Rucete ✏ AP Chemistry In a Nutshell
11. Thermodynamics — Practice Questions
This chapter introduces thermodynamic principles including energy types, state functions, enthalpy, entropy, Gibbs free energy, Hess's law, and criteria for thermodynamically favored reactions.
(Multiple Choice — Click to Reveal Answer)
1. Which of these must be negative for a reaction to be thermodynamically favored?
(A) ∆G°
(B) ∆H°
(C) ∆G°, ∆S°, and ∆G
(D) ∆G
Answer
(A) — A negative value for ΔG° indicates a thermodynamically favorable reaction.
2. Why is ∆H° a state function?
(A) It has the required superscript zero.
(B) Its value does not depend on the path between the initial and final states.
(C) It requires exactly 1 molar concentrations of all reactants.
(D) Its value is equal to the initial state minus the final state.
Answer
(B) — Its value does not depend on the path between initial and final states.
3. Scientists divide energy into two broad groups. They are
(A) chemical and solar
(B) electrical and nuclear
(C) potential and kinetic
(D) thermal and electromagnetic
Answer
(C) — Energy is broadly divided into potential and kinetic forms.
4. How can you tell if a chemical reaction or a physical process is endothermic or exothermic?
(A) Use the ∆G° and ∆S° for the reaction.
(B) Observe how temperature affects the products.
(C) Use Hess’s law and heats of formation.
(D) All of the above.
Answer
(D) — All listed methods are valid.
5. When 0.400 g of CH₄ is burned in a calorimeter with C = 3245 J/°C and ΔT = 6.795 °C, what is q and ΔH?
(A) 220 kJ mol⁻¹ and –220 kJ mol⁻¹
(B) −882 kJ and 882 kJ
(C) 477 kJ and –477 kJ
(D) –22.05 kJ and –22.05 kJ
Answer
(D) — q = –(3245)(6.795) = –22.05 kJ; heat released.
6. In a frictionless U-ramp diagram, which is correct as a ball rolls back and forth?
(A) Point 4 is the highest velocity, and point 3 is the lowest velocity.
(B) Point 1 is the highest potential energy, and point 3 is the highest kinetic energy.
(C) Point 1 is the highest kinetic energy, and point 3 is the lowest potential energy.
(D) Point 1 is the highest kinetic energy, and point 3 is the lowest velocity.
Answer
(B) — Point 1 is highest in potential energy, point 3 highest in kinetic energy.
7. Which system CANNOT be thermodynamically favored?
(A) ΔH° is positive, and ΔS° is negative.
(B) ΔH° is positive, and ΔS° is positive.
(C) ΔH° is negative, and ΔS° is negative.
(D) ΔH° is negative, and ΔS° is positive.
Answer
(A) — Positive ΔH° and negative ΔS° always give a positive ΔG° (not favored).
8. When KCl(s) dissolves and condensation is seen outside the beaker, what does this indicate?
(A) The condensation is extra information.
(B) The condensation informs about heat; the dissociation informs entropy.
(C) Only heat of solution can be inferred.
(D) The condensation tells us it is exothermic; entropy always increases.
Answer
(B) — Condensation informs about the heat change; dissolution informs entropy.
9. The reaction with the greatest entropy decrease is
(A) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
(B) CH₄(l) + 2O₂(g) → CO₂(g) + 2H₂O(g)
(C) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
(D) CH₄(g) + 2O₂(g) → CO₂(s) + 2H₂O(g)
Answer
(C) — Fewer gas molecules after reaction means lower entropy.
10. Water boils at 100 °C with ΔHvap = +43.9 kJ/mol. What is ΔS when H₂O condenses?
(A) Cannot solve; ΔG° needed.
(B) Cannot solve; not a chemical reaction.
(C) −439 J/K
(D) −118 J/K
Answer
(D) — ΔS = –43900 J / 373 K = –118 J/K.
11. Which statement is NOT correct about the use of the free energy diagram?
(A) If Q is left of the minimum, no reaction occurs.
(B) The sign of ΔG determines thermodynamic favorability.
(C) The minimum indicates Q = K.
(D) If Q is right of the minimum, products convert to reactants.
Answer
(A) — Q will shift to reach equilibrium; the statement is false.
12. A gas expands from a high-pressure state. Estimate the work done.
(A) +30.0 L·atm
(B) +15.0 L·atm
(C) –10.0 L·atm
(D) Data not sufficient
Answer
(C) — Expansion results in negative work (system does work).
13. What is needed to convert L·atm to joules?
(A) Avogadro’s constant and Planck’s constant
(B) R in L·atm units
(C) R in J/mol·K units
(D) Both (B) and (C)
Answer
(D) — Both forms of R are used to convert energy units.
