being energetically
“favorable” because energy is released, but favorable
does not necessarily mean spontaneous.
Is every exothermic process spontaneous? Is every
endothermic process nonspontaneous?
Non-spontaneous processes do not
violate the first law of thermodynamics,
we need another law of thermodynamics
to predict spontaneity: the Second Law
of Thermodynamics
18 | 11
Second Law of Thermodynamics
Second Law of Thermodynamics:
The total entropy of the universe increases
in any spontaneous process
ΔSuniverse = ΔSsystem + ΔSsurroundings > 0 (spontaneous proc.)
Entropy (S):
A measure of the distribution of energy
in a system at a specific temperature
Energy distribution affected by molecular motion,
volume
18 | 12
Entropy and the Second Law of
Thermodynamics
The second law of thermodynamics states that
the total entropy of a system and its
surroundings always increases for a
spontaneous process.
The net change in entropy of the system, DS ,
equals the sum of the entropy created (or
destroyed) during the spontaneous process and
the change in entropy associated with the heat
flow.
18 | 13
13
Entropy and the Second Law of
Thermodynamics
Entropy and Molecular Disorder
Entropy is essentially related to energy
dispersal. The entropy of a molecular
system may be concentrated in a few
energy states and later dispersed among
many more energy states. The entropy of
such a system increases.
In the case of the cup of hot coffee, as heat
moves from the hot coffee, molecular
motion becomes more disordered. In
becoming more disordered, the energy is
more dispersed.
18 | 14
Entropy and Microstates
The motion of molecules is quantized:
Different molecular states related to molecular
motion are separated by specific energies
Energy state or energy level:
An allowed value of energy
Microstate:
A unique distribution of particles among energy
levels
18 | 15
Types of Molecular Motion
Three types of motion:
Translational—movement through
space
Rotational—spinning motion around
axis ⊥ to bond
Vibrational—movement of atoms
toward/away from each other
As temperature increases, the amount
of motion increases
18 | 16
Quantized Energy States
Quantized rotational
states.
Quantized vibrational
states of O2 molecule.
18 | 17
Microstates: Energy Distribution
The number of times
a molecule occupies an
accessible energy level
follows a Boltzmann
distribution:
Number of energy states
increases as volume
increases
18 | 18
Microstates: Energy Distribution
Increasing volume
“stretches” the energy
distribution:
Accessible energy levels
move closer together.
Number of accessible
energy states increases.
18 | 19
Statistical Entropy
Boltzmann Equation:
S = k ln W
S = entropy
W = # of microstates (Ω)
k = Boltzmann constant (1.38 × 10-23 J/K)
This equation indicates that entropy increases
as the number of microstates increases.
18 | 20
Second Law of Thermodynamics
Statistical interpretation of Second Law
insulated,
rigid
containers
→
low entropy
→
high entropy
Spontaneous
insulated,
rigid
containers
high entropy
→
low entropy
18 | 21
Not Spontaneous
Statistical entropy
These are energetically
equivalent states for the
expansion of a gas
It doesn’t matter, in terms of
potential energy, whether the
molecules are all in one flask, or
evenly distributed
But one of these states is more
probable than the other two
18 | 22
Tro: Chemistry: A Molecular Approach, 2/e
Macrostates → Microstates
These microstates
all have the same
macrostate
So there are six
different particle
arrangements that
result in the same
macrostate
18 | 23
Tro: Chemistry: A Molecular Approach, 2/e
Macrostates and Probability
There is only one possible
arrangement that gives State A
and one that gives State B
There are six possible arrangements
that give State C
Therefore State C has higher
entropy than either State A or
State B
The macrostate with the highest
entropy also has the greatest
dispersal of energy, which is State C
18 | 24
Tro: Chemistry: A Molecular Approach, 2/e
Thermodynamic Entropy
Isothermal Process:
A process that takes place at constant T.
Reversible Process:
A process that can be run in the reverse direction
with no net heat flow into or out of the system.
For an isothermal process: ΔSsys= qrev/T
qrev = flow of heat for reversible process.
