CHEM134 C003 Lesson 6 Quiz |
Questions and Answers – spring
2026 | 100% Correct - American
Public University.
SECTION 1: THERMODYNAMICS
1. The enthalpy of vaporization of isopropanol is 44.00 kJ/mol at its boiling point
(82.3°C). Calculate the value of ΔS when 1.00 mole of isopropanol is vaporized at
82.3°C and 1.00 atm.
A) -1.24 × 10² J/K·mol
B) 1.24 × 10² J/K·mol
C) -5.35 × 10² J/K·mol
, D) 0 J/K·mol
E) 5.35 × 10² J/K·mol
Correct Answer: B
Rationale: At the boiling point, ΔG = 0 for vaporization, so ΔS = ΔH/T. T = 82.3 + 273.15
= 355.45 K. ΔH = 44,000 J/mol. ΔS = 44,000 J/mol / 355.45 K = 1.24 × 10² J/K·mol. The
positive sign indicates increasing entropy during vaporization .
2. Arrange the following substances in the order of increasing entropy at 25°C:
HF(g), NaF(s), SiF₄(g), SiH₄(g), Al(s)
A) Al(s) < HF(g) < NaF(s) < SiF₄(g) < SiH₄(g)
B) HF(g) < Al(s) < NaF(s) < SiF₄(g) < SiH₄(g)
C) Al(s) < NaF(s) < HF(g) < SiH₄(g) < SiF₄(g)
D) NaF(s) < Al(s) < HF(g) < SiF₄(g) < SiH₄(g)
E) SiF₄(g) < SiH₄(g) < NaF(s) < HF(g) < Al(s)
Correct Answer: C
Rationale: Entropy generally increases with phase: solids < liquids < gases. Al(s) and
NaF(s) are solids (lowest entropy). Among gases, SiF₄(g) has higher entropy than SiH₄(g)
due to greater molecular complexity, and HF(g) has intermediate values .
, 3. Sodium reacts violently with water according to the equation: 2Na(s) + 2H₂O(l)
→ 2NaOH(aq) + H₂(g). The resulting solution has a higher temperature than the
water prior to the addition of sodium. What are the signs of ΔH° and ΔS° for this
reaction?
A) ΔH° is negative and ΔS° is negative.
B) ΔH° is negative and ΔS° is positive.
C) ΔH° is positive and ΔS° is negative.
D) ΔH° is positive and ΔS° is positive.
Correct Answer: B
Rationale: The reaction is exothermic (temperature increases), so ΔH° < 0. The reaction
produces gas (H₂) and aqueous ions, increasing disorder, so ΔS° > 0 .
4. A two-bulbed flask contains 5 particles. What is the probability of finding all 5
particles on the left side?
A) 2.24%
B) 2.50%
C) 6.25%
D) 0.20%
E) 3.13%
Correct Answer: E
, Rationale: Each particle has two possible locations (left or right). Total possible
arrangements = 2⁵ = 32. Only one arrangement has all particles on the left. Probability =
1/32 = 0.03125 = 3.125% .
5. Which of the following shows a decrease in entropy?
A) Melting ice
B) A burning piece of wood
C) Two of these
D) Precipitation
E) Gaseous reactants forming a liquid
Correct Answer: D
Rationale: Precipitation involves ions in solution forming a solid, which significantly
decreases disorder (entropy). Melting increases entropy; combustion increases entropy; gas
to liquid decreases entropy but precipitation is the most clear example .
6. Which electron on an atom of copper would have the highest value of W in the
Boltzmann formula?
A) 4p
B) 3d
C) 3s
D) 4s
Questions and Answers – spring
2026 | 100% Correct - American
Public University.
SECTION 1: THERMODYNAMICS
1. The enthalpy of vaporization of isopropanol is 44.00 kJ/mol at its boiling point
(82.3°C). Calculate the value of ΔS when 1.00 mole of isopropanol is vaporized at
82.3°C and 1.00 atm.
A) -1.24 × 10² J/K·mol
B) 1.24 × 10² J/K·mol
C) -5.35 × 10² J/K·mol
, D) 0 J/K·mol
E) 5.35 × 10² J/K·mol
Correct Answer: B
Rationale: At the boiling point, ΔG = 0 for vaporization, so ΔS = ΔH/T. T = 82.3 + 273.15
= 355.45 K. ΔH = 44,000 J/mol. ΔS = 44,000 J/mol / 355.45 K = 1.24 × 10² J/K·mol. The
positive sign indicates increasing entropy during vaporization .
2. Arrange the following substances in the order of increasing entropy at 25°C:
HF(g), NaF(s), SiF₄(g), SiH₄(g), Al(s)
A) Al(s) < HF(g) < NaF(s) < SiF₄(g) < SiH₄(g)
B) HF(g) < Al(s) < NaF(s) < SiF₄(g) < SiH₄(g)
C) Al(s) < NaF(s) < HF(g) < SiH₄(g) < SiF₄(g)
D) NaF(s) < Al(s) < HF(g) < SiF₄(g) < SiH₄(g)
E) SiF₄(g) < SiH₄(g) < NaF(s) < HF(g) < Al(s)
Correct Answer: C
Rationale: Entropy generally increases with phase: solids < liquids < gases. Al(s) and
NaF(s) are solids (lowest entropy). Among gases, SiF₄(g) has higher entropy than SiH₄(g)
due to greater molecular complexity, and HF(g) has intermediate values .
, 3. Sodium reacts violently with water according to the equation: 2Na(s) + 2H₂O(l)
→ 2NaOH(aq) + H₂(g). The resulting solution has a higher temperature than the
water prior to the addition of sodium. What are the signs of ΔH° and ΔS° for this
reaction?
A) ΔH° is negative and ΔS° is negative.
B) ΔH° is negative and ΔS° is positive.
C) ΔH° is positive and ΔS° is negative.
D) ΔH° is positive and ΔS° is positive.
Correct Answer: B
Rationale: The reaction is exothermic (temperature increases), so ΔH° < 0. The reaction
produces gas (H₂) and aqueous ions, increasing disorder, so ΔS° > 0 .
4. A two-bulbed flask contains 5 particles. What is the probability of finding all 5
particles on the left side?
A) 2.24%
B) 2.50%
C) 6.25%
D) 0.20%
E) 3.13%
Correct Answer: E
, Rationale: Each particle has two possible locations (left or right). Total possible
arrangements = 2⁵ = 32. Only one arrangement has all particles on the left. Probability =
1/32 = 0.03125 = 3.125% .
5. Which of the following shows a decrease in entropy?
A) Melting ice
B) A burning piece of wood
C) Two of these
D) Precipitation
E) Gaseous reactants forming a liquid
Correct Answer: D
Rationale: Precipitation involves ions in solution forming a solid, which significantly
decreases disorder (entropy). Melting increases entropy; combustion increases entropy; gas
to liquid decreases entropy but precipitation is the most clear example .
6. Which electron on an atom of copper would have the highest value of W in the
Boltzmann formula?
A) 4p
B) 3d
C) 3s
D) 4s
