Class 11 · Chemistry · Chapter 5

Thermodynamics

This chapter builds the language of energy changes in chemical and physical processes — system, surroundings, work, heat, internal energy, enthalpy, and entropy — and uses the first and second laws to predict whether a reaction gives out or takes in energy, and whether it happens on its own. It is the direct foundation for electrochemistry and equilibrium in Class 12 and for JEE/NEET physical chemistry.

14 sections12 NCERT questions24 board questions15-question quiz

Before you startWhat the board asks from this chapter

No board exam in Class 11, but this chapter is heavily tested in the school final/pre-board and is foundational for Class 12 boards (thermodynamics ideas reappear in electrochemistry and equilibrium) and for JEE/NEET, where it is one of the highest-yield physical chemistry chapters.

Question typeMarksUsually asked
MCQ1Sign conventions, state vs path functions, spontaneity, and identifying q, w, ΔU for a given process.
Assertion–Reason1Often tests whether a negative ΔG always means a fast reaction, or whether ΔH = ΔU for all reactions.
Very short answer2Define a term (state function, isochoric process) or state a law (first law, Hess's law) with its equation.
Short answer3Calculate ΔU, ΔH, w or q for an expansion or a reaction; or derive the relation ΔH = ΔU + ΔnRT.
Long answer5Full numerical using Hess's law or bond enthalpies, or a derivation such as Cp − Cv = R, combined with a conceptual part on ΔG and spontaneity.
Case-based4A passage on a real process (engine, dissolution, reaction) followed by questions on system type, sign of ΔH, and spontaneity.

Hot topics, in order

  1. First law of thermodynamics and sign convention (ΔU = q + w)
  2. Enthalpy, ΔH = ΔU + ΔnRT, and Hess's law numericals
  3. Enthalpies of reaction — formation, combustion, atomization, bond enthalpy
  4. Spontaneity, entropy, and Gibbs energy ΔG = ΔH − TΔS
  5. ΔG and equilibrium constant, ΔG° = −RT ln K
  6. Extensive vs intensive properties, state vs path functions
  7. Isothermal and adiabatic reversible expansion of an ideal gas
  8. Second and third laws of thermodynamics (conceptual)

The whole chapter, in orderNotes

01 5.1 Thermodynamic Terms — System, Surroundings, Boundary

Thermodynamics deals with energy changes as a reaction or process occurs, without worrying about how fast it happens or the mechanism — that is the job of chemical kinetics. To study energy change we first mark off the part of the universe we are interested in: the system. Everything else is the surroundings, and the real or imaginary surface separating them is the boundary. Heat and work are exchanged across this boundary.

Systems are classified by what can cross the boundary. An open system can exchange both matter and energy with the surroundings (hot tea in an open cup). A closed system exchanges only energy, not matter (tea in a sealed but conducting flask). An isolated system exchanges neither matter nor energy (tea in a perfect thermos flask). Most laboratory reactions in a stoppered flask are treated as closed systems.

The state of a system is described by macroscopic, measurable properties — pressure, volume, temperature, and amount — called state variables. When these fix the state completely, the system is said to be in a definite state. Any property whose value depends only on the initial and final state, not on how the change was carried out, is a state function (e.g. internal energy, enthalpy). A quantity that depends on the path taken is a path function (e.g. heat, work). Examiners frequently ask you to classify a list of quantities as state or path functions.

Exam tip 'State function' and 'path function' come up almost every year as a one-mark or VSA question — memorise the definition word for word and be ready to classify U, H, S, G (state) vs q, w (path).

02 5.1.1 The State of the System — Extensive and Intensive Properties

Properties of a system fall into two classes based on how they behave when the size of the system changes. An extensive property depends on the amount of matter present in the system — mass, volume, internal energy, enthalpy, entropy, and Gibbs energy are all extensive; double the amount of substance and these quantities double. An intensive property does not depend on the amount of matter — temperature, pressure, density, and molar quantities (like molar heat capacity) are intensive; they stay the same whether you have 1 g or 100 g of the substance.

