75 free MCQs on Thermodynamics with worked answers and explanations. Study energy changes in chemical reactions. Understand enthalpy, entropy, Gibbs free energy, and the laws of thermodynamics that decide whether a reaction will occur spontaneously or not.
Take the timed Thermodynamics chapterwise test →Below are 75 practice questions on Thermodynamics, sorted Easy → Hard. Tap “Show answer & explanation” under any question to check yourself. Want the full theory first? Read the Thermodynamics notes.
Isothermal expansion follows a gentler curve (heat flows in to keep T constant) while adiabatic expansion drops in pressure more steeply (no heat exchange, so internal energy and temperature fall as the gas does work).
Q1.
Which of the following is an intensive property?
Answer: A. Temperature
Why: Intensive properties (temperature, pressure, density) do not depend on the amount of substance; internal energy, volume and enthalpy are extensive.
Q2.
According to the equation ΔU = q + w, when work is done on the system, the sign of w is:
Answer: B. Positive
Why: In the IUPAC convention, work done on the system increases internal energy and is taken as positive; work done by the system is negative.
Q3.
Which of the following is a path function?
Answer: C. Work
Why: Work (and heat) depend on the path taken between states, so they are path functions. Enthalpy, entropy and internal energy are state functions.
Q4.
In NCERT thermodynamics, the standard state of a substance is defined at a pressure of:
Answer: B. 1 bar
Why: The current IUPAC/NCERT standard state is defined at a pressure of exactly 1 bar (the older convention used 1 atm).
Q5.
For a complete cyclic process, the change in internal energy (ΔU) is:
Answer: D. Zero
Why: Internal energy is a state function, so over a complete cycle the system returns to its initial state and ΔU = 0.
Q6.
A reaction that releases heat to its surroundings is called:
Answer: B. Exothermic
Why: Exothermic reactions release heat, making the surroundings warmer (e.g., combustion, neutralisation).
Q7.
The SI unit of heat energy is:
Answer: B. Joule
Why: The SI unit of all energy, including heat, is the joule (J).
Q8.
Which law of thermodynamics states that energy cannot be created or destroyed?
Answer: B. First law
Why: The first law (conservation of energy) states the total energy of the universe remains constant.
Q9.
Enthalpy change at constant pressure equals:
Answer: C. Heat absorbed or released
Why: At constant pressure, ΔH = qp, the heat absorbed or released by the system.
Q10.
A reaction is spontaneous when Gibbs free energy change (ΔG) is:
Answer: C. Negative
Why: A negative ΔG means the reaction proceeds spontaneously in the forward direction.
Q11.
The standard enthalpy of formation of any element in its most stable state is:
Answer: C. Zero
Why: By definition, ΔHf° for elements in their standard state (e.g., O₂ gas, C-graphite) is zero.
Q12.
In an adiabatic process:
Answer: C. No heat exchange occurs
Why: Adiabatic means no heat is transferred between system and surroundings (q = 0).
Q13.
The unit of entropy is:
Answer: C. J K⁻¹ mol⁻¹
Why: Entropy (S) has units of joules per kelvin per mole (J K⁻¹ mol⁻¹).
Q14.
Gibbs free energy is defined as:
Answer: B. G = H - TS
Why: Gibbs free energy G = H - TS, combining enthalpy and entropy to predict spontaneity.
Q15.
Which thermodynamic process occurs at constant temperature?
Answer: D. Isothermal
Why: Isothermal means constant temperature. Isobaric = constant pressure, isochoric = constant volume.
Q16.
The branch of science dealing with energy changes accompanying chemical and physical processes is:
Answer: B. thermodynamics
Why: Thermodynamics studies energy changes; kinetics deals with reaction rates.
Q17.
A system that exchanges both energy and matter with its surroundings is a(n):
Answer: A. open system
Why: An open system exchanges both matter and energy - for example, water boiling in an open beaker.
Q18.
