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Free MCAT-style practice, 5E

MCAT Thermodynamics Practice Questions

By StudyStore · Published 2026-10-30 · 9 original questions · about 14 min

Nine free MCAT-style thermodynamics practice questions on Hess's law, bond energies, entropy, Gibbs free energy, temperature and spontaneity, and K, with full explanations.

Quick concept primer

Enthalpy is a state function, so the ΔH of a reaction depends only on its start and end points. That is why Hess's law works: reverse a step and flip the sign of its ΔH, scale a step and scale its ΔH by the same factor, then add. Bond dissociation energies give an estimate of the same quantity: energy spent breaking reactant bonds minus energy released forming product bonds.

Spontaneity is decided by ΔG = ΔH − TΔS, not by ΔH alone. When ΔH and ΔS share a sign, temperature decides, and the crossover sits at T = ΔH/ΔS. At standard conditions ΔG° = −RT ln K, so a negative ΔG° means K is greater than 1. Away from standard conditions, ΔG = ΔG° + RT ln Q tells you which way a mixture will shift. Cells exploit additivity too: an unfavorable step runs when it shares an intermediate with a strongly favorable one, such as ATP hydrolysis.

Traps: thermodynamics says nothing about rate, so a negative ΔG does not mean a fast reaction; convert ΔS from J to kJ before combining it with ΔH; and the second law concerns the entropy of the universe, not of the system alone. The 5E Thermodynamics and Kinetics pack adds flashcards and more practice across the whole category, and you can try a free sampler first.

9 practice questions

Try each question before opening the explanation. Click "Show answer and explanation" to check yourself. All questions are original and written for study practice.

Question 1

A chemist measures two steps at constant pressure:

StepReactionΔH, kJ/mol
1A + 2B → C−210
2C → 2D+80

What is ΔH for the reaction 2D → A + 2B?

  1. −130 kJ/mol
  2. +130 kJ/mol
  3. −290 kJ/mol
  4. +290 kJ/mol
Show answer and explanation

Answer: B. +130 kJ/mol

Adding steps 1 and 2 gives A + 2B → 2D with ΔH = −210 + 80 = −130 kJ/mol. The requested reaction is the reverse, so its sign flips to +130 kJ/mol. Choice A forgets to reverse, and choices C and D subtract 80 instead of adding it.

Question 2

Use these average bond energies for a hypothetical gas-phase reaction X2 + Y2 → 2XY: X-X 440 kJ/mol, Y-Y 240 kJ/mol, X-Y 430 kJ/mol. What is the estimated ΔH of the reaction?

  1. −180 kJ/mol
  2. +180 kJ/mol
  3. +250 kJ/mol
  4. −250 kJ/mol
Show answer and explanation

Answer: A. −180 kJ/mol

ΔH is bonds broken minus bonds formed: (440 + 240) − 2(430) = 680 − 860 = −180 kJ/mol. Choice B reverses the subtraction, and choices C and D count only one X-Y bond formed even though two form.

Question 3

A phase change has ΔH = +36 kJ/mol and ΔS = +120 J/(mol K), both roughly constant with temperature. Which statement about spontaneity is correct?

  1. It is spontaneous at every temperature
  2. It is spontaneous only below 300 K
  3. It is spontaneous only above 300 K
  4. It is spontaneous only above about 3.3 K
Show answer and explanation

Answer: C. It is spontaneous only above 300 K

With ΔH and ΔS both positive, ΔG = ΔH − TΔS becomes negative only when TΔS exceeds ΔH, above T = 36,000 J/mol ÷ 120 J/(mol K) = 300 K. Choice B reverses the direction, and choice D inverts the ratio and drops the unit conversion.

Question 4

A reaction has an equilibrium constant of 1.0 × 103 at 298 K. Using R = 8.314 J/(mol K), what is ΔG° for the reaction?

