AP Chemistry : Ch. 16 Notes
Review of Ch 6 :
Mr. Ferwerda - Tecumseh High School
5/20/08
_____ (T/F) Heat is an intensive property.
_____ (T/F) Volume is an extensive property.
_____ (T/F) Density is an intensive property.
_____ (T/F) Heat is a state function.
_____ (T/F) Entropy is a state function.
Review : Hess’s law :
a. C(graphite) + O2(g)  CO2(g)
b.
H2(g) + 1/2O2(g)  H2O(l)
c. C2H2(g) + 5 O2 (g)  4 CO2(g) + 2 H2O(l)
73.2 kJ/mol
Review of calc of ΔHºf : 2 B5H9 (l) +
0.0 kJ/mol
-1263 kJ/mol
12 O2(g)
---> 5 B2O3(s)
ΔH = -393.5 kJ
ΔH = -285.8 kJ
ΔH = -2598.8 kJ____
-258.8 kJ/mol
+
9 H2O(l)
Which of the following have enthalpies of formation equal to zero?
_____ H2(l)
_____H2O(l)
_____ S(s)
_____ Cl2(g)
_____ Br2(l)
Writing equations for standard enthalpy of formation :
for H3PO4(s):
for ZnBrO3(s)
What is the enthalpy of formation of HCl given the following? H2(g) + Cl2(g) --> 2 HCl
ΔH = -131 kJ
AP Chemistry : Ch. 16 Notes
Mr. Ferwerda - Tecumseh High School
5/20/08
A. Spontaneity, Entropy, and Free Energy
1. Spontaneous Processes and Entropy
a. Spontaneous reactions occur under the given conditions without external stimuli.
- spontaneous processes may be fast or slow
- thermodynamics tell us only if a reaction will occur, not how fast (kinetics tells us how fast)
- thermodynamics considers only the initial and final states, the pathway is relevant only to
kinetics
- thermodynamics and kinetics are needed to fully describe a reaction
- examples - ice melting at T>0 ºC, NaCl dissolving in water, iron rusting
________________________________________________ reactions occur without outside intervention.
_____ (T/F) Thermodynamics tells us if a reaction will occur, not how fast it will occur.
_____ (T/F) Thermodynamics considers only the initial and final states of substances.
b. Entropy (S) - a measure of disorder
- a state function (independent of pathway)
- the driving force behind spontaneous reactions (if Suniv is positive for a process it will be
spontaneous and irreversible)
- entropy is a thermodynamic function that describes the number of possible arrangements that are
available to a system
- nature proceeds spontaneously towards the states that have the highest probability of existing
(greatest positional probability)
- positional probability increases from solid to liquid to gas (i.e. gases have greatest entropy)
- a solution of a solid has more positional probability than the solid sitting in a beaker of water
- the more complex the molecule, the greater its entropy
- due to greater degrees of freedom or forms of motion
- vibrational motion or rotational motion (spinning of atoms on their axes)
_____________ is the driving force for spontaneous reactions in the universe.
Spontaneous?
_____ (T/F) Entropy is a state function.
AP Chemistry : Ch. 16 Notes
Mr. Ferwerda - Tecumseh High School
5/20/08
_______________________________ probability depends on the number of possible configurations in space.
_____ (T/F) A change from a liquid to a gas increases the number of possible configurations in space.
_____(T/F) Nature spontaneously moves toward those states that have the highest probabilities of existing (i.e.
highest entropy).
____ (T/F) If Suniv is positive for a change it will be spontaneous.
If you roll two dice, what number (sum)is most likely?
Which one, in each of the following pairs, has the highest entropy?
1 mol H2O(g)
N2 gas at .01 atm
He gas at 12 K
C12H26
1 mol H2O(l)
N2 gas at 4 atm
He gas at 273 K
C6H14
2. Entropy and the Second Law of Thermodynamics
a. Second Law : In any __________________________ process there is always an _______________
in the entropy of the________________________.
