Chapter 19: Chemical
Thermodynamics
Tyler Brown
Hailey Messenger
Shiv Patel
Agil Jose
19.1 Spontaneous Processes
19.1 Spontaneous Processes
Spontaneous Process:
• A process that proceeds on its own without any
outside assistance
• Occurs in a definite direction
“Processes that are spontaneous in one direction are nonspontaneous
in the other”
(Pg 805 Brown and LeMay)
19.1 Spontaneous Processes
Reversible Process:
• A process in which the system is changed in a way
that the system and surroundings can be restored
to their original state by exactly reversing the
change (can be completely restored to original
condition)
• Reversible processes reverse direction whenever
an infinitesimal change is made in some property
of the system
19.1 Spontaneous Processes
Irreversible Process:
• Cannot be reversed to return system and
surroundings to original state
19.1 Spontaneous Processes
Isothermal Process:
• A process with a constant temperature
19.1 Spontaneous Processes
• First Law of Thermodynamics: Energy is
conserved
• “Physical and chemical processes have a
directional character” (Pg 804)
• I.e. Sodium and Chlorine come together on their
own to make NaCl, but it does not decompose of
its own accord.
19.1 Spontaneous Processes
• Experimental conditions help to determine
if a process is spontaneous
• I.e. When the temperature > 0°C in ordinary
atmospheric pressure, ice melting is spontaneous.
» In these conditions, liquid water turning in to ice is
not spontaneous
19.1 Spontaneous Processes
• Entropy: The ratio of heat delivered to the
temperature at which it is delivered
• “A reversible change produces the
maximum amount of work that can be
achieved by the system on the
surroundings (wrev= wmax)” (Page 807)
• All spontaneous processes are
IRREVERSIBLE
19.1 Spontaneous Processes
Tell whether or not the following process is
spontaneous
Ice melts at -1 degree Celsius
19.1 Spontaneous Processes
Tell whether or not the following process is
spontaneous
Ice melts at -1 degree Celsius
This would not be spontaneous because Ice does
not melt under 0 degrees Celsius. This reaction
would be spontaneous in the reverse reaction
19.1 Spontaneous Processes
• 19.1 Problems: 7-18
19.2 Entropy and the Second
Law of Thermodynamics
19.3 The Molecular
Interpretation of Entropy
Molecular motion
• Translational Motion – entire molecule
moves in one direction (like throwing a
baseball)
• Vibrational Motion – atoms in the molecule
periodically move toward and away from
one another
• Rotational Motion – molecules spin like a
top
What is Statistical
Thermodynamics?
• It’s the use of tools of statistics and
probability to provide the link between
macroscopic and microscopic worlds
• We molecules in bulk in a microstate: a
single possible arrangement of the
positions and kinetic energies of gas
molecules in a specific thermodynamic
state
So what equation do we use?
• Boltzmann created this beautiful equation:
S = k lnW
K is Boltzmann’s constant which is: 1.38 x 10-23J/K
-This means entropy is the measure of how many
microstates are associated with a particular
macroscopic state
Relationship between microstates
and entropy
• The more microstates, the more entropy
• Increasing volume, temperature, and
number of molecules increases entropy
due to larger number of microstates
GENERALLY SPEAKING
Entropy increases when:
- Gases are formed from either solids or
liquids
- Liquids or solutions are formed from solids
- The number of gas molecules increases
during a chemical reaction
The Third Law of Thermodynamics
• The entropy of a pure crystalline
substance at absolute zero is zero
S(0 K) = 0
- Makes sense because when there is no
molecular motion (temp is 0 K), there is
only ONE microstate
Trends
• Sharp increase of entropy at melting points
• Sharp increase of entropy at boiling points
Entropy of solids < Entropy of liquids < Entropy of gas
Example
• What will the sign of change of entropy be
for the following reactions?