An ideal thermos flask, exchanging neither matter nor energy, is a(n):
Answer: C. isolated system
Why: An isolated system exchanges neither matter nor energy with its surroundings.
Q19.
Which of the following is a state function?
Answer: C. enthalpy
Why: Enthalpy depends only on the state of the system, not the path; work and heat are path functions.
Q20.
The first law of thermodynamics is a statement of the conservation of:
Answer: B. energy
Why: The first law states that energy can be neither created nor destroyed: ΔU = q + w.
Q21.
For an exothermic reaction, the enthalpy change ΔH is:
Answer: B. negative
Why: Exothermic reactions release heat, so the products have lower enthalpy and ΔH is negative.
Q22.
The SI unit of enthalpy is the:
Answer: A. joule
Why: Enthalpy is a form of energy, measured in joules (J).
Q23.
At constant pressure, the heat exchanged by a system equals its change in:
Answer: B. enthalpy
Why: At constant pressure, q_p = ΔH by definition of enthalpy.
Q24.
The internal energy of a system is represented by the symbol:
Answer: B. U
Why: Internal energy is denoted U (H is enthalpy, S entropy, G Gibbs energy).
Q25.
A process carried out at constant temperature is described as:
Answer: B. isothermal
Why: Isothermal means constant temperature; isobaric is constant pressure and isochoric is constant volume.
Q26.
For H<sub>2</sub>(g) + Cl<sub>2</sub>(g) → 2HCl(g), using bond enthalpies H–H = 435, Cl–Cl = 242 and H–Cl = 431 kJ/mol, ΔH is:
Answer: C. −185 kJ
Why: ΔH = bonds broken − bonds formed = (435 + 242) − 2(431) = 677 − 862 = −185 kJ.
Q27.
A system absorbs 700 J of heat and does 300 J of work on the surroundings. Its change in internal energy is:
Answer: D. +400 J
Why: ΔU = q + w = (+700) + (−300) = +400 J, since work done by the system is negative.
Q28.
Given ΔHf(CO<sub>2</sub>) = −393.5 and ΔHf(CO) = −110.5 kJ/mol, the enthalpy change for CO(g) + ½O<sub>2</sub>(g) → CO<sub>2</sub>(g) is:
Answer: B. −283 kJ
Why: ΔH = ΔHf(CO<sub>2</sub>) − ΔHf(CO) = −393.5 − (−110.5) = −283 kJ.
Q29.
The enthalpy of vaporisation of water is 40.8 kJ/mol at its boiling point of 373 K. The entropy of vaporisation is about:
Answer: A. 109 J/K/mol
Why: At the boiling point ΔS = ΔH/T = 40800/373 ≈ 109 J/K/mol.
Q30.
For which of these reactions is ΔH equal to ΔU (i.e. Δn<sub>g</sub> = 0)?
Answer: C. H<sub>2</sub>(g) + Cl<sub>2</sub>(g) → 2HCl(g)
Why: ΔH = ΔU + Δn<sub>g</sub>RT. Only H<sub>2</sub> + Cl<sub>2</sub> → 2HCl has Δn<sub>g</sub> = 2 − 2 = 0, so ΔH = ΔU.
Q31.
For a reaction with ΔH = -200 kJ and ΔS = +100 J/K at 300 K, ΔG is:
Answer: B. -230 kJ
Why: ΔG = ΔH - TΔS = -200,000 - (300 × 100) = -230,000 J = -230 kJ.
Q32.
In an isothermal reversible expansion of an ideal gas, ΔU is:
Answer: C. Zero
Why: For an ideal gas, internal energy depends only on temperature. At constant T, ΔU = 0.
Q33.
Bond enthalpy (bond dissociation energy) is always:
Answer: B. Positive
Why: Breaking bonds always requires energy input (endothermic), so bond dissociation enthalpy is always positive.
Q34.
Hess's law states that the enthalpy change of a reaction:
Answer: B. Is independent of the path
Why: Enthalpy is a state function, so ΔH depends only on initial and final states, not on the intermediate steps.