18 | 25
Entropy Change For Phase Transition
Consider phase change reversible process occurring
at constant T and P
q
ΔS =
(reversible process, constant T)
T
q is given by enthalpy change of phase transition
T is given by the temperature of the phase
transition
18 | 26
Second Law of Thermodynamics
Thermodynamic interpretation of Second Law
ΔSuniverse = ΔSsystem + ΔSsurroundings
a process will be spontaneous if |q/T| transferred to or
from the surroundings is greater than the |q/T|
transferred from or to the surroundings (irreversible
process, constant T)
exothermic process: q is negative, system decreases its
entropy, surroundings increase its entropy
endothermic process: q is positive, system increases its
entropy, surroundings decrease its entropy
18 | 27
Entropy of Cold Packs
ΔHsoln = positive, (heat lost
by surroundings → ΔSsurr =
negative)
ΔSsys = positive, due to
increased freedom of
movement of dissolved ions
Net: ΔSuniverse > 0
18 | 28
Entropy of the Universe
2nd Law: “Entropy of the universe increases for a
spontaneous process”
ΔSuniv = ΔSsys + ΔSsurr
Consider an ice cube melting at 0.0°C (273 K) on a
counter top at room temperature (293 K)
• Heat flows into ice cube: ΔSsys = qrev/273
• Heat flows out of counter top: ΔSsurr = −qrev/293
• ΔSuniv= (qrev/273) + (−qrev/293) > 0 spontaneous!
18 | 29
Factors Affecting Entropy
Temperature:
Entropy increases as T increases
Volume:
Entropy increases as volume increases
Number of particles:
Entropy increases as the number of particles increases
Can use these observations to predict entropy changes
associated with reactions
18 | 30
Ways to increase entropy, DS > 0
Increase # of Particles
Increase Energy of Particles
Linear Molecule to Branched
Increase the Volume
Decompose Particles
18 | 31
Entropy and Temperature
Entropy increases as temperature increases
Increases in kinetic energy increase
the number of accessible microstates
Decreasing temperature decreases entropy
At what temperature does all molecular motion
cease, and entropy equal zero?
18 | 32
Third Law of Thermodynamics
Third Law of Thermodynamics:
The entropy of a perfect crystal is zero at absolute zero
Absolute Entropy:
The entropy of a substance at some temperature
above 0 Kelvin
Calculated from measurement of molar heat capacities
as a function of temperature
Standard Molar Entropy (S°):
The absolute entropy of 1 mole of a substance in its
standard state at 298 K and 1 bar of pressure
18 | 33
Trends in Entropies
Ssolid < Sliquid < Sgas
Example:
H2O at 298 K
H2O(l) = 69.9 J/(mol∙K)
H2O(g) =188.8 J/(mol∙K)
ΔSvap = 119 J/(mol∙K)
18 | 34
Other Trends
Entropy increases as the complexity of
molecular structure increases
More bonds, more opportunities for internal motion
(more microstates)
S° (J/(mol∙K):
186
230
270
18 | 35
310
Entropy Differences in Allotropes
Rigid network of covalent
More flexibility and range of
bonds; less range of motion motion in planar rings
= less entropy
= more entropy
18 | 36
Calculating Entropy Changes
Entropy change for the system (ΔSrxn°):
ΔS°rxn = ΣnproductsS°products − ΣnreactantsS°reactants
Where n = coefficients of the products/reactants in the
balanced equation
Entropy change for the universe:
Heat gained/lost by system affects entropy
of the surroundings
ΔSsurr° = −qsys / T
18 | 37
(qsys = ΔHrxn°)
Standard Molar Entropies at 25 °C and 1 atm
Substance
S° (J/(mol·K))
Substance
S° (J/(mol·K))
Br2(g)
245.5
CH4(g)
186.2
Br2(l)
152.2
C2H6(g)
229.5
Cdiamond(s)
2.4
CH3OH(g)
239.9
Cgraphite(s)
5.7
CH3OH(l)
126.8
CO(g)
197.7
CH3CH2OH(g)
282.6
CO2(g)
213.8
CH3CH2OH(l)
160.7
H2(g)
130.6
CH3CH2CH3(g)
269.9
N2(g)
191.5
CH3(CH2)2CH3(g)
310.0
O2(g)
205.2
CH3(CH2)2CH3(l)
231.0
P4(s, red)
22.8
CH3(CH2)6CH3(g)
466.7
P4(s, white)
41.1
CH3(CH2)6CH3(l)
361.1
H2O(g)
188.8
C6H6(g)
269.2
H2O(l)
69.9
C6H6(l)
172.9
NH3(g)
192.5
C12H22O11(s)
360.2
18 | 38
Practice: Calculating ΔS°
1. For the following reaction, calculate ΔS°rxn.
2. If ΔH°comb for one mole of C2H6 is -1560 kJ at
25 °C, what is ΔS°univ?
