1
General Chemistry II Jasperse
Entropy, Spontaneity, and Free Energy. Extra Practice Problems .
General Types/Groups of problems:
Evaluating Relative Molar Entropy for Chemicals
p1
Calculating ΔG for Reactions (Math)
p5
Evaluating ΔS for Reactions (non-math)
p2
ΔG, ΔH, ΔS, Equilibrium, and Temperature
p6
Calculating ΔS for Reactions (Math)
p2
Answers
p7
Entropy/Enthalpy and Spontaneity.
p4
Key Equations Given for Test:
For weak acids alone in water:
[H
+
] =
€
K
a
x WA
[ ]
For weak bases alone in water:
[OH
-
] =
€
K
b
x WB
[ ]
pZ= -logZ
General definition for p of anything
pH + pOH = 14
[H
+
][HO
-
] = 1.00 x 10
-14
K
a
K
b
= 1.00 x 10
-14
for conjugate acid/base pair
For Buffer: pH = pK
a
+ log[base]/[acid]
Henderson-Hasselbalch Equation
∆S˚ = S˚ (products) – S˚ (reactants)
∆G˚ = G˚ (products) – G˚ (reactants)
∆G˚ = ∆H˚ –T∆S˚ (T in Kelvin)
EVALUATING/RANKING STANDARD MOLAR ENTROPY (S°) FOR CHEMICALS (non-math)
1. Which of the following is in the correct order of standard state entropy?
I.
Liquid water < gaseous water
II.
Liquid water < solid water
III.
NH
3
< H
2
a.
I only
d.
I and II only
b.
II only
e.
I and III only
c.
III only
2. Which of the following will have the greatest standard molar entropy (S°)?
a.
NH
3
(g)
d.
H
2
O(l)
b.
He(g)
e.
CaCO
3
(s)
c.
C(s, graphite)
3. Indicate which of the following has the lowest standard molar entropy (S°).
a.
CH
4
(g)
d.
Na(s)
b.
CH
3
CH
2
OH()
e.
He(g)
c.
H
2
O(s)
4. Indicate which of the following has the highest entropy at 298 K.
a.
0.5 g of HCN
b.
1 mol of HCN
c.
2 kg of HCN
d.
2 mol of HCN
e.
All of the above have the same entropy at 298 K.
T
T
T
1. GAS > Liquid > Solid
-gas wins over any other
2. Molecular Size
3. Number of Molecules
4. Dissolved > undissolved
5. Temperature
2
EVALUATING ΔS FOR REACTIONS (non-math recognition)
5. Indicate which one of the following reactions result in a positive ΔS
sys
.
a.
AgNO
3
(aq) + NaCl(aq) ! AgCl(s) + NaNO
3
(aq)
b.
H
2
O (g) + CO
2
(g) ! H
2
CO
3
(aq)
c.
H
2
(g) + I
2
(g) ! 2
Hl(g)
d.
C
2
H
2
O
2
(g) ! 2
CO(g) + H
2
(g)
e.
H
2
O(g) ! H
2
O(l)
6. Indicate which one of the following reactions results in a negative ΔS
sys
.
a.
H
2
O(g) ! H
2
O(s)
b.
CaCO
3
(s) ! CaO(s) + CO
2
(g)
c.
CuSO
4
(H
2
O)
5
(s) ! CuSO
4
(s) + 5H
2
O(g)
d.
14O
2
(g) + 3NH
4
NO
3
(s) + C
10
H
22
(l) → 3N
2
(g) + 17H
2
O(g) + 10CO
2
(g)
e.
CO
2
(aq) ! CO
2
(g)
7. Which of the processes A–D will lead to a positive change in the entropy of the system? If all of these processes lead
to a positive change in the entropy of the system, select E.
a.
Sodium chloride crystals form as saltwater evaporates.
b.
Helium gas escapes from the hole in a balloon.
c.
Stalactites form in a cave.
d.
Water freezes in a freezer.
e.
All of these lead to a positive change in entropy of the system, as they are all spontaneous.
8. Which of the following processes will lead to a decrease in the entropy of the system?
a.
Salt crystals dissolve in water.
b.
Air escapes from a hole in a balloon.
c.
Iron and oxygen react to form rust.
d.
Ice melts in your hand.
e.
None of these lead to a negative change in the entropy of the system, as they are all
spontaneous.
CALCULATING ΔS FOR REACTIONS (Math)
Substance
S° (J/mol · K)
H
2
(g)
130.58
I
2
(g)
116.73
HI(g)
206.3
a.
–41.10 J/mol · K
d.
+165.29 J/mol · K
b.
–165.29 J/mol · K
e.
+41.10 J/mol · K
c.