Q35.
The entropy change for melting of ice at 0°C (ΔH_fusion = 6 kJ/mol) is:
Answer: B. 21.98 J K⁻¹ mol⁻¹
Why: ΔS = ΔH/T = 6000 J / 273 K = 21.98 J K⁻¹ mol⁻¹.
Q36.
A reaction is spontaneous only at high temperatures when:
Answer: B. ΔH > 0, ΔS > 0
Why: When ΔH > 0 and ΔS > 0, ΔG = ΔH - TΔS becomes negative only at high T. Below T = ΔH/ΔS, non-spontaneous.
Q37.
The standard Gibbs energy change relates to the equilibrium constant by:
Answer: B. ΔG° = -RT ln K
Why: ΔG° = -RT ln K. A large K means a large negative ΔG°, confirming the reaction is product-favoured.
Q38.
Work done during isothermal reversible expansion of an ideal gas is:
Answer: B. w = -nRT ln(V₂/V₁)
Why: w = -nRT ln(V₂/V₁). For expansion V₂ > V₁, the system does work on surroundings (work done on system is negative).
Q39.
For an ideal gas undergoing free expansion into a vacuum, the values of q and w are:
Answer: A. Both zero
Why: Free expansion into a vacuum occurs against zero external pressure, so w = 0, and since it happens in an isolated system, q = 0 as well.
Q40.
For the reaction N<sub>2</sub>(g) + 3H<sub>2</sub>(g) -> 2NH<sub>3</sub>(g), how is the standard enthalpy of formation of NH<sub>3</sub> related to the standard reaction enthalpy?
Answer: A. Standard reaction enthalpy equals twice the standard enthalpy of formation of NH<sub>3</sub>
Why: Since 2 moles of NH<sub>3</sub> are formed from the elements in their standard states, the standard reaction enthalpy equals 2 times the standard enthalpy of formation of NH<sub>3</sub>.
Q41.
For the isothermal expansion of an ideal gas, the change in internal energy ΔU is:
Answer: C. zero
Why: Internal energy of an ideal gas depends only on temperature, so ΔU = 0 in any isothermal process.
Q42.
In an adiabatic process, the heat exchanged with the surroundings (q) is:
Answer: B. zero
Why: An adiabatic process occurs with no heat exchange, so q = 0 and ΔU = w.
Q43.
In ΔH = ΔU + ΔnₘRT, the term Δnₘ is the change in the number of moles of:
Answer: C. gases
Why: Only gaseous species change the PV term appreciably, so Δnₘ counts moles of gaseous products minus reactants.
Q44.
Which of these has the highest standard molar entropy at 298 K?
Answer: C. water vapour
Why: Gases have far more positional disorder than liquids or solids, so water vapour has the highest entropy.
Q45.
Hess’s law of constant heat summation is a direct consequence of enthalpy being a:
Answer: B. state function
Why: Because enthalpy is a state function, the total ΔH is the same regardless of the route taken.
Q46.
The enthalpy change when one mole of a compound forms from its elements in their standard states is the standard enthalpy of:
Answer: B. formation
Why: This is the definition of the standard enthalpy of formation, ΔH_f°.
Q47.
For the reaction N₂(g) + 3H₂(g) → 2NH₃(g), the value of Δnₘ (gaseous) is:
Answer: A. −2
Why: Δnₘ = moles of gaseous products − reactants = 2 − 4 = −2.
Q48.
The heat change measured at constant volume is equal to the change in:
Answer: B. internal energy
Why: At constant volume no PV work is done, so q_v = ΔU.
Q49.
For one mole of an ideal gas, the difference between the molar heat capacities Cp and Cv equals:
Answer: A. R
Why: From ΔH = ΔU + Δ(PV) for an ideal gas, Cp − Cv = R (Mayer’s relation).
Q50.