14. Which is LEAST likely in a combustion reaction?
(A) ΔG° is large and negative
(B) ΔS° is large and negative
(C) ΔH° is large and negative
(D) Keq is large and positive
Answer
(B) — Combustion usually increases entropy, not decreases it.
15. Which molecular-level diagram shows the most entropy for argon gas?
(A) Atoms far apart
(B) Argon is a liquid
(C) Most gas atoms and states to occupy
(D) Argon atoms strongly attract
Answer
(C) — Entropy increases with more particles and microstates.
16. Given heats of formation: CH₃CH₂OH(l) = –258 kJ/mol, CO₂(g) = –393.5 kJ/mol, H₂O(g) = –241.8 kJ/mol. What is ΔH° for combustion of ethanol?
(A) –5021.6 kJ
(B) –2510.8 kJ
(C) –1254.4 kJ
(D) +5021.6 kJ
Answer
(C) — Apply Hess’s law using formation data.
17. When is the rate of reaction fastest?
(A) ΔG° is a large negative number
(B) ΔS° is large and negative
(C) ΔH° is large and negative
(D) None of the above
Answer
(D) — Thermodynamics doesn't determine rate; kinetics does.
18. From thermochemical data, what is the correct ΔH°?
(A) –171.07 kJ
(B) –55.21 kJ
(C) +55.21 kJ
(D) +171.07 kJ
Answer
(A) — Apply Hess’s law and proper sign handling.
19. What can change the value of ΔG°?
(A) Total pressure
(B) Reactant pressure
(C) Reactant concentration
(D) Temperature in °C
Answer
(D) — Only temperature affects standard Gibbs free energy.
20. Which is NOT an example of kinetic ↔ potential energy conversion without loss?
(A) Vibrating spring
(B) Bouncing ball on bench
(C) Pendulum
(D) Ball bouncing down stairs
Answer
(D) — Ball loses energy permanently with each stair impact.
21. Given: ΔHf of SO₃(g) = –396 kJ/mol; S° of S(s) = 31.8, O₂(g) = 205.0, SO₃(g) = 256 J/mol·K. What is ΔG° for 2SO₃(g) → 2S(s) + 3O₂(g)?
(A) –446 kJ
(B) –346 kJ
(C) +396 kJ
(D) +742 kJ
Answer
(D) — Use ΔG = ΔH – TΔS with proper mole scaling.
22. To calculate ΔH° for: 2C₆H₆(l) + 15O₂(g) → 12CO₂(g) + 6H₂O(l), you need
(A) Heat of combustion of carbon
(B) Heat of combustion of hydrogen
(C) Heat of formation of C₆H₆(l)
(D) All of the above
Answer
(D) — All are required to calculate overall ΔH°.
23. The evaporation of any liquid has
(A) Positive ΔH and negative ΔS
(B) Negative ΔH and negative ΔS
(C) Positive ΔH and positive ΔS
(D) Negative ΔH and positive ΔS
Answer
(C) — Evaporation absorbs heat and increases disorder.
24. Which is most likely to be true?
(A) Combustion has negative ΔH°.
(B) Positive ΔG° means favored.
(C) Positive ΔS° always means favored.
(D) Favored reactions go to completion.
Answer
(A) — Combustion is typically exothermic (ΔH° < 0).
25. Which affects reaction direction (Q vs K), not favorability?
(A) ∆H
(B) ∆S
(C) ∆G°
(D) Q vs K value
Answer
(D) — Reaction direction is determined by Q compared to K.
26. Which equation allows calculation of free energy change at non-standard conditions?
(A) ΔG = ΔH – TΔS
(B) ΔG° = –RT ln Q
(C) ΔG = ΔG° + RT ln Q
(D) ΔG° = –RT ln K
Answer
(C) — This is the proper equation for non-standard Gibbs free energy.
27. What does it mean when ΔG = 0 for a reaction?
(A) Reaction is not spontaneous
(B) Reaction is at equilibrium
(C) Reaction is endothermic
(D) Reaction will proceed in forward direction
Answer
(B) — At equilibrium, ΔG = 0 and there is no net change.
28. A chemical reaction has ΔH° = +100 kJ and ΔS° = +300 J/K. At what temperature does it become thermodynamically favored?
(A) 100 K
(B) 273 K
(C) 373 K
(D) Above 333 K
Answer
(D) — ΔG = ΔH – TΔS; solve 0 = 100 – T(0.3) → T > 333 K.