+398.75 J/mol · K
10. Determine ΔS for N
2
O
4
(g) ! 2NO
2
(g) given the following information.
Substance
S° (J/mol · K)
N
2
O
4
(g)
304.3
NO
2
(g)
240.45
a.
+176.7 J/mol · K
d.
–50.7 J/mol · K
b.
–63.8 J/mol · K
e.
–176.7 J/mol · K
c.
+63.8 J/mol · K
T
T
T
1. GAS > Liquid > Solid
-gas wins over any other
2. Molecular Size
3. Number of Molecules
4. Dissolved > undissolved
5. Temperature
Any side with more gas has
more entropy
ΔS = S (products) - S (reactants)
Note: Sº for elements is NOT zero
Factor # of moles
9. Determine ΔS for H2(g) + I2(g) !! 2HI(g) given the following information.
3
11. What is the entropy change to make 1 mole of SO
3
for the reaction SO
2
(g) + 1/2 O
2
(g) → SO
3
(g)
Substance
S° (J/mol · K)
SO
2
(g)
248.2
O
2
(g)
205.0
SO
3
(g)
256.8
a.
–196.4 J/K
d.
+93.9 J/K
b.
+196.4 J/K
e.
+401.4 J/K
c.
–93.9 J/K
12. NO gas is converted to NO
2
gas according to the following reaction, NO(g) + 1/2 O
2
(g) → NO
2
(g)
What is the standard entropy change when 0.5 mol of NO gas reacts with 0.5 mol of O
2
gas?
Substance
S° (J/mol · K)
NO(g)
210.7
O
2
(g)
205.0
NO
2
(g)
240.0
a.
–36.6 J/K
d.
+83.4 J/K
b.
–175.7 J/K
e.
+36.6 J/K
c.
–83.4 J/K
13. If 3.500 g of Ni (58.69 g/mol)are reacted with excess oxygen to form nickel oxide (NiO) under standard state
conditions, what is the entropy change for the reaction?
2Ni(s) + O
2
! 2NiO(s)
Substance
S° (J/mol · K)
Ni
182.1
O
2
205.0
NiO
37.99
a.
–49.3 J/K
d.
+49.3 J/K
b.
–24.7 J/K
e.
–10.4 J/K
c.
–14.7 J/K
14. What is the entropy change if 4.500 g of CaCO
3
(s) is placed in a container and allowed to decompose to CaO(s) and
CO
2
(g) according to the following reaction?
CaCO
3
(s) ! CaO(s) + CO
2
(g)
Substance
S° (J/mol · K)
CaCO
3
(s)
92.88
CaO(s)
39.75
CO
2
(g)
213.6
a.
+7.2 J/K
d.
+160.5 J/K
b.
–160.5 J/K
e.
+3.57 J/K
c.
+35.7 J/K
15. What is the standard entropy change when 10.0 g of methane reacts with 10.0 g of oxygen?
CH
4
(g) + 2O
2
(g) → CO
2
(g) + 2H
2
O()
Substance
S° (J/mol · K)
CH
4
(g)
186.2
O
2
(g)
205.0
H
2
O(l)
70.0
CO
2
(g)
213.6
a.
–121 J/K
d.
–154.4 J/K
b.
–37.9 J/K
e.
–16.8 J/K
c.
–242.6 J/K
T
T
T
ΔS = S (products) - S (reactants)
Note: Sº for elements is NOT zero
Factor # of moles
Notice importance of a BALANCED reaction, and how
we needed to factor in the 1/2 coefficient
Balanced reaction needed
1. Factoring in which chemical is limiting.
2. Find ΔSº on a per-mole basis first
3. Then adjust for how many moles were
actually used.
1.Find ΔSº on a per-mole basis first
2.Then adjust for how many moles
were actually used.
1.Find ΔS on a per-mole basis first
2.Then adjust for how many moles
were actually used.
1. Factoring in which chemical is limiting.
2. Find ΔS on a per-mole basis first
3. Then adjust for how many moles were
actually used.
Note: you should normally be
able to predict whether ΔS is
positive or negative. Use this
to eliminate implausible
answers.
4
CHANGES IN ENTROPY OF UNIVERSE VS SYSTEM. Evaluating Spontaneity Considering both
Entropy and Enthalpy
16. In a spontaneous process, which of the following always increases?
a.
the entropy of the system
b.
the entropy of the surroundings
c.
the entropy of the universe
d.
the entropy of the system and the universe
e.
the entropy of the system, surroundings and the universe
17. Processes are always spontaneous, regardless of temperature, when __________ (H and S refer to the system).
a.
ΔH > 0 and ΔS < 0
b.
ΔH < 0 and ΔS < 0
c.
ΔH > 0 and ΔS > 0
d.
