NEET UG Chemistry · Chapter 5

Chemical Thermodynamics

System and surroundings, first law, enthalpy changes, Hess’s law, entropy, Gibbs energy and spontaneity.

NCERT Class 11NEET: 1–2 Qs per paper 13 min read

1. Chapter at a Glance

Chemical thermodynamics deals with energy changes in chemical reactions and tells us whether a reaction can happen on its own (spontaneity). It uses a few big ideas: the system and surroundings, internal energy and enthalpy, the first law, Hess's law, entropy and Gibbs energy. NEET asks one or two questions every year — usually on ΔH–ΔU relation, Hess's law, bond enthalpies, work done in expansion, or the sign of ΔG.

What NEET asks most
  • State vs path functions; intensive vs extensive properties.
  • First law ΔU = q + w; work in isothermal expansion.
  • ΔH = ΔU + ΔngRT.
  • Hess's law, enthalpies of formation, combustion and bond enthalpies.
  • Entropy, ΔG = ΔH − TΔS and ΔG° = −RT ln K.

2. Basic Terms

  • System: the part of the universe we study; surroundings: everything else.
  • Open system exchanges energy and matter; closed exchanges only energy; isolated exchanges neither (an ideal thermos flask).
  • State functions depend only on the initial and final states: U, H, S, G, p, V, T. Path functions depend on the route taken: heat q and work w.
  • Extensive properties depend on the amount of matter (mass, volume, U, H, heat capacity). Intensive properties do not (T, p, density, molar heat capacity).
  • Processes: isothermal (T constant), adiabatic (q = 0), isobaric (p constant), isochoric (V constant), reversible (infinitely slow, always near equilibrium).

3. First Law of Thermodynamics

ΔU = q + w    (IUPAC sign convention: q and w positive when given to the system)
  • Work of expansion against constant external pressure: w = −pextΔV.
  • Free expansion (into vacuum, pext = 0): w = 0.
  • Isothermal reversible expansion of an ideal gas: w = −2.303 nRT log(V₂/V₁). Since ΔU = 0 at constant T, q = −w.
  • At constant volume, ΔU = qV; at constant pressure, ΔH = qp.

4. Enthalpy and Heat Capacity

H = U + pV    ΔH = ΔU + ΔngRT    Cp − CV = R (per mole of ideal gas)
  • Δng = moles of gaseous products − moles of gaseous reactants. If Δng = 0 (e.g. H₂ + Cl₂ → 2HCl), ΔH = ΔU.
  • Heat capacity: q = C ΔT; for n moles, q = nCmΔT; for mass m, q = mcΔT.
  • ΔU is measured in a bomb calorimeter (constant volume); ΔH in a coffee-cup calorimeter (constant pressure).
  • Exothermic reactions have ΔH < 0; endothermic have ΔH > 0.
H reactants products ΔH < 0exothermic reactants products ΔH > 0endothermic
In an exothermic reaction the products have less enthalpy than the reactants, and heat is released. In an endothermic reaction heat is absorbed.

5. Enthalpies of Reaction and Hess's Law

Enthalpy changeDefinitionNEET Note
Standard enthalpy of formation (ΔfH°)1 mol of compound from its elements in their standard statesZero for elements in standard states (O₂(g), C(graphite))
Enthalpy of combustion (ΔcH)Complete burning of 1 mol in oxygenAlways negative
Enthalpy of neutralisation1 mol of water from acid and base−57.1 kJ mol⁻¹ for strong acid + strong base
Bond dissociation enthalpyBreaking 1 mol of a bond in the gas phaseAlways positive
Enthalpy of atomisation1 mol of gaseous atoms from the substance—
Lattice enthalpy1 mol of ionic solid into gaseous ionsBorn–Haber cycle
Enthalpy of fusion / vaporisation / sublimationPhase changesΔsubH = ΔfusH + ΔvapH
  • Hess's law: the total enthalpy change is the same whether a reaction takes place in one step or many. So equations can be added and subtracted like algebra, along with their ΔH values.
  • ΔrH° = ΣΔfH°(products) − ΣΔfH°(reactants).
  • From bond enthalpies (gas phase): ΔrH = Σ(bonds broken) − Σ(bonds formed).

6. Entropy and the Second Law

  • Entropy (S) measures disorder or randomness. ΔS = qrev/T.
  • Entropy increases: solid → liquid → gas; dissolving a solid; increasing temperature; reactions that increase the number of gas molecules.
  • Second law: the total entropy of the universe increases in every spontaneous process: ΔStotal = ΔSsystem + ΔSsurroundings > 0.
  • At a phase transition, ΔS = ΔHtransition/T.
  • Third law: the entropy of a perfect crystal at 0 K is zero.

