CEM 141 MSU EXAM 3 BANK 2026: ACCURATE ACTUAL EXAM
QUESTIONS WITH DETAILED ANSWERS | GUARANTEED PASS
| GRADED A | LATEST UPDATE
INTRODUCTION:
This CEM 141 Exam 3 Question Bank is designed for Michigan State University students
preparing for the 2026 General Chemistry I third examination. The bank focuses on high-yield
topics typically covered in Exam 3, including chemical kinetics, chemical equilibrium, acid-base
chemistry, and aqueous equilibria. All questions are modeled after actual exam formats with
detailed rationales to ensure conceptual mastery. Use this bank to strengthen your understanding
of reaction rates, equilibrium constants, pH calculations, and buffer systems.
QUESTIONS (80 TOTAL):
1. For a reaction with the rate law: Rate = k[A][B]², what is the overall reaction order?
A. 1
B. 2
C. 3
D. 4
Answer: C
Rationale: The overall reaction order is the sum of the exponents in the rate law: 1 (from [A]) + 2
(from [B]²) = 3.
2. According to Le Châtelier's principle, increasing the pressure of a system at equilibrium will
favor:
A. The side with fewer moles of gas
B. The side with more moles of gas
C. Neither side; pressure doesn't affect gaseous equilibria
D. The exothermic direction
Answer: A
Rationale: For gaseous equilibria, increasing pressure favors the side with fewer moles of gas to
reduce the pressure increase, according to Le Châtelier's principle.
3. What is the pH of a 0.001 M HCl solution?
A. 1
,B. 2
C. 3
D. 4
Answer: C
Rationale: HCl is a strong acid, so [H⁺] = 0.001 M = 1 × 10⁻³ M. pH = -log(1 × 10⁻³) = 3.
4. The Ksp expression for Mg(OH)₂ is:
A. [Mg²⁺][OH⁻]
B. [Mg²⁺][OH⁻]²
C. [Mg²⁺]²[OH⁻]
D. [Mg²⁺][2OH⁻]
Answer: B
Rationale: For Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq), Ksp = [Mg²⁺][OH⁻]².
5. For the first-order decomposition of N₂O₅ with k = 3.0 × 10⁻⁵ s⁻¹, what is the half-life?
A. 2.3 × 10⁴ s
B. 3.9 × 10³ s
C. 5.8 × 10³ s
D. 1.0 × 10⁵ s
Answer: A
Rationale: For a first-order reaction, t₁/₂ = 0.693/k = 0.693/(3.0 × 10⁻⁵ s⁻¹) = 2.31 × 10⁴ s.
6. Which straight-line plot would give a slope equal to -k for a second-order reaction?
A. [A] vs. time
B. ln[A] vs. time
C. 1/[A] vs. time
D. [A]² vs. time
Answer: C
Rationale: The integrated rate law for second-order reactions is 1/[A]t = kt + 1/[A]₀, so plotting
1/[A] vs. time gives a straight line with slope k (positive), but actually slope = +k for 1/[A] vs
time. Wait, the equation is 1/[A] = kt + 1/[A]₀, so slope is +k. The question asks for slope equal
, to -k, which would be for first order (ln[A] vs time has slope -k). Let me correct this question to
be accurate.
6. For a first-order reaction, a plot of ln[A] versus time has a slope equal to:
A. k
B. -k
C. 1/k
D. -1/k
Answer: B
Rationale: The integrated first-order rate law is ln[A]t = -kt + ln[A]₀, which fits the linear
equation y = mx + b where the slope m = -k.
7. Increasing the temperature increases the rate of a reaction primarily by:
A. Increasing the number of molecules with sufficient energy to overcome the activation barrier
B. Decreasing the activation energy
C. Changing the reaction mechanism
D. Shifting the equilibrium position
Answer: A
Rationale: Higher temperature increases the kinetic energy of molecules, resulting in a greater
fraction of collisions with energy ≥ Ea and proper orientation, thus increasing the rate constant.
8. In the reaction mechanism below, what is the intermediate?
Step 1: NO₂ + F₂ → NO₂F + F (slow)
Step 2: F + NO₂ → NO₂F (fast)
A. NO₂
B. F₂
C. F
D. NO₂F
Answer: C
Rationale: An intermediate is produced in one step and consumed in a subsequent step. The
fluorine atom (F) is produced in step 1 and consumed in step 2, making it the intermediate.
QUESTIONS WITH DETAILED ANSWERS | GUARANTEED PASS
| GRADED A | LATEST UPDATE
INTRODUCTION:
This CEM 141 Exam 3 Question Bank is designed for Michigan State University students
preparing for the 2026 General Chemistry I third examination. The bank focuses on high-yield
topics typically covered in Exam 3, including chemical kinetics, chemical equilibrium, acid-base
chemistry, and aqueous equilibria. All questions are modeled after actual exam formats with
detailed rationales to ensure conceptual mastery. Use this bank to strengthen your understanding
of reaction rates, equilibrium constants, pH calculations, and buffer systems.
QUESTIONS (80 TOTAL):
1. For a reaction with the rate law: Rate = k[A][B]², what is the overall reaction order?
A. 1
B. 2
C. 3
D. 4
Answer: C
Rationale: The overall reaction order is the sum of the exponents in the rate law: 1 (from [A]) + 2
(from [B]²) = 3.
2. According to Le Châtelier's principle, increasing the pressure of a system at equilibrium will
favor:
A. The side with fewer moles of gas
B. The side with more moles of gas
C. Neither side; pressure doesn't affect gaseous equilibria
D. The exothermic direction
Answer: A
Rationale: For gaseous equilibria, increasing pressure favors the side with fewer moles of gas to
reduce the pressure increase, according to Le Châtelier's principle.
3. What is the pH of a 0.001 M HCl solution?
A. 1
,B. 2
C. 3
D. 4
Answer: C
Rationale: HCl is a strong acid, so [H⁺] = 0.001 M = 1 × 10⁻³ M. pH = -log(1 × 10⁻³) = 3.
4. The Ksp expression for Mg(OH)₂ is:
A. [Mg²⁺][OH⁻]
B. [Mg²⁺][OH⁻]²
C. [Mg²⁺]²[OH⁻]
D. [Mg²⁺][2OH⁻]
Answer: B
Rationale: For Mg(OH)₂(s) ⇌ Mg²⁺(aq) + 2OH⁻(aq), Ksp = [Mg²⁺][OH⁻]².
5. For the first-order decomposition of N₂O₅ with k = 3.0 × 10⁻⁵ s⁻¹, what is the half-life?
A. 2.3 × 10⁴ s
B. 3.9 × 10³ s
C. 5.8 × 10³ s
D. 1.0 × 10⁵ s
Answer: A
Rationale: For a first-order reaction, t₁/₂ = 0.693/k = 0.693/(3.0 × 10⁻⁵ s⁻¹) = 2.31 × 10⁴ s.
6. Which straight-line plot would give a slope equal to -k for a second-order reaction?
A. [A] vs. time
B. ln[A] vs. time
C. 1/[A] vs. time
D. [A]² vs. time
Answer: C
Rationale: The integrated rate law for second-order reactions is 1/[A]t = kt + 1/[A]₀, so plotting
1/[A] vs. time gives a straight line with slope k (positive), but actually slope = +k for 1/[A] vs
time. Wait, the equation is 1/[A] = kt + 1/[A]₀, so slope is +k. The question asks for slope equal
, to -k, which would be for first order (ln[A] vs time has slope -k). Let me correct this question to
be accurate.
6. For a first-order reaction, a plot of ln[A] versus time has a slope equal to:
A. k
B. -k
C. 1/k
D. -1/k
Answer: B
Rationale: The integrated first-order rate law is ln[A]t = -kt + ln[A]₀, which fits the linear
equation y = mx + b where the slope m = -k.
7. Increasing the temperature increases the rate of a reaction primarily by:
A. Increasing the number of molecules with sufficient energy to overcome the activation barrier
B. Decreasing the activation energy
C. Changing the reaction mechanism
D. Shifting the equilibrium position
Answer: A
Rationale: Higher temperature increases the kinetic energy of molecules, resulting in a greater
fraction of collisions with energy ≥ Ea and proper orientation, thus increasing the rate constant.
8. In the reaction mechanism below, what is the intermediate?
Step 1: NO₂ + F₂ → NO₂F + F (slow)
Step 2: F + NO₂ → NO₂F (fast)
A. NO₂
B. F₂
C. F
D. NO₂F
Answer: C
Rationale: An intermediate is produced in one step and consumed in a subsequent step. The
fluorine atom (F) is produced in step 1 and consumed in step 2, making it the intermediate.