A useful test: if you divide a system into two equal halves, an extensive property is halved in each part, while an intensive property stays unchanged. This distinction matters when combining thermodynamic quantities — for instance, dividing an extensive property (like total enthalpy) by an extensive property (like moles) always gives an intensive property (molar enthalpy).

Exam tip A quick trick examiners like: mass ÷ volume = density. Extensive ÷ extensive = intensive. Use this to classify any unfamiliar quantity.

03 5.2 Applications — Work, Heat, Internal Energy, and the First Law PYQ CBSE 2020

Internal energy (U) is the total energy stored in a system — the sum of translational, rotational, vibrational, and bond energies of all the molecules. We can never measure U in absolute terms, only its change, ΔU, between two states; since U depends only on the state, ΔU is the same regardless of the path taken.

Internal energy of a system changes when energy enters or leaves as heat (q) or as work (w). The sign convention used in Class 11 NCERT (IUPAC convention): heat absorbed by the system is positive (q > 0), heat given out is negative (q < 0); work done on the system is positive (w > 0), work done by the system is negative (w < 0).

The first law of thermodynamics is the law of conservation of energy applied to a system: energy can neither be created nor destroyed, though it can change form. Mathematically, ΔU = q + w. This single equation is the working tool of the whole chapter.

Work of expansion: when a gas expands against an external pressure pext by a volume ΔV, the work done by the gas is pextΔV, so the work done on the system is w = −pextΔV. For an ideal gas expanding into vacuum (free expansion), pext = 0, so w = 0. For a reaction at constant volume (ΔV = 0), w = 0, so ΔU = qV — the heat measured in a bomb calorimeter directly gives ΔU.

For a reversible isothermal expansion of an ideal gas from V1 to V2 at temperature T, the gas does maximum work: w = −2.303 nRT log(V2/V1) = −2.303 nRT log(p1/p2). Since ΔU depends only on T for an ideal gas and T is constant here, ΔU = 0, so q = −w = 2.303 nRT log(V2/V1).

ΔU = q + w
ΔUchange in internal energyqheat absorbed by the system (+) or released (−)wwork done on the system (+) or by the system (−)
w = −pextΔV
pextexternal (opposing) pressureΔVVfinal − Vinitial
wrev = −2.303 nRT log(V2/V1)
nmoles of gasRgas constant, 8.314 J K−1 mol−1Tabsolute temperature (K)
Exam tip Always state which sign convention you are using at the start of a numerical — NCERT/CBSE uses q, w positive when the system gains energy. Marks are lost purely for sign slips.

04 5.3 The First Law of Thermodynamics — Applications to Isothermal and Adiabatic Processes

An isothermal process occurs at constant temperature (dT = 0); since U of an ideal gas depends only on T, ΔU = 0 for any isothermal change of an ideal gas, so q = −w. An adiabatic process occurs with no heat exchange with the surroundings (q = 0); the boundary is thermally insulated, so ΔU = w — all the work done on/by the system changes its internal energy, and the temperature changes.

In an adiabatic expansion the gas does work at the expense of its own internal energy, so it cools (w is negative, ΔU decreases, T falls). In an adiabatic compression, work is done on the gas, ΔU increases, and T rises. Contrast this with an isothermal expansion, where heat flows in from the surroundings exactly fast enough to keep T — and hence U — constant.

An isochoric process occurs at constant volume (ΔV = 0), so w = 0 and ΔU = qV. An isobaric process occurs at constant pressure; here qp = ΔU + pΔV, which is exactly how enthalpy is defined in the next section.

Isothermal (ideal gas): ΔU = 0, q = −w
Adiabatic: q = 0, ΔU = w
Isochoric: ΔV = 0, w = 0, ΔU = qV
Exam tip A common trap: 'adiabatic' does not mean 'no temperature change' — it means no heat exchange. Temperature does change in an adiabatic process because work is converted to/from internal energy.