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)
18 | 39
Free Energy
Free energy (G):
The energy available to do useful work
Free energy change (ΔG):
The change in free energy of a process occurring
at constant temperature and pressure
For spontaneous processes, ΔG < 0
18 | 40
ΔG and Entropy of the Universe
For a spontaneous process:
Δ Suniverse = Δ Ssys + Δ Ssurroundings > 0
Change in S of system = ΔSsys
Change in S of surroundings = −ΔHsys/T
Change in S of universe: ΔSuniverse = ΔSsys − ΔHsys/T > 0
Multiply by -T: −TΔSuniverse = ΔHsys − TΔSsys
Plug in ΔGsys = −TΔSuniverse
ΔGsys = ΔHsys − TΔSsys
18 | 41
Energy Conditions to Favor Spontaneity
Free energy change (ΔG) relates enthalpy, entropy, and
temperature for a process:
ΔG = ΔH − TΔS
A negative ΔG will be a spontaneous process, so the
conditions that help to favor a spontaneous process are:
1. The formation of low energy products: ΔHrxn < 0
2. The formation of products that have greater entropy
than the reactants: ΔSrxn > 0
However, not all spontaneous processes follow these conditions…
18 | 42
Effects of ΔH°, ΔS°, and T on ΔG°
ΔG = ΔH − TΔS
ΔH° ΔS°
ΔG°
-
+
-
+
-
+
-
+
Description*
ΔG° Always (-)
Spontaneous at all temperatures
ΔG° Always (+)
Nonspontaneous at all temperatures
Spontaneous at lower temperature
Nonspontaneous at higher temperature
+
+
-
Nonspontaneous at lower temperature
Spontaneous at higher temperature
+
*Lower and Higher Temperatures are relative terms dependent on when ΔG° = 0
18 | 43
ΔGo as a Criteria for Spontaneity
The following rules are useful in judging the
spontaneity of a reaction.
DGo
Reaction
When DGo is a large negative number, the reaction is
spontaneous in the forward direction, and the reactants
transform almost entirely to products when equilibrium is
reached.
More
negative
than -10 kJ
Products are
favored at
equilibrium
When DGo is a large positive number, the reaction is
nonspontaneous in the forward direction, and reactants do
not give significant amounts of product at equilibrium.
More
positive
than +10 kJ
Reactants are
favored at
equilibrium
When DGo is a small negative or positive, the reaction gives
an equilibrium mixture with significant amounts of both
reactants and products.
Between
-10 kJ and
+10 kJ
Mixture of
reactants and
products
Rule
18 | 44
Free Energy
Free Energy = energy available to do useful work (w)
Change in internal energy of system can be used to
perform work: ΔE = q + w
Some energy lost as heat to the surroundings (q)
Some energy lost due to entropy (−TΔS)
Efficiency = (work done)/(energy consumed)
18 | 45
Temperature and Spontaneity
Consider:
H2O(s) → H2O(l) at 273 K
ΔH = positive
ΔS = positive
ΔG = ΔH − TΔS
= negative above a certain
temp, when TΔS > ΔH
18 | 46
Practice: Transition Temperature
Given the data below, calculate the temperature at
which the reaction becomes spontaneous:
N2(g) + 3H2(g) ⇌ 2NH3(g)
(ΔH° = −92 kJ/mol and ΔS° = −199 J/mol K)
18 | 47
Calculating Free-Energy Changes
Standard Free Energy of Formation (ΔGf°):
Change in free energy associated with formation of
1 mole of a compound from its constituent
elements
Standard free energy of formation of most stable
form of element in standard state = 0
Free-energy change for a reaction:
ΔG°rxn = Σ(nprodΔG°f,prod) − Σ(nreactΔG°f,react)
18 | 48
Standard Free Energies at 25 °C and 1 atm
Substance
ΔG°f (kJ/mol)
Substance
ΔG°f (kJ/mol)
Br2(g)
3.1
CH4(g)
-50.8
Br2(l)
0.0
C2H6(g)
-32.9
Cdiamond(s)
2.9
CH3OH(g)
-162.0
Cgraphite(s)
0.0
CH3OH(l)
-166.4
CO(g)
-137.2
CH3CH2OH(g)
-168.6
CO2(g)
-394.4
CH3CH2OH(l)
-174.9
H2(g)
0.0
CH3CH2CH3(g)
-23.5
N2(g)
0.0
CH3(CH2)2CH3(g)
-15.7
O2(g)
0.0
CH3(CH2)2CH3(l)
-15.0
P4(s, red)
-12.1
CH3(CH2)6CH3(g)
16.4
P4(s, white)
0.0
CH3(CH2)6CH3(l)
6.4
H2O(g)
-228.6
C6H6(g)
129.7
H2O(l)
-237.2
C6H6(l)
124.5
NH3(g)
-16.5
C12H22O11(s)
-1543.8
18 | 49
Practice: Free-Energy Change
Calculate the free-energy change for the
following reaction using the ΔGf° values in the
appendix:
C12H22O11(s) + 12O2 (g) → 12CO2 (g) + 11H2O (l)
18 | 50
Driving the Human Engine
In living systems:
Breaking down food, generating heat
ΔG < 0; spontaneous
Metabolic processes (building muscle, etc.)
