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Joe DiLosa
Hannah Robinson
Justin Whetzel
Emily Penn
Casey White
Spontaneous Process – a process that
proceeds on its own without any outside
Reversible Process – A process that can go
back and forth between states along exactly
the same path
Irreversible Process – A process that is not
Isothermal Process – A process that occurs
at a constant temperature
Occur in a definite direction
Ex: When a brick is dropped and it falls down
(gravity is always doing work on the brick)
 Opposite direction (reverse) would be nonspontaneous
(takes work to go against gravity)
If the surroundings must do work on the
system to return it to its original state, the
process is irreversible
Any spontaneous process is irreversible
A spontaneous reaction does not mean
the reaction occurs very fast
 Iron rusting is spontaneous
Experimental conditions are important in
determining if a reaction is spontaneous
(temperature and pressure)
To determine spontaneity, distinguish
between the reversible and irreversible
paths between states
Which of the following processes are
spontaneous and which are
nonspontaneous: (a) the melting of ice
cubes at -5°C and 1 atm pressure, (b)
dissolution of sugar in a cup of hot coffee,
(c) the reaction of nitrogen atoms to form
N2 molecules at 25°C (d) alignment of iron
filings in a magnetic field, (e) formation of
CH4 and O2 molecules from CO2 and H2O at
room temperature and 1 atm of pressure?
(a) is nonspontaneous because the ice will not
change state
(b) is spontaneous because to get the sugar back,
the temperature of the system must be altered.
(c) is spontaneous because work must be done to
form N2 and it is an irreversible process
(d) is spontaneous because the system must do
work against the magnetic field to go back to its
original state
(e) is nonspontaneous because a change in
temperature and work must be done to break the
products up and form the original reactants
Practice Problems - #7, 8, and 9
Entropy – A thermodynamic function
associated with the number of different
equivalent energy states or spatial
arrangements in which a system may be
Second Law of Thermodynamics – in
general, any irreversible process results in
an overall increase in entropy whereas a
reversible process results in no overall
change in entropy
Infinitesimal difference - an extremely
small difference
Entropy = q/T = S
A reversible change produces the maximum
amount of work that can be achieved by the
system on the surroundings (wrev = wmax)
Reversible processes are those that reverse
direction whenever an infinitesimal change
is made in some property of the system
Example of Reversible Process
◦ Flow of heat between a system and its
Example of Irreversible Process
◦ Gas that has been limited to part of a container
expands to fill the entire container when the
barrier is removed
∆S = Sfinal – Sinitial
∆S = qrev/T (where T = constant, isothermal
qrev = ∆Hfusion when calculating ∆Sfusion for a
phase change
∆Ssystem = qrev/T
∆Ssurroundings= -qrev/T
∆Stotal = ∆Suniv = ∆Ssystem + ∆Ssurroundings
For a reversible process ∆Stotal = 0
For an irreversible process ∆Stotal > 0
The normal boiling point of methanol (CH3OH) is
64.7˚C and its molar enthalpy of vaporization is
∆Hvap=71.8 kJ/mol. When CH3OH boils at its normal
boiling point does its entropy increase or decrease?
Calculate the value of ∆S when 1.00 mol CH3OH (l) is
vaporized at 64.7 ˚C
Increases, the process is endothermic
∆S = qrev/T = ∆Hvap/T
64.7 ˚C + 273 = 337.7 K
(1 mol)(71.8 kJ/mol)/(337.7 K)= 0.213 kJ/K = 213 J/K
entropy increases
Practice Problems- # 20 and 21
Translational Motion – movement in which an
entire molecule moves in a definite direction
Vibrational Motion – the atoms in the molecule
move periodically toward and away from one
Rotational Motion – the molecules are
spinning like a top
Microstate – the state of a system at a particular
instant, one of many possible states of the system
Third Law of Thermodynamics – the entropy of a
pure crystalline substance at absolute zero is zero
Img src = wps.prenhall.com
S = k ln W (W = number of microstates)
k = Boltzmann’s constant = 1.38 x 10-23 J/K
Entropy is a measure of how many
microstates are associated with a particular
macroscopic state
∆S= k ln Wfinal - k ln Winitial = k ln Wfinal/Wintial
Entropy increase with the number of
microstates of the system
The # of microstates available to a system
increases with
Increase in volume
Increase in temperature
Increase in the number of molecules
Because any of these changes increases the
possible positions and energies of the
molecules of the system
Generally expect the entropy of a system to
increase for processes in which
Gases are formed from either solids or
Liquids or solutions are formed from
The # of gas molecules increases during a
chemical reaction
At absolute zero there is no thermal motion
so there is only one microstate
As the temperature increases the molecules
gain energy in the form of vibrational
When the molecules increase their motion
they have a greater number of microstates
Practice Problems- # 28, 30, 38, 40
Standard Molar Entropy – The molar entropy
value of substances in their standard states
The standard molar entropies of elements
are not 0 at 298K
The standard molar entropies of gases are
greater than those of liquids and solids
standard molar entropies generally
increase with increasing molar mass
Standard molar entropies generally increase
with an increasing number of atoms in the
formula of a substance
Pictures taken from p820 of the Tenth Edition of Chemistry: The Natural Science textbook written by Brown, LeMay,
and Bursten
∆S˚= ΣnS˚(products) - ΣmS˚(reactants)
∆Ssurr= -qsys/T
For a reaction at constant pressure qsys= ∆H
Calculate ∆S˚ values for the following reaction:
C2H4 (g) + H2 (g)  C2H6 (g)
∆S˚= ΣnS˚(products) - ΣmS˚(reactants)
= 229.5 J/molK – (219.4 J/molK + 130.58
= -120.5 J/molK
Practice Problems - # 48
Gibbs free energy- a thermodynamic state
function that combines entropy and
Standard free energies of formation- the
change in free energy associated with the
formation of a substance from its elements
under standard conditions
Free energy (G) is defined by:
G = H-TS
Where T is absolute temperature, S is
entropy, and H is enthalpy
When temperature is constant the equation
for the change in free energy for the system
If T and P are constant then the sign of ΔG
and spontaneity of a reaction are related.
