THERMODYNAMICS
Review of Energy and Enthalpy Changes (Ch. 5)
For each pair, which state is more stable?
Pencil on desk vs. raised in the air
Skier at top of mountain vs. bottom
Toaster at 25oC vs. hot toaster
Liquid water at 25oC vs. ice at 25oC
Salt water vs. pure water in contact with solid salt
MgO vs. Mg metal in contact with O2
In each case, how was energy transferred in going to the
more stable state?
Energy Changes: Heat and Work
Heat = q = energy transferred due to a difference in
temperature.
+q means heat is added to the system
Work = w = action of force through a distance (often P∆V)
+w means work is done on the system
Find one pair in which energy was transferred as heat.
Find one pair in which energy was transferred as work.
What force was involved?
Find a second pair in which a different force was involved.
STATE FUNCTIONS
A state function is a quantity that only does not depend
on the process by which the system was prepared
Example: Your altitude (height above sea level) does not
depend on the route you took to class this morning.
State functions are written as uppercase letters (E, H, P, V,
T, S…)
Changes in state functions are path-independent:
reactants
1
2
products
q and w are not state functions
but ∆E (= q + w) is a state function
∆E
Energy Changes
∆E = Efinal state - Einitial state
∆E (kJ/mol)
O2 (g)  2 O atoms
+498.3
Water  ice at 25oC
-6.0
Water + salt  salty water
-3.9
Si (s) + O2 (g) SiO2 (s)
-908
Which set of reactants/products has the biggest difference in
energy?
Which set has the lowest absolute energy?
If we assign O atoms an energy of zero, what is the energy of O2?
Is the lowest energy state always most stable?
First Law of Thermodynamics
∆Euniverse = 0
Energy is conserved
∆Esystem + ∆Esurroundings = ∆Euniverse
so
∆Esystem = - ∆Esurroundings
Heat and work:
∆Esystem = q + w
and for PV work at const. pressure,
∆Esystem = q – P∆V
Energy Changes - Heat and Work
²V
∆ =0
w = P²V
∆
Dry ice (CO2) is
heated to room
temperature at
const. P or at const. V
+q
+q
What are the signs of q and w for each?
Does Esystem increase or decrease? How do you know?
Which system has higher E at the end?
When natural gas burns:
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
Reaction produces heat and light
Is energy conserved?
What are the signs of q and w for the overall reaction?
Where does the energy come from?
Is energy stored or released when bonds are broken?
Enthalpy (H)
∆H = heat transferred at const. P
If (+), endothermic (need to add heat)
If (-), exothermic (heat is given off)
Classify as endo- or exothermic:
Ice melting
Water boiling
Wood burning
H is a state function – changes are path-independent
H = E + PV (sums and products of state functions are
also state functions)
Standard Enthalpy of Formation
∆Hof:
∆H for making a compound from elements in their
standard states
Standard state is the most stable form (pure solid, pure
liquid, or gas at P = 1 atm)
For solutes in solution, standard state is usually 1 M
There are tables of ∆Hof
∆Horxn =  ∆Hof (products) –  ∆Hof (reactants)
Standard Enthalpy of Formation of Oxides
Reaction
∆Hof (kJ/mol)
2 Li + 1/2 O2  Li2O
-598
2 Na + 1/2 O2  Na2O
-414
Ca + 1/2 O2  CaO
-635
Mg + 1/2 O2  MgO
-602
2 Al + 3/2 O2  Al2O3
-1676
2 Fe + 3/2 O2  Fe2O3
-824
2 Cr + 3/2 O2  Cr2O3
-1128
2 Ag + 1/2 O2  Ag2O
-30
2 Cu + 1/2 O2 Cu2O
-167
C + O2  CO2
-394
1/2 N2 + O2  NO2
+34
Cl2 + 1/2 O2 Cl2O
+80
What are the least and most stable oxides in the table?
What periodic trends do you see for the ∆Hof’s of the
oxides?
How do oxidation states relate to ∆Hof?
Predict ∆Hof for Au2O, K2O, and SrO.
Why does Mg burn in dry ice?
SPONTANEOUS REACTIONS
A spontaneous reaction is one that can proceed in the
forward direction under a given set of conditions.
Note: spontaneity has nothing to do with the rate at which a
reaction occurs.
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) Spontaneous
CO2(g)  C(s) + O2(g)
Not Spontaneous
2 Fe2O3(s)  4 Fe(s) + 3 O2(g)
Not Spontaneous
What determines whether these reactions are spontaneous?