2Na(s) + Cl2 (g)
2NaCl (s)
2H2 (g) + O2 (g)
2H2O (l)
Answers
• Because of Boltzmann’s equation, we
know that the more particles there are in a
system, the more microstates there are, so
the more entropy there is, so:
• The reaction to form NaCl has a negative
sign for change in entropy, since there are
less particles after the reaction
• The reaction to form water also has a
negative sign for change in entropy since
there are less particles
Example
• In a chemical reaction two gases combine
to form a solid. What do you expect for the
sign of change in S? For which of the
processes does the entropy of the system
increase? A)The melting of ice cubes at -5
degrees Celsius and 1 atm pressure B)
dissolution of sugar in a cup of hot coffee
C) the reaction of nitrogen atoms to form
N2
Answer
• Entropy decreases when two gases form a
solid, so negative sign
• A) Positive because the ice is melting from
a structured solid to unstructured liquid
• B) Positive because the sugar went from
being structured solids to numerous
separated particles
• C) Negative, there is a huge decrease in
the number of gas particles (by 1/2 to be
exact)
Practice, practice, practice!
• Page 838 #’s 19.27 – 19.40
• A closer look on page 810
19.4 Entropy Changes in
Chemical Reactions
How do we do it?
• We can’t measure change in entropy for a
reaction, but we CAN find the absolute
value of the entropy for many substances
at any temperature
• Standard molar entropies are denoted by
S° (pronounced S knot) for any pure
substance in their standard state at 1 atm
Observations on Standard Entropy
• Standard molar entropies (SME) of
elements at reference temperature 298 K
are not zero
• SME of gases greater than liquids
• SME generally increase with molar mass
• SME generally increase with number of
atoms in molecular formula
So what’s the equation we use?
ΔS° = ΣnS(products) – ΣnS(reactants)
In english
Change in entropy equals sum of entropies
of products minus the sum of entropies of
reactants
Entropy Changes in the
Surroundings
• The change in entropy of the surroundings will
depend on how much heat is absorbed or given
off by the system
• Isothermal process equation:
ΔSsurr=-qsys/T
- At constant pressure, qsys is the enthalpy
change of the system, ΔH
Example
• Calculate change in standard entropy for
the following reaction
• 2Na(s) + Cl2 (g)
2NaCl (s)
Answer
• 1. The standard entropy change is equal to the standard
entropy of products minus the reactants
2. So we must first find the standard entropy of the products(using
a standard entropy table), and multiply them by the number of
moles in the reaction
Standard entropy of NaCl = 72.1 J/K-mol, so 2(72.1)= 144.2
3. Then we find the standard entropy of reactants
Standard entropy of Na = 51.2 J/K-mol, so 2(51.2)= 102.4 J/K-mol
Standard entropy of Cl2 = 223.1 J/K-mol, 223.1 J/K-mol
4. Now we subtract product and reactants
144.2 J/K-mol – (102.4 J/K-mol + 223.1 J/K-mol) = - 181.3 J/K-mol
Example
• Cyclopropane and propylene isomers both
have the formula C3H6. Based on the
molecular structures shown, which of
these isomers would you expect to have
the higher standard molar entropy at 25
degrees Celsius?
Cyclopropane
Propylene
Answer
• Propylene because it is much less
organized compared to Cyclopropane
We love to practice!
• Page 839 #’s 19.41 – 19.48
• Example on page 821, figure 19.5
19.5 Gibbs Free Energy
19.5 Gibbs Free Energy
• Standard Free Energies of Formation:
Used to calculate standard free-energy
change for chemical processes
19.5 Gibbs Free Energy
• Equation for Standard Free-Energy
Change:
ΔG° = ΣnΔG°f (products) - ΣmΔG°f
(reactants)
19.5 Gibbs Free Energy
• G = H – TS
• T is the absolute temperature.
• The change in free energy of a system
(ΔG) is given by: ΔG= ΔH – TΔS
19.5 Gibbs Free Energy
• If ΔG is negative, the reaction is
spontaneous
• If ΔG is zero, then reaction is at
equilibrium
• If ΔG is positive, the forward reaction is
not spontaneous, but the reverse reaction
is.
19.5 Gibbs Free Energy
• For a certain chemical reaction, ΔH = 35.4 kJ and ΔS = - 85.5 J/K. Calculate ΔG
for the reaction at 298 K.