7. Gibbs Energy and Spontaneity

ΔG = ΔH − TΔS    ΔG° = −RT ln K = −2.303 RT log K    ΔG° = −nFE°
ΔHΔSΔGSpontaneity
−+Always −Spontaneous at all temperatures
+−Always +Never spontaneous
−−− at low TSpontaneous below T = ΔH/ΔS
++− at high TSpontaneous above T = ΔH/ΔS
  • ΔG < 0 spontaneous; ΔG > 0 non-spontaneous; ΔG = 0 at equilibrium.
  • ΔG° < 0 ⇒ K > 1 (products favoured).
  • Spontaneous does not mean fast: diamond turning into graphite has ΔG < 0 but is extremely slow.

8. Formula Master Table

QuantityFormula
First lawΔU = q + w
Work (constant pext)w = −pextΔV
Isothermal reversible workw = −2.303nRT log(V₂/V₁)
EnthalpyΔH = ΔU + ΔngRT
Heat capacitiesCp − CV = R
Heatq = nCmΔT = mcΔT
From formation enthalpiesΔrH° = ΣΔfH°(P) − ΣΔfH°(R)
From bond enthalpiesΔrH = ΣBE(broken) − ΣBE(formed)
Entropy changeΔS = qrev/T
Gibbs energyΔG = ΔH − TΔS
Standard Gibbs energy and KΔG° = −2.303RT log K
Crossover temperatureT = ΔH/ΔS

9. Master Summary Table

ConceptKey ideaNEET Note
State functionPath-independentq and w are not state functions
First lawEnergy conservedΔU = 0 for isothermal ideal gas
ΔH vs ΔUDiffer by ΔngRTEqual when Δng = 0
Hess's lawAdd equationsConsequence of H being a state function
EntropyDisorderGas > liquid > solid
Gibbs energyΔG < 0 spontaneousFour sign cases
ΔG° and K−RT ln KK > 1 when ΔG° < 0

10. Mind Map

Chemical Thermodynamics
Basics
  • System types
  • State functions
  • Processes
First law
  • ΔU = q + w
  • Work
  • ΔH = ΔU + ΔngRT
Thermochemistry
  • ΔfH, ΔcH
  • Hess's law
  • Bond enthalpy
Spontaneity
  • Entropy
  • ΔG = ΔH − TΔS
  • ΔG° = −RT ln K

11. Tips and Tricks

Shortcuts
  • Use R = 8.314 J K⁻¹ mol⁻¹ with ΔH in joules; convert kJ carefully (divide by 1000).
  • 2.303RT at 298 K ≈ 5.71 kJ mol⁻¹, so ΔG° = −5.71 log K kJ mol⁻¹.
  • Only gases count in Δng; ignore liquids and solids.
  • To decide the sign of ΔS for a reaction, compare the moles of gas on both sides.
Common NEET traps
  • Using the old sign convention (w positive for work done by the system).
  • Including water as a gas in Δng when it is formed as a liquid.
  • Saying exothermic reactions are always spontaneous.
  • Forgetting that ΔfH° of an element in its standard state is zero.

12. Important One-liners

  • Heat and work are path functions; internal energy and enthalpy are state functions.
  • In an isolated system, energy remains constant.
  • In an adiabatic process, q = 0, so ΔU = w.
  • The enthalpy of neutralisation of a strong acid by a strong base is about −57.1 kJ mol⁻¹.
  • Graphite, not diamond, is the standard state of carbon.
  • A reaction is spontaneous when ΔG is negative.
  • The entropy of a perfectly crystalline substance at 0 K is zero.

13. Solved Numericals: Average to Advanced

Five average and five advanced numericals. Take R = 8.314 J K⁻¹ mol⁻¹. Click to reveal each answer and working.

Average level

N1Average

2 mol of an ideal gas expands isothermally and reversibly at 300 K from 10 L to 20 L. Find the work done and the heat absorbed.

Show answer & solution

Concept: w = −2.303nRT log(V₂/V₁); ΔU = 0 for an isothermal ideal gas.

  1. w = −2.303 × 2 × 8.314 × 300 × log 2.
  2. = −2.303 × 4988.4 × 0.301 ≈ −3458 J.
  3. q = −w = +3458 J (heat absorbed).