- the entropy change of the universe equals the sum of the changes of entropy in the system and
the surroundings
ΔSuniv = ΔSsys + ΔSsurr
_____ (T/F) If ΔSuniv is positive (increasing entropy) the process is spontaneous (and irreversible).
_____ (T/F) If ΔSuniv is negative (decreasing entropy) the process is nonspontaneous.
_____ (T/F) If a system has a positive change in entropy (increase) it will always be spontaneous.
_____ (T/F) If a change in a system occurs so the change in entropy in the universe is positive the reaction will
be spontaneous.
_____ (T/F) Entropy changes in the surroundings are determined primarily by the heat flow.
Sign of ΔSuniv determines direction of heat flow.
Magnitude of ΔSuniv is determined by temperature.
AP Chemistry : Ch. 16 Notes
Mr. Ferwerda - Tecumseh High School
5/20/08
3. The Effect of Temperature on Spontaneity
a. both ΔSsys and ΔSsurr need to be considered
- if both ΔSsys and ΔSsurr are positive, then ΔSuniv is positive (spontaneous)
- if both ΔSsys and ΔSsurr are negative, then ΔSuniv is negative (nonspontaneous)
- in an exothermic process heat flows from the system to the surroundings and ΔSsurr
increases(positive) and ΔSsys decreases (negative)
- in an endothermic process heat flows from the surroundings to the system and ΔSsys increases
(positive) and ΔSsurr decreases (negative)
- in both of the above cases one entropy change favors an increase in ΔSuniv and one entropy
change is unfavorable towards and increase in ΔSuniv.
- In these cases the temperature is the factor which determines if ΔSuniv is positive or not.
- Consider phase changes in water. Above 100 ºC, at 1 atm, the change from the liquid state to
the gaseous state is spontaneous. Below 100 ºC at 1 atm, condensation is the spontaneous
process.
- Entropy changes in the surroundings are primarily determined by heat flow.
- The magnitude of ΔSsurr is temperature dependent.
- The significance of exothermicity is temperature dependent.
- The impact of heat transfer to or from a system to the surroundings will be greater at lower
temperatures.
Predicting sign of ΔSsurr :
Endothermic processes - heat flows out of surroundings and therefore ΔSsurr is negative and vice-versa.
condensation of water ?
melting of ice?
AP Chemistry : Ch. 16 Notes
Mr. Ferwerda - Tecumseh High School
5/20/08
To Summarize :
- The sign of ΔSsurr depends on the direction of heart transfer.
- The magnitude of ΔSsurr depends on the temperature.
Signs of Entropy Changes
Spontaneous Process?
ΔSsys
ΔSsurr
ΔSuniv
+
+
+
Yes
No (will be spontaneous in
opposite direction)
+
?
Yes, if ΔSsys has a greater
magnitude than ΔSsurr
+
?
Yes, if ΔSsurr has a greater
magnitude than ΔSsys
- At constant temperature :
S surr  
H
T
Note temp. dependence of ΔSsurr :
If T is large - small change in ΔS
If T is small - large change in ΔS
- where : T is Kelvin temperature
ΔH is the enthalpy of system (negative if exothermic)
- ΔSsys will be equal in magnitude and opposite in sign of ΔSsurr
ex. Calculate the change in entropy of the surroundings for the following reaction at 25 ºC
Sb2S3(s) + 3Fe(s) ----> 2 Sb(s)
+ 3 FeS(s)
ΔH = -125 kJ
4. Free Energy (G) - can be used to determine if a reaction is spontaneous or not at constant T and P
G  H  TS
- if ΔG is negative, the reaction is spontaneous (at constant T and P)
- if ΔG is positive, the reaction is nonspontaneous (at constant T and P)
- if ΔG = 0, the reaction is at equilibrium
a. Also, at constant T and P
S univ  
G
T
:
AP Chemistry : Ch. 16 Notes
Mr. Ferwerda - Tecumseh High School
Calculation of ΔG at constant T and P :
5/20/08
ΔG = ΔH +ΔTS (T in Kelvin)(Energy units must be alike).
