1
UNE General Chemistry II Midterm Exam
Official Practice Exam Actual Exam
2026/2027 with Detailed Rationales |
Complete Exam-Style Questions | Pass
Guaranteed – A+ Graded
TABLE OF CONTENTS
Section 1 | Kinetics & Chemical Equilibrium | Q1 – Q10
Section 2 | Acid-Base Chemistry | Q11 – Q20
Section 3 | Solubility & Thermodynamics | Q21 – Q30
Section 4 | Electrochemistry | Q31 – Q40
Section 5 | NGN-Style Integrated Case Analysis | Q41 – Q50
Instructions: Choose the single best answer. Pass: 37 in 90 minutes.
══════════════════════════════════════
SECTION 1: KINETICS & CHEMICAL EQUILIBRIUM Q1 – Q10
══════════════════════════════════════
Question 1 of 50
A graduate student is studying the gas-phase decomposition of dinitrogen
pentoxide in a sealed reaction vessel at 298 K. In three separate experiments, the
initial rate of disappearance of N₂O₅ is measured at different initial concentrations:
Experiment | [N₂O₅]₀ (M) | Initial Rate (M/s)
1 | 0.010 | 2.4 × 10⁻⁵
2 | 0.020 | 4.8 × 10⁻⁵
3 | 0.030 | 7.2 × 10⁻⁵
What is the rate law for this reaction?
,2
A. Rate = k[N₂O₅]²
B. Rate = k[N₂O₅] ✓ CORRECT
C. Rate = k[N₂O₅]⁰
D. Rate = k[N₂O₅]¹·⁵
Correct Answer: B
Rationale: When the concentration of N₂O₅ doubles from experiment 1 to 2, the
initial rate also doubles, indicating a first-order dependence on [N₂O₅]. Choice A is
incorrect because a second-order rate law would predict the rate quadruples when
the concentration doubles, which is not observed in the data.
Question 2 of 50
A pharmaceutical chemist monitors the degradation of an antibiotic at 25°C and
finds that its concentration drops from 0.80 M to 0.20 M in 120 minutes. The
degradation follows first-order kinetics. What is the half-life of this antibiotic?
A. 30 minutes
B. 45 minutes
C. 90 minutes
D. 60 minutes ✓ CORRECT
Correct Answer: D
Rationale: For a first-order reaction, the integrated rate law shows that the
concentration drops by a factor of four in two half-lives, so each half-life must be
60 minutes. Choice C incorrectly assumes zero-order kinetics where the half-life
would depend on initial concentration.
Question 3 of 50
An industrial chemist needs to determine the activation energy for the gas-phase
reaction between NO and Cl₂. The rate constant is 4.5 × 10⁻⁵ M⁻¹s⁻¹ at 300 K and
3.6 × 10⁻⁴ M⁻¹s⁻¹ at 320 K. Using the Arrhenius equation, what is the approximate
activation energy?
A. 83 kJ/mol ✓ CORRECT
B. 52 kJ/mol
, 3
C. 35 kJ/mol
D. 120 kJ/mol
Correct Answer: A
Rationale: The two-point Arrhenius equation ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂) yields
approximately 83 kJ/mol when the given rate constants and temperatures are
substituted. Choice B results from forgetting to take the natural logarithm of the
rate constant ratio and instead using the ratio directly.
Question 4 of 50
In a materials science lab, a sealed container holds the equilibrium system CO(g) +
Cl₂(g) ⇌ COCl₂(g) at 500 K. If additional chlorine gas is injected into the container
at constant temperature, what happens to the system?
A. The value of Kc increases and more phosgene forms
B. The equilibrium shifts left, consuming COCl₂
C. The equilibrium shifts right, producing more COCl₂ ✓ CORRECT
D. The partial pressure of CO increases while Kc decreases
Correct Answer: C
Rationale: Le Chatelier's principle states that adding a reactant (Cl₂) to a system at
equilibrium causes the reaction to shift in the direction that consumes the added
species, thereby forming more product. Choice A is incorrect because Kc depends
only on temperature and remains unchanged when concentration is altered at
constant temperature.
Question 5 of 50
A research team is studying the endothermic decomposition of calcium carbonate:
CaCO₃(s) ⇌ CaO(s) + CO₂(g) ΔH° = +178 kJ/mol. Which change will increase the
amount of CaO present at equilibrium?