ΔG > 0, nonspontaneous
Process of life increases the entropy of the
universe by converting chemical energy into
heat
18 | 51
Breakdown of Glucose
Glucose:
Used by living systems for
energy
Glycolysis:
Series of reactions to convert
glucose into pyruvate (major
path for metabolism of
glucose)
18 | 52
Phosphorylation
Phosphorylation reaction: results in addition of a
phosphate group to an organic molecule
ΔGrxn° = + 13.8 kJ/mol, nonspontaneous
18 | 53
Adenosine Triphosphate (ATP)
ATP:
Produced as a result of the breaking down
(metabolizing) of food
Can be used to drive nonspontaneous cellular
reactions
18 | 54
Hydrolysis of ATP to ADP
ΔGrxn° = −30.5 kJ/mol,
spontaneous
18 | 55
Coupled Reactions
Phosphate group from ATP
used in phosphorylation
of glucose
Net process:
ΔGsum° = −16.7 kJ (Spontaneous)
18 | 56
ₒ
Relation ΔG to the Equilibrium Constant
The thermodynamic equilibrium constant, K, is the
equilibrium constant in which the concentrations of
gases are expressed as partial pressures in atmospheres
and the concentrations of solutes in solutions are
expressed in molarities.
If only gases are present, K = Kp
If only solutes in liquid solution are present, K = Kc
18 | 57
ₒ
Relation ΔG to the Equilibrium Constant
Standard free energy change is related to equilibrium. If
not at equilibrium, we can use the following equation to
solve for the free energy change, ΔG, using the reaction
quotient, Q, at a temperature, T (in K):
ΔG = ΔG˚ + RTlnQ
where ΔG˚ is the standard free energy change at
equilibrium and R is 8.314 J/mol∙K.
• If Q < K, ΔG will be negative, and the reaction will
proceed towards the products to get to equilibrium.
• If Q > K, ΔG will be positive, and the reaction will
proceed towards the reactants to get to equilibrium.
18 | 58
ₒ
Relation ΔG to the Equilibrium Constant
If we are at equilibrium, then Q = K, and ΔG will equal 0
K
0
ΔG = ΔG˚ + RTlnQ
So the equation becomes:
0 = ΔG˚ + RTlnK
Which can be rearranged to:
ΔG˚ = -RTlnK
18 | 59
Practice
For the reaction at 25 ˚C:
2NH3(g) + CO2(g) ⇌ NH2CONH2(aq) + H2O(l); ΔG˚ = –13.6 kJ
1. Find the value of the equilibrium constant K at 25 ˚C
(298 K)
2. What is ΔG at 25 ˚C if the [NH3] = 1.2 M, [CO2] = 2.3 M,
and [NH2CONH2] = 4 M
18 | 60
Solution (Part 1)
2NH3(g) + CO2(g) ⇌ NH2CONH2(aq) + H2O(l); ΔG°= –13.6 kJ
We are asked about the equilibrium constant, K, so we
can use the equation:
ΔG˚ = −RTlnK
If we are solving for K, we can rearrange the equation:
∆𝐺°
−13600 𝐽
𝑙𝑛𝐾 =
=
𝐽
−𝑅𝑇
−(8.314
)(298𝐾)
𝑚𝑜𝑙𝐾
lnK = 5.49
elnK = e5.49
K = e5.49 = 242
18 | 61
Solution (Part 2)
2NH3(g) + CO2(g) ⇌ NH2CONH2(aq) + H2O(l); ΔG°= –13.6 kJ
In Part 2, we may not be at equilibrium, so we use:
ΔG = ΔG˚ + RTlnQ
We have ΔG˚, R, and T, so to solve for Q:
[𝑁𝐻2 𝐶𝑂𝑁𝐻2 ]
(4)
𝑄=
=
= 1.208
2
2
[𝑁𝐻3 ] [𝐶𝑂2 ]
(1.2) (2.3)
Since Q < K, we expect the reaction to proceed in the
forward direction, so ΔG should be negative:
ΔG = -13.6 kJ + (0.008314 kJ/molK)(298K)ln(1.208)
ΔG = -13.1 kJ
18 | 62
Change of Free Energy with Temperature
ΔG˚ or K at another temperature can be
ascertained by a simpler method. In this
method, assume that ΔH˚ and ΔS˚ are essentially
constant with respect to temperature
The value of ΔG˚T at any temperature T can be
determined by substituting the values of ΔH˚
and ΔS˚ at 25 ˚C into the following equation:
ΔG˚T = ΔH˚ − TΔS˚
18 | 63
(approximation for ΔG˚T)
Practice
1. What is the ΔG˚ at 600 ˚C for the following reaction
given the following thermochemical information? Is
the reaction spontaneous at this temperature and 1
atm?