05 5.4 Measurement of ΔU and ΔH — Calorimetry

ΔU is measured using a bomb calorimeter: the reaction (usually combustion) is carried out in a sealed, rigid steel vessel immersed in a known mass of water, so the volume is constant. No expansion work is possible (w = 0), so all the heat released or absorbed shows up as a temperature change of the water and calorimeter: qV = ΔU = −C × ΔT, where C is the heat capacity of the calorimeter assembly. This value is reported per mole after scaling to the amount reacted.

ΔH, the enthalpy change at constant pressure, is measured in a calorimeter open to the atmosphere (like a coffee-cup calorimeter for solution reactions) where the system is free to expand or contract against constant atmospheric pressure: qp = ΔH.

qV = ΔU (bomb calorimeter, constant volume)
qp = ΔH (constant-pressure calorimeter)
Exam tip If a numerical says 'sealed vessel' or 'bomb calorimeter', use ΔU directly from the heat measured; if it says 'open vessel' or 'constant pressure', use ΔH directly.

06 5.5 Enthalpy, H — a Useful New State Function PYQ CBSE 2019 · CBSE 2023

Most reactions in the laboratory are carried out in open vessels, at constant atmospheric pressure, not constant volume. It is convenient to define a new state function, enthalpy, H = U + pV, so that the heat absorbed at constant pressure equals ΔH directly, without needing to separately track the pV work.

Derivation of ΔH = ΔU + pΔV (at constant pressure): Starting from the first law, ΔU = qp + w. For expansion work at constant external pressure p, w = −pΔV. So ΔU = qp − pΔV, which gives qp = ΔU + pΔV. Since H = U + pV, at constant pressure ΔH = ΔU + pΔV. Comparing the two, qp = ΔH — the heat absorbed at constant pressure is exactly the enthalpy change.

Relation between ΔH and ΔU for gaseous reactions: for a reaction in which the moles of gas change, using pV = nRT at constant T and p, pΔV = ΔngRT, where Δng is the difference between moles of gaseous products and moles of gaseous reactants. So ΔH = ΔU + ΔngRT. If Δng = 0 (equal moles of gas on both sides, or no gas involved), ΔH = ΔU.

By convention, ΔH is negative for an exothermic reaction (heat is released to the surroundings, enthalpy of products is lower than reactants) and positive for an endothermic reaction (heat is absorbed, enthalpy of products is higher).

H = U + pV
HenthalpyUinternal energyppressureVvolume
ΔH = ΔU + ΔngRT
Δngmoles of gaseous products − moles of gaseous reactantsR8.314 J K−1 mol−1Tabsolute temperature
Exam tip The ΔH = ΔU + ΔnRT derivation is one of the most repeated 2–3 mark questions. Practice writing it in the exact three steps: first law → definition of H → substitute pV = nRT.

07 5.5.1 Extensive and Intensive Properties, Heat Capacity

The heat capacity of a system is the heat needed to raise its temperature by 1 K (or 1 °C): C = q/ΔT. It depends on the size and nature of the system, and on the conditions of measurement — heat capacity at constant volume, CV, and heat capacity at constant pressure, Cp, are different because at constant pressure some of the heat also does expansion work.

Relation between Cp and CV for an ideal gas — derivation: At constant volume, qV = ΔU = CVΔT. At constant pressure, qp = ΔH = CpΔT. For one mole of an ideal gas, ΔH = ΔU + Δ(pV) = ΔU + RΔT (since pV = RT for one mole). Substituting, CpΔT = CVΔT + RΔT. Dividing throughout by ΔT gives Cp − CV = R. Cp is always greater than CV because heating at constant pressure must also supply energy for the gas to expand and do work, in addition to raising its internal energy.

Molar heat capacity is the heat capacity per mole, an intensive quantity, with units J K−1 mol−1. Specific heat capacity is heat capacity per gram (J g−1 K−1). Water's high specific heat (4.18 J g−1 K−1) is why it is used as the medium in calorimeters.

Cp − CV = R (per mole of ideal gas)
Cpmolar heat capacity at constant pressureCVmolar heat capacity at constant volumeRgas constant
q = m × c × ΔT
mmasscspecific heat capacityΔTtemperature change
Exam tip The Cp − CV = R derivation is a favourite 3-mark question — write every step, do not skip to the answer.