Note : Reaction is spontaneous when ΔG is negative (only at constant T and P).
_____(T/F) A reaction that is spontaneous has a negative ΔG and favors the formation of the products.
_____(T/F) A reaction that has a positive ΔG favors the formation of the reactants and is not spontaneous in
the forward direction.
_____ (T/F) The value of ΔG of a reaction at equilibrium is zero.
- Note : A negative enthalpy value and a positive entropy value favor a negative free energy value and
a spontaneous reaction.
ΔH
ΔS
ΔG
Negative
Positive
Positive
Positive
Negative
Negative
positive
Negative
For the reaction A + B - C + D :
Given :
ΔH = -233.4 kJ
ΔS = -34.5 J/K x mol
ΔH = + 678 kJ
ΔS = - 35.7J/K x mol
ΔH = -455.5 kJ
ΔS = + 12.2 J/ K x mol
ΔG will be :
Calculate ΔG for the following reaction from the given data: 3 C(graphite) + 2O3(g)  3 CO2(g)
Substance S (J/K·mol) H (kJ/mol)
C(graphite)
5.694
1.9
O3(g)
238
142.0
CO2(g)
213.6
-393.5
At what temperature would a reaction with a ΔH of 25 kJ and ΔS of 5.0 J/K be spontaneous?
AP Chemistry : Ch. 16 Notes
Mr. Ferwerda - Tecumseh High School
5/20/08
At what temperature would a reaction be spontaneous (constant T and P) given the following values?
ΔH = + 25 kJ
ΔS = - 5.0 J/K
If ΔG is negative, the reaction is ________________________________.
If ΔG is positive, the reaction is _________________________________.
If ΔG is 0 the reaction must be at ________________________________.
For mercury, the enthalpy of vaporization is 58.51 kJ/mol and the entropy of vaporization is 92.92 J/K. What
is the normal boiling point of mercury?
Day
Review :
Which of the following are state functions?
- entropy
- heat
- work
- enthalpy
- free energy
- volume
- mass
The first law of thermodynamics states that _________________ can neither be __________________ nor
________________________.
The second law of thermodynamics states that in any _____________________ change the _______________
of the _______________________ will ______________________.
The third law states that the _____________ of a ____________ _____________ at 0 K is _____________.
Which of the following indicates a spontaneous reaction?
a. positive ΔSuniv
f. negative ΔSsys
b. positive ΔSsurr
g. positive ΔG at constant T and P
c. positive ΔSsys
h. negative ΔG at constant T and P
d. negative ΔSuniv
e. negative ΔSsurr
Which of the following would be spontaneous?
a. ΔSsurr = 30 J/K , ΔSsys = 50 J/K
b. ΔSsurr = -20 J/K, ΔSsys = 15 J/K
c. ΔSsurr = 30 J/K , ΔSsys = -40 J/K
AP Chemistry : Ch. 16 Notes
Mr. Ferwerda - Tecumseh High School
Which of the following shows an increase in entropy?
a. H2O(l) --> H2O(g)
b. C3H8(g) + 5O2(g) --> 3CO2(g)
c. 4 Na(s) + O2(g) --> 2Na2O(s)
5/20/08
+ 4H2O(g)
The following reaction has a negative change in entropy. What is a plausible reason for this change in entropy?
HX(aq) <===> H+(aq)
+ X-(aq)
Which of the following favors a spontaneous reaction? (ΔG = ΔH – TΔS)
+ ΔH or
- ΔH
+ ΔS or - ΔS
Is the dissociation of chlorine gas spontaneous or nonspontaneous? : Cl2(g) --> 2 Cl(g)
Consider the following :
- sign of ΔH :
- sign of ΔS :
- sign of ΔG :
For any given reaction such as : A + B  C + D , where ΔG = X,
_____ (T/F) A positive free energy favors the reactants.
_____ (T/F) A negative free energy favors the products.
_____ (T/F) At ΔG = 0 the system will be at equilibrium.
ΔH = 16.6 kJ/mol
ΔS = 50.2 J/K mol
ΔH = -12.2 kJ/mol
ΔS = 80.0 J/K mol
A(s) <====> A(l)
How would you calculate the melting temperature of the above compound?