Answer: w ≈ −3.46 kJ; q ≈ +3.46 kJ

N2Average

For N₂(g) + 3H₂(g) → 2NH₃(g), ΔU = −87.4 kJ at 298 K. Find ΔH.

Show answer & solution

Concept: ΔH = ΔU + ΔngRT.

  1. Δng = 2 − 4 = −2.
  2. ΔngRT = −2 × 8.314 × 298 = −4955 J ≈ −4.96 kJ.
  3. ΔH = −87.4 − 4.96 ≈ −92.4 kJ.

Answer: about −92.4 kJ

N3Average

How much heat is needed to raise the temperature of 36 g of water from 25 °C to 75 °C? (Molar heat capacity of water = 75.3 J K⁻¹ mol⁻¹)

Show answer & solution

Concept: q = nCmΔT.

  1. n = 36/18 = 2 mol; ΔT = 50 K.
  2. q = 2 × 75.3 × 50 = 7530 J ≈ 7.53 kJ.

Answer: 7.53 kJ

N4Average

For a reaction, ΔH = +30 kJ mol⁻¹ and ΔS = +100 J K⁻¹ mol⁻¹. At what temperature is it at equilibrium? Above or below this temperature will it be spontaneous?

Show answer & solution

Concept: at equilibrium ΔG = 0 ⇒ T = ΔH/ΔS.

  1. T = 30000/100 = 300 K.
  2. Both ΔH and ΔS are positive, so TΔS wins at higher T: spontaneous above 300 K.

Answer: 300 K; spontaneous above 300 K

N5Average

A system absorbs 500 J of heat and does 200 J of work on the surroundings. Find the change in its internal energy.

Show answer & solution

Concept: ΔU = q + w with IUPAC signs.

  1. q = +500 J (absorbed); w = −200 J (done by the system).
  2. ΔU = 500 − 200 = +300 J.

Answer: +300 J

Advanced level

N6Advanced

The enthalpies of combustion of C(graphite), H₂(g) and CH₄(g) are −393.5, −285.8 and −890.3 kJ mol⁻¹. Find the standard enthalpy of formation of methane.

Show answer & solution

Concept: Hess's law: C + 2H₂ → CH₄ = (burn C) + 2(burn H₂) − (burn CH₄).

  1. ΔfH = ΔcH(C) + 2ΔcH(H₂) − ΔcH(CH₄).
  2. = −393.5 + 2(−285.8) − (−890.3).
  3. = −393.5 − 571.6 + 890.3 = −74.8 kJ mol⁻¹.

Answer: −74.8 kJ mol⁻¹

N7Advanced

Using bond enthalpies H–H = 436, Cl–Cl = 242 and H–Cl = 431 kJ mol⁻¹, find ΔH for H₂(g) + Cl₂(g) → 2HCl(g). Is ΔU different from ΔH here?

Show answer & solution

Concept: ΔH = Σ bonds broken − Σ bonds formed.

  1. Broken: 436 + 242 = 678 kJ.
  2. Formed: 2 × 431 = 862 kJ.
  3. ΔH = 678 − 862 = −184 kJ. Δng = 2 − 2 = 0, so ΔU = ΔH.

Answer: −184 kJ; ΔU = ΔH

N8Advanced

The equilibrium constant of a reaction at 300 K is 10. Find ΔG°. Is the forward reaction favoured under standard conditions?

Show answer & solution

Concept: ΔG° = −2.303RT log K.

  1. ΔG° = −2.303 × 8.314 × 300 × log 10.
  2. = −5744 J ≈ −5.74 kJ mol⁻¹.
  3. ΔG° < 0 and K > 1, so products are favoured.

Answer: about −5.74 kJ mol⁻¹; yes

N9Advanced

The enthalpy of vaporisation of water is 40.79 kJ mol⁻¹ at its boiling point, 373 K. Find the entropy change of vaporisation.

Show answer & solution

Concept: at a phase transition, ΔS = ΔH/T.

  1. ΔS = 40790/373.
  2. ΔS ≈ 109.4 J K⁻¹ mol⁻¹ (positive: liquid → gas increases disorder).

Answer: about 109.4 J K⁻¹ mol⁻¹

N10Advanced

For the synthesis of ammonia, ΔH° = −92.4 kJ mol⁻¹ and ΔS° = −198.3 J K⁻¹ mol⁻¹. Find ΔG° at 298 K, and the temperature above which the reaction stops being spontaneous.