1) If ΔG < 0, reaction proceeds forward
2) If ΔG = 0, reaction is at equilibrium
3) If ΔG > 0, the forward reaction is not
spontaneous because work must be supplied
from the surroundings but the reverse reaction is
be spontaneous.
In any spontaneous process at constant
temperature and pressure the free energy
always decreases
Conditions of Standard free energies of
formation (implied)
 1 atm pressure for gases
 pure solids
 pure liquids
 1M solutions
 25˚ C (normally)
The free energy of elements is 0 in their
standard states.
ΔGf˚ = standard free energies of formation
Calculated the same way as ΔH
ΔG˚ = ΣnΔGf˚ (products) - ΣmΔGf˚ (reactants)
ΔG = -wmax
Practice Problems - #49, 51, 53, 54, 56, 58,
63, 66
For a certain reaction ΔH = -35.4 kJ and
ΔS = -85.5 J/K. Calculate ΔGo for the
reaction at 298 K. Is the reaction
spontaneous at 298 K?
ΔG = ΔH – ΔS*T
ΔG = -35.4KJ – (-85.5 J/K * 298K)
ΔG = -9920 J
No new vocabulary
When ΔH and –TΔS are negative, ΔG is
negative and the process is spontaneous
When ΔH and –TΔS are positive, ΔG is positive
and the process is non-spontaneous
When ΔH and –TΔS are opposite signs, ΔG will
depend on the magnitudes
Picture taken from p828 of the Tenth Edition of Chemistry: The Natural Science textbook written by Brown, LeMay, and Bursten
For a certain reaction ΔH = -35.4 kJ and
ΔS = -85.5 J/K. is the reaction exothermic
or endothermic? Does the reaction lead to
an increase or decrease of disorder in the
system? Calculate ΔGo for the reaction at
298 K. Is the reaction spontaneous at 298
A) The reaction is exothermic because ΔH is
B) It leads to a decrease in the disorder of
the system (increase in the order of the
system) because ΔS is negative
C) ΔG = ΔH – ΔS*T
ΔG = -35.4KJ – (-85.5 J/K * 298K)
ΔG = -9920 J
D) Yes, if all reactants are in their
standards states because ΔG is negative
Practice problems (same as 19.5) - #49, 51,
53, 54, 56, 58, 63, 66
Nonstandard conditions – conditions other
than the standard conditions
ΔG = ΔGo + RT ln Q
Where R = 8.314 J/mol-K, T = absolute
temperature, Q = reaction quotient
ΔGo = ΔG under standard conditions
Because Q = 1 and lnQ = 0
ΔGo = -RT ln K
At equilibrium, ΔG = 0
K = e –(delta G standard)/RT
Smaller (more negative ΔG0), the larger the
K value
Direction of Reaction
Toward forming more products >1 negative
Toward forming more reactants <1 positive
Products and reactants equal
#71 – Explain Quantitatively how ΔG
changes for each of the following reactions
as the partial pressure of O2 is increased:
A) 2 CO(g) + O2(g) → 2 CO2 (g)
2 CO(g) +
-137.2 kJ/mol
O2(g) →
0 kJ/mol
2 CO2 (g)
-394.4 kJ/mol
ΔG = products – reactants
ΔG = -394.4 kJ/mol + 137.2 kJ/mol
ΔG = -257.2 kJ/mol
Practice Problems: #71,76,79
#82, 87, 88, 89, 90, 92

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