Transcript Slide 1

General Chemistry
Principles & Modern Applications
9th Edition
Petrucci/Harwood/Herring/Madura
Chapter 19
Spontaneous Change: Entropy and Free
Energy
Dr. Travis D. Fridgen
Memorial University of Newfoundland
© 2007 Pearson Education
Which of the following processes would you expect to result in a greater
positive change in entropy?
1. H2O(l, 25 oC)  H2O(l, 75 oC)
2. CO2 (s, 101 kPa)  CO(g, 2 kPa)
3. CH3OH(l, 101 kPa)  CH3OH(g, 2 kPa)
4. Fe(s)
 Fe(l)
5. None of these processes involve positive
entropy changes.
Which of the following processes would you expect to result in a greater
positive change in entropy?
1. H2O(l, 25 oC)  H2O(l, 75 oC)
2. CO2 (s, 101 kPa)  CO(g, 2 kPa)
3. CH3OH(l, 101 kPa)  CH3OH(g, 2 kPa)
4. Fe(s)
 Fe(l)
5. None of these processes involve positive
entropy changes.
Which of the following statements is incorrect?
1. The absolute entropy for gaseous C2H6 is lower
than that for gaseous C3H8 at the same
temperature.
2. The entropy of formation for gaseous C3H8 is negative.
3. The entropy of formation of gaseous C3H8 is more negative
than the entropy of formation for C2H6.
4. The absolute entropy for gaseous C3H8 decreases when the temperature
is increased.
5. DfusS (solid to liquid) for H2O is greater than that for C3H8.
Which of the following statements is incorrect?
1. The absolute entropy for gaseous C2H6 is lower
than that for gaseous C3H8 at the same
temperature.
2. The entropy of formation for gaseous C3H8 is negative.
3. The entropy of formation of gaseous C3H8 is more negative
than the entropy of formation for C2H6.
4. The absolute entropy for gaseous C3H8 decreases when the temperature
is increased.
5. DfusS (solid to liquid) for H2O is greater than that for C3H8.
The entropy change for the decomposition
of ozone forming diatomic oxygen,
2O3 (g)  3O2 (g)
DH =  285 kJ mol-1
1. is positive because two moles of gas are
forming three moles of gas.
2. is close to zero because there are the same number of
atoms on each side of the equation.
3. is negative because energy is released as the reaction proceeds.
4. is close to zero because both ozone and oxygen are in the gas phase.
5. None of the above answers makes sense.
The entropy change for the decomposition
of ozone forming diatomic oxygen,
2O3 (g)  3O2 (g)
DH =  285 kJ mol-1
1. is positive because two moles of gas are
forming three moles of gas.
2. is close to zero because there are the same number of
atoms on each side of the equation.
3. is negative because energy is released as the reaction proceeds.
4. is close to zero because both ozone and oxygen are in the gas phase.
5. None of the above answers makes sense.
Which of the following reactions, occurring when
propane is burned, would you expect to be the most
entropically favored (most positive change in
entropy)?
1. C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O(g)
2. C3H 8 (g) +
7
O2 (g)
2
 3CO(g) + 4H 2O(g)
3. C3H8 (g) + 2O2 (g)  3C(graph) + 4H2O(g)
Which of the following reactions, occurring when
propane is burned, would you expect to be the most
entropically favored (most positive change in
entropy)?
1. C3H8 (g) + 5O2 (g)  3CO2 (g) + 4H2O(g)
2. C3H 8 (g) +
7
O2 (g)  3CO(g) + 4H 2O(g)
2
3. C3H8 (g) + 2O2 (g)  3C(graph) + 4H2O(g)
In which of the following processes involving
CO2 do you think the standard free energy
change not make sense?
1. MgCO3 (s)  MgO(s) + CO2 (g)
DS = 175 J K-1 mol-1
2. CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O(g)
3. 2CO(g)  C(graph) + CO2 (g)
DS = -4 J K-1 mol-1
DS =  175 J K-1 mol-1
4. C2H5OH(g) + 3O2 (g)  2CO2 (g) + 3H2O(l)
DS = 260 J K-1 mol-1
5. C6H12O6 (s) + 6O2 (g)  6CO2 (g) + 6H2O(g)
DS = 976 J K-1 mol-1
In which of the following processes involving
CO2 do you think the standard free energy
change not make sense?
1. MgCO3 (s)  MgO(s) + CO2 (g)
DS = 175 J K-1 mol-1
2. CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O(g)
3. 2CO(g)  C(graph) + CO2 (g)
DS = -4 J K-1 mol-1
DS =  175 J K-1 mol-1
4. C2 H 5OH(g) + 3O2 (g)  2CO2 (g) + 3H 2O(l)
DS = 260 J K -1 mol -1
5. C6H12O6 (s) + 6O2 (g)  6CO2 (g) + 6H2O(g)
DS = 976 J K-1 mol-1
Which of the following compounds
would obey Trouton’s rule most
closely?
1.
D vapS =
2.
D vap H
Tb
= 87 J K-1 mol-1
3.
Which of the following compounds
would obey Trouton’s rule most
closely?
1.
D vapS =
2.
D vap H
Tb
= 87 J K-1 mol-1
3.
At room temperature ~290 K the reaction of H2 and O2 to
form water:
2H 2 (g) + O2 (g)
2H 2O(l)
DH =  285.8 kJ mol-1
DS =  327 J K-1 mol-1
1. is spontaneous because it is exothermic.
2. is non-spontaneous because DS is negative.
3. is spontaneous because DS is negative.
4. is spontaneous because DG is negative.
5. is spontaneous because DG is positive.
J. W. Gibbs
Edgar Fahs Smith Collection
University of Pennsylvania Library
At room temperature ~290 K the reaction of H2 and O2 to
form water:
2H 2 (g) + O2 (g)
2H 2O(l)
DH =  285.8 kJ mol-1
DS =  327 J K-1 mol-1
1. is spontaneous because it is exothermic.
2. is non-spontaneous because DS is negative.
3. is spontaneous because DS is negative.
4. is spontaneous because DG is negative.
5. is spontaneous because DG is positive.
J. W. Gibbs
Edgar Fahs Smith Collection
University of Pennsylvania Library
Which of the following processes would you expect to be spontaneous at all
temperatures?
1. 2SO2 (g) + O2 (g)  2SO3 (g)
2.
3.

