1 Line-by-Line Walkthrough
Read each line, peek at the "Easy idea" for a friendly version, then move to the tables.
1
Thermodynamics is the branch of science that deals with energy changes accompanying physical and chemical transformations. It does not deal with rate or mechanism — only with the energy bookkeeping.
Easy ideaIt's the accountant of energy — counts how much energy moves around but doesn't watch the clock or the journey.
2
A thermodynamic system is the part of the universe under study; the rest is the surroundings. The boundary separates them. Systems are open (exchange both matter and energy), closed (energy only), or isolated (neither).
Easy ideaOpen coffee mug = open system. Sealed flask = closed system. Perfectly insulated thermos = isolated.
3
State of a system is defined by macroscopic properties like p, V, T and n. A state function depends only on the current state, not on how the system got there. Examples: U, H, S, G, T, P, V.
Easy ideaLike your height — it doesn't matter what path you took to grow up. State function only cares about start and end.
4
Path functions depend on the route taken between two states. The main examples are heat (q) and work (w). Different paths between the same two states give different q and w.
Easy ideaLike the petrol you use on a trip — depends on whether you go via highway or scenic route.
5
Internal energy U is the total energy stored inside a system — sum of all molecular kinetic and potential energies. It is a state function. For an ideal gas, U depends only on T.
Easy ideaThe system's "energy bank balance". Only the current balance matters, not how money moved in and out.
6
First law of thermodynamics: energy can be neither created nor destroyed, only converted. Mathematically: ΔU = q + w. The change in internal energy equals heat added to the system plus work done on it.
Easy ideaEnergy is like money — it doesn't appear or vanish. If your balance changed, money must have moved in (q) or out (w).
7
Sign convention (IUPAC): q is positive if heat is absorbed BY system; negative if released. w is positive if work is done ON system; negative if done BY system. Compression ⇒ w > 0; expansion ⇒ w < 0.
Easy ideaEnergy flowing INTO the system is +. Energy flowing OUT is −.
8
Work of expansion against an external pressure: w = −pext · ΔV. For reversible isothermal expansion of an ideal gas: w = −nRT ln(V₂/V₁) — this gives maximum work.
Easy ideaTo squeeze the most work out of a gas, you do it gently (reversibly) so it pushes against the highest possible matching pressure all the way.
9
Enthalpy H is defined as H = U + pV. For a process at constant pressure, ΔH equals heat absorbed: qp = ΔH. For constant volume: qv = ΔU. Relationship between them for gases: ΔH = ΔU + Δng · RT.
Easy ideaAt constant pressure, the heat going in shows up as ΔH. The "pV" extra term accounts for the gas pushing out as it warms.
10
Heat capacity: C = q/ΔT. At constant volume: Cv = (∂U/∂T)V. At constant pressure: Cp = (∂H/∂T)p. For an ideal gas: Cp − Cv = R (Mayer's relation). For a monoatomic gas Cv = (3/2)R; diatomic ≈ (5/2)R.
Easy ideaHow much heat you need per degree of warming. Hotter at constant p needs a bit more because the gas also pushes outward.
11
Hess's law of constant heat summation: ΔH of a reaction is the same whether the reaction takes place in one step or several. This is a direct consequence of H being a state function. It allows calculation of ΔH for reactions that are difficult to measure directly.
Easy ideaJust like total altitude gain on a hike depends only on start and end, regardless of the trail you take.
12
Standard enthalpy changes: ΔH°_f (formation: 1 mole compound from elements in standard state), ΔH°_c (combustion in O₂), ΔH°_neut (acid + base → salt + water), ΔH°_sol (1 mole in large excess solvent). Standard state = 1 bar, specified T (usually 298 K).
Easy idea"Standard" means same agreed-upon conditions so everyone's tables match.
13
Bond enthalpy is the average energy required to break one mole of a particular bond in a gas-phase molecule. ΔH of reaction ≈ Σ BE(bonds broken) − Σ BE(bonds formed). Useful estimate when calorimetric data is unavailable.
Easy ideaBonds are like little springs. Total energy = energy to snap old springs − energy released when new springs form.
14
Spontaneous process is one that takes place by itself without outside help (e.g., water flowing downhill, heat flowing from hot to cold). Spontaneity is NOT about speed — only about direction.
Easy ideaSpontaneous = "wants to happen". A rusty nail is spontaneous (just slow). The reverse is non-spontaneous.
15
Entropy S is a measure of disorder or number of microscopic arrangements. Boltzmann: S = kB · ln W. Second law: for a spontaneous process, ΔSuniverse > 0. Gas > liquid > solid in entropy.
Easy ideaDisorder always tends to win in the universe overall. Tidy rooms become messy by themselves — that's the second law.
16
Gibbs free energy G = H − TS. At constant T and p, the criterion for spontaneity is ΔG < 0. At equilibrium ΔG = 0. Relationship to equilibrium constant: ΔG° = −RT ln K = −2.303 RT log K.
Easy ideaGibbs energy combines tendency to lower energy (ΔH) with tendency to spread out (TΔS). When both push the same way, the process is spontaneous.
17
Third law of thermodynamics: the entropy of a perfect crystalline substance is zero at absolute zero (0 K). This gives an absolute reference for entropy, unlike enthalpy (which only has relative values).
Easy ideaAt 0 K, atoms freeze into perfect order — only one arrangement possible — so disorder is exactly zero.
