AP Chemistry - Problem Drill 23: Thermodynamics
Question No. 1 of 10
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1. An exothermic reaction releases heat to the surroundings, H < 0. For an
exothermic process where entropy increases, it is spontaneous at ____.
Question
(A)
(B)
(C)
(D)
All temperatures
Low temperatures only
High temperatures only
Never
A. Correct!
Good job! Exothermic process (H < 0) and entropy increase (S > 0) are both
favorable for free energy change (G) to be negative (for a spontaneous reaction)
at any temperature, since T in Kelvin is always positive or zero.
B. Incorrect.
Exothermic process (H < 0) and entropy increase (S > 0) are both favorable for
free energy change (G) to be negative (for a spontaneous reaction) at any
temperature.
C. Incorrect.
Feedback
Exothermic process (H < 0) and entropy increase (S > 0) are both favorable for
free energy change (G) to be negative (for a spontaneous reaction) at any
temperature.
D. Incorrect.
Exothermic process (H < 0) and entropy increase (S > 0) are both favorable for
free energy change (G) to be negative (for a spontaneous reaction) at any
temperature.
Exothermic: Energy is released. ∆H = Entropy Increase: ∆S = +
Gibb’s Free Energy: G = H - TS
When H < 0 and S > 0, the G would be negative regardless the temperature.
If G < 0, the process will be spontaneous at all temperatures.
Solution
The correct answer is (A).
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Question No. 2 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
2. The Second Law of Thermodynamics states that a natural process that occurs in
an isolated system is spontaneous with an increase in entropy of the system.
Entropy is a measure of disorder. Liquids are more disordered than solids and gases
are more disordered than liquids. If a reaction occurs in the same phase and the
number of particles in a system increases, the entropy then increases, and vice
versa. Which of the following has an incorrect pairing of the reaction with the sign
of the entropy change?
Question
(A)
(B)
(C)
(D)
I2(g)  I2(s)
H2O(l)  H2O(s)
CH3OH(g) + 3/2 O2(g)  CO2(g) + 2H2O(l)
2O2(g) + 2SO(g)  2SO3(g)
∆S = +
∆S = ∆S = ∆S = -
A. Correct!
Good job! Moving from gas to solid is a decrease in disorder. This is a mis-match in
change in entropy sign.
B. Incorrect.
Moving from a liquid to a solid is a decrease in disorder. The entropy change is
indeed to be negative.
C. Incorrect.
Feedback
Moving from 5/2 gas molecules to 1 gas and 2 liquid molecules is a decrease in
disorder. The entropy change is indeed to be negative.
D. Incorrect.
Moving from 4 gas molecules to 2 gas molecules is a decrease in disorder. The
entropy change is indeed to be negative.
Entropy is defined as the disorder or randomness.
+ ∆S is an increase in disorder.
- ∆S is a decrease in disorder.
A: Gas  solid is a decrease in disorder. The sign is incorrect.
B: Liquid  solid is a decrease in disorder. The sign is correct.
C: 5/2 of gas molecules  1 gas and 2 liquid molecules is a decrease in disorder.
The sign is correct.
Solution
D: 4 gas molecules  2 gas molecules is a decrease in disorder. The sign is
correct.
The correct answer is (A).
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Question No. 3 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
3. The Gibbs free energy of a system is defined as the enthalpy of the system
minus the product of the temperature times the entropy of the system, i.e. G = H TS. For 2N2O5(g) ⇄ 4NO2(g) + O2(g) at 25C. Use the data below and calculate the
∆G of this reaction.
Question
∆Hf
S
N2O5
11.29 kJ/mol
655.3 J/K*mol
NO2
33.15 kJ/mol
239.9 J/K*mol
O2
0 kJ/mol
204.8 J/K*mol
(A)
(B)
(C)
(D)
154 kJ/mol
43678 kJ/mol
84.6 kJ/mol
-154 kJ/mol
A. Correct!
Good job! First calculate the H and S using their formations of reactions, H =
Hf (products) - Hf (reactants) and S = S (products) - S (reactants).
Apply the free energy equation G = H - TS.
B. Incorrect.
First calculate the H and S using their formations of reactions, H = Hf
(products) - Hf (reactants) and S = S (products) - S (reactants). Apply
the free energy equation G = H - TS.
Feedback
C. Incorrect.
First calculate the H and S using their formations of reactions, H = Hf
(products) - Hf (reactants) and S = S (products) - S (reactants). Apply
the free energy equation G = H - TS.
D. Incorrect.
First calculate the H and S using their formations of reactions, H = Hf
(products) - Hf (reactants) and S = S (products) - S (reactants). Apply
the free energy equation G = H - TS.
2N2O5(g) ⇄ 4NO2(g) + O2(g) ….. T = 25°C+273 = 298 K, ∆G = ?
H   H f prod   H f react
S   S prod   S react
G  H  TS
H  4  33.15  1 0  2  11.29 = 110.02 kJ/mol
S  4  239.9  1 204.8  2  655.3 = -146.2 J/Kmol
Solution

