AP Chemistry Thermodynamics — Worked Answer Explanations

Unit 6 · 12 questions explained

Below is a complete answer key for our AP Chemistry Thermodynamics practice questions. For each question you'll find the correct choice, a full written explanation of how to get there, and — for every wrong answer — a short note on exactly why it's tempting and where it goes wrong. Reading these straight through is one of the fastest ways to find the gaps in a unit before exam day.

Prefer to test yourself first? Take the timed Thermodynamics practice test and come back here to review, or head back to the Thermodynamics unit overview.

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Question 1 · Easy
When a hot metal block is placed in cool water and the system reaches thermal equilibrium, which statement is correct?
- A
  Heat flows from the water to the metal until their temperatures equalize.
  Why not A: Heat flows from hot to cold, not cold to hot.
- B
  The heat lost by the metal equals the heat gained by the water (assuming no heat loss to surroundings).Correct
- C
  The metal loses more heat than the water gains because metal is denser.
  Why not C: Conservation of energy requires $q_{metal} = - q_{water}$ ; density is irrelevant.
- D
  The temperature changes of the metal and water must be equal.
  Why not D: Temperature changes differ because mass and specific heat differ.
Explanation
Conservation of energy: $q_{metal} + q_{water} = 0 \Rightarrow q_{metal} = - q_{water}$ . Heat released by the hot metal equals heat absorbed by the cool water. The final temperature lies between the two initial temperatures.

Key takeaway
In a closed system, heat lost by the hot object equals heat gained by the cool object: $q_1 = -q_2$.
Question 2 · Easy
Which of the following best describes an endothermic reaction?
- A
  Products have lower potential energy than reactants; $Δ H < 0$ .
  Why not A: That describes an exothermic reaction.
- B
  The system absorbs heat from the surroundings; $Δ H > 0$ .Correct
- C
  The reaction releases energy as light.
  Why not C: Releasing light is a separate phenomenon; endothermic refers to heat absorbed from surroundings.
- D
  Temperature of the surroundings increases during the reaction.
  Why not D: If the system absorbs heat from surroundings, the surroundings cool (temperature decreases).
Explanation
Endothermic: the system takes in heat from the surroundings, so $Δ H_{rxn} > 0$ and the surroundings cool. Products are at higher enthalpy than reactants.

Key takeaway
Endothermic: system absorbs heat, $\Delta H > 0$, surroundings cool. Exothermic: system releases heat, $\Delta H < 0$, surroundings warm.
Question 3 · Easy
A student burns $1.00 g$ of ethanol in a calorimeter containing $500 g$ of water. The temperature rises from $22.5° C$ to $35.7° C$ . What is the heat absorbed by the water? ( $c_{water} = 4.18 J/ (g \cdot °C)$ )
- A
  $2.76 kJ$
  Why not A: Off by a factor of 10; did not complete the multiplication correctly.
- B
  $27.6 kJ$ Correct
- C
  $55.2 kJ$
  Why not C: Used $Δ T = 26.4° C$ (doubled the temperature change).
- D
  $0.276 kJ$
  Why not D: Off by a factor of 100; likely used 5 g instead of 500 g.
Explanation
$q = m c Δ T = (500 g) (4.18 J/ (g \cdot °C)) (35.7 - 22.5) ° C = 500 \times 4.18 \times 13.2 = 27, 588 J \approx 27.6 kJ$

Key takeaway
Calorimetry: $q = mc\Delta T$. Use mass in grams, specific heat in J/(g·°C), and $\Delta T$ in °C or K.
Question 4 · Easy
What is the difference between heat ( $q$ ) and temperature ( $T$ )?
- A
  Heat and temperature are the same thing measured in different units.
  Why not A: They are fundamentally different concepts; heat is energy in transit, temperature is a measure of average kinetic energy.
- B
  Temperature measures the average kinetic energy of particles; heat is the energy transferred due to a temperature difference.Correct
- C
  Heat is a property of objects; temperature is a form of energy.
  Why not C: It is the reverse: temperature is a property of objects; heat is energy in transit.
- D
  A large object always has more heat than a small object at the same temperature.
  Why not D: This conflates total thermal energy with heat flow; the question asks about the conceptual distinction.
Explanation
Temperature is an intensive property that measures the average translational kinetic energy of molecules ( $\overset{ˉ}{K E} \propto T$ ). Heat is an extensive quantity describing energy transferred between objects at different temperatures; it depends on path.