08 5.6 Measurement of ΔU and ΔH — Enthalpy Change of a Reaction, Enthalpy of Reaction

The enthalpy of a reaction, ΔrH, is the enthalpy change when the number of moles of reactants shown in the balanced equation react completely to give products, under standard conditions unless stated otherwise. It equals the sum of the standard enthalpies of formation of the products minus that of the reactants, each multiplied by its stoichiometric coefficient: ΔrH° = ΣaΔfH°(products) − ΣbΔfH°(reactants).

Standard conditions (indicated by the ° symbol) mean 1 bar pressure and a specified temperature, usually 298 K, with substances in their standard states (the most stable physical state of the substance at that temperature and 1 bar).

Different named enthalpy changes are simply ΔrH for a reaction of a specific type:

  • Standard enthalpy of formation, ΔfH° — enthalpy change when 1 mole of a compound is formed from its elements in their standard states. ΔfH° of an element in its standard state is zero by definition.
  • Standard enthalpy of combustion, ΔcH° — enthalpy change when 1 mole of a substance is completely burnt in excess oxygen; always negative (exothermic).
  • Enthalpy of atomization, ΔaH° — enthalpy needed to break a substance into gaseous atoms, e.g. CH4(g) → C(g) + 4H(g).
  • Enthalpy of sublimation, fusion, and vaporization — enthalpy changes for solid → gas, solid → liquid, and liquid → gas at constant T.
  • Enthalpy of solution and dilution — heat change when a solute dissolves in a solvent, or when a solution is further diluted.
ΔrH° = Σ ΔfH°(products) − Σ ΔfH°(reactants)
ΔfH°standard enthalpy of formation, each term × its stoichiometric coefficient
Exam tip Enthalpy of formation of an element in its most stable standard state (O2(g), C(graphite), Br2(l)) is always zero — a common one-mark question.

09 5.7 Enthalpy Change of a Reaction — Reaction Enthalpy, Hess's Law PYQ CBSE 2022

Hess's law of constant heat summation states that the total enthalpy change for a reaction is the same whether it occurs in one step or in a series of steps, because enthalpy is a state function and ΔH depends only on the initial and final states, not the path.

This lets us calculate the enthalpy of a reaction that cannot be measured directly by combining the enthalpies of reactions that can be measured, using ordinary algebra: reverse an equation and reverse the sign of its ΔH; multiply an equation by a factor and multiply its ΔH by the same factor; add equations that share common intermediates so they cancel, and add their ΔH values.

Classic example: enthalpy of formation of CO(g) from C(graphite) and O2(g) cannot be measured directly because burning carbon in a limited supply of oxygen always gives some CO2 as well. Instead we use: (i) C(graphite) + O2(g) → CO2(g), ΔH1 = −393.5 kJ mol−1, and (ii) CO(g) + ½O2(g) → CO2(g), ΔH2 = −283.0 kJ mol−1. Reversing (ii) and adding to (i): C(graphite) + ½O2(g) → CO(g), ΔH = ΔH1 − ΔH2 = −393.5 − (−283.0) = −110.5 kJ mol−1.

Hess's law numericals are also solved using a Born–Haber-cycle-style energy diagram, drawing the target reaction directly between reactants and products, and the known steps as an alternative path, then equating the sum of ΔH along both routes.

ΔHdirect = ΣΔHsteps (path independence)
Exam tip In a Hess's law numerical, always check whether an equation needs to be reversed (flip the sign of ΔH) or scaled (multiply ΔH by the same factor) before adding — this is where most marks are lost.

10 5.7.1 Enthalpies for Different Types of Reactions

Bond enthalpy, ΔbondH°, is the enthalpy required to break one mole of a particular covalent bond in a gaseous molecule to give gaseous atoms/fragments. For a diatomic molecule this equals the bond dissociation enthalpy directly. For a polyatomic molecule with several identical bonds (e.g. the four C–H bonds in methane), the mean bond enthalpy is used, since each successive bond-breaking step actually requires a slightly different amount of energy.