19.5 Gibbs Free Energy
• For a certain chemical reaction, ΔH = 35.4 kJ and ΔS = - 85.5 J/K. Calculate ΔG
for the reaction at 298 K (19.51 C)
ΔG = ΔH –TΔS
ΔG = - 35.4 kJ – (298K)(-85.5 J/K)
ΔG = - 35.4 kJ + 25500 J
ΔG = -35.4 kJ + 25.5 kJ
ΔG = -9.9 kJ
19.5 Gibbs Free Energy
• 19.5 Problems: 49 -70
19.6 Free Energy and
Temperature
19.6 Free Energy and Temperature
•The equation ΔG=ΔH+(–TΔS) in which
ΔH is the enthalpy term and –TΔS is the
entropy term, can be used to
demonstrate how free energy is affected
by the change in temperature.
•The value of –TΔS depends directly on
the absolute temperature, T, ΔG will vary
with the temperature.
19.6 Free Energy and Temperature
• T is a positive number at all temperatures
other than absolute zero.
• Both the enthalpy and entropy can be
either positive or negative.
• When ΔS is positive, which means greater
randomness than the original state, then
the TΔS is negative. When the ΔS is
negative, then the TΔS is positive.
19.6 Free Energy and Temperature
• ΔG will always be negative when the
process is spontaneous, and the opposite
is true when the process is not
spontaneous.
19.6 Free Energy and Temperature
• H2O(s) → H2O(l) ΔH>0 ΔS>0
• This process is endothermic, meaning the
ΔH is positive. Entropy also increases with
this process, meaning that ΔS is positive
as well, which makes –TΔS a negative
value. This means that the –TΔS term
dominates, making ΔG negative. A
negative ΔG means that this process is
spontaneous at T>0ºC
19.6 Free Energy and Temperature
• Calculate ΔG for 2NO2(g) → N2O4(g) at
298K using appendix C.
19.6 Free Energy and Temperature
• ΔH 2NO2(g) → N2O4(g) = -58.02 kJ/mol
• ΔS 2NO2(g) → N2O4(g) = -175.79 J/K
•
19.6 Free Energy and Temperature
•
•
•
•
•
•
•
ΔGº =
-58.02kJ – (298K)(-175.79J/K)(1kJ/1000J)
-58.02kJ – (298K)(-0.176kJ/K)
-58.02kJ – (-52.45kJ)
ΔGº = -5.57kJ
This means that at 298K,
2NO2(g) → N2O4(g) is a spontaneous
reaction.
19.6 Free Energy and Temperature
• Textbook problems: chapter 19, 65,66,79.
80, and 87
19.7 Free Energy and the
Equilibrium Constant
Most reactions occur in
nonstandard conditions
• To find the free energy in nonstandard
reactions, we use the standard free energy
change to calculate
ΔG = ΔG° + RT ln(Q)
R is ideal gas constant = 8.314 J/mol-K
T is absolute temperature
Q is reaction quotient for that particular
mixture of interest
Some more info
• At equilibrium ΔG = 0
• Q = equilibrium constant K
Trends
• The more negative ΔG°, is the larger K is
• The more positive ΔG° is, the smaller K
is
Example
• 2CO(g) + O2 (g)
2CO2 (g)
• Calculate the change in free energy for the
above reaction at 298K, for a reaction
mixture that consists of 1.0 atm O2 , 2.0
atm CO , and 0.75 atm CO2
Answer
1. First we must find Q in order to use the equation provided
earlier. So we solve for the reaction quotient:
Q=(PCO2 )2 / (PCO)2(PO2 )=(.75 atm)2 / (2.0 atm)2 (1.0 atm) = .14atm
2. Now calculate standard change in free energy for reaction
using tables
Products – Reactants, so 2(-394.4kJ/mol) – (2(-137.2kJ/mol) + 0)
Change in standard free energy is -514.4 kJ/mol
3. Now we can finally use ΔG = ΔG° + RT ln(Q)
ΔG = (-514.4 kJ/mol) + (8.314 J/mol-K)(298 K) (1 kJ/1000J)ln(.14)
= (-514.4 kJ/mol) – 4.871 kJ/mol = -519.271 = - 520 kJ/mol
Practice Problems
• Page 841
• Problems 19.71 – 19.95
• You can find answers on pages A-24 thru
A-25
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