ThermodynamicsNEET MCQs with solutions
Thermodynamics covers the laws of thermodynamics, enthalpy, Hess's law, entropy, Gibbs free energy and spontaneity. NEET questions involve enthalpy calculations (bond enthalpy, formation enthalpy), predicting spontaneity using ΔG = ΔH − TΔS, and applying Hess's law for indirect enthalpy determination.
- Class
- 11 Chemistry
- Free on this page
- 24 questions
- In the RankUp app
- 551 questions
- ELITE questions
- 234
- NCERT topics
- 12
Practise 24 questions
Tap an option to check it. Questions from every NCERT topic in this chapter, from easy to hard.
Q1Important PYQ Mixed Concepts
Using combustion enthalpies ΔH°c[C(s)] = −94 kcal mol⁻¹, ΔH°c[H₂(g)] = −68 kcal mol⁻¹ and ΔH°c[CH₄(g)] = −213 kcal mol⁻¹, the standard enthalpy of formation of CH₄(g) is:
Not quite — the answer is B.
By Hess's law: ΔH°f(CH₄) = ΔH°c(C) + 2ΔH°c(H₂) − ΔH°c(CH₄) = −94 + 2(−68) − (−213) = −94 − 136 + 213 = −17 kcal mol⁻¹. Option B is correct.
Q2Important PYQ Mixed Concepts
A closed, insulated container holds a liquid stirred by a paddle until its temperature rises. Using ΔU = q + w, which statement is correct?
Not quite — the answer is C.
The container is insulated, so no heat exchange occurs: q = 0. The paddle does work on the system, so w > 0. By the first law, ΔU = q + w = w > 0. Options A and D require q ≠ 0, which contradicts insulation. Option B requires w = 0, which contradicts paddle action.
Q3Grand Test
For the reaction C(s)+2H₂(g)→CH₄(g), the heats of combustion of C, H₂ and CH₄ are −94, −68 and −213 kcal mol⁻¹ respectively. The standard enthalpy of formation of CH₄(g) is:
Not quite — the answer is B.
By Hess's law, ΔHf(CH₄) = ΔHc(C) + 2ΔHc(H₂) − ΔHc(CH₄) = −94 + 2(−68) − (−213) = −94 − 136 + 213 = −17 kcal mol⁻¹. Options A, C and D are numerically incorrect.
Q4Grand Test
A closed insulated container contains a liquid stirred by a paddle and its temperature rises. Which thermodynamic statement is correct?
Not quite — the answer is C.
The container is insulated, so heat exchange q=0. The paddle does mechanical work on the system (w>0), so by the first law ΔU = q + w = 0 + w = w. Options A, B and D are each inconsistent with these conditions.
Q5Mixed Revision
Find the heat released when 35.0 g of CO₂ is formed from carbon by combustion. Given ΔHc(C)=−390 kJ mol⁻¹.
Not quite — the answer is C.
Molar mass of CO₂=44 g mol⁻¹. Moles=35.0/44=0.7955 mol. Heat released=0.7955×390≈310 kJ. Option B is the value for exactly 1 mol. Options A and D are numerically incorrect.
Q6Mixed Revision
At constant volume, 1 mol of gas is heated from 298 K to 308 K and 500 J of heat is supplied. Which relation is correct?
Not quite — the answer is B.
At constant volume ΔV=0, so w=−PextΔV=0. By the first law ΔU=q+w=500+0=500 J. Option A confuses adiabatic with constant-volume conditions. Option C assigns non-zero work. Option D incorrectly sets ΔU=0.
Q7Internal Energy and First Law of Thermodynamics
When heat is absorbed and work is done BY the system, the correct IUPAC signs are:
Not quite — the answer is B.
IUPAC: heat absorbed by system → q positive; work done BY system on surroundings → w negative (system loses energy as work). Both signs reflect the system's perspective.
Q8Internal Energy and First Law of Thermodynamics
At constant volume, which of the following is correct?
Not quite — the answer is D.