Estimating reaction enthalpy from bond enthalpies: ΔrH° = Σ(bond enthalpies of bonds broken in reactants) − Σ(bond enthalpies of bonds formed in products), because breaking bonds always requires energy (endothermic) and forming bonds always releases energy (exothermic). This method gives an approximate value since it uses average bond enthalpies, not the exact bond enthalpy in that specific molecule.

Enthalpy of lattice formation, ΔlatticeH° — enthalpy change when one mole of an ionic solid is formed from its gaseous ions; it is large and negative because of the strong electrostatic attraction between oppositely charged ions. It cannot be measured directly and is obtained using a Born–Haber cycle, which applies Hess's law to a sequence of steps: sublimation of the metal, ionization, dissociation and electron gain of the non-metal, and lattice formation, connecting them to the directly measurable enthalpy of formation of the compound.

ΔrH° = Σ(bond enthalpy of reactant bonds) − Σ(bond enthalpy of product bonds)
Exam tip Bond-enthalpy calculations give only an estimate of ΔH, not the exact value — a favourite assertion-reason trap.

11 5.8 Spontaneity — Is a Chemical Reaction Feasible? PYQ CBSE 2021

A spontaneous process is one that occurs on its own, without continuous external intervention, once started. It may be fast or slow — spontaneity says nothing about the rate. A spontaneous process cannot reverse on its own; it needs external work to be reversed (e.g. water flows downhill spontaneously; pumping it back up needs work). Non-spontaneous does not mean impossible — it means the process will not occur by itself under the given conditions.

Early thinking assumed all exothermic reactions (ΔH < 0) are spontaneous, since systems tend towards lower energy. But some spontaneous processes are endothermic — ice melting above 0°C, or NH4Cl dissolving in water, both absorb heat yet occur on their own. So enthalpy alone cannot decide spontaneity; a second factor, entropy, is needed.

Exam tip Never write 'exothermic reactions are always spontaneous' — this is a very commonly penalised misconception; give a counter-example (ice melting, salt dissolving) if asked to justify.

12 5.9 Entropy and Spontaneity PYQ CBSE 2020

Entropy (S) is a state function that measures the degree of randomness or disorder of a system. The second law of thermodynamics states that the total entropy of the universe (system + surroundings) always increases in a spontaneous process: ΔStotal = ΔSsystem + ΔSsurroundings > 0 for a spontaneous process, and = 0 at equilibrium.

For a reversible process at temperature T, the entropy change of the system is ΔS = qrev/T. Entropy generally increases (ΔS > 0) when a solid melts, a liquid vaporizes, a gas expands, moles of gas increase in a reaction, or a solid dissolves to give ions in solution; entropy decreases in the reverse of these changes.

The third law of thermodynamics states that the entropy of a perfectly crystalline substance is taken as zero at absolute zero (0 K), since there is then only one way of arranging the molecules — this gives an absolute reference point for calculating standard molar entropies, unlike enthalpy where only changes can be measured.

ΔS = qrev/T
ΔSentropy changeqrevheat absorbed reversiblyTabsolute temperature
ΔStotal = ΔSsys + ΔSsurr ≥ 0
>0spontaneous=0equilibrium
Exam tip When asked to predict the sign of ΔS for a reaction, count moles of gas on each side — more moles of gas on the product side means ΔS is positive.

13 5.10 Gibbs Energy and Spontaneity PYQ CBSE 2023

Since spontaneity needs the total entropy change of the universe, but we usually only know about the system, Gibbs defined a new state function combining ΔH and ΔS of the system alone: Gibbs energy, G = H − TS. At constant temperature and pressure, ΔG = ΔH − TΔS.

Derivation of the criterion: ΔStotal = ΔSsys + ΔSsurr. At constant p and T, the surroundings absorb −ΔHsys of heat reversibly, so ΔSsurr = −ΔHsys/T. Substituting, ΔStotal = ΔSsys − ΔHsys/T. Multiplying through by −T: −TΔStotal = ΔHsys − TΔSsys = ΔGsys. Since ΔStotal ≥ 0 for a spontaneous or equilibrium process and T is positive, ΔGsys ≤ 0 for the same process.