ΔH < 0 and ΔS > 0
e.
None of these is true, as temperature must always be taken into account.
18. The dissolution of ammonium nitrate in water is a spontaneous endothermic process. It is spontaneous because the
system undergoes __________
a.
a decrease in enthalpy.
d.
a decrease in entropy.
b.
an increase in entropy.
e.
an increase in free energy.
c.
an increase in enthalpy.
19. Which of the following must be true for a spontaneous exothermic process?
a.
only that ΔS
sys
< 0
b.
only that ΔS
sys
> 0
c.
both ΔS
sys
< 0 and the magnitude of ΔS
sys
< the magnitude of ΔS
surr
d.
both ΔS
sys
< 0 and the magnitude of ΔS
sys
> the magnitude of ΔS
surr
e.
either ΔS
sys
> 0, or else ΔS
sys
< 0 but the magnitude of ΔS
sys
< the magnitude of ΔS
surr
20. Suppose a chemical reaction is found to be spontaneous, but with ΔS
sys
< 0. Which of the following statements must
be TRUE?
a.
ΔS
surr
< 0 and its magnitude is < ΔS
sys
. In other words, the system loses entropy and the surroundings also lose
entropy. The loss by the surroundings is less than the loss by the system.
b.
ΔS
surr
< 0 and its magnitude is > ΔS
sys
. In other words, the system loses entropy and the surroundings also lose
entropy. The loss by the surroundings is greater than the loss by the system.
c.
ΔS
surr
> 0 and its magnitude is < ΔS
sys
. In other words, the system loses entropy but the surroundings gain
entropy. The gain by the surroundings is less than the loss by the system.
d.
ΔS
surr
> 0 and its magnitude is > ΔS
sys
. In other words, the system loses entropy but the surroundings gain
entropy, and the gain by the surroundings outweighs the loss by the system.
e.
an error has been made, as S
sys
> 0 by necessity for a spontaneous process.
T
T
T
ΔS
universe
depends on both ΔS
system
and
ΔS
surroundings.
ΔS
system
is reflected by ΔSº,
(system entropy) while ΔS
surroundings
is
reflected by ΔH
system
(system enthlpy)
If you know it's spontaneous, then either entropy
or enthalpy (or both) must be favorable.
If you know one is NOT favorable, then the other
must certainly be favorable.
Since Enthalpy is favorable (exothermic), then
entropy could be either favorable or unfavorable.
But if it's unfavorable, it must not be bad enough to
win over the good enthalpy.
5
FREE ENERGY AND CALCULATING ΔG FOR REACTIONS (Math)
21. Any reaction will be spontaneous if __________
a.
ΔG
sys
> 0
d.
ΔS
sys
< 0
b.
ΔG
sys
< 0
e.
ΔH
sys
< 0
c.
ΔS
sys
> 0
From ΔG
formationn
22. What is the ΔG
rxn
for the reaction given:
CH
4
(g) + 2O
2
(g) ! CO
2
(g) + 2H
2
O(g)
Substance
ΔG
form
(kJ/mol)
CH
4
(g)
50.8
CO
2
(g)
394.4
H
2
O(g)
–228.57
a.
–50.8 kJ/mol
d.
–115 kJ/mol
b.
–751 kJ/mol
e.
–807 kJ/mol
c.
–113 kJ/mol
23. Determine ΔG
rxn
for C
4
H
10
(l) + 13/2 O
2
(g) ! 4CO
2
(g) + 5H
2
O(g) given the following.
Substance
ΔG
form
(J/mol · K)
C
4
H
10
(l)
–15.0
CO
2
(g)
–394.4
H
2
O(g)
–228.57
a.
–2705 kJ/mol
d.
–3457 kJ/mol
b.
–608.0 kJ/mol
e.
+608.0 kJ/mol
c.
–1791 kJ/mol
24. Given the following data, determine the molar free energy of combustion for propane gas, C
3
H
8
.
ΔG (C
3
H
8
, g)
–23.5 kJ/mol
ΔG (CO
2
, g)
–394.4 kJ/mol
ΔG (H
2
O, g)
–105.6 kJ/mol
a.
–1629.1 kJ/mol
d.
+476.5 kJ/mol
b.
–1582.1 kJ/mol
e.
+1582.1 kJ/mol
c.
–476.5 kJ/mol
From ΔH° and ΔS°
25. Hydrogen reacts with nitrogen to form ammonia (NH
3
) according to the reaction
3H
2
(g) + N
2
(g) ! 2NH
3
(g)
The value of ΔH° is –92.38 kJ/mol, and that of ΔS° is –198.2 J/mol · K. Determine ΔG° at 25°C.
a.