2 Concepts to know for NEET MCQs
| Concept | Key idea | NEET focus |
|---|---|---|
| System types | Open / closed / isolated | "Hot tea in open cup = ?" style MCQs |
| State vs path function | State: only start/end matter. Path: route matters | "Which is NOT a state function?" |
| Sign convention | q, w positive when heat/work flows INTO system (IUPAC) | Setting up first-law equations |
| First law ΔU = q + w | Energy conservation in any process | Calculating ΔU from q and w |
| Enthalpy H | H = U + pV; ΔH = qp | Heat of reaction at constant p |
| ΔH vs ΔU | ΔH = ΔU + Δng RT for gases | Common numerical pattern |
| Heat capacity | Cp − Cv = R for ideal gas; Cv depends on f | Adiabatic calculations |
| Hess's law | ΔH path-independent ⇒ can add/subtract steps | Indirect ΔH calculations |
| Bond enthalpy | ΔH ≈ Σ BE(broken) − Σ BE(formed) | Estimate ΔH from bond data |
| Spontaneity | ΔG < 0 at constant T, p | Predicting direction of change |
| Entropy | Measure of disorder; gas > liquid > solid | Sign of ΔS in given process |
| ΔG and K | ΔG° = −RT ln K | Connecting thermodynamics to equilibrium |
3 Formulas Bank
| Quantity / Process | Formula | Notes |
|---|---|---|
| First law | ΔU = q + w | IUPAC sign convention |
| Work (expansion/compression) | w = −pext · ΔV | Irreversible against constant pext |
| Reversible isothermal (ideal gas) | w = −nRT ln(V₂/V₁) = −nRT ln(p₁/p₂) | Maximum work |
| Heat at constant V | qv = ΔU = nCv ΔT | For ideal gas |
| Heat at constant p | qp = ΔH = nCp ΔT | For ideal gas |
| Enthalpy definition | H = U + pV | State function |
| ΔH vs ΔU for gases | ΔH = ΔU + Δng · R · T | Δng = mol gaseous products − mol gaseous reactants |
| Mayer's relation | Cp − Cv = R | Per mole, ideal gas |
| Adiabatic ideal gas | pV^γ = constant; TVγ−1 = constant | γ = Cp/Cv |
| Adiabatic work (ideal) | w = nCv(T₂ − T₁) = (p₁V₁ − p₂V₂)/(γ − 1) | q = 0 |
| Hess's law | ΔHoverall = Σ ΔHsteps | Path independent |
| Bond enthalpy | ΔHrxn = Σ BE(broken) − Σ BE(formed) | Estimate only |
| Entropy change (rev) | ΔS = qrev/T | For constant-T process |
| Phase change entropy | ΔSfus = ΔHfus/Tm; ΔSvap = ΔHvap/Tb | At constant T (mp or bp) |
| Boltzmann entropy | S = kB · ln W | W = microstates |
| Second law | ΔSuniv = ΔSsys + ΔSsurr ≥ 0 | > 0 for spontaneous |
| Gibbs energy | G = H − TS; ΔG = ΔH − TΔS | At constant T |
| Spontaneity (T, p) | ΔG < 0 ⇒ spontaneous; ΔG = 0 ⇒ equilibrium | The most important criterion |
| ΔG and K | ΔG° = −RT ln K = −2.303 RT log K | Connects thermo to equilibrium |
| ΔG and ΔG° | ΔG = ΔG° + RT ln Q | Q = reaction quotient |
| Carnot efficiency | η = 1 − Tcold/Thot | Best possible heat engine |
4 Facts you must remember
| Fact | Why it matters for NEET |
|---|---|
| For an ideal gas, U depends only on T | Isothermal ΔU = 0; free expansion ΔT = 0 |
| Standard state: 1 bar pressure, specified T (usually 298 K) | Definition trap in PYQs |
| ΔH°_f of element in standard state = 0 | Reference point for all formation enthalpies |
| Enthalpy of neutralisation of strong acid + strong base ≈ −57.1 kJ/mol | Standard NEET fact |
| Heat of fusion of ice = 6.01 kJ/mol; vaporisation of water = 40.66 kJ/mol | Phase-change PYQ fodder |
| For 1 mole gas: Cp − Cv = R | Used in nearly every adiabatic problem |
| Monoatomic: Cv = (3/2)R, γ = 5/3; Diatomic: Cv ≈ (5/2)R, γ = 7/5 | Common gas-property questions |
| For cyclic process: ΔU = ΔH = ΔG = ΔS = 0 | Identify path/state function quickly |
| Free expansion of ideal gas into vacuum: q = w = ΔU = 0 | Classic conceptual MCQ |
| Trouton's rule: ΔSvap ≈ 88 J/(K·mol) for normal liquids | Approximation question |
| Born-Haber cycle is used to find lattice enthalpy | Definition question |
| Carnot efficiency: η = 1 − Tc/Th; only depends on temperatures | Engine problems |
5 Controversial / Confusing Points
| Confusion | Clarification |
|---|---|
| "Sign convention for w" | IUPAC: w > 0 when work done ON system. Older convention: w > 0 when work done BY system. NEET uses IUPAC ⇒ wcompression > 0, wexpansion < 0. |
| "ΔH vs ΔU — when are they equal?" | For reactions with no change in moles of gas (Δng = 0). Also for purely solid/liquid reactions. |
| "Is q a state function?" | No. q and w are path functions. But qp = ΔH and qv = ΔU are special cases where heat equals a state function. |
| "Spontaneous = fast?" | No! Spontaneity says nothing about rate. Diamond → graphite is spontaneous but unimaginably slow. |
| "ΔG < 0 always means it will happen?" | Only that it's thermodynamically allowed. Activation energy may make it kinetically frozen. |
| "ΔSsystem can be negative even for spontaneous process?" | Yes — as long as ΔSuniverse (= ΔSsys + ΔSsurr) is positive. Water freezing below 0 °C: ΔSsys < 0 but ΔSsurr more positive. |
| "ΔG° vs ΔG" | ΔG° is under standard conditions (1 bar, 298 K). ΔG is under actual conditions. Related by ΔG = ΔG° + RT ln Q. |
| "At equilibrium ΔG = 0 but ΔG° ≠ 0" | ΔG = 0 at the actual equilibrium state. ΔG° = −RT ln K can be positive, negative or zero depending on K. |
6 Assumptions in this chapter
| Assumption | When it's invoked |
|---|---|