5. Entropy Changes in Chemical Reactions
a. Fewer molecules in products indicates a decrease in entropy.
b. A change of state also indicates a change in entropy. The change in positional entropy is dominated
by gaseous reactants or products.
c. Third Law of Thermodynamics : The entropy of a perfect crystal at 0 K is zero.
AP Chemistry : Ch. 16 Notes
Mr. Ferwerda - Tecumseh High School
d. Calculation of entropy from standard entropy values :
5/20/08
ΔSº reaction   Sº products   Sº reactants
- generally, the more complex the molecule, the higher the standard entropy (more rotational and
vibrational positions)
- the standard enthalpies of elements in their standard states are not zero (unlike standard
enthalpies and standard free energies)
6. Free Energy Change and Chemical Reactions
a. Standard free energy change (ΔGº) - the free energy change that occurs in a chemical reaction when
all reactants and products are in their standard states.
b. Calculation of standard free energy change :
- free energy is a state function
ΔGº reaction   Gº products   Gº reactants
or :
ΔGº  Hº-TSº
or by adding ΔGº values from chemical reactions as in Hess's law for ΔH
7. The Dependence of Free Energy on Pressure
_____ (T/F) An increase in pressure would decrease the volume of a gas and therefore decrease its entropy.
a. The equilibrium position of a reaction system represents the lowest free energy value available.
- reason why some reactions will not go to completion
b. For an ideal gas :
- enthalpy is independent of pressure
- entropy is pressure-dependent (change in volume changes positional probabilities)
G  G º  RT ln( Q)
- where : R = universal gas constant (8.3145J/K∙mol)
T = Kelvin temperature
Q = reaction quotient
1.0 atm
1.5 atm 2.0 atm
ex. of calc. of Q : 2A(g) + B(s) ---> 3C(g) + 2D(g)
Q=
Ex. Calculate ΔG for the following reaction at 25 ºC :
2 H2S(g) + SO2(g) <===> 3S(s) + 2 H2O(g)
Substance
H2S(g)
SO2(g)
3S(s)
H2O(g)
Pressure (atm)
1.0 x 10-4
1.0 x 10-2
--3.0 x 10-2
G (kJ/mol)
-34
-300.
0.0
-229
AP Chemistry : Ch. 16 Notes
Mr. Ferwerda - Tecumseh High School
5/20/08
8. Free Energy and Equilibrium
a. The equilibrium point is at the lowest free energy value available to the system.
- therefore, Gprod = Greact, or ΔG = Gprod - Greact = 0
-ΔG
and ΔGº = -RTln(K) or K =e RT
b. Equilibrium constant (K) :
For the reaction 2A + 3 B <===> 4C
K=
C4
(A 2 )(B3 )
c. Relationship between ΔGº and K :
ΔGº
ΔGº = 0
ΔGº < 0
ΔGº > 0
K
K=1
K>1
K<1
Favors
Example: Solve for K at 298 K (see table for values of G)
2 H2S(g) + SO2(g) <===> 3S(s) + 2 H2O(
Substance
H2S(g)
SO2(g)
3S(s)
H2O(g)
G (kJ/mol)
-34
-300.
0.0
-229
9. Free Energy and Work
a. The change in free energy, at constant temperature and pressure, is equal to the maximum possible
amount of useful work obtainable from that system.
-
wmax  G
- According to second law, this amount of work is not really available - some energy will be lost
in any real pathway
- reversible processes are really hypothetical since energy is lost in the changes (it really takes
more work to restore the original condition)
- all real processes are therefore irreversible
Energy is _________________________________________
If a certain process has a ΔG value of -56.7 kJ, what is the maximum amount of work that can come out of the
process?
Is the above process real or theoretical?
All real processes are__________________________________.
b. Summary of first two laws of thermodynamics from Henry Bent :
- First law : You can't win, you can only break even.
- Second Law : You can't break even.