+5.897 × 10
4
kJ/mol
d.
–16.66 kJ/mol
b.
+297.8 kJ/mol
e.
+49.5 kJ/mol
c.
–33.32 kJ/mol
26. Hydrochloric acid (HCl) reacts with sodium hydroxide (NaOH) to form sodium chloride (NaCl) and water. If ΔH° = –
56.13 kJ/mol and ΔS° = 79.11 J/mol · K, what is ΔG° for this reaction at 20°C?
a.
–79.31 kJ/mol
d.
79.31 kJ/mol
b.
–77.73 kJ/mol
e.
–1638 kJ/mol
c.
–2.324 × 10
4
kJ/mol
T
T
T
ΔGº = Gº (products) - Gº (reactants)
G for elements is ZERO
Factor Coefficients
Be patient and systematic! With all
the plus/minuses, and multiplying,
easy to goof up!
1. Balance first.
2. Then solve
1. 25ºC = 298K
2. Entropy units in J; enthalpy in kJ,
so need to factor by a thousand.
Enthalpy good, entropy bad,
overall good, so enthalpy won
6
FREE ENERGY, ENTROPY, ENTHALPY, EQUILIBRIUM, and TEMPERATURE
27. A reaction is at equilibrium at a given temperature and constant pressure when __________
a.
ΔS
rxn
= 0.
d.
ΔG = 0.
b.
ΔS = 0.
e.
ΔH
rxn
= 0.
c.
ΔG
rxn
= 0.
28. Which of the following statements about equilibrium are true?
I.
ΔG
sys
= 0
II.
ΔS
sys
= 0
III.
ΔS
universe
= 0
a.
I only
d.
Both I and II
b.
II only
e.
Both I and III
c.
III only
29. A reaction with a low enthalpy of reaction value is not spontaneous at low temperature but becomes spontaneous at
high temperature. What are the signs for ΔH° and ΔS°, respectively?
a.
+, –
b.
–, –
c.
–, +
d.
+, +
e.
Insufficient data is provided to answer this question.
30. The enthalpy of fusion for benzene (C
6
H
6
, 78.0 g/mol) is 127.40 kJ/kg, and its melting point is 5.5°C. What is the
entropy change when 1 mole of benzene melts at 5.5°C?
a.
9.95 kJ/K
d.
1.81 J/K
b.
35.7 J/K
e.
127.40 kJ/K
c.
1809 J/K
31. The entropy of vaporization of water is 109.0 J/mol · K. What is the enthalpy of vaporization of water at its normal
boiling point of 100°C?
a.
+10.90 kJ/mol
d.
+40.66 kJ/mol
b.
–40.66 kJ/mol
e.
–10.90 kJ/mol
c.
+3.42 kJ/mol
32. The enthalpy and entropy of vaporization of ethanol are 38.6 kJ/mol and 109.8 J/mol · K, respectively. What is the
boiling point of ethanol, in °C?
a.
352°C
b.
78.5°C
c.
2.84°C
d.
624°C
e.
Not enough information is given to answer the question.
33. Dinitrogen tetroxide (N
2
O
4
) decomposes to nitrogen dioxide (NO
2
). If ΔH° = 58.02 kJ/mol and ΔS° = 176.1 J/mol · K,
at what temperature are reactants and products in their standard states at equilibrium?
a.
+56.5°C
d.
+25.0°C
b.
+329.5°C
e.
+98.3°C
c.
–272.7°C
T
T
T
1. Entropy factor increases at higher temp
2. A temperature "flip" means entropy and enthalpy disagree.
a. If so, high Temp = entropy control,
b. low temp = enthalpy control.
ΔGº = Gº (products) - Gº (reactants)
ΔGº = ΔHº - TΔSº
1. At equilibrium ΔGº = 0
2. So, at eq, ΔHº = TΔSº
3. Phase changes are equilibrium
Oops! Extra hard
because enthalpy per kg
rather than per mole
Note: since entropy favorable,
(positive), enthalpy must be
unfavorable (also positive).
Otherwise vaporization favorable at
ALL temps, not just high ones.
1. Convert S in J to kJ
2. Convert Temp answer in K to ºC (subtract 273)
1. Convert S in J to kJ
2. Convert Temp answer in K to ºC (subtract 273)
7
General Chemistry II Jasperse ANSWERS
Entropy, Spontaneity, and Free Energy. Extra Practice Problems
1. A
2. A
3. D
4. C
5. D
6. A
7. B
8. C
9. D
10. A
11. C
12. A
13. C
14. A
15. B
16. C
17. D
18. B
19. E
20. D
21. B
22. C
23. A
24. B
25. C
26. A
27. C
28. E
29. D
30. B
31. D
32. B
33. A