  1. +17.1 kJ/mol
  2. −7.4 kJ/mol
  3. −1.7 kJ/mol
  4. −17.1 kJ/mol
Show answer and explanation

Answer: D. −17.1 kJ/mol

ΔG° = −RT ln K = −(8.314)(298)(ln 1000) = −(8.314)(298)(6.91) ≈ −17,100 J/mol, or −17.1 kJ/mol. The negative sign matches K greater than 1. Choice B uses log base 10 instead of the natural log, and choice A has the wrong sign.

Question 5

Which process has a negative entropy change for the system?

  1. 2SO2(g) + O2(g) → 2SO3(g)
  2. N2O4(g) → 2NO2(g)
  3. Ice melting at 0 °C
  4. Solid CO2 subliming at 1 atm
Show answer and explanation

Answer: A. 2SO2(g) + O2(g) → 2SO3(g)

Three moles of gas become two moles of gas, so the number of ways to arrange the particles drops and ΔS is negative. Choice B doubles the moles of gas, and choices C and D move from a solid to a less ordered phase, all of which raise entropy.

Question 6

A reaction has ΔG° < 0 at 298 K. All of the following must be true EXCEPT:

  1. K at 298 K is greater than 1
  2. The reverse reaction has ΔG° > 0 at 298 K
  3. The reaction reaches equilibrium within seconds
  4. ln K at 298 K is positive
Show answer and explanation

Answer: C. The reaction reaches equilibrium within seconds

ΔG° is a thermodynamic quantity and carries no information about activation energy or rate, so a favorable reaction can still be very slow. A negative ΔG° does force K above 1 and ln K above zero, and reversing a reaction flips the sign of ΔG°.

Question 7

A student tabulates approximate values for four reactions, assuming each value stays constant with temperature:

ReactionΔH, kJ/molΔS, J/(mol K)
P−50−100
R−30+15
S+20+60
T+10−5

Which reaction is spontaneous at every temperature?

  1. P
  2. R
  3. S
  4. T
Show answer and explanation

Answer: B. R

Only reaction R has negative ΔH and positive ΔS, so both terms of ΔG = ΔH − TΔS are negative at any temperature. Reaction P is spontaneous only at low temperature, S only at high temperature, and T, with ΔH positive and ΔS negative, never.

Question 8

In a cell, phosphorylating a sugar with inorganic phosphate has ΔG°′ of about +13.8 kJ/mol, and ATP hydrolysis to ADP and phosphate has ΔG°′ of about −30.5 kJ/mol. Using these values, what is ΔG°′ for the coupled reaction sugar + ATP → sugar phosphate + ADP?

  1. +44.3 kJ/mol
  2. −44.3 kJ/mol
  3. +16.7 kJ/mol
  4. −16.7 kJ/mol
Show answer and explanation

Answer: D. −16.7 kJ/mol

Coupled reactions add, and so do their free energies: +13.8 + (−30.5) = −16.7 kJ/mol, so the overall reaction is favorable. Choices A and B subtract the values instead of adding them, and choice C has the wrong sign.

Question 9

At 300 K, a process occurs spontaneously in a closed system while the system's entropy falls by 20 J/K. Which statement must be true?

  1. The process violates the second law of thermodynamics
  2. The system absorbed at least 6.0 kJ of heat
  3. The entropy of the universe stays exactly constant
  4. The system released more than 6.0 kJ of heat to the surroundings
Show answer and explanation

Answer: D. The system released more than 6.0 kJ of heat to the surroundings

A spontaneous process requires the total entropy change to be positive, so the surroundings must gain more than 20 J/K. At 300 K that means more than (20 J/K)(300 K) = 6000 J of heat flowed into the surroundings. Choice A ignores the surroundings, choice B gets the heat direction backward, and choice C describes a reversible process.

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Written by StudyStore from standard scientific facts as a study aid. These are not official exam questions and are not affiliated with or endorsed by any testing organization. Always check values against the reference provided with your exam or course.

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