AP Chemistry Practice Quiz: Introduction to Entropy
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 9 questions to check your progress.
Question 1 of 9
All Questions (9)
A) H₂O(l) → H₂O(s) (Freezing)
B) CO₂(g) → CO₂(s) (Deposition)
C) I₂(s) → I₂(g) (Sublimation)
D) N₂(g) → N₂(l) (Condensation)
Correct Answer: C
Sublimation is the transition from a solid to a gas. According to the provided content, entropy increases when matter becomes more dispersed. The gas phase is significantly more dispersed than the solid phase, resulting in a positive entropy change. The other options represent processes where matter becomes less dispersed (more ordered), leading to a negative entropy change.
A) Entropy decreases because the gas particles move faster and collide more frequently.
B) Entropy increases because the distribution of kinetic energies among the gas particles broadens.
C) Entropy remains constant because the number of moles and the volume do not change.
D) Entropy decreases because the increase in temperature leads to a more ordered state of motion.
Correct Answer: B
The provided content states that as temperature increases, the distribution of kinetic energies broadens (per Kinetic Molecular Theory), and the entropy of the system increases. This is because the thermal energy becomes more dispersed among the available motional states of the particles.
A) 2 H₂(g) + O₂(g) → 2 H₂O(g)
B) N₂(g) + 3 H₂(g) → 2 NH₃(g)
C) CaCO₃(s) → CaO(s) + CO₂(g)
D) Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
Correct Answer: C
Entropy increases when the total moles of gas increase in a reaction. In option C, a solid reactant produces a solid and a gaseous product. The formation of a gas from a solid represents a large increase in the dispersal of matter, leading to a positive entropy change. In options A and B, the moles of gas decrease. In option D, a solid is formed from aqueous ions, which is a decrease in dispersal.
A) Positive, because the gas particles are more dispersed in a larger volume.
B) Negative, because the pressure of the gas decreases.
C) Zero, because the temperature remains constant.
D) Negative, because the gas particles are farther apart.
Correct Answer: A
The provided content states that entropy increases when the volume of a gas increases. This is because the gas particles have more available positions to occupy, leading to a greater dispersal of matter. Therefore, the entropy change (ΔS) is positive.
A) Melting 1 mole of ice at 0°C: H₂O(s) → H₂O(l)
B) Raising the temperature of 1 mole of liquid water from 25°C to 50°C.
C) Boiling 1 mole of water at 100°C: H₂O(l) → H₂O(g)
D) Increasing the pressure on 1 mole of O₂(g) from 1 atm to 2 atm at 25°C.
Correct Answer: C
The magnitude of the entropy change is related to the degree of increased dispersal. The transition from a liquid to a gas (boiling) involves the largest increase in the dispersal of matter, as particles go from being relatively close together to being far apart and moving randomly in a much larger volume. This change is significantly larger in magnitude than the entropy change for melting or a small temperature increase. Increasing pressure (option D) would decrease volume and thus decrease entropy.
A) Positive, because the total moles of gas increase from 6 to 7.
B) Negative, because the reactants are more complex molecules than the products.
C) Approximately zero, because all major substances are in the gas phase.
D) Negative, because the total moles of gas decrease.
Correct Answer: A
The content states that entropy increases when the total moles of gas increase in a reaction. In this reaction, there are 6 moles of gaseous reactants (1 mole of C₃H₈ and 5 moles of O₂) and 7 moles of gaseous products (3 moles of CO₂ and 4 moles of H₂O). Since the number of moles of gas increases, the matter becomes more dispersed, and the entropy change (ΔS) is expected to be positive.
A) A < C < B
B) C < A < B
C) B < A < C
D) A = C < B
Correct Answer: D
Entropy is an extensive property, meaning it depends on the amount of substance. At the same temperature and volume, more particles lead to more possible arrangements and thus higher entropy. Therefore, Flask B (2 moles) has higher entropy than Flask A (1 mole). Comparing Flask A and Flask C, both have 1 mole of a noble gas at the same temperature and volume. While Ne is more massive than He, the primary driver of entropy in this comparison is the number of particles. For ideal gases under these conditions, the entropy is primarily dependent on the number of moles, not the identity of the gas. Thus, the entropy of A and C are approximately equal. The correct order is A = C < B.
A) The molecules in gaseous nitrogen have a broader distribution of kinetic energies than in liquid nitrogen.
B) The molecules in gaseous nitrogen are much more dispersed and occupy a significantly larger volume than in the liquid state.
C) The conversion from liquid to gas is an endothermic process, and the absorption of energy itself is the entropy.
D) Gaseous nitrogen has weaker intermolecular forces, which allows for more stored potential energy.
Correct Answer: B
The primary reason for the large entropy increase during vaporization is the massive increase in the dispersal of matter. In the liquid state, molecules are close together. In the gaseous state, the same molecules occupy a volume that is orders of magnitude larger, leading to a much more disordered and dispersed arrangement. This corresponds to a much higher entropy.
A) ΔS is positive.
B) ΔS is negative.
C) ΔS is zero.
D) The sign of ΔS cannot be determined without knowing the temperature change.
Correct Answer: B
Entropy is a measure of dispersal or disorder. The problem states that the system becomes more ordered and that the moles of gas, which have high entropy, decrease. Both of these factors point to a decrease in the dispersal of matter and energy. Therefore, the entropy change for the system (ΔS) must be negative.