SPONTANEITY AND WORK
Useful work can be extracted from a spontaneous process
2 H2 + O2  2 H2O
can be used to drive a rocket
Water in a tower 
Water on the ground
Can do work
(drive a turbine)
Work must be done to drive a non-spontaneous process
2 H2 O  2 H2 + O2
Energy (electrolysis work) must be put in.
Water on the ground  Water in a tower
Work is done to pump water
Spontaneity and H
Not all spontaneous reactions are exothermic
Examples:
At +10°C
H2O(s)  H2O(l)
H>0
Ba(OH)2•8H2O(s) + 2NH4SCN(s) 
Ba(SCN)2(aq) + 2NH3(aq) + 10H2O(l)
H>0
NH4Cl(s) + H2O(l)  NH4Cl(aq)
H>0
QUESTIONS
Why are some endothermic processes spontaneous?
Evaporation of water
Dissolution of NH4+NO3-
What makes an ideal gas expand into a vacuum
(q = 0, w = 0)?
What distinguishes "past" from "future" in chemistry
and physics?
ENTROPY
A thermodynamic parameter that measures the
disorder or randomness in a system.
The more disordered a system, the greater its
entropy.
Entropy is a state function -- its value depends only
upon the state of the system (not how it got there).
We are usually concerned with the change in entropy
(S ) during a process such as a chemical reaction.
S = S final - S initial
Which "reactions" have S > 0?
Cards arranged in order  Random order after
shuffling the deck
Messy dorm room  Cleaned up room
CO2 + H2O + Minerals  Tree
1 mole of gas
in a 1 L flask
 1 mole of gas
in a 2 L flask
NH4Cl(s)  NH4+(aq) + Cl-(aq)
ENTROPY
Gases have a lot more entropy than solids or liquids.
Reactions that form gases usually have S > 0
S (+ or - ?)
H2O (l, 25oC)  H2O(g)
CaCO3 (s)  CaO(s) + CO2(g)
N2(g) + 3 H2(g)  2 NH3(g)
N2(g) + O2(g)  2 NO(g)
Au(s) at 298K  Au(s) at 1000K
Ag+(aq) + Cl-(aq)  AgCl(s)
Adding heat increases entropy
Entropy and heat: S = qrev/T
e.g., for melting ice, S = Hfus/273 K
(endothermic) 
+
+ m.p.= 0oC
Entropy units: J/mol-K (same units as specific heat)
S(1 mole HCl(g)) >> S(1 mole NaCl(s))
why?
S(2 moles HCl(g)) = 2 S(1 mole HCl(g))
“
S(1 mole HCl(g)) > S(1 mole Ar(g))
“
S(1 mole N2(g) at 300K) > S(1 mole N2(g) at 200K)
“
ENTROPY and PHASE CHANGES
Sgas >> Sliq > Ssolid
Svap = qrev = Hvap
T
T
During phase changes temperature and pressure are
constant. (Heat transfer is reversible, so H = q = qrev)
Calculate the entropy change when 1 mole of liquid
water evaporates at 100oC (Hvap = +44 kJ/mol)
2nd Law of Thermodynamics:
The total entropy in the universe is increasing.
Suniverse > 0
Suniverse = Ssystem + Ssurroundings > 0
3rd Law:
The entropy of every pure substance at 0K
(absolute zero temperature) is zero.
S = 0 at 0 K
3rd LAW ENTROPY
Entropy is a state function - its value depends only
on the initial and final states.
And S = 0 at T = 0 K (3rd Law)
This means we can measure absolute entropy S
(not just ∆S)
So (rxn) = So (products) - So (reactants)
N2(g) + 3 H2(g) 2 NH3(g)
So (25oC) = 191.5
130.58
192.5 J/mol-K
So (rxn) = (2x192.5) - (191.5 + 3x130.58)
= -198.3 J/mol-K
CRITERIA FOR SPONTANEITY
Hsystem < 0
Exothermic reactions are usually spontaneous.
S
Reactions are always spontaneous if
Ssystem + Ssurroundings > 0 (2nd Law)
We need a new state function (G) that can predict spontaneity
from just the system. This is called the Free Energy:
G = H - TS (T = absolute temperature (in K))
Gsystem predicts rxns at constant T and P:
G < 0
Spontaneous
G > 0
Not spontaneous
G = 0
Reaction at equilibrium
EFFECT OF TEMPERATURE ON SPONTANEITY
G = H - TS
H and S do not change much with temperature,
but ∆G does.