2. What is the value of Kp at 600 ˚C for this reaction?
3. What is the equilibrium partial pressure of SO3 if
the equilibrium pressures of SO2 and O2 are atm?
2SO2(g) + O2(g) ⇌ 2SO3(g)
ΔH˚f (kJ/mol)
S˚ (J/mol∙K)
18 | 64
2SO2(g) +
-296.83
248.2
O2(g) ⇌ 2SO3(g)
0
-395.72
205.2
256.76
Solution (Part 1)
2SO2(g) + O2(g) ⇌ 2SO3(g)
ΔH˚f (kJ/mol) -296.83
0
-395.72
S˚ (J/mol∙K)
248.2
205.2 256.76
ΔH˚rxn = Σ(nprodΔH°f,prod) − Σ(nreactΔH°f,react)
ΔH˚rxn = (2 mol)(-395.72 kJ/mol) − [(2 mol)(-296.83 kJ/mol)
+ (1 mol)(0 kJ/mol)] = -197.78 kJ
ΔS˚rxn = Σ(nprodS°prod) − Σ(nreactS°react)
ΔS˚rxn = (2 mol)(256.76 J/molK) − [(2 mol)(248.2 J/molK)
+ (1 mol)(205.2 J/molK)] = -188.08 J/K = -0.1881 kJ/K
ΔG˚T = -197.78 kJ – (873 K)(-0.1881 kJ/K) = -33.6 kJ
Since the value of ΔG˚T is negative, the reaction should be
18 | 65
spontaneous at 600 ºC and 1 atm
Solution (Part 2)
What is the value of Kp at 600 ˚C for this reaction?
ΔG˚T = −RTlnK = -33.6 kJ
∆𝐺°
−33600 𝐽
𝑙𝑛𝐾 =
=
𝐽
−𝑅𝑇
−(8.314
)(873𝐾)
𝑚𝑜𝑙𝐾
lnK = 4.63
elnK = e4.63
K = e4.63 = 102.2
18 | 66
Solution (Part 3)
What is the equilibrium partial pressure of SO3 if the
equilibrium pressures of SO2 and O2 are atm?
2SO2(g) + O2(g) ⇌ 2SO3(g); Kp = 102.2
𝑃𝑆𝑂3 2
𝑃𝑆𝑂3 2
𝐾𝑝 =
=
= 102.2
2
2
(0.5) (0.5)
𝑃𝑆𝑂2 𝑃𝑂2
𝑃𝑆𝑂3 2 = 12.78085
PSO3 = 3.58 atm
18 | 67
Learning Objectives
● Entropy and Spontaneity
○ state the second law of thermodynamics
○ understand what a spontaneous process is
○ understand the Third Law of Thermodynamics
○ comprehend and explain the relationship of spontaneity and entropy of universe
○ recognize entropy trends based on molecular structure and properties
○ predict ∆Srxn for a chemical or physical change
○ calculate entropy changes
○ understand how temperature and other properties can affect spontaneity
● Free Energy and Equilibrium
○ Comprehend the relationship between free energy change for a reaction and the
reaction quotient for a reaction
○ comprehend relationship between standard state free energy change and equilibrium
constant
○ relate the sign of the change of free energy with the work available and to the
spontaneity of a process in the forward direction
○ calculate ∆Grxn for a chemical or physical change