For the isothermal reversible expansion of an ideal gas from V₁ to V₂, the work done by the gas is:
Answer: B. nRT ln(V₂/V₁)
Why: Reversible isothermal work = nRT ln(V₂/V₁), obtained by integrating P dV with P = nRT/V.
Q51.
Given ΔHf(CO<sub>2</sub>) = −393.5, ΔHf(H<sub>2</sub>O, l) = −285.8 and ΔHf(C<sub>2</sub>H<sub>5</sub>OH, l) = −277.0 kJ/mol, the standard enthalpy of combustion of ethanol is:
Answer: A. −1367 kJ/mol
Why: C<sub>2</sub>H<sub>5</sub>OH + 3O<sub>2</sub> → 2CO<sub>2</sub> + 3H<sub>2</sub>O: ΔH = [2(−393.5) + 3(−285.8)] − (−277.0) = −1644.4 + 277 = −1367 kJ/mol.
Q52.
A gas expands from 2 L to 6 L against a constant external pressure of 2 atm. The work done by the gas is about (1 L·atm = 101.3 J):
Answer: A. −810 J
Why: w = −P<sub>ext</sub>ΔV = −(2 atm)(4 L) = −8 L·atm = −8 × 101.3 ≈ −810 J (work done by the gas is negative).
Q53.
For a reaction with Δn<sub>g</sub> = +2 carried out at 300 K, the value of (ΔH − ΔU) is (R = 8.314 J/K/mol):
Answer: B. +4.99 kJ
Why: ΔH − ΔU = Δn<sub>g</sub>RT = 2 × 8.314 × 300 = 4988 J ≈ +4.99 kJ.
Q54.
Assertion (A): In the isothermal reversible expansion of an ideal gas, q = −w. Reason (R): The internal energy of an ideal gas depends only on temperature, so ΔU = 0 in an isothermal change.
Answer: C. Both A and R true; R correctly explains A
Why: For an ideal gas ΔU depends only on T, so isothermally ΔU = 0; then q + w = 0, i.e. q = −w. R correctly explains A.
Q55.
An exothermic reaction releases 100 kJ of heat to the surroundings at 300 K. The entropy change of the surroundings is approximately:
Answer: D. +333 J/K
Why: ΔS<sub>surr</sub> = +q<sub>surr</sub>/T = +100000/300 ≈ +333 J/K (surroundings gain heat, so their entropy rises).
Q56.
At what temperature does a reaction with ΔH = +80 kJ/mol and ΔS = +200 J/K/mol become spontaneous?
Answer: A. 400 K
Why: ΔG = 0 when T = ΔH/ΔS = 80,000/200 = 400 K. Above 400 K, ΔG < 0 (spontaneous).
Q57.
For C(s) + O₂(g) → CO₂(g), the relationship between ΔH and ΔU is:
Answer: C. ΔH = ΔU
Why: Δng = 1 gas product - 1 gas reactant = 0. Since ΔH = ΔU + ΔngRT and Δng = 0, ΔH = ΔU.
Q58.
Given ΔHc(C) = -393.5, ΔHc(H₂) = -285.8, ΔHc(CH₄) = -890 kJ/mol, the enthalpy of formation of CH₄ is:
Answer: A. -74.8 kJ/mol
Why: ΔHf(CH₄) = ΔHc(C) + 2ΔHc(H₂) - ΔHc(CH₄) = -393.5 + 2(-285.8) - (-890) = -75.1 ≈ -74.8 kJ/mol.
Q59.
For a reaction at 298 K with ΔG° = -20 kJ/mol, the equilibrium constant K is approximately:
Answer: A. 3200
Why: ΔG° = -RT ln K → ln K = 20000/(8.314×298) = 8.07 → K = e<sup>8.07</sup> ≈ 3200.
Q60.
The Kirchhoff equation d(ΔH)/dT = ΔCp is used to calculate:
Answer: B. Enthalpy change at different temperatures
Why: Kirchhoff's equation accounts for how ΔH varies with temperature based on heat capacity differences (ΔCp).