| Gas behaves ideally (pV = nRT) | All ideal-gas calculations (work, ΔU, ΔH) |
| For an ideal gas, U depends only on T | Used to set ΔU = 0 in isothermal processes |
| Reversible processes are infinitely slow (quasi-static) | Maximum-work derivations |
| Adiabatic means perfectly insulated (q = 0) | All adiabatic equations |
| Standard state is at 1 bar pressure (and specified T) | All standard ΔH°, ΔG° values |
| Bond enthalpies are averages over many molecules | Bond-enthalpy estimates of ΔH |
| Heat capacities Cp, Cv are independent of T | Simple integration of heat capacity |
| Perfect crystal exists at 0 K (third law) | Absolute entropy reference |
| Surroundings are effectively infinite at constant T | ΔSsurr = −qsys/T well-defined |
7 Exceptions to remember
| General rule | Exception |
|---|---|
| ΔH and ΔU are equal | Only when Δng = 0 (no change in moles of gas). Otherwise they differ by Δng RT. |
| For real gases, U = U(T) only | False for real gases — U depends on V too (only ideal gases have U = U(T)). |
| Free expansion is always isothermal | True for ideal gas only. Real gases cool slightly (Joule-Thomson effect). |
| Joule-Thomson coefficient is non-zero | Zero for ideal gas — H depends only on T, so no cooling on throttling. |
| Bond-enthalpy ΔH equals true reaction ΔH | Only an average estimate; actual values can deviate by 10–20 kJ/mol. |
| Trouton's rule applies to all liquids | Fails for H-bonded liquids (water, ethanol) due to abnormal structure in liquid. |
| For elements ΔH°_f = 0 | Only for standard state. Diamond ΔH°_f ≠ 0 because graphite is the standard form of carbon. |
| ΔS positive ⇒ spontaneous | True only when also considering ΔSuniverse. ΔSsystem alone is insufficient. |
| Third law: S = 0 at 0 K | Holds only for perfect crystals. Glassy/amorphous solids have residual entropy at 0 K. |
8 Scientists and years
| Scientist | Year | Contribution |
|---|---|---|
| Germain Henri Hess | 1840 | Hess's law of constant heat summation |
| James Prescott Joule | 1843–1850 | Mechanical equivalent of heat; conservation of energy |
| Rudolf Clausius | 1850, 1865 | Formulated first and second laws; coined the word entropy |
| William Thomson (Lord Kelvin) | 1851 | Absolute temperature scale; alternative statement of second law |
| Sadi Carnot | 1824 | Theoretical maximum efficiency of heat engines (Carnot cycle) |
| Hermann von Helmholtz | 1882 | Helmholtz free energy; A = U − TS |
| Josiah Willard Gibbs | 1873–1878 | Gibbs free energy G = H − TS; criterion for spontaneity |
| Ludwig Boltzmann | 1877 | Statistical definition of entropy: S = kB ln W |
| Walther Nernst | 1906 | Third law of thermodynamics (Nernst heat theorem) |
| James Joule & William Thomson | 1852 | Joule-Thomson effect (real-gas cooling on expansion) |
9 NEET traps to avoid
| Trap | How to avoid it |
|---|---|
| Mixing up sign conventions for w | NEET uses IUPAC. Heat IN, work IN ⇒ positive. wexpansion is negative. |
| Forgetting Δng in ΔH−ΔU | Always check whether moles of gas change. If only solids/liquids: ΔH ≈ ΔU. |
| Claiming ΔSsys > 0 ⇒ spontaneous | You need ΔSuniverse > 0. Or at const T, p use ΔG < 0. |
| Calling ΔG° = 0 at equilibrium | ΔG = 0 at equilibrium. ΔG° = −RT ln K which is generally non-zero. |
| Heat of formation = heat of combustion? | True only for H₂O(l) from H₂(g) and ½O₂(g). Not general. |
| "Spontaneous = fast" misconception | Spontaneity says nothing about rate. ΔG < 0 only means thermodynamically allowed. |
| Cyclic process gives non-zero ΔH | Cyclic ⇒ ΔX = 0 for any state function X. Heat and work can still be non-zero. |
| Adiabatic always means temperature constant | No — adiabatic means no heat exchange. Temperature usually changes. |
| Free expansion of ideal gas reduces T | For ideal gas, T is unchanged in free expansion (Joule expansion). |
| Negative ΔG means K < 1 | Opposite: ΔG° < 0 ⇒ K > 1 (products favoured). |
📊 Diagrams
a) Types of thermodynamic systems
Open exchanges matter and energy; closed exchanges only energy; isolated exchanges neither.
b) First law and sign convention
Energy flowing INTO the system (q, w) is taken as positive (IUPAC convention).
c) p–V curves for key processes
Isothermal (solid) is shallower; adiabatic (dashed) is steeper since γ > 1.
d) Enthalpy diagram for an exothermic reaction
Products sit lower in enthalpy than reactants ⇒ ΔH negative.
e) Spontaneity zones (ΔG = ΔH − TΔS)
Four quadrants depending on signs of ΔH and ΔS. Two are temperature-dependent.
f) Hess's law cycle
Direct ΔH equals the sum of step-wise ΔH values along any alternate path.
📝 50 PYQs — 25 NEET + 25 JEE Main
Each card shows a question-specific Easy idea, then Given / To find / Formula / Solution / Answer.
Q1NEET 2019For the process at 298 K, the value of ΔSsurroundings (in J/K) is
A+200
B−200
C+298
D0
Easy idea
Heat that flows OUT of the system flows INTO the surroundings, so the surroundings' entropy goes up.
Given
ΔHsystem = −59.6 kJ released to surroundings at T = 298 K.
To find
ΔSsurroundings in J/K.
Formula
ΔSsurr = −ΔHsystem / T
Solution
ΔSsurr = +59600 / 298 ≈ +200 J/K. Positive because heat flows out of system into surroundings.
Answer
(A)
Q2NEET 2020For an ideal gas undergoing isothermal expansion, ΔU is
A+ve
B−ve
CZero
DEqual to w
Easy idea
Internal energy of an ideal gas depends only on temperature. Isothermal means constant T, so U doesn't change.
Given
Ideal gas; isothermal process (ΔT = 0).
To find
Change in internal energy ΔU.
Formula
Uideal = f(T) only; ΔU = n·Cv·ΔT
Solution
ΔT = 0 ⇒ ΔU = 0. All heat absorbed equals work done (q = -w).
Answer
(C)
Q3NEET 2018The first law of thermodynamics states
Aenergy is conserved
Bentropy of universe always increases
Centropy of pure crystal at 0 K is zero
Dheat is path independent
Easy idea
First law = energy bookkeeping. Energy can move around (heat, work) but the total never changes.
Given
Statement identification.
To find
Identify the first law.
Formula
ΔU = q + w
Solution
First law = law of conservation of energy. The change in internal energy equals heat added plus work done on the system.
Answer
(A)
Q4NEET 2021Which of the following is NOT a state function?
AInternal energy U
BEnthalpy H
CWork w
DEntropy S
Easy idea
State functions only care about start and end. Work and heat depend on HOW you get there (the path).
Given
Definition of state vs path function.
To find
Pick the path function.
Formula
State function: ΔX depends only on initial & final states.
Solution
Work (and heat) depend on the path. U, H, S, G, T, P, V are all state functions.
Answer
(C)
Q5NEET 2017For an exothermic reaction, ΔH is
A+ve
B−ve
CZero
DCannot be predicted
Easy idea
Exothermic = heat coming OUT. The system loses energy, so its enthalpy goes down.
Given
Exothermic reaction.
To find
Sign of ΔH.
Formula
ΔH = Hproducts − Hreactants; exothermic ⇒ products less energy than reactants
Solution
ΔH is negative (system loses heat to surroundings).
Answer
(B)
Q6NEET 2022Cp − Cv for one mole of an ideal gas is
AR
B2R
CR/2
D0
Easy idea
When you heat at constant pressure, the gas also pushes its surroundings out (does work). That extra work shows up as the difference Cp − Cv.
Given
1 mole ideal gas.
To find
Difference Cp − Cv.
Formula
Mayer's relation: Cp − Cv = R (for ideal gas, 1 mole)
Solution
Cp − Cv = R = 8.314 J/(K·mol).
Answer
(A)
Q7NEET 2018For the reaction N2(g) + 3H2(g) → 2NH3(g) at 298 K, ΔH − ΔU equals
A−RT
B+RT
C−2RT
D+2RT
Easy idea
Going from 4 moles of gas to 2 moles drops Δn by 2. Since ΔH = ΔU + Δn·RT, and Δn = −2, ΔH is less than ΔU by 2RT.
Given
Δn(gas) = 2 − (1 + 3) = −2; T = 298 K.
To find
Value of ΔH − ΔU.
Formula
ΔH = ΔU + Δn·R·T
Solution
ΔH − ΔU = Δn·R·T = (−2)·R·T = −2RT.
Answer
(C)
Q8NEET 2020Which has the highest entropy?
ASolid H2O
BLiquid H2O
CGaseous H2O
DAll equal
Easy idea
Gas molecules move freely all over — that's a lot of disorder. Solid is very ordered, liquid in between.
Given
Three phases of water.
To find
Phase with highest entropy.
Formula
Entropy: gas > liquid > solid (more disorder ⇒ more entropy)
Solution
Gaseous water has the highest entropy due to high molecular randomness.
Answer
(C)
Q9NEET 2019A process will be spontaneous at constant T and p if
AΔH < 0 only
BΔS > 0 only
CΔG < 0
DΔG > 0
Easy idea
Spontaneous = wants to happen on its own. The rule at constant T and p is: Gibbs energy must drop.
Given
Constant T and p.
To find
Criterion for spontaneity.
Formula
ΔG = ΔH − TΔS; spontaneous when ΔG < 0
Solution
A process is spontaneous at constant T, p when ΔG < 0.
Answer
(C)
Q10NEET 2021The enthalpy change for the reaction H2(g) + ½O2(g) → H2O(l) is called
AEnthalpy of formation
BEnthalpy of combustion
CEnthalpy of neutralisation
DBoth (a) and (b)
Easy idea
Making H2O from elements is formation. Burning H2 in O2 is also combustion. Here both definitions apply because both reactants are elements in standard states.
Given
Equation: H2(g) + ½O2(g) → H2O(l).
To find
Type of enthalpy.
Formula
Formation: 1 mole product from elements in standard states; Combustion: complete oxidation
Solution
It is BOTH the standard enthalpy of formation of H2O AND the standard enthalpy of combustion of H2.
Answer
(D)
Q11NEET 2017For the reversible isothermal expansion of 1 mole of an ideal gas from V1 to V2 at temperature T, the work done by the gas is
A−RT ln(V2/V1)
B+RT ln(V2/V1)
CRT(V2 − V1)
D0
Easy idea
For maximum work in an isothermal expansion, the gas pushes against the highest possible opposing pressure at every instant — that's reversible expansion.
Given
1 mole ideal gas; isothermal reversible expansion from V1 to V2.
To find
Work done BY the gas.
Formula
wby = RT ln(V2/V1)
Solution
Work done by gas = nRT ln(V2/V1) = RT ln(V2/V1) for 1 mole. Maximum possible work.
Answer
(B)
Q12NEET 2022For 1 mole of an ideal monoatomic gas, Cv is
A(3/2)R
B(5/2)R
CR
D2R
Easy idea
Monoatomic gas only has translation (3 directions). Each direction contributes (1/2)R to Cv, so 3 × (1/2)R = (3/2)R.
Given
Monoatomic ideal gas; 1 mole.
To find
Heat capacity at constant volume Cv.
Formula
Cv = (degrees of freedom / 2) × R; monoatomic has f = 3
Solution
Cv = (3/2)R = 12.47 J/(K·mol).
Answer
(A)
Q13NEET 2018Bond dissociation enthalpy of H−H bond is 436 kJ/mol. The enthalpy of formation of 1 mole of H atoms is
A+218 kJ/mol
B−218 kJ/mol
C+436 kJ/mol
D−436 kJ/mol
Easy idea
Breaking ½H2 into one H atom takes half the bond energy. Breaking absorbs energy, so ΔH is positive.
Given
Bond enthalpy of H−H = 436 kJ/mol.
To find
ΔHf of 1 mole H atoms.
Formula
½H2(g) → H(g); ΔH = ½ × BDE(H−H)
Solution
ΔHf (H atom) = +436/2 = +218 kJ/mol.
Answer
(A)
Q14NEET 2020ΔS for a cyclic process is
A+ve
B−ve
CZero
DAlways equal to ΔH
Easy idea
Going back to where you started means all state functions return to their original value. Net change is zero.
Given
Cyclic process (returns to initial state).
To find
Total change in entropy of the system.
Formula
ΔXcycle = 0 for any state function X.
Solution
S is a state function; in a cyclic process the system returns to its initial state ⇒ ΔS = 0.
Answer
(C)
Q15NEET 2019Which process is irreversible?
ASudden expansion of gas into vacuum
BVery slow compression
CQuasi-static heating
DReversible isothermal expansion
Easy idea
Reversible = infinitely slow, near equilibrium. Sudden / quick = irreversible. Expansion into vacuum is the classic irreversible example (free expansion).
Given
Four processes listed.
To find
Identify the irreversible process.
Formula
Reversible ⇒ infinitesimal driving force at every instant.
Solution
Sudden free expansion into vacuum cannot be reversed by an infinitesimal change ⇒ irreversible.
Answer
(A)
Q16NEET 2017The enthalpy of neutralisation of a strong acid with a strong base is approximately
A−13.7 kcal/mol
B−57.1 kJ/mol
C+57.1 kJ/mol
DBoth (a) and (b)
Easy idea
When strong acid meets strong base, the only chemical change is H+ + OH- → H2O. That reaction is the same regardless of which acid/base, so the heat released is fixed.
Given
Strong acid + strong base.
To find
Enthalpy of neutralisation.
Formula
Net: H+(aq) + OH-(aq) → H2O(l); ΔH ≈ −57.1 kJ/mol ≈ −13.7 kcal/mol
Solution
Standard value is −57.1 kJ/mol = −13.7 kcal/mol. Both options describe the same value.
Answer
(D)
Q17NEET 2021Boltzmann's entropy formula is
AS = k·ln W
BS = k/W
CS = k·W
DS = k·ln(1/W)
Easy idea
Entropy is a measure of disorder. The more ways (W) a system can be arranged, the higher its entropy. Boltzmann found S grows with the LOG of W.
Given
Boltzmann's statistical mechanics.
To find
Boltzmann entropy formula.
Formula
S = kB · ln W; kB = 1.38×10-23 J/K, W = microstates
Solution
S = kB ln W. Found on Boltzmann's grave in Vienna.
Answer
(A)
Q18NEET 2022For an ideal gas, the work done in free expansion into vacuum is
A+ve
B−ve
CZero
DEqual to ΔH
Easy idea
Pushing against NOTHING means doing no work. External pressure is zero, so w = 0.
Given
Ideal gas; free expansion into vacuum.
To find
Work done by gas.
Formula
w = −pext · ΔV; pext = 0 for vacuum
Solution
pext = 0 ⇒ w = 0. Also q = 0 (insulated) ⇒ ΔU = 0 (isothermal for ideal gas).
Answer
(C)
Q19NEET 2018ΔG° for a reaction at equilibrium is
A+ve
B−ve
CZero
DDepends on K
Easy idea
At equilibrium, no further net change is happening, and the Gibbs energy has hit its minimum. So ΔG° relates to K through ΔG° = −RT ln K, but ΔG at the actual equilibrium state is zero.
Given
Reaction at equilibrium.
To find
Sign of ΔG at equilibrium.
Formula
ΔG = ΔG° + RT ln Q; at equilibrium Q = K so ΔG = 0
Solution
ΔG = 0 at equilibrium. (ΔG° is not zero, ΔG = 0.)
Answer
(C)
Q20NEET 2020The standard enthalpy of formation of an element in its standard state is
APositive
BNegative
CZero
DDepends on element
Easy idea
Forming an element from itself doesn't change anything, so the enthalpy change is zero — by definition.
Given
Element in standard state.
To find
Standard enthalpy of formation.
Formula
ΔH°_f (element in standard state) = 0
Solution
By convention, ΔH°_f of an element in its standard state is exactly zero.
Answer
(C)
Q21NEET 2017Which of the following is an intensive property?
AInternal energy
BEnthalpy
CVolume
DTemperature
Easy idea
Intensive = doesn't change when you take more. Cut a hot iron bar in half — each half is still the same temperature.
Given
Property classification.
To find
Identify intensive property.
Formula
Intensive: independent of amount of substance
Solution
Temperature is intensive. U, H, V are all extensive (depend on amount).
Answer
(D)
Q22NEET 2021The bond enthalpy of N≡N is 941 kJ/mol. The energy needed to break 0.5 mole of N2 is
A235.5 kJ
B470.5 kJ
C941 kJ
D1882 kJ
Easy idea
1 mole costs 941 kJ. Half a mole costs half of that.
Given
Bond enthalpy of N≡N = 941 kJ/mol; 0.5 mole.
To find
Energy needed.
Formula
E = n × Bond enthalpy
Solution
E = 0.5 × 941 = 470.5 kJ.
Answer
(B)
Q23NEET 2019The work done in an isothermal reversible expansion of 2 mole of an ideal gas at 300 K from 1 L to 10 L is
A−9.97 kJ
B−11.5 kJ
C+11.5 kJ
D+9.97 kJ
Easy idea
Reversible isothermal expansion is the textbook log formula. The gas does work ON surroundings, so w (on system) is negative.
Given
n = 2 mol, T = 300 K, V1 = 1 L, V2 = 10 L, R = 8.314 J/(K·mol).
To find
Work done w.
Formula
w = −nRT ln(V2/V1)
Solution
w = −2 × 8.314 × 300 × ln 10 = −4988.4 × 2.303 = −11488 J ≈ −11.5 kJ.
Answer
(B)
Q24NEET 2022For the reaction A → B, ΔH = +20 kJ/mol and ΔS = +50 J/(K·mol). The temperature above which this reaction becomes spontaneous is
A100 K
B400 K
C200 K
D250 K
Easy idea
ΔG = ΔH − TΔS. For the reaction to be spontaneous, ΔG must be negative. Set ΔG = 0 to find the crossover temperature.
Given
ΔH = +20000 J/mol; ΔS = +50 J/(K·mol).
To find
Temperature above which spontaneous (ΔG < 0).
Formula
ΔG < 0 ⇒ T > ΔH/ΔS
Solution
Tcross = 20000/50 = 400 K. Above 400 K the reaction becomes spontaneous.
Answer
(B)
Q25NEET 2018Third law of thermodynamics states that
Aentropy of any pure crystal at absolute zero is zero
Benergy is conserved
Centropy of universe always increases
DΔG = 0 at equilibrium
Easy idea
At absolute zero, all atoms in a perfect crystal sit still and orderly — only one arrangement is possible, so disorder (entropy) is zero.
Given
Statement identification.
To find
Identify the third law.
Formula
Third law: limT→0 S(crystal) = 0
Solution
Entropy of any pure perfect crystal is zero at absolute zero (0 K).
Answer
(A)
Q26JEE Main 2020For an adiabatic process,
Aq = 0
BΔU = 0
Cw = 0
DΔH = 0
Easy idea
Adiabatic means perfectly insulated — no heat in, no heat out. So q is the one that's zero.
Given
Adiabatic process definition.
To find
Identify what equals zero.
Formula
Adiabatic: no heat exchange with surroundings
Solution
q = 0 (by definition of adiabatic).
Answer
(A)
Q27JEE Main 2019The internal energy of 1 mole of an ideal monoatomic gas at 300 K is
A1247 J
B3742 J
C6235 J
D4988 J
Easy idea
Internal energy of an ideal gas only depends on temperature. For a monoatomic gas, U = (3/2)RT per mole.
Given
1 mole monoatomic ideal gas, T = 300 K, R = 8.314 J/(K·mol).
To find
Internal energy U.
Formula
U = (3/2)nRT for monoatomic ideal gas
Solution
U = 1.5 × 1 × 8.314 × 300 = 3741.3 ≈ 3742 J.
Answer
(B)
Q28JEE Main 2021For a reversible adiabatic expansion of an ideal gas, the relation is
ApV = constant
BpV^γ = constant
Cp/V = constant
Dp^γ V = constant
Easy idea
Reversible adiabatic = no heat exchange + done very slowly. The pressure-volume relation includes γ (ratio of heat capacities).
Given
Reversible adiabatic process.
To find
Equation of state.
Formula
Adiabatic equation: pV^γ = const where γ = Cp/Cv
Solution
pV^γ = constant for reversible adiabatic process of ideal gas.
Answer
(B)
Q29JEE Main 2018Hess's law states that
AΔH of a reaction is independent of path
BΔG = 0 at equilibrium
Centropy is path dependent
Denergy is conserved
Easy idea
Hess's Law is a direct consequence of enthalpy being a state function — the change depends only on start and end, not the route.
Given
Statement identification.
To find
What does Hess's law state?
Formula
Hess: ΔHtotal = Σ ΔHsteps regardless of path
Solution
The enthalpy change of a reaction is independent of the path (number of steps) taken.
Answer
(A)
Q30JEE Main 2022The work done by 1 mole of an ideal gas in an isothermal expansion from 10 atm to 1 atm at 298 K against constant external pressure of 1 atm is
A−2.227 kJ
B−5.74 kJ
C+2.227 kJ
D−22.27 kJ
Easy idea
Constant external pressure expansion is irreversible. Work = -pext × ΔV. Find final volume from p2V2 = nRT, initial from p1V1 = nRT.
Given
n = 1, T = 298, pext = 1 atm, gas goes from 10 atm to 1 atm.
To find
Work w done by gas (against constant pext).
Formula
w = −pext(V2 − V1); Vi = nRT/pi
Solution
V1 = 1·0.0821·298/10 = 2.45 L; V2 = 24.5 L. w = −1 atm × (24.5 − 2.45) L = −22.05 L·atm × 101.3 J/(L·atm) = −2234 J ≈ −2.227 kJ.
Answer
(A)
Q31JEE Main 2019ΔG° for the reaction A → B at 298 K with Keq = 10 is
A−5.71 kJ/mol
B+5.71 kJ/mol
C−2.30 kJ/mol
D−11.42 kJ/mol
Easy idea
Larger K means more product at equilibrium, which means the reaction strongly favors products, so ΔG° is negative.
Given
T = 298 K, K = 10, R = 8.314 J/(K·mol).
To find
ΔG°.
Formula
ΔG° = −RT ln K = −2.303 RT log K
Solution
ΔG° = −2.303 × 8.314 × 298 × log 10 = −5705 J ≈ −5.71 kJ/mol.
Answer
(A)
Q32JEE Main 2020Standard enthalpy of combustion of carbon is −393.5 kJ/mol and of CO is −283.0 kJ/mol. Standard enthalpy of formation of CO is
A−110.5 kJ/mol
B+110.5 kJ/mol
C−676.5 kJ/mol
D+676.5 kJ/mol
Easy idea
C + O2 → CO2 gives −393.5. CO + ½O2 → CO2 gives −283.0. Subtract to find C + ½O2 → CO.
Given
ΔHcomb(C) = −393.5 kJ/mol; ΔHcomb(CO) = −283.0 kJ/mol.
To find
ΔHf(CO).
Formula
Use Hess's law: C + ½O2 → CO has ΔH = ΔHcomb(C) − ΔHcomb(CO)
Solution
ΔHf(CO) = −393.5 − (−283.0) = −110.5 kJ/mol.
Answer
(A)
Q33JEE Main 2018For an ideal gas in adiabatic free expansion,
AΔU = 0
Bq = +ve
Cw = +ve
DΔH = +ve
Easy idea
Free expansion: no work (vacuum), no heat (adiabatic). So ΔU = 0. For ideal gas, T also doesn't change.
Given
Adiabatic free expansion of ideal gas.
To find
What is zero?
Formula
q = 0 (adiabatic); w = 0 (no external pressure)
Solution
ΔU = q + w = 0 + 0 = 0. T constant for ideal gas.
Answer
(A)
Q34JEE Main 2021If at 298 K, ΔG° = −5.7 kJ/mol for A ⇌ B, the equilibrium constant K is approximately
A10
B100
C1
D0.1
Easy idea
ΔG° = −5.7 kJ/mol ≈ −2.303 RT × 1, which means log K = 1, so K = 10.
Given
ΔG° = −5700 J/mol; T = 298 K.
To find
Equilibrium constant K.
Formula
ΔG° = −RT ln K ⇒ log K = −ΔG°/(2.303RT)
Solution
log K = 5700/(2.303 × 8.314 × 298) = 5700/5705 ≈ 1; K ≈ 10.
Answer
(A)
Q35JEE Main 2019Heat of combustion of methane is −890 kJ/mol. The heat released on combustion of 16 g of methane is
A890 kJ
B445 kJ
C1780 kJ
D222.5 kJ
Easy idea
16 g of methane is exactly 1 mole (since molar mass = 16). So heat released = heat per mole.
Given
ΔHcomb(CH4) = −890 kJ/mol; mass = 16 g; M(CH4) = 16 g/mol.
To find
Heat released.
Formula
Heat = n × |ΔHcomb|; n = mass / M
Solution
n = 16/16 = 1 mol. Heat = 1 × 890 = 890 kJ released.
Answer
(A)
Q36JEE Main 2022The Carnot cycle efficiency η of an engine working between 500 K and 300 K is
A40%
B60%
C20%
D100%
Easy idea
Carnot efficiency only depends on the two temperatures. The bigger the gap, the more efficient.
Given
Thot = 500 K, Tcold = 300 K.
To find
Efficiency η.
Formula
η = 1 − Tcold/Thot
Solution
η = 1 − 300/500 = 0.4 = 40%.
Answer
(A)
Q37JEE Main 2020Which has zero standard enthalpy of formation?
AO2(g)
BH2O(l)
CCO2(g)
DNH3(g)
Easy idea
Elements in their standard state have ΔH°_f = 0 by definition. Oxygen exists as O2(g) at standard state.
Given
Standard state of various species.
To find
Species with ΔH°_f = 0.
Formula
ΔH°_f of an element in its standard state = 0
Solution
O2(g) is the standard state of oxygen ⇒ ΔH°_f = 0. The others are compounds.
Answer
(A)
Q38JEE Main 2018Bond enthalpies (kJ/mol): C−H = 414, H−H = 436, C−C = 347. The ΔH for the reaction CH3−CH3 → CH3· + CH3· is
A+347 kJ
B−347 kJ
C+414 kJ
D+693 kJ
Easy idea
Breaking one C−C bond requires exactly its bond enthalpy.
Given
C−C bond enthalpy = 347 kJ/mol; reaction breaks one C−C bond.
To find
ΔH of the reaction.
Formula
ΔH = Σ (BE bonds broken) − Σ (BE bonds formed)
Solution
Only one C−C bond is broken, no bonds formed. ΔH = +347 kJ/mol.
Answer
(A)
Q39JEE Main 2021Enthalpy of vaporisation of water is 40.66 kJ/mol at 373 K. The entropy change is
A+109 J/(K·mol)
B−109 J/(K·mol)
C+218 J/(K·mol)
D+40.66 J/(K·mol)
Easy idea
Phase change at constant T means ΔS = ΔH/T. Vapor has higher entropy than liquid, so ΔS positive.
Given
ΔHvap = 40660 J/mol; T = 373 K.
To find
ΔSvap.
Formula
ΔS = ΔH / T at constant T
Solution
ΔS = 40660 / 373 = 109 J/(K·mol). Positive: liquid → vapour has more disorder.
Answer
(A)
Q40JEE Main 2019The relation between ΔG and ΔG° is
AΔG = ΔG° + RT ln Q
BΔG = ΔG° − RT ln Q
CΔG = ΔG° + ln Q
DΔG = ΔG° − ln Q
Easy idea
ΔG° is fixed (under standard conditions). When you're NOT at standard, you add an extra term RT ln Q to get the actual ΔG.
Given
Relationship between ΔG and ΔG°.
To find
Correct relation.
Formula
Standard derivation from chemical potential
Solution
ΔG = ΔG° + RT ln Q. At equilibrium Q = K and ΔG = 0, giving ΔG° = −RT ln K.
Answer
(A)
Q41JEE Main 2022The molar entropy change for fusion of ice at 273 K is 22 J/(K·mol). The molar enthalpy of fusion is
A6.006 kJ/mol
B2.273 kJ/mol
C22 kJ/mol
D273 kJ/mol
Easy idea
Phase change at constant T: ΔH = T × ΔS.
Given
ΔSfus = 22 J/(K·mol); T = 273 K.
To find
Molar ΔHfus.
Formula
ΔH = T × ΔS at constant T (phase change)
Solution
ΔH = 273 × 22 = 6006 J/mol = 6.006 kJ/mol. Close to actual value 6.01 kJ/mol.
Answer
(A)
Q42JEE Main 2020For a process to be spontaneous at all temperatures, the requirements are
AΔH > 0, ΔS > 0
BΔH < 0, ΔS > 0
CΔH < 0, ΔS < 0
DΔH > 0, ΔS < 0
Easy idea
ΔG = ΔH − TΔS. If ΔH is negative AND ΔS is positive, then ΔG is always negative regardless of T.
Given
Spontaneity at all T.
To find
Conditions for always-spontaneous.
Formula
ΔG < 0 must hold for all T
Solution
ΔH < 0 and ΔS > 0. Then ΔG = (−) − T(+) = always negative.
Answer
(B)
Q43JEE Main 2018Joule-Thomson coefficient μ = (∂T/∂p)H. For an ideal gas this equals
A0
BCp
CCv
DCp/Cv
Easy idea
For an ideal gas, enthalpy depends only on temperature. So changing pressure at constant enthalpy doesn't change T.
Given
Ideal gas.
To find
Joule-Thomson coefficient μ.
Formula
μ = (∂T/∂p)H = 0 for ideal gas (H = H(T))
Solution
For an ideal gas H depends only on T, so μ = 0 (no Joule-Thomson cooling).
Answer
(A)
Q44JEE Main 20211 mole of ideal gas expands isothermally and reversibly from 22.4 L to 224 L at 273 K. Heat absorbed q is
A+5226 J
B−5226 J
C+2613 J
D0
Easy idea
Isothermal for ideal gas: ΔU = 0, so all heat absorbed equals work done by gas. Expansion does positive work, so q must be positive.
Given
n = 1, T = 273, V1 = 22.4 L, V2 = 224 L, reversible isothermal.
To find
Heat absorbed q.
Formula
Isothermal ideal gas: ΔU = 0, q = −w = nRT ln(V2/V1)
Solution
q = 1 × 8.314 × 273 × ln 10 = 2269 × 2.303 = 5226 J. Positive (heat absorbed).
Answer
(A)
Q45JEE Main 2019For a perfect gas in adiabatic reversible process, work done is
AnCv(T2 − T1)
BnCv(T1 − T2)
CnR(T2 − T1)
DnCp(T2 − T1)
Easy idea
For adiabatic q = 0, so ΔU = w. And ΔU = nCv ΔT for any ideal gas process.
Given
Adiabatic reversible process of ideal gas.
To find
Work done on the system.
Formula
Adiabatic: q = 0 ⇒ ΔU = w; ΔU = nCv ΔT
Solution
w = ΔU = nCv(T2 − T1). If expansion (T2<T1), w negative.
Answer
(A)
Q46JEE Main 2022Born-Haber cycle is used to determine
ALattice enthalpy
BBond enthalpy
CCombustion enthalpy
DHydration enthalpy
Easy idea
Lattice energies can't be measured directly, so we use Hess's law to find them via a thermodynamic cycle.
Given
Born-Haber cycle application.
To find
What does it determine?
Formula
Cycle of: sublimation + IE + EA + ½ bond + ΔHf = lattice enthalpy
Solution
Used to find lattice enthalpy of ionic compounds indirectly using known steps.
Answer
(A)
Q47JEE Main 2020For the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g), at constant T and P, ΔS will be
A+ve
B−ve
CZero
DCannot be predicted
Easy idea
1 mole of gas becomes 2 moles of gas. More gas particles means more disorder.
Given
Reaction: 1 mole gas → 2 moles gas.
To find
Sign of ΔS.
Formula
Δn(gas) > 0 ⇒ ΔS > 0 typically
Solution
Increase in moles of gas (Δn = +1) ⇒ ΔS positive.
Answer
(A)
Q48JEE Main 2018Trouton's rule states that ΔSvap is approximately
A88 J/(K·mol)
B22 J/(K·mol)
C10 J/(K·mol)
D200 J/(K·mol)
Easy idea
Most liquids have about the same change in disorder when going to gas — Trouton spotted this regularity.
Given
Trouton's empirical rule.
To find
Approximate value of ΔSvap.
Formula
ΔSvap ≈ 88 J/(K·mol) for normal liquids
Solution
Trouton: ΔSvap ≈ 88 J/(K·mol). Water and H-bonded liquids deviate.
Answer
(A)
Q49JEE Main 2021Δ_r G° for the cell reaction Zn + Cu2+ → Zn2+ + Cu with E°_cell = 1.10 V (n = 2) is
A−212.3 kJ/mol
B+212.3 kJ/mol
C−106.15 kJ/mol
D−424.6 kJ/mol
Easy idea
Cell reaction connects to free energy via ΔG° = −nFE°.
Given
n = 2; F = 96500 C/mol; E°_cell = 1.10 V.
To find
Δ_r G° in kJ/mol.
Formula
Δ_r G° = −n·F·E°_cell
Solution
Δ_r G° = −2 × 96500 × 1.10 = −212300 J = −212.3 kJ/mol.
Answer
(A)
Q50JEE Main 2019The standard enthalpy of reaction is the change in enthalpy when reactants and products are in their
AMost stable form
BStandard state at 1 bar
CGaseous form
DLiquid form
Easy idea
Standard means agreed-upon reference: 1 bar pressure (formerly 1 atm) at specified T (usually 298 K).
Given
Definition of standard enthalpy.
To find
Conditions for standard state.
Formula
Standard state: 1 bar pressure, specified T (usually 298 K), pure form
Solution
Reactants and products in their standard states (1 bar, specified T).
Answer
(B)