At constant volume, no PV work is done (w = 0), so first law gives ΔU = q. This heat at constant volume is denoted qv, measured directly by bomb calorimetry. Option A is only true when Δng = 0.
Q9Bond Enthalpy and Calculations
Which of the following correctly defines bond enthalpy?
Not quite — the answer is A.
Bond enthalpy is the energy required to break one mole of a specific type of bond in gaseous-phase molecules. It is always defined for the gaseous state, not liquid or solution. Option C is wrong because bond formation releases, not absorbs, energy.
Q10Bond Enthalpy and Calculations
Which of the following thermodynamic quantities is always positive?
Not quite — the answer is C.
Bond breaking always requires energy input, making bond enthalpy always endothermic and hence always positive. Enthalpy of formation can be negative (e.g., H₂O), combustion is always negative, and neutralization of strong acid-base is always negative (−57 kJ/mol).
Q11Entropy and Spontaneity
Which of the following is the most accurate definition of entropy?
Not quite — the answer is A.
Entropy (S) quantifies the degree of randomness or disorder; it is a thermodynamic state function. Heat content is enthalpy (H), not entropy. Internal energy is a separate state function.
Q12Entropy and Spontaneity
The correct mathematical expression for entropy change in a reversible process is:
Not quite — the answer is C.
For a reversible process, ΔS = qrev/T, where qrev is heat exchanged reversibly and T is absolute temperature. Option A inverts the relationship. Options B and D use ΔH and ΔG, which apply to different expressions.
527 more questions on this chapter are waiting in the app
Every one with a detailed explanation, plus flashcards and chapter tests.
Get RankUp on Google PlayQ13Gibbs Free Energy and Spontaneity
Which of the following is the correct definition of Gibbs free energy (G)?
Not quite — the answer is A.
Gibbs free energy is defined as G = H − TS, where H is enthalpy, T is absolute temperature, and S is entropy. Option C (G = U − TS) defines Helmholtz free energy (A), a common NEET-level confusion. Option D reverses the sign, giving a physically meaningless quantity.
Q14Gibbs Free Energy and Spontaneity
Which of the following correctly identifies the condition for a spontaneous process at constant T and P?
Not quite — the answer is B.
At constant T and P, a process is spontaneous when ΔG < 0 — the system loses free energy. ΔG = 0 represents equilibrium, not spontaneity. ΔG > 0 means the reverse process is spontaneous. ΔG = ΔH would only hold if TΔS = 0.
Q15Basic Concepts & Terminology
Which of the following correctly defines an isolated system?
Not quite — the answer is D.
An isolated system has no exchange of matter or energy with surroundings. Option B describes a closed system — the most common wrong choice here.
Q16Basic Concepts & Terminology
Which type of system allows exchange of energy but not matter?
Not quite — the answer is B.
A closed system permits energy exchange (heat/work) but not matter. Open systems allow both; isolated systems allow neither. Adiabatic is a process property, not a system type.
Q17Enthalpy and Enthalpy Changes
Which of the following correctly defines enthalpy (H)?
Not quite — the answer is A.
Enthalpy is defined as H = U + PV, where U is internal energy and PV is the pressure-volume term. This definition accounts for the energy stored in the system plus the work done by the system against external pressure. H = q + w is the first law expression for internal energy change, not enthalpy.
Q18Enthalpy and Enthalpy Changes
Under which condition is heat exchanged by a system equal to its enthalpy change?
Not quite — the answer is B.
At constant pressure (isobaric process), no PV work other than expansion work occurs, so qp = ΔH. At constant volume, qv = ΔU. Adiabatic means q = 0, so ΔH ≠ q. This is the thermodynamic basis of coffee-cup calorimetry.
Q19Work Heat and Sign Convention
Which of the following correctly defines heat in thermodynamics?
Not quite — the answer is B.
Heat is energy in transit between system and surroundings driven by a temperature gradient — it is not stored. Internal energy is stored; temperature measures average kinetic energy. All three are distinct concepts.
Q20Work Heat and Sign Convention
Which of the following correctly represents work done BY a gas during expansion against constant external pressure?
Not quite — the answer is D.
IUPAC: w = −PextΔV for work done BY the system. During expansion ΔV > 0, making w negative — the system loses energy as work. Option A gives the magnitude without sign, causing errors in ΔU calculations.
Q21Thermochemical Equations and Hess Law
Which of the following is a thermochemical equation?
Not quite — the answer is B.
A thermochemical equation must include: (1) a balanced equation, (2) physical state symbols for all species, and (3) the enthalpy change (ΔH) value. Options A, C, and D lack ΔH values and state symbols — they are ordinary chemical equations, not thermochemical equations.
Q22Thermochemical Equations and Hess Law
Which of the following must be specified in a thermochemical equation?
Not quite — the answer is B.
Physical states must be specified because enthalpy depends on state — H₂O(l) and H₂O(g) have different enthalpies differing by the enthalpy of vaporisation (≈44 kJ mol⁻¹). Changing a state symbol changes the ΔH value of the equation. Colour, time, and catalyst do not affect ΔH.
Q23Second Law of Thermodynamics
Which of the following is the most precise statement of the Second Law of Thermodynamics?
Not quite — the answer is B.
The Second Law states that for any spontaneous process, the total entropy of the universe (system + surroundings) increases. Option C is incorrect — entropy of the system alone can decrease (e.g., freezing of water) as long as entropy of surroundings increases more.
Q24Second Law of Thermodynamics
Which of the following correctly states the change in entropy of the universe for a spontaneous process?
Not quite — the answer is C.
For spontaneous processes, ΔS_universe = ΔS_system + ΔS_surroundings > 0. At equilibrium, ΔS_universe = 0. ΔS_universe < 0 is impossible for any real process (it would violate the Second Law). Option D is a trap — surroundings entropy must also be included.
ELITE question · AIR under 50 level
This chapter has 234 ELITE questions for students aiming at the very top. They are only in the app.
Unlock ELITE questions in the appKey Thermodynamic Relations
Quick revision: most questions in this chapter test these facts.
| Concept | Key Formula / Fact |
|---|---|
| First law | ΔU = q + w; for ideal gas isothermal expansion: ΔU = 0 |
| Enthalpy | ΔH = ΔU + ΔngRT; at constant pressure: qp = ΔH |
| Hess's law | ΔH is path-independent; sum of steps = overall enthalpy change |
| Entropy | ΔS = qrev/T; increases with disorder; S(gas) > S(liquid) > S(solid) |
| Gibbs free energy | ΔG = ΔH − TΔS; spontaneous if ΔG < 0 |
| Spontaneity conditions | ΔH < 0, ΔS > 0 → always spontaneous; ΔH > 0, ΔS < 0 → never spontaneous |
What the app covers in this chapter
551 questions in total, each with a detailed explanation.
| Important PYQ Mixed Concepts | 137 |
| Grand Test | 58 |
| Mixed Revision | 58 |
| Internal Energy and First Law of Thermodynamics | 40 |
| Bond Enthalpy and Calculations | 40 |
| Entropy and Spontaneity | 40 |
| Gibbs Free Energy and Spontaneity | 40 |
| Basic Concepts & Terminology | 39 |
| Enthalpy and Enthalpy Changes | 39 |
| Work Heat and Sign Convention | 20 |
| Thermochemical Equations and Hess Law | 20 |
| Second Law of Thermodynamics | 20 |
Questions students ask
Is Thermodynamics important for NEET?
Yes — it is a high-weightage Physical Chemistry chapter. Enthalpy calculations, Hess's law application and Gibbs free energy spontaneity prediction are tested every year.
Which topics should I revise first?
Focus on the first law with work calculations, enthalpy change types (formation, combustion, bond), Hess's law problems, and ΔG = ΔH − TΔS for predicting spontaneity at different temperatures.
How many questions from this chapter are on RankUp?
The RankUp app has 551 questions on Thermodynamics, including 234 ELITE questions. Every question has a detailed explanation.