Q61.
For an adiabatic reversible expansion, which relation holds?
Answer: A. TV<sup>γ-1</sup> = constant
Why: For adiabatic reversible processes: TV<sup>γ-1</sup> = constant (and also PV^γ = constant), where γ = Cp/Cv.
Q62.
The efficiency of a Carnot engine operating between 500 K and 300 K is:
Answer: A. 40%
Why: Carnot efficiency = 1 - (Tc/Th) = 1 - 300/500 = 1 - 0.6 = 0.4 = 40%.
Q63.
Which of the following has a negative entropy change (ΔS < 0)?
Answer: B. Dissolving a gas in liquid
Why: Dissolving a gas in liquid reduces disorder (gas → dissolved species), so ΔS < 0. Other options all increase disorder.
Q64.
For the reaction N<sub>2</sub>(g) + 3H<sub>2</sub>(g) -> 2NH<sub>3</sub>(g) at 300 K, if the internal energy change (Delta-U) is -100 kJ, what is the enthalpy change Delta-H (R = 8.314 J/K/mol)?
Answer: C. -105.0 kJ
Why: Delta-n(gas) = 2 - 4 = -2; Delta-H = Delta-U + Delta-n(g)RT = -100000 + (-2 x 8.314 x 300) = -100000 - 4988 = -104988 J, approximately -105.0 kJ.
Q65.
One mole of an ideal gas is compressed isothermally and reversibly from 10 L to 1 L at 300 K. What is the work done on the gas (R = 8.314 J/K/mol)?
Answer: B. +5744 J
Why: For isothermal reversible compression, w = -nRT ln(Vf/Vi) = -1 x 8.314 x 300 x ln(1/10) = +5744 J done on the gas.
Q66.
The second law of thermodynamics states that for any spontaneous process the total entropy of the universe:
Answer: B. increases
Why: A spontaneous change increases the total entropy of the universe (system + surroundings).
Q67.
The Gibbs free energy change is defined by the equation ΔG =:
Answer: B. ΔH − TΔS
Why: ΔG = ΔH − TΔS; a negative ΔG marks a spontaneous process at constant T and P.
Q68.
A reaction is spontaneous at all temperatures when:
Answer: B. ΔH < 0 and ΔS > 0
Why: With ΔH negative and ΔS positive, ΔG = ΔH − TΔS is negative at every temperature.
Q69.
At equilibrium, the Gibbs free energy change ΔG for the reaction is:
Answer: B. zero
Why: At equilibrium there is no net drive in either direction, so ΔG = 0.
Q70.
The relationship between the standard free energy change and the equilibrium constant is ΔG° =:
Answer: B. −RT ln K
Why: ΔG° = −RT ln K, so a large K corresponds to a large negative ΔG°.
Q71.
A reaction has ΔH = +50 kJ/mol and ΔS = +100 J/K·mol. It becomes spontaneous above a temperature of about:
Answer: B. 500 K
Why: Spontaneous when T > ΔH/ΔS = 50000/100 = 500 K.
Q72.
The entropy change for the melting of ice at 0 °C is:
Answer: B. positive
Why: Melting turns an ordered solid into a more disordered liquid, so ΔS is positive.
Q73.
By the third law of thermodynamics, the entropy of a perfect crystal at absolute zero is:
Answer: B. zero
Why: A perfect crystal at 0 K has only one microstate, so its entropy is zero.
Q74.
For a reversible adiabatic process involving an ideal gas, which relation holds?
Answer: B. PVᵞ = constant
Why: For a reversible adiabatic change, PVᵞ = constant, where γ = Cp/Cv.
Q75.
A Carnot engine operating between 500 K and 300 K has an efficiency of:
Answer: A. 40%
Why: Carnot efficiency = 1 − T_cold/T_hot = 1 − 300/500 = 0.40 = 40%.