Key takeaway
Temperature = measure of average kinetic energy (intensive). Heat = energy transferred due to temperature difference (extensive, path-dependent).
Question 5 · Medium
Using the following standard enthalpies of formation, calculate $Δ H^{\circ}$ for $C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)$ :

$Δ H_{f}^{\circ} [C H_{4} (g)] = - 74.8 kJ/mol$

$Δ H_{f}^{\circ} [C O_{2} (g)] = - 393.5 kJ/mol$

$Δ H_{f}^{\circ} [H_{2} O (l)] = - 285.8 kJ/mol$
- A
  $- 604.5 kJ/mol$
  Why not A: Only included one water molecule instead of two.
- B
  $- 890.3 kJ/mol$ Correct
- C
  $+ 890.3 kJ/mol$
  Why not C: Reversed the sign — combustion is exothermic.
- D
  $- 965.1 kJ/mol$
  Why not D: Did not subtract the enthalpy of formation of CH $_{4}$ .
Explanation
$Δ H^{\circ} = \sum Δ H_{f}^{\circ} (products) - \sum Δ H_{f}^{\circ} (reactants)$

$= [(- 393.5) + 2 (- 285.8)] - [(- 74.8) + 2 (0)]$

$= (- 393.5 - 571.6) - (- 74.8) = - 965.1 + 74.8 = - 890.3 kJ/mol$

Key takeaway
$\Delta H^\circ_{\text{rxn}} = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants})$. Elements in standard state have $\Delta H_f^\circ = 0$.
Question 6 · Medium
At $25° C$ , the enthalpy of combustion of hydrogen is $- 241.8 kJ/mol$ (as gas) when $H_{2} O$ is formed as a vapor. The enthalpy of vaporization of water is $44.0 kJ/mol$ . What is the enthalpy of combustion when liquid water is formed?
- A
  $- 197.8 kJ/mol$
  Why not A: Added $Δ H_{vap}$ instead of subtracting (condensation releases heat, not absorbs).
- B
  $- 285.8 kJ/mol$ Correct
- C
  $- 241.8 kJ/mol$
  Why not C: This is the value for gaseous water product; liquid formation releases more energy.
- D
  $+ 44.0 kJ/mol$
  Why not D: This is only the vaporization enthalpy, not the total combustion enthalpy.
Explanation
Form gaseous water ( $- 241.8 kJ$ ) then condense it to liquid (condensation = $- 44.0 kJ$ ):

$Δ H = - 241.8 + (- 44.0) = - 285.8 kJ/mol$

This is also the standard enthalpy of formation of liquid water.

Key takeaway
Hess's Law applies to phase changes too: $\Delta H_{\text{liq}} = \Delta H_{\text{gas}} - \Delta H_{\text{vap}}$ (condensation releases energy).
Question 7 · Medium
The average bond enthalpy of an O=O double bond is $498 kJ/mol$ and that of an O–H bond is $464 kJ/mol$ . Using bond enthalpies, estimate $Δ H$ for:

$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (g)$

( $D (H - H) = 436 kJ/mol$ )
- A
  $- 478 kJ$
  Why not A: Did not account for all four O–H bonds formed.
- B
  $- 484 kJ$ Correct
- C
  $+ 484 kJ$
  Why not C: Reversed the sign; forming stronger bonds releases more energy than breaking weaker bonds.
- D
  $- 928 kJ$
  Why not D: Added all bond energies rather than subtracting bonds formed from bonds broken.
Explanation
Bonds broken: $2 \times 436 (H - H) + 1 \times 498 (O = O) = 872 + 498 = 1370 kJ$

Bonds formed: $4 \times 464 (O - H) = 1856 kJ$

$Δ H \approx 1370 - 1856 = - 486 kJ$

(Closest answer choice: $- 484 kJ$ , reflecting rounding in bond enthalpies used.)

Key takeaway
$\Delta H \approx \sum D(\text{broken}) - \sum D(\text{formed})$. Bond-enthalpy estimates are approximate due to bond-context variation.
Question 8 · Medium
A reaction at constant pressure releases $150 kJ$ of heat and does $30 kJ$ of work on the surroundings. What is $Δ H$ for this process?
- A
  $- 120 kJ$
  Why not A: Subtracted the work from the heat; but $Δ H = q_{P}$ — work done is already accounted for at constant pressure.
- B
  $- 180 kJ$
  Why not B: Added heat and work instead of recognizing that $Δ H = q_{P}$ .
- C
  $- 150 kJ$ Correct
- D
  $+ 150 kJ$
  Why not D: Sign error: heat released by system is negative from the system's perspective.
Explanation
At constant pressure, $Δ H = q_{P}$ (heat at constant pressure). The system releases $150 kJ$ , so $q_{P} = - 150 kJ$ and $Δ H = - 150 kJ$ . The work done ( $30 kJ$ ) is already incorporated in the definition of enthalpy ( $H = U + P V$ ) at constant $P$ .