G  110.2kJ/mol  298K  0.1462 kJ
K*mol

∆G = 154 kJ/mol …… Not spontaneous
The correct answer is (A).
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Question No. 4 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
4. Entropy is a quantitative measure of disorder in a system. The less disordered
phase in solid has lower entropy than more disordered liquid and gas phases. In a
reaction of a closed system, the direction of the increasing entropy goes to the
greater number of molecules (particles) involved in the process. All of the following
reactions illustrate an increase in entropy except _____.
Question
(A)
(B)
(C)
(D)
N2O4 (g)  2 NO2 (g)
C6H6 (l)  C6H6 (g)
2 KClO3 (s)  3 O2 (g) + 2 KCl (s)
3 Fe (s) + 2 O2 (g)  Fe3O4 (s)
A. Incorrect.
This is an increase in disorder. In this all gaseous reaction, the forward reaction
goes from 1 molecule to 2 molecules with the increase in entropy.
B. Incorrect.
This is an increase in disorder. The forward reaction goes from liquid phase to gas
phase, an increase in entropy.
C. Incorrect.
Feedback
This is an increase in disorder. The forward reaction starts in solid phase and ends
with a product in gas phase, with higher entropy.
D. Correct!
Good job! This is a decrease in disorder. The forward reaction goes from some gas
phase to all solid phase, a decrease in entropy.
Entropy is disorder or randomness. The direction toward more molecules or toward
gas phase will increase the entropy since the disorder will increase.
A. 1 gas molecule  2 gas molecules. This is an increase in disorder
B. 1 liquid molecule  1 gas molecule. This is an increase in disorder
C. 2 solid molecules  3 gas molecules and 2 solid molecules. This is an
increase in disorder.
Solution
D. 3 solid and 2 gas molecules  1 solid molecule. This is a decrease in
disorder.
The correct answer is (D).
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Question No. 5 of 10
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on paper as needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
5. For exothermic or endothermic reactions, the heat absorbed or released is
directly proportional to the moles of products produced according to reaction
stoichiometry. If 1000 kJ of energy is released, how many grams of water are
produced?
B2H6 (g) + 3 O2 (g)  B2O3 (s) + 2 H2O (g) … ∆H = -2035 kJ
Question
(A)
(B)
(C)
(D)
8.86 g
73.34 g
-73.34 g
17.71 g
A. Incorrect.
Keep in mind that 2 moles of H2O produced will release 2035 kJ of heat. Apply
stoichiometric ratio of the balanced reaction to calculate the moles of the water
generated from the total energy released. Convert to the grams of water produced.
B. Incorrect.
Keep in mind that 2 moles of H2O produced will release 2035 kJ of heat. Apply
stoichiometric ratio of the balanced reaction to calculate the moles of the water
generated from the total energy released. Convert to the grams of water produced.
C. Incorrect.
Feedback
Keep in mind that 2 moles of H2O produced will release 2035 kJ of heat. Apply
stoichiometric ratio of the balanced reaction to calculate the moles of the water
generated from the total energy released. Convert to the grams of water produced.
D. Correct!
Good job! Keep in mind that 2 moles of H2O produced will release 2035 kJ of heat.
Apply stoichiometric ratio of the balanced reaction to calculate the moles of the
water generated from the total energy released. Convert to the grams of water
produced.
Use the energy change of the reaction along with the mole ratio from the balanced
equation.
-2035 kJ = 2 mole H2O
(molar mass of H2O) 1 mole H2O = 18.02 g H2O
Set up the dimensional analysis:
-1000 kJ
2 mole H2O
-2035 kJ
18.02 g H2O
1 mole H2O
Solution
The correct answer is (D).
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= __17.71___ g
H2O
Question No. 6 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
6. Standard Enthalpy (Heat) of Reaction, H, is the difference between the sum of
standard enthalpies (heats) of formation of the products and that of the reactants.
For 2N2O5(g) ⇄ 4NO2(g) + O2(g), find heat of reaction if the following data is given.
N2O5
NO2
O2
Question
(A)
(B)
(C)
(D)
∆Hf
11.29 kJ/mole
33.15 kJ/mole
0 kJ/mole
-110.02 kJ
110.02 kJ
21.86 kJ
-21.86 kJ
A. Incorrect.
The heat of reaction is calculated as: Heat of reaction = [the sum of all heats of
formation of all products] - [the sum of all heats of formation of all reactants].
B. Correct.
Good job! The heat of reaction is calculated as: Heat of reaction = [the sum of all
heats of formation of all products] - [the sum of all heats of formation of all
reactants].
C. Incorrect.
Feedback
The heat of reaction is calculated as: Heat of reaction = [the sum of all heats of
formation of all products] - [the sum of all heats of formation of all reactants].
D. Incorrect.
The heat of reaction is calculated as: Heat of reaction = [the sum of all heats of
formation of all products] - [the sum of all heats of formation of all reactants].
To find heat of reaction from heat of formation, find the sum of the products and
subtract the sum of the reactants.
Standard Heat (Enthalpy) of Formation, Hf, of any compound is the enthalpy
change of the reaction by which it is formed from its elements, reactants and
products all being in a given standard state.
H rxn 
Solution
H 