A. Decreasing the volume of the container at constant temperature
B. Increasing the temperature of the system ✓ CORRECT
C. Adding more solid CaCO₃ to the reaction vessel
D. Introducing an inert gas at constant volume
UNE General Chemistry II Midterm Exam
Official Practice Exam Actual Exam
2026/2027 with Detailed Rationales |
Complete Exam-Style Questions | Pass
Guaranteed – A+ Graded
TABLE OF CONTENTS
Section 1 | Kinetics & Chemical Equilibrium | Q1 – Q10
Section 2 | Acid-Base Chemistry | Q11 – Q20
Section 3 | Solubility & Thermodynamics | Q21 – Q30
Section 4 | Electrochemistry | Q31 – Q40
Section 5 | NGN-Style Integrated Case Analysis | Q41 – Q50
Instructions: Choose the single best answer. Pass: 37 in 90 minutes.
══════════════════════════════════════
SECTION 1: KINETICS & CHEMICAL EQUILIBRIUM Q1 – Q10
══════════════════════════════════════
Question 1 of 50
A graduate student is studying the gas-phase decomposition of dinitrogen
pentoxide in a sealed reaction vessel at 298 K. In three separate experiments, the
initial rate of disappearance of N₂O₅ is measured at different initial concentrations:
Experiment | [N₂O₅]₀ (M) | Initial Rate (M/s)
1 | 0.010 | 2.4 × 10⁻⁵
2 | 0.020 | 4.8 × 10⁻⁵
3 | 0.030 | 7.2 × 10⁻⁵
What is the rate law for this reaction?
,2
A. Rate = k[N₂O₅]²
B. Rate = k[N₂O₅] ✓ CORRECT
C. Rate = k[N₂O₅]⁰
D. Rate = k[N₂O₅]¹·⁵
Correct Answer: B
Rationale: When the concentration of N₂O₅ doubles from experiment 1 to 2, the
initial rate also doubles, indicating a first-order dependence on [N₂O₅]. Choice A is
incorrect because a second-order rate law would predict the rate quadruples when
the concentration doubles, which is not observed in the data.
Question 2 of 50
A pharmaceutical chemist monitors the degradation of an antibiotic at 25°C and
finds that its concentration drops from 0.80 M to 0.20 M in 120 minutes. The
degradation follows first-order kinetics. What is the half-life of this antibiotic?
A. 30 minutes
B. 45 minutes
C. 90 minutes
D. 60 minutes ✓ CORRECT
Correct Answer: D
Rationale: For a first-order reaction, the integrated rate law shows that the
concentration drops by a factor of four in two half-lives, so each half-life must be
60 minutes. Choice C incorrectly assumes zero-order kinetics where the half-life
would depend on initial concentration.
Question 3 of 50
An industrial chemist needs to determine the activation energy for the gas-phase
reaction between NO and Cl₂. The rate constant is 4.5 × 10⁻⁵ M⁻¹s⁻¹ at 300 K and
3.6 × 10⁻⁴ M⁻¹s⁻¹ at 320 K. Using the Arrhenius equation, what is the approximate
activation energy?
A. 83 kJ/mol ✓ CORRECT
B. 52 kJ/mol
, 3
C. 35 kJ/mol
D. 120 kJ/mol
Correct Answer: A
Rationale: The two-point Arrhenius equation ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂) yields
approximately 83 kJ/mol when the given rate constants and temperatures are
substituted. Choice B results from forgetting to take the natural logarithm of the
rate constant ratio and instead using the ratio directly.
Question 4 of 50
In a materials science lab, a sealed container holds the equilibrium system CO(g) +
Cl₂(g) ⇌ COCl₂(g) at 500 K. If additional chlorine gas is injected into the container
at constant temperature, what happens to the system?
A. The value of Kc increases and more phosgene forms
B. The equilibrium shifts left, consuming COCl₂
C. The equilibrium shifts right, producing more COCl₂ ✓ CORRECT
D. The partial pressure of CO increases while Kc decreases
Correct Answer: C
Rationale: Le Chatelier's principle states that adding a reactant (Cl₂) to a system at
equilibrium causes the reaction to shift in the direction that consumes the added
species, thereby forming more product. Choice A is incorrect because Kc depends
only on temperature and remains unchanged when concentration is altered at
constant temperature.
Question 5 of 50
A research team is studying the endothermic decomposition of calcium carbonate:
CaCO₃(s) ⇌ CaO(s) + CO₂(g) ΔH° = +178 kJ/mol. Which change will increase the
amount of CaO present at equilibrium?
A. Decreasing the volume of the container at constant temperature
B. Increasing the temperature of the system ✓ CORRECT
C. Adding more solid CaCO₃ to the reaction vessel
D. Introducing an inert gas at constant volume