Key takeaway
Enthalpy change $\Delta H = q_P$, the heat flow at constant pressure. Separate PV work is implicit in $H$.
Question 9 · Medium
The specific heat of aluminum is $0.900 J/ (g \cdot °C)$ and of water is $4.18 J/ (g \cdot °C)$ . Equal masses of aluminum and water absorb equal amounts of heat. Which statement is correct?
- A
  Aluminum undergoes a greater temperature increase.Correct
- B
  Water undergoes a greater temperature increase.
  Why not B: Lower specific heat (aluminum) means more temperature change per joule — aluminum heats faster.
- C
  Both undergo the same temperature increase.
  Why not C: Same heat absorbed but different specific heats → different $Δ T$ .
- D
  Water absorbs more total heat because it has a higher specific heat.
  Why not D: The question states equal amounts of heat are absorbed by both.
Explanation
$q = m c Δ T \Rightarrow Δ T = q / (m c)$ . With equal $q$ and equal $m$ , larger $c$ means smaller $Δ T$ . Water ( $c = 4.18$ ) warms less; aluminum ( $c = 0.900$ ) warms $\sim 4.6 \times$ more.

Key takeaway
A material with lower specific heat undergoes a larger temperature change for the same amount of heat absorbed.
Question 10 · Hard
Given:

(1) $C (s) + O_{2} (g) \to C O_{2} (g)$ , $Δ H_{1} = - 393.5 kJ$

(2) $2 CO (g) + O_{2} (g) \to 2 C O_{2} (g)$ , $Δ H_{2} = - 566.0 kJ$

Use Hess's Law to find $Δ H$ for: $2 C (s) + O_{2} (g) \to 2 CO (g)$
- A
  $- 959.5 kJ$
  Why not A: Added the two equations directly without reversing equation (2).
- B
  $- 221.0 kJ$ Correct
- C
  $+ 221.0 kJ$
  Why not C: Correct magnitude but wrong sign; the net process is exothermic.
- D
  $- 172.5 kJ$
  Why not D: Did not double equation (1) to account for 2 mol C.
Explanation
Target: $2 C + O_{2} \to 2 CO$

Step: Use $2 \times$ Eq.(1): $2 C (s) + 2 O_{2} (g) \to 2 C O_{2} (g)$ , $Δ H = 2 (- 393.5) = - 787.0 kJ$

Reverse Eq.(2): $2 C O_{2} (g) \to 2 CO (g) + O_{2} (g)$ , $Δ H = + 566.0 kJ$

Sum: $2 C + O_{2} \to 2 CO$ , $Δ H = - 787.0 + 566.0 = - 221.0 kJ$

Key takeaway
Hess's Law: $\Delta H$ is a state function — add or reverse steps as needed, adjusting signs accordingly.
Question 11 · Hard
Which of the following thermochemical properties can be calculated from a Born-Haber cycle?
- A
  Rate constant of an ionic reaction
  Why not A: Rate constants are kinetic properties; Born-Haber cycles involve thermodynamic state functions.
- B
  Lattice energy of an ionic compoundCorrect
- C
  Equilibrium constant for a dissolution reaction
  Why not C: Equilibrium constants relate to $Δ G$ , not directly to the Born-Haber cycle (though related).
- D
  Bond order of an ionic compound
  Why not D: Bond order applies to covalent bonds; ionic compounds have electrostatic lattice interactions.
Explanation
The Born-Haber cycle is a Hess's Law application connecting ionization energies, electron affinities, sublimation enthalpy, bond dissociation enthalpy, and enthalpy of formation to calculate the lattice energy of an ionic solid — which is otherwise very difficult to measure directly.

Key takeaway
Born-Haber cycle: a Hess's Law thermodynamic cycle used to calculate lattice energies of ionic compounds.
Question 12 · Hard
In a bomb calorimeter experiment, $1.00 g$ of glucose ( $M = 180 g/mol$ ) causes a temperature rise of $2.00° C$ in a calorimeter with heat capacity $C_{cal} = 7.78 kJ/°C$ . What is the molar heat of combustion of glucose?
- A
  $- 15.6 kJ/g$
  Why not A: This is the heat per gram, not per mole.
- B
  $- 2800 kJ/mol$ Correct
- C
  $- 5330 kJ/mol$
  Why not C: Used molar mass of sucrose (342 g/mol) instead of glucose (180 g/mol).
- D
  $- 156 kJ/mol$
  Why not D: Off by a factor of 18; divided by 10 instead of multiplying by 180 g/mol.
Explanation
$q = C_{cal} \times Δ T = 7.78 kJ/°C \times 2.00° C = 15.56 kJ$ per gram

Molar heat $= 15.56 kJ/g \times 180 g/mol = 2, 801 \approx 2800 kJ/mol$

$Δ H_{comb} = - 2800 kJ/mol$ (negative because combustion releases heat).

Key takeaway
Bomb calorimetry: $q = C_{\text{cal}} \Delta T$; molar heat of combustion = (heat per gram) × (molar mass).