f prod 
H rxn  4  33.15 kJ
mole
H 
f react
 1 0 kJ mole 2 11.29 kJ mole
Hrxn = 110.02 kJ/mole
The correct answer is (B).
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Question No. 7 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
7. Gibbs free energy G combines enthalpy H and entropy S into one. The change of
free energy ΔG is equal to the sum of its enthalpy plus the product of the
temperature and entropy of the system. ΔG can be used to predict the direction of
a chemical reaction. If ΔG is positive then the reaction is non-spontaneous. If it is
negative, the reaction is spontaneous. For the following, N2 + 3 H2  2 NH3, ∆H = 92 kJ/mol and ∆S = -199 J/Kmol. Which interval G will fall at 200 K?
Question
(A)
(B)
(C)
(D)
Less than -100 kJ/mol
Between -100 kJ/mol and -40 kJ/mol
Between -40 kJ/mol and +40 kJ/mol
Between +40 kJ/mol and +100 kJ/mol
A. Incorrect.
Find change in free energy by multiplying temperature by entropy and subtracting
that value from enthalpy.
B. Correct!
Good job! Apply the definition of Gibbs Free Energy and use the formula to
calculate the G.
C. Incorrect.
Feedback
Find change in free energy by multiplying temperature by entropy and subtracting
that value from enthalpy.
D. Incorrect.
Find change in free energy by multiplying temperature by entropy and subtracting
that value from enthalpy.
Gibbs Free Energy:
G  H  TS

G  92kJ/mol  200K  0.199 kJ
K mol

G =-92+40 = -52 kJ/mol
The correct answer is (B).
Solution
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Question No. 8 of 10
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on paper as needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
8. In a chemical reaction, solids can be heated to the point where the molecules
holding their bonds together break apart and form a liquid. This process is better
known as melting, or heat of fusion. A 200. g solid is allowed to melt in 400. g of
water. In the process, the water temperature decreases from 85.1°C to 30.0°C.
What is the heat of fusion for this solid in J/g?
Question
(A)
(B)
(C)
(D)
115 J/g
-115 J/g
460 J/g
-460 J/g
A. Incorrect.
Apply the equality using the law of conservation of energy and establish the
equation to solve the heat of fusion. The temperature remains constant during the
process of fusion (melting).
B. Incorrect.
Apply the equality using the law of conservation of energy and establish the
equation to solve the heat of fusion.
C. Correct!
Feedback
Good job! Apply the equality using the law of conservation of energy and establish
the equation to solve the heat of fusion. The temperature remains constant during
the process of fusion (melting).
D. Incorrect!
Apply the equality using the law of conservation of energy and establish the
equation to solve the heat of fusion.
msolid = 200.g = mass of the solid; mwater = 400. g = mass of the water
T1 (water) = 85.1°C; T2 (water) = 30.0°C; Cp (water) = 4.18 J/g°C
Hfus solid = ? J/g; ∆Hwater = mCp∆T; ∆Hsolid = mHfus
Apply the law of conservation of energy: ∆Hwater = -∆Hsolid
(mCp∆T)water = (mHfus)solid
400.g  4.18 J
400.g  4.18 J
Solution
g C