DH = -200 kJ mol-1
DH = 6 kJ mol-1
DH = -5 kJ mol-1
4. None are spontaneous at all temperatures.
5. All are spontaneous irrespective of the temperature.
Which of the following processes would you expect to be spontaneous at all
temperatures?
1. 2SO2 (g) + O2 (g)  2SO3 (g)
2.
3.

DH = -200 kJ mol-1
DH = 6 kJ mol-1
DH = -5 kJ mol-1
4. None are spontaneous at all temperatures.
5. All are spontaneous irrespective of the temperature.
The transformation between ice and water:
H 2O(s)
H 2O(l)
DH = 6.0 kJ mol-1
DS = 22 J K-1 mol-1
1. Is spontaneous in the directions written because
DS is positive.
2. Is spontaneous in the backwards direction
because DH is positive.
3. Is spontaneous in the forward direction above 273
K and spontaneous in the backward direction
below 273 K.
4. All three answers are correct.
5. There is not enough information provided to answer this question.
The transformation between ice and water
H 2O(s)
H 2O(l)
DH = 6.0 kJ mol-1
DS = 22 J K-1 mol-1
1. Is spontaneous in the directions written because
DS is positive.
2. Is spontaneous in the backwards direction
because DH is positive.
3. Is spontaneous in the forward direction above 273
K and spontaneous in the backward direction
below 273 K.
4. All three answers are correct.
5. There is not enough information provided to answer this question.
A mixture of H2 and O2 can sit in a flask almost indefinitely at 298 K without
reacting.
2H 2 (g) + O2 (g)  2H 2O(l)
DH =  285.8 kJ mol-1
DS =  327 J K-1 mol-1
What is the best explanation for the absence of observable reaction?
1. A significant energy barrier hinders the start of the reaction.
2. The reaction is not spontaneous at this temperature.
3. The reaction is entropically unfavorable.
4. All three of these factors contribute.
5. None of the above answers is correct.
?
A mixture of H2 and O2 can sit in a flask almost indefinitely at 298 K without
reacting.
2H 2 (g) + O2 (g)  2H 2O(l)
DH =  285.8 kJ mol-1
DS =  327 J K-1 mol-1
What is the best explanation for the absence of observable reaction?
1. A significant energy barrier hinders the start of the reaction.
2. The reaction is not spontaneous at this temperature.
3. The reaction is entropically unfavorable.
4. All three of these factors contribute.
5. None of the above answers is correct.
?
There is enough energy in lightning bolts that O2
and N2 in the atmosphere are decomposed into N
and O atoms. The reaction of N and O to form NO,
N(g) + O(g)

