Chemical Thermodynamics — NEET Chemistry MCQs with answers

Chemical Thermodynamics is a Class 11 Chemistry chapter in the NEET (UG) syllabus. NEET720 has 797 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.

Topics covered

System, state functions and processes · First law, internal energy and enthalpy · Thermochemistry and Hess law · Thermochemistry · Entropy and second law · Second law and entropy · Gibbs energy and spontaneity · Gibbs free energy · System and surroundings · Enthalpy and internal energy · Spontaneity · Enthalpy of reaction · Work · Heat capacity · Basic Concepts · Spontaneity and entropy · Reversible isothermal work · Hess's law and bond enthalpy · P-V work · Entropy of phase transitions · Adiabatic process · State functions · Calorimetry · Enthalpy vs internal energy · Bond enthalpy · Gibbs free energy and equilibrium constant · Cp and Cv · Third law of thermodynamics · Gibbs free energy vs temperature · Hess's law · Second law of thermodynamics · Entropy change with temperature · Free expansion · Kirchhoff's law · Isothermal vs adiabatic work · Exothermicity and spontaneity · Standard enthalpy of formation · Extensive and intensive properties · Bomb calorimeter · Trouton's rule

8 free Chemical Thermodynamics practice questions with answers

Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.

1. Question 1 · easy · System, state functions and processes

   Which of the following is an intensive property of a system?

   • A.Total volume
   • B.Total mass
   • C.Total internal energy
   • D.Density
   Show answer and explanation

   Answer: D. Density

   Intensive properties do not depend on the amount of matter present; density (mass/volume) stays constant regardless of sample size, unlike volume, internal energy or mass which all scale with amount.

   An intensive property is one whose value is independent of the size or amount of the system, e.g. temperature, pressure, density, molar volume, refractive index. An extensive property depends on the quantity of matter, e.g. mass, volume, total internal energy, total enthalpy, total entropy. Density = mass/volume; if you double the mass you also double the volume, so the ratio (density) stays the same — it is intensive. Volume, internal energy and mass all increase if you take more of the substance, so they are extensive.

   Common mistake: Assuming any single measurable quantity of the system is automatically intensive.

2. Question 2 · medium · System, state functions and processes

   A gas is compressed by suddenly increasing the external pressure and letting the system reach equilibrium with its surroundings at every stage rejected as 'infinitely slow'. If instead the compression happens instantaneously against a constant high external pressure with no intermediate equilibrium states, the process is best described as:

   • A.Reversible and quasi-static
   • B.Adiabatic and reversible
   • C.Isothermal and reversible
   • D.Irreversible, with the system passing only through non-equilibrium intermediate states
   Show answer and explanation

   Answer: D. Irreversible, with the system passing only through non-equilibrium intermediate states

   A reversible process must proceed through a continuous series of equilibrium states with only an infinitesimal difference between system and external pressure at every step. Sudden compression against a constant finite external pressure skips these intermediate equilibrium states, making it irreversible.

   Reversibility requires the process to occur infinitesimally slowly (quasi-statically), so that the system is always in equilibrium with its surroundings and the direction can be reversed by an infinitesimal change in conditions. A sudden, one-step compression against a constant (finite) external pressure does not allow the system to equilibrate at each intermediate stage — the system passes through non-equilibrium states, and the process cannot be exactly reversed without net changes elsewhere. Such a process is irreversible, regardless of whether it is isothermal or adiabatic (those describe heat exchange, not reversibility).

   Common mistake: Believing any process that starts and ends at defined states is automatically reversible.

3. Question 3 · easy · System, state functions and processes

   A litre of pure water is divided equally into two 500 mL beakers at the same temperature. Which property of the water changes on splitting it into the two portions?

   • A.Temperature
   • B.Density
   • C.Total mass of each portion
   • D.Refractive index
   Show answer and explanation

   Answer: C. Total mass of each portion

   Splitting a sample changes only extensive (amount-dependent) properties like total mass or volume of each portion; intensive properties such as temperature, density, and refractive index remain unchanged since they don't depend on the quantity of matter.

   Temperature, density, and refractive index are intensive properties: their values are fixed by the nature and state of the substance, not by how much of it you have. Dividing 1 L of water into two 500 mL portions does not alter these properties in either half. However, the total mass (and total volume) of each portion is now half of the original — mass is extensive, so it changes when the sample is divided.

   Common mistake: Assuming that splitting a sample changes intensive properties as well as extensive ones.

4. Question 4 · medium · Entropy and second law

   Entropy is a state function. This means that for a gas taken from state (P1, V1, T1) to state (P2, V2, T2),

   • A.ΔS is different if the change is carried out reversibly versus irreversibly between the same two states
   • B.ΔS can only be calculated if the process is reversible
   • C.ΔS depends only on the initial and final states, not on the path taken
   • D.ΔS is zero whenever the initial and final temperatures are equal
   Show answer and explanation

   Answer: C. ΔS depends only on the initial and final states, not on the path taken

   As a state function, the entropy change ΔS between two states depends only on the initial and final states, never on the path (reversible or irreversible) connecting them.

   Unlike q and w, which are path functions, S is a state function. So ΔS for a given initial and final state is fixed, whether the actual process is carried out reversibly or irreversibly — this is exactly why ΔS for an irreversible process is computed via any convenient reversible path between the same two states (ruling out A and B). D is wrong because even at constant T, a volume/pressure change (e.g., isothermal expansion) produces a nonzero ΔS. Hence C.