H
S
-
+
-
-
+
+
+
-
G
Spontaneous?
Is this reaction spontaneous at room temp? At 1100 oC?
CaCO3(s)  CaO(s) + CO2(g)
So
92.88
39.75
213.6
1207.1
635.3
393.5
J/mol K
 Hfo
(kJ/mol)
Calculate the boiling point of bromine
Hovap = 31.0 (kJ/mol)
Sovap = 92.9 (J/mol-K)
STANDARD FREE ENERGY OF FORMATION
Gf = Free energy change in forming one mole of a
compound from its elements, each in their standard states.
Standard States:
Solid
Liquid
Gas
Solution
Elements
Temperature
Pure solid
Pure liquid
P = 1 atm
1M solution
Gf = 0
Usually 25C
Grxn = Gf(prods) - Gf(reactants)
Units: kJ/mole.
COMBUSTION OF SUCROSE
C12H22O11(s) + 8 KClO3(s) 12 CO2(g) + 11 H2O(l) + 8 KCl(s)
Is this reaction spontaneous?
∆Gof(sucrose)
∆Gof(KClO3, s)
∆Gof(CO2, g)
∆Gof(H2O, l)
∆Gof(KCl, s)
= -1544.3 kJ/mol
= -289.9 ”
= -394.4 "
= -237.1 "
= -408.3 "
How does H2SO4 affect the reaction rate?
What are two possible reasons for using high T to carry
out a reaction?
EXTENT OF REACTIONS
So far we have considered only standard conditions
and ∆Ho, ∆So, ∆Go.
What happens under other conditions?
G = G + RTlnQ = G + 2.303RTlog10Q
aA + bB  cC + dD
Q = ([C]c[D]d) = REACTION
([A]a[B]b) QUOTIENT
Example: 2NO2 (g)  N2O4 (g)
Go298 = 5.4 kJ/mol
Partial pressure of NO2 is 0.25 atm and partial pressure
of N2O4 is 0.6 atm. What is ∆G?
RELATIONSHIP BETWEEN G AND G
How do concentrations affect Q and G?
Q
Add more reactants
Take away products
Take away reactants
G ( or )
ANALOGY BETWEEN POTENTIAL ENERGY
AND FREE ENERGY
Go
slope = 0
At equilibrium point, G = 0 for interconverting
reactants  products
Note: This does NOT mean Go = 0
Nitrogen Fixation
N2(g) + 3 H2(g) = 2 NH3(g)
∆Gof:
0
0
-16.7 kJ/mol
∆Gorxn =
What is the sign of ∆So?
∆Ho?
Would higher or lower P favor products?
"
"
"
"
T
"
"
?
RELATIONSHIP BETWEEN
GAND Keq
At equilibrium, G = 0:
G = 0 = G + 2.303RTlogQ
= G + 2.303RTlogKeq
G = -2.303RTlogKeq
G > 0
G < 0
G = 0
Keq < 1
Keq > 1
Keq = 1
Solubility Equilibrium
Guess the solubilites of these salts in water:
KClO3(s)  K+(aq) + ClO3-(aq)
∆Go > 0
NaF(s)  Na+(aq) + F-(aq)
∆Go ≈ 0
NaCl(s)  Na+(aq) + Cl-(aq)
∆Go < 0
What is the solubility product Ksp of AgBr?
AgBr(s) Ag+(aq) + Br(aq)
Gfo Ag+
Gfo Br
Gfo AgBr
77.1 kJ/mol
104 kJ/mol
96.9 kJ/mol
Spontaneity and Equilibrium
At 10°C
H2O(l)  H2O(s) Spontaneous
Can get the system to do work!
At + 10°C
H2O(l)  H2O(s) Non- Spontaneous
Must do work to get this to go.
But at 0°C H2O(l)
H2O(s)
Heat in and out is reversible
Equilibrium
q = qrev
When a chemical system is at equilibrium, reactants and
products can interconvert reversibly.
RELATIONSHIP BETWEEN
G AND WORK
For a spontaneous process,
G = Wmax = The maximum work that can be obtained
from a process at constant T and P.
For a non-spontaneous process,
G = Wmin = The minimum work that must be done to
make a process go at constant T and P.
Calculating W from G
What is the maximum work available from the
oxidation of 1 mole of octane under standard
conditions (P = 1 atm)?
C8H18(l) + 12.5 O2(g) 8CO2(g) + 9H2O(l)
Gfo
17.3
kJ/mol
0
394.4
237.1