NO(g)
1. is spontaneous at all temperatures.
2. is non-spontaneous at all temperatures.
3. is spontaneous at high temperatures but nonspontaneous at low temperatures.
4. is spontaneous at low temperatures but nonspontaneous at high temperatures.
5. It is impossible to choose between the above
responses without thermochemical data.
There is enough energy in lightning bolts that O2
and N2 in the atmosphere are decomposed into N
and O atoms. The reaction of N and O to form NO,
N(g) + O(g)

NO(g)
1. is spontaneous at all temperatures.
2. is non-spontaneous at all temperatures.
3. is spontaneous at high temperatures but nonspontaneous at low temperatures.
4. is spontaneous at low temperatures but nonspontaneous at high temperatures.
5. It is impossible to choose between the above
responses without thermochemical data.
Below is some thermochemical data for
diamond and graphite.
C(diamond)
C(graphite)
Df H o / kJ mol-1
2
0
So / J K -1 mol-1
2
6
True or false, equilibrium favors graphite
in the above reaction.
1. True
2. False
3. The proportion of diamond and graphite are equal.
4. Cannot determine from the data given.
Below is some thermochemical data for
diamond and graphite.
C(diamond)
C(graphite)
Df H o / kJ mol-1
2
0
So / J K -1 mol-1
2
6
True or false, equilibrium favors graphite
in the above reaction.
1. True
2. False
3. The proportion of diamond and graphite are equal.
4. Cannot determine from the data given.
1. The equilibrium constant is 1.
2. The system has reached equilibrium.
3. The reaction will be too slow to observe.
4. All of the above are correct.
System free energy, G
When DG = 0,
5. 1 and 2 above are both correct.
Reactants
Products
1. The equilibrium constant is 1.
2. The system has reached equilibrium.
3. The reaction will be too slow to observe.
4. All of the above are correct.
System free energy, G
When DG = 0,
Q<K
DGo
Q>K
Q=K
equilibrium
5. 1 and 2 above are both correct.
Reactants
(std state)
Products
(std state)
DG = G(P)-G(R) = 0
The equilibrium constant for the decomposition of
colorless dinitrogen tetroxide to form brown nitrogen
dioxide
N2O4 (g)
2NO2 (g)
1. favors reactants at all temperatures.
2. favors products at all temperatures.
3. favors reactants high temperatures only
4. favors products at high temperatures only.
5. It is impossible to choose between the above
responses without thermochemical data.
The equilibrium constant for the decomposition of
colorless dinitrogen tetroxide to form brown nitrogen
dioxide
N2O4 (g)
2NO2 (g)
1. favors reactants at all temperatures.
2. favors products at all temperatures.
3. favors reactants high temperatures only
4. favors products at high temperatures only.
5. It is impossible to choose between the above
responses without thermochemical data.
CO binds to iron in hemoglobin more
favorably than does oxygen, which is why
CO is poisonous.
Hgb + CO
Hgb + O2
HgbCO
HgbO2
DGo   80 kJ mol-1
DGo   70 kJ mol-1
What is the free energy change for O2 replacing
CO bound to iron of hemoglobin?
1. 150 kJ mol-1
4. -150 kJ mol-1
2. 10 kJ mol-1
5. 5600 kJ mol-1
3. -10 kJ mol-1
CO binds to iron in hemoglobin more
favorably than does oxygen, which is why
CO is poisonous.
Hgb + CO
Hgb + O2
HgbCO
HgbO2
DGo   80 kJ mol-1
DGo   70 kJ mol-1
What is the free energy change for O2 replacing
CO bound to iron of hemoglobin?
1. 150 kJ mol-1
4. -150 kJ mol-1
2. 10 kJ mol-1
5. 5600 kJ mol-1
3. -10 kJ mol-1
The White Cliffs of Dover, England are
made of chalk, CaCO3, which obviously
has a very low solubility, Ksp is ~1 x 10-13
at 298 K. DGo for the dissolution of
CaCO3 is:
1. ~ 70 kJ mol-1
2. ~ 10 kJ mol-1
3. ~ 1 kJ mol-1
4. ~ -10 kJ mol-1
5. ~ -70 kJ mol-1
From http://www.dover-web.co.uk/
The White Cliffs of Dover, England are
made of chalk, CaCO3, which obviously
has a very low solubility, Ksp is ~1 x 10-13
at 298 K. DGo for the dissolution of
CaCO3 is:
1. ~ 70 kJ mol-1
2. ~ 10 kJ mol-1
3. ~ 1 kJ mol-1
4. ~ -10 kJ mol-1
5. ~ -70 kJ mol-1
From http://www.dover-web.co.uk/