 30.0  C  85.1 C  200.g  H fus


 30.0  C  85.1 C
g C
 H fus
 200.g
Hfus = 460. J/g
Please take the time to truly understand this problem and its solution. It uses
several important concepts!
The correct answer is (C).
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Question No. 9 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
9. To melt or vaporize a substance, energy must be added to separate the sample
into its component pieces which may involve breaking intermolecular forces. For
the following diagram, which answer below best explains the difference in length
between b & c and between d & e?
Question
(A) The strength of individual intermolecular forces is stronger between b & c
than between d & e.
(B) The strength of individual intermolecular forces is weaker between b & c
than between d & e.
(C) There are fewer intermolecular forces broken between b & c than between d
& e.
(D) There are more intermolecular forces broken between b & c than between d
& e.
A. Incorrect.
The strength of the individual intermolecular forces is not a function of what state
the matter is in, but the type of molecule. The strength of individual intermolecular
forces doesn’t change.
B. Incorrect.
The strength of the individual intermolecular forces is not a function of what state
the matter is in, but the type of molecule. The strength of individual intermolecular
forces doesn’t change.
C. Correct!
Feedback
Good job! There are fewer intermolecular forces broken between b & c (melting)
than between d & e (boiling) and therefore it takes less added energy at a lower
temperature.
D. Incorrect.
There are fewer intermolecular forces broken from solid to liquid than from liquid to
gas.
Solution
During phase changes, the energy being added to the system is used to break
intermolecular forces.
Since it’s the same molecule during both intervals, the strength of intermolecular
forces will be the same.
Since the d-e range requires more energy to be added in order to complete the
phase change but the strength of the intermolecular forces is the same during both
changes, there must be more intermolecular forces broken between d & e.
The correct answer is (C).
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Question No. 10 of 10
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needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed.
10. The combustion of H2 gas yields water and heat. In an isolated system, 2.00
moles of H2 and 1.00 moles of O2 at 25.0C react to produce H2O. The heat of
formation (Enthalpy of Formation Hf) of this reaction is -57.8 kcal/mol. The release
of heat from the reaction will raise the water temperature. Determine the final
temperature. The specific heat capacity of water is 1.00 cal/g°C (i.e. 4.18 J/g°C).
Question
(A)
(B)
(C)
(D)
3240°C
1032°C
25°C
1000°C
A. Correct!
Good job! The heat released from the reaction will all be used to warm up the water
(100%). Set the equality of Hf(reaction) = -Hwater. Convert the units and calculate
the final temperature.
B. Incorrect.
The heat released from the reaction will all be used to warm up the water (100%).
Set the equality of Hf(reaction) = -Hwater. Convert the units and calculate the final
temperature.
C. Incorrect.
Feedback
The heat released from the reaction will all be used to warm up the water (100%).
Set the equality of Hf(reaction) = -Hwater. Convert the units and calculate the final
temperature.
D. Incorrect.
The heat released from the reaction will all be used to warm up the water (100%).
Set the equality of Hf(reaction) = -Hwater. Convert the units and calculate the final
temperature.
The combustion of H2 with O2 yields H2O and heat:
(1/2)H2 + O2  H2O + Heat
Since the heat of formation Hf = -57.8 kcal/mol, this means that for every mole of
the product generated, there will be 57.8 kcal of heat released to the system.
Since there are total of 2.00 moles of H2 and 1.00 moles of O2, if react completely,
there will be 2.00 moles of H2O produced. The total heat released = 2.00 moles x
57.8 kcal/moles = 115.6 kcal. This amount of heat will warm up the water from
initial 25C to final temperature (Tf).
Solution
Conservation of Heat: Hf(reaction) = -Hwater
Hwater= m x Cp x (Tf – Ti)
-115.6 kcal x 1000 cal/kcal = - (2.00 moles x 18.0 g/mol) x (1.00 Cal/g°C) x (Tf –
25.0)
Tf = 3240°C (3 significant figures).
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The Correct Answer is (A).
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