   Common mistake: Believing ΔS can only be computed for reversible processes, or that ΔS = 0 whenever ΔT = 0

5. Question 5 · medium · First law, internal energy and enthalpy

   Combustion of 0.5 mol of a fuel in a bomb calorimeter (heat capacity 2.5 kJ K⁻¹) raises the calorimeter temperature by 4.0 K. The molar internal energy change of combustion is:

   • A.−10 kJ mol⁻¹
   • B.−20 kJ mol⁻¹
   • C.+20 kJ mol⁻¹
   • D.−5 kJ mol⁻¹
   Show answer and explanation

   Answer: B. −20 kJ mol⁻¹

   Heat released = 2.5 × 4.0 = 10 kJ for 0.5 mol, so per mole ΔU = −10/0.5 = −20 kJ mol⁻¹ (bomb calorimeter measures q_v = ΔU).

   A bomb calorimeter operates at constant volume, so the heat measured is q_v = ΔU. Heat absorbed by the calorimeter = C × ΔT = 2.5 kJ K⁻¹ × 4.0 K = 10 kJ; the reaction released this, so ΔU(total) = −10 kJ for 0.5 mol. Molar value: −10 kJ / 0.5 mol = −20 kJ mol⁻¹.

   Common mistake: Not scaling the measured heat to one mole

6. Question 6 · medium · First law, internal energy and enthalpy

   An ideal gas expands from 10 L to 30 L against a constant external pressure of 1 bar. If the same expansion were done reversibly and isothermally, which statement about the magnitude of work is correct? (1 L·bar = 100 J)

   • A.Irreversible work = 2 kJ; reversible work would be smaller in magnitude
   • B.Both works are exactly equal because the end states are the same
   • C.Irreversible work = 2 kJ; reversible work would be larger in magnitude
   • D.Irreversible work = 3 kJ; reversible work would be larger in magnitude
   Show answer and explanation

   Answer: C. Irreversible work = 2 kJ; reversible work would be larger in magnitude

   Irreversible: |w| = P_extΔV = 1 bar × 20 L = 20 L·bar = 2 kJ. Reversible isothermal expansion between the same states does the maximum work, so its magnitude exceeds 2 kJ.

   For the irreversible step, |w| = P_extΔV = (1 bar)(30 − 10 L) = 20 L·bar = 2000 J = 2 kJ (w = −2 kJ for the gas). In a reversible expansion the gas pushes against an external pressure only infinitesimally smaller than its own pressure at every stage, so it extracts the maximum possible work; hence |w_rev| > |w_irrev| for the same expansion. Work is a path function — identical end states give different work along different paths.

   Common mistake: Believing work is the same for all paths between the same states

7. Question 7 · hard · First law, internal energy and enthalpy

   5 mol of an ideal diatomic gas is heated at constant pressure from 300 K to 350 K. Calculate ΔU, ΔH and w done on the gas for the process (R = 8.314 J mol⁻¹ K⁻¹). Which option lists all three correctly (in kJ)?

   • A.ΔU = +5.20, ΔH = +7.27, w = −2.08
   • B.ΔU = +7.27, ΔH = +5.20, w = +2.08
   • C.ΔU = +5.20, ΔH = +5.20, w = 0
   • D.ΔU = +5.20, ΔH = +7.27, w = +2.08
   Show answer and explanation

   Answer: A. ΔU = +5.20, ΔH = +7.27, w = −2.08

   Diatomic: C_v = (5/2)R, C_p = (7/2)R. ΔU = nC_vΔT = 5(20.785)(50) = 5196 J ≈ +5.20 kJ; ΔH = nC_pΔT = 5(29.099)(50) = 7275 J ≈ +7.27 kJ; w = ΔU − ΔH ≈ −2.08 kJ.

   Step 1: ΔT = 350 − 300 = 50 K, n = 5 mol. Step 2: For a diatomic gas, C_v = (5/2)R = 20.785 J mol⁻¹K⁻¹ and C_p = (7/2)R = 29.099 J mol⁻¹K⁻¹. Step 3: ΔU = nC_vΔT = 5 × 20.785 × 50 = 5196.25 J ≈ +5.20 kJ. Step 4: ΔH = nC_pΔT = 5 × 29.099 × 50 = 7274.75 J ≈ +7.27 kJ. Step 5: Since ΔH = ΔU − w (work done on the gas is negative for expansion), w = ΔU − ΔH = 5196.25 − 7274.75 = −2078.5 J ≈ −2.08 kJ, confirming the gas expands and does work on the surroundings while being heated at constant pressure.

   Common mistake: Swapping Cp and Cv, or mis-signing the work term

8. Question 8 · hard · First law, internal energy and enthalpy

   A reaction has ΔH° = −250.0 kJ mol⁻¹ at 300 K with Δn_g = −3. At 600 K (assume ΔH and Δn_g are unchanged for this calculation), what is ΔU° at 600 K (R = 8.314 J mol⁻¹ K⁻¹)?

   • A.−235.03 kJ mol⁻¹
   • B.−264.97 kJ mol⁻¹
   • C.−250.00 kJ mol⁻¹
   • D.−242.52 kJ mol⁻¹
   Show answer and explanation

   Answer: A. −235.03 kJ mol⁻¹

   ΔU = ΔH − Δn_gRT = −250.0 − (−3)(8.314)(600)/1000 = −250.0 + 14.97 = −235.03 kJ mol⁻¹.

   Step 1: Δn_gRT at T = 600 K: (−3)(8.314)(600) = −14,965.2 J = −14.965 kJ. Step 2: ΔU = ΔH − Δn_gRT = −250.0 − (−14.965) = −250.0 + 14.965 = −235.035 ≈ −235.03 kJ mol⁻¹. This tests careful substitution of the correct temperature (600 K, not 300 K) into the Δn_gRT term, since the question explicitly asks for the value at 600 K.

   Common mistake: Using the wrong temperature value in the RT term

Questions about Chemical Thermodynamics for NEET

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