So at constant temperature and pressure: ΔG < 0 — reaction is spontaneous (exergonic); ΔG > 0 — reaction is non-spontaneous in the forward direction (endergonic), though the reverse reaction is spontaneous; ΔG = 0 — system is at equilibrium.

Because ΔG = ΔH − TΔS, four combinations of sign are possible: ΔH < 0, ΔS > 0 gives ΔG < 0 always (spontaneous at all T); ΔH > 0, ΔS < 0 gives ΔG > 0 always (non-spontaneous at all T); ΔH < 0, ΔS < 0 is spontaneous only at low T (below T = ΔH/ΔS); ΔH > 0, ΔS > 0 is spontaneous only at high T (above T = ΔH/ΔS). This is why some endothermic processes (like ice melting, or many dissolution reactions) are spontaneous — the −TΔS term outweighs the positive ΔH once T is large enough.

ΔG = ΔH − TΔS
ΔGGibbs energy changeΔHenthalpy changeTabsolute temperatureΔSentropy change of the system
ΔG° = −RT ln K = −2.303 RT log K
ΔG°standard Gibbs energy changeKequilibrium constant
Exam tip The sign-combination table (ΔH, ΔS, ΔG at low/high T) is asked almost every year as a 2–3 mark question — draw it as a table in the answer, examiners reward the structure.

14 5.11 Gibbs Energy Change and Equilibrium

Gibbs energy links thermodynamics to chemical equilibrium (studied in the next chapter) through ΔG° = −RT ln K = −2.303 RT log K, where K is the equilibrium constant of the reaction at temperature T. A large negative ΔG° gives a large K (reaction goes almost to completion); a large positive ΔG° gives a very small K (reaction barely proceeds); ΔG° = 0 gives K = 1.

It is important to distinguish ΔG (at the actual, non-equilibrium concentrations, which determines the direction a reaction will shift) from ΔG° (the fixed standard value used to calculate K). Once the system reaches equilibrium, ΔG = 0, even though ΔG° for the reaction is not necessarily zero.

ΔG° = −2.303 RT log K
R8.314 J K−1 mol−1Tabsolute temperatureKequilibrium constant
Exam tip This equation is the bridge to Class 12's electrochemistry (ΔG° = −nFE°cell) and equilibrium chapters — examiners like combined numericals across chapters at the JEE/NEET level.

VocabularyKey terms

SystemThe part of the universe chosen for thermodynamic study, separated from the rest (surroundings) by a real or imaginary boundary.
Open systemA system that can exchange both matter and energy with its surroundings.
Closed systemA system that can exchange only energy, not matter, with its surroundings.
Isolated systemA system that can exchange neither matter nor energy with its surroundings.
State functionA property whose value depends only on the initial and final state of the system, not on the path taken, e.g. U, H, S, G.
Path functionA quantity whose value depends on the path taken between two states, e.g. heat (q) and work (w).
Extensive propertyA property that depends on the amount of matter in the system, e.g. mass, volume, internal energy, enthalpy.
Intensive propertyA property that does not depend on the amount of matter, e.g. temperature, pressure, density, molar heat capacity.
Internal energy (U)The total energy stored within a system; only ΔU can be measured, not the absolute value.
Enthalpy (H)A state function defined as H = U + pV; the heat absorbed or released at constant pressure equals ΔH.
First law of thermodynamicsEnergy can neither be created nor destroyed, only transformed; ΔU = q + w for a system.
Isothermal processA process occurring at constant temperature (dT = 0).
Adiabatic processA process in which no heat is exchanged with the surroundings (q = 0).
Isochoric processA process occurring at constant volume, so no expansion work is done (w = 0).
Standard enthalpy of formation (Δ<sub>f</sub>H°)The enthalpy change when 1 mole of a compound is formed from its elements in their standard states; zero for elements in their standard state