Show answer & solution

Concept: ΔG = ΔH − TΔS (use the same units).

  1. ΔG° = −92.4 − 298 × (−0.1983) = −92.4 + 59.1 = −33.3 kJ mol⁻¹ (spontaneous).
  2. Both ΔH and ΔS are negative, so it is spontaneous only at low T.
  3. T = ΔH/ΔS = 92400/198.3 ≈ 466 K; above this, ΔG becomes positive.

Answer: about −33.3 kJ mol⁻¹; above about 466 K

14. NEET-pattern Questions with Detailed Answers

These questions are original and based on concepts that NEET UG asks again and again.

Q1Easy

Which of the following is NOT a state function?

  1. Internal energy
  2. Enthalpy
  3. Work
  4. Entropy
Show solution

Work depends on the path taken between two states, so it is a path function. Internal energy, enthalpy and entropy depend only on the initial and final states.

Answer: (C)

Q2Easy

For which reaction is ΔH equal to ΔU?

  1. N₂(g) + 3H₂(g) → 2NH₃(g)
  2. H₂(g) + I₂(g) → 2HI(g)
  3. PCl₅(g) → PCl₃(g) + Cl₂(g)
  4. 2SO₂(g) + O₂(g) → 2SO₃(g)
Show solution

ΔH = ΔU when Δng = 0. In reaction (B), Δng = 2 − 2 = 0. For the others, Δng is −2, +1 and −1.

Answer: (B)

Q3Medium

A reaction has ΔH < 0 and ΔS < 0. It will be spontaneous:

  1. At all temperatures
  2. At low temperatures only
  3. At high temperatures only
  4. At no temperature
Show solution

ΔG = ΔH − TΔS. Here −TΔS is positive, and it grows with T. So ΔG is negative only when T is small enough (below ΔH/ΔS).

Answer: (B)

Q4Medium

The enthalpy of formation is zero for:

  1. O₃(g)
  2. C(diamond)
  3. Br₂(l)
  4. H₂O(l)
Show solution

Only elements in their standard states have ΔfH° = 0. Liquid bromine is the standard state of bromine. O₃ is not the standard form of oxygen, diamond is not the standard form of carbon, and water is a compound.

Answer: (C)

Q5Medium

In an adiabatic process:

  1. ΔU = 0
  2. q = 0
  3. w = 0
  4. ΔT = 0
Show solution

An adiabatic process allows no heat exchange, so q = 0 and ΔU = w. ΔU = 0 and ΔT = 0 describe an isothermal process for an ideal gas.

Answer: (B)

Q6Medium

Entropy increases in which change?

  1. Freezing of water
  2. Condensation of steam
  3. Sublimation of iodine
  4. 2NO₂(g) → N₂O₄(g)
Show solution

Sublimation turns a solid into a gas, which increases disorder a great deal. Freezing and condensation make the system more ordered. In (D), the moles of gas fall from 2 to 1, so entropy decreases.

Answer: (C)

Q7Hard

If ΔG° for a reaction at 298 K is −11.4 kJ mol⁻¹, its equilibrium constant is about (2.303RT = 5.7 kJ mol⁻¹):

  1. 10
  2. 100
  3. 0.01
  4. 2
Show solution

log K = −ΔG°/(2.303RT) = 11.4/5.7 = 2, so K = 100. (C) comes from a sign error, and (D) confuses log K with K.

Answer: (B)

Q8Hard

An ideal gas expands freely into a vacuum under adiabatic conditions. Which set is correct?

  1. q = 0, w = 0, ΔU = 0
  2. q = 0, w < 0, ΔU < 0
  3. q > 0, w = 0, ΔU > 0
  4. q = 0, w > 0, ΔU > 0
Show solution

The process is adiabatic, so q = 0. Against zero external pressure w = 0, so ΔU = 0 and the temperature of an ideal gas stays the same. (B) would be true for an adiabatic expansion against a pressure.

Answer: (A)

15. Quick Revision Summary

Remember before the exam
  • ΔU = q + w; w = −pextΔV; isothermal reversible w = −2.303nRT log(V₂/V₁).
  • ΔH = ΔU + ΔngRT; Cp − CV = R.
  • Hess's law; ΔrH = ΣΔfH(P) − ΣΔfH(R); bonds broken − bonds formed.
  • ΔS = qrev/T; ΔStotal > 0 for spontaneous change.
  • ΔG = ΔH − TΔS; ΔG° = −2.303RT log K; four sign cases.
All Chemistry chaptersNEET Mock Tests