29. A system has ΔH = –150 kJ and ΔS = –200 J/K. What temperature will make ΔG = 0?
(A) 300 K
(B) 400 K
(C) 500 K
(D) 750 K
Answer
(C) — Use ΔG = ΔH – TΔS = 0 → T = ΔH / ΔS = 150,000 / 200 = 750 K.
30. When calculating ΔG° = –RT ln K, if K = 1, what is ΔG°?
(A) 0
(B) 1
(C) RT
(D) –RT
Answer
(A) — ln(1) = 0, so ΔG° = 0 at equilibrium.
31. A chemical system with more gas particles on the product side than reactant side tends to have
(A) lower entropy
(B) zero entropy change
(C) increased entropy
(D) decreased enthalpy
Answer
(C) — More gas particles → more microstates → higher entropy.
32. What is the best method to determine the entropy change (ΔS) experimentally?
(A) Bomb calorimetry
(B) Measuring temperature rise
(C) Using ΔG = ΔH – TΔS after measuring ΔH and ΔG
(D) Reacting elements in standard state
Answer
(C) — Measure ΔH and ΔG, then calculate ΔS.
33. Which of the following is NOT a state function?
(A) Enthalpy (H)
(B) Work (w)
(C) Entropy (S)
(D) Gibbs Free Energy (G)
Answer
(B) — Work depends on the path taken; it is not a state function.
34. Which factor directly determines whether a reaction proceeds spontaneously under constant T and P?
(A) Change in entropy only
(B) Heat released
(C) Sign of ΔG
(D) Activation energy
Answer
(C) — ΔG < 0 means the reaction is spontaneous at constant T, P.
35. For a process to be both exothermic and entropy increasing, which is true?
(A) ΔH > 0 and ΔS < 0
(B) ΔH < 0 and ΔS > 0
(C) ΔH > 0 and ΔS > 0
(D) ΔH < 0 and ΔS < 0
Answer
(B) — Negative enthalpy and positive entropy ensures ΔG < 0 at all T.
36. Define what it means for a chemical reaction to be thermodynamically favored.
Answer
ΔG° < 0 — A reaction is thermodynamically favored if it proceeds spontaneously under standard conditions, indicated by a negative ΔG°.
37. What are the two main types of energy in thermodynamics?
Answer
Kinetic and potential energy — Kinetic energy relates to motion; potential energy relates to position or stored energy in bonds.
38. What does the equation ΔG° = –RT ln K tell us about a reaction?
Answer
It relates free energy to equilibrium — If K > 1, ΔG° is negative, meaning the reaction is product-favored at equilibrium.
39. Why is work not considered a state function?
Answer
It depends on the path — Unlike enthalpy or entropy, work depends on how the process occurs, not just initial and final states.
40. What does a positive ΔS value suggest about a process?
Answer
Disorder increases — A positive entropy change indicates the system becomes more disordered or has more possible microstates.
41. Describe what the term “standard state” means in thermodynamics.
Answer
1 atm, 25 °C, and 1 M concentration — Standard state refers to specified conditions used to report thermodynamic values like ΔH°, ΔG°, and ΔS°.
42. What is the formula used to calculate the amount of heat absorbed or released by a substance?
Answer
q = mcΔT — Where q = heat, m = mass, c = specific heat, ΔT = change in temperature.
43. What does it mean if ΔG is positive for a chemical process?
Answer
The reaction is not spontaneous — A positive ΔG means energy input is required for the reaction to proceed in the forward direction.
44. Under what condition does ΔG = 0?
Answer
At equilibrium — No net reaction occurs; the system is thermodynamically balanced.
45. Why do heats of formation for elements in their standard state equal zero?
Answer
No formation occurs — The element already exists in its most stable form at standard conditions, so ΔH°f = 0 by definition.
46. What is the significance of the sign of ΔH in a reaction?
Answer
It shows heat flow — ΔH < 0 means exothermic (heat released); ΔH > 0 means endothermic (heat absorbed).
47. If a reaction has ΔH° > 0 and ΔS° > 0, under what temperature conditions is it spontaneous?
Answer
At high temperatures — Because ΔG = ΔH – TΔS, a large T makes –TΔS more negative, favoring spontaneity.
48. What happens to entropy when a gas condenses into a liquid?
Answer
Entropy decreases — The molecules become more ordered and restricted in motion.
49. State the First Law of Thermodynamics.
Answer
Energy is conserved — It cannot be created or destroyed, only transformed between forms like heat and work.
50. How does increasing temperature affect the value of ΔG for a reaction with both ΔH > 0 and ΔS > 0?
Answer
It makes ΔG more negative — This favors spontaneity since the –TΔS term becomes more dominant.