CHEM 103 General Chemistry I – Module 7 Exam 2026 |
Thermochemistry & Energy Changes in Chemical Reactions |
Comprehensive Practice Examination | High-Yield Questions, Answers
& Detailed Rationales | Enthalpy, Calorimetry, Specific Heat, Hess’s
Law & Energy Diagrams | Complete Exam Prep Guide
Question 1
Which statement correctly describes the relationship between a system and its surroundings in
thermodynamics?
A. The system and surroundings together constitute the universe, and energy can be transferred
between them
B. The system is always isolated from its surroundings in thermodynamic studies
C. The surroundings are defined as everything outside the system, but energy cannot cross this
boundary
D. The system and surroundings must be at the same temperature for thermodynamic analysis
Correct Answer: A
Explanation: The system and surroundings together comprise the thermodynamic universe. Energy
can be transferred across the boundary between them through heat or work, while matter transfer
depends on the system type (open, closed, or isolated).
Question 2
A chemical reaction absorbs 85 kJ of heat from its surroundings while performing 45 kJ of work on the
surroundings. What is the change in internal energy (ΔE) of the system?
,A. +130 kJ
B. -130 kJ
C. +40 kJ
D. -40 kJ
Correct Answer: C
Explanation: Using the first law: ΔE = q + w. Heat absorbed is positive (+85 kJ), work done by the
system is negative (-45 kJ). Therefore, ΔE = 85 + (-45) = +40 kJ. The system's internal energy increased
by 40 kJ.
Question 3
Which of the following is an example of an intensive property?
A. Enthalpy
B. Internal energy
C. Temperature
D. Heat capacity
Correct Answer: C
Explanation: Intensive properties are independent of the amount of substance present. Temperature
is intensive because it remains constant regardless of sample size, whereas enthalpy, internal energy,
and heat capacity are extensive properties that depend on the quantity of matter.
Question 4
,Calculate the work done when a gas expands from 2.0 L to 8.0 L against a constant external pressure
of 1.5 atm. Express your answer in joules (1 L·atm = 101.3 J).
A. +911.7 J
B. -911.7 J
C. +607.8 J
D. -607.8 J
Correct Answer: B
Explanation: Work = -P_ext × ΔV = -1.5 atm × (8.0 - 2.0) L = -9.0 L·atm. Converting: -9.0 L·atm × 101.3
J/L·atm = -911.7 J. The negative sign indicates work done by the system during expansion.
Question 5
A system undergoes a process where ΔE = -250 J and the system absorbs 100 J of heat. Calculate the
work done by or on the system.
A. 350 J of work done by the system
B. 350 J of work done on the system
C. 150 J of work done by the system
D. 150 J of work done on the system
Correct Answer: A
Explanation: From ΔE = q + w, we have -250 = 100 + w, so w = -350 J. The negative sign means work is
done BY the system (350 J). The system lost internal energy through both heat absorption and work
output.
Question 6
, Which thermodynamic quantity is defined as H = E + PV?
A. Gibbs free energy
B. Helmholtz free energy
C. Enthalpy
D. Entropy
Correct Answer: C
Explanation: Enthalpy (H) is defined as the sum of internal energy (E) and the product of pressure and
volume (PV). This state function is particularly useful for constant-pressure processes, which are
common in open laboratory conditions.
Question 7
For a chemical reaction carried out in a bomb calorimeter at constant volume, which of the following
statements is true?
A. ΔH = q_v (heat at constant volume)
B. ΔE = q_v (heat at constant volume)
C. ΔH = ΔE + RTΔn
D. Both B and C are correct
Correct Answer: D
Explanation: In a bomb calorimeter at constant volume, no work is done (w = 0), so ΔE = q_v.
Additionally, the relationship between enthalpy and internal energy is ΔH = ΔE + RTΔn (for ideal
gases), making both statements B and C correct.
Thermochemistry & Energy Changes in Chemical Reactions |
Comprehensive Practice Examination | High-Yield Questions, Answers
& Detailed Rationales | Enthalpy, Calorimetry, Specific Heat, Hess’s
Law & Energy Diagrams | Complete Exam Prep Guide
Question 1
Which statement correctly describes the relationship between a system and its surroundings in
thermodynamics?
A. The system and surroundings together constitute the universe, and energy can be transferred
between them
B. The system is always isolated from its surroundings in thermodynamic studies
C. The surroundings are defined as everything outside the system, but energy cannot cross this
boundary
D. The system and surroundings must be at the same temperature for thermodynamic analysis
Correct Answer: A
Explanation: The system and surroundings together comprise the thermodynamic universe. Energy
can be transferred across the boundary between them through heat or work, while matter transfer
depends on the system type (open, closed, or isolated).
Question 2
A chemical reaction absorbs 85 kJ of heat from its surroundings while performing 45 kJ of work on the
surroundings. What is the change in internal energy (ΔE) of the system?
,A. +130 kJ
B. -130 kJ
C. +40 kJ
D. -40 kJ
Correct Answer: C
Explanation: Using the first law: ΔE = q + w. Heat absorbed is positive (+85 kJ), work done by the
system is negative (-45 kJ). Therefore, ΔE = 85 + (-45) = +40 kJ. The system's internal energy increased
by 40 kJ.
Question 3
Which of the following is an example of an intensive property?
A. Enthalpy
B. Internal energy
C. Temperature
D. Heat capacity
Correct Answer: C
Explanation: Intensive properties are independent of the amount of substance present. Temperature
is intensive because it remains constant regardless of sample size, whereas enthalpy, internal energy,
and heat capacity are extensive properties that depend on the quantity of matter.
Question 4
,Calculate the work done when a gas expands from 2.0 L to 8.0 L against a constant external pressure
of 1.5 atm. Express your answer in joules (1 L·atm = 101.3 J).
A. +911.7 J
B. -911.7 J
C. +607.8 J
D. -607.8 J
Correct Answer: B
Explanation: Work = -P_ext × ΔV = -1.5 atm × (8.0 - 2.0) L = -9.0 L·atm. Converting: -9.0 L·atm × 101.3
J/L·atm = -911.7 J. The negative sign indicates work done by the system during expansion.
Question 5
A system undergoes a process where ΔE = -250 J and the system absorbs 100 J of heat. Calculate the
work done by or on the system.
A. 350 J of work done by the system
B. 350 J of work done on the system
C. 150 J of work done by the system
D. 150 J of work done on the system
Correct Answer: A
Explanation: From ΔE = q + w, we have -250 = 100 + w, so w = -350 J. The negative sign means work is
done BY the system (350 J). The system lost internal energy through both heat absorption and work
output.
Question 6
, Which thermodynamic quantity is defined as H = E + PV?
A. Gibbs free energy
B. Helmholtz free energy
C. Enthalpy
D. Entropy
Correct Answer: C
Explanation: Enthalpy (H) is defined as the sum of internal energy (E) and the product of pressure and
volume (PV). This state function is particularly useful for constant-pressure processes, which are
common in open laboratory conditions.
Question 7
For a chemical reaction carried out in a bomb calorimeter at constant volume, which of the following
statements is true?
A. ΔH = q_v (heat at constant volume)
B. ΔE = q_v (heat at constant volume)
C. ΔH = ΔE + RTΔn
D. Both B and C are correct
Correct Answer: D
Explanation: In a bomb calorimeter at constant volume, no work is done (w = 0), so ΔE = q_v.
Additionally, the relationship between enthalpy and internal energy is ΔH = ΔE + RTΔn (for ideal
gases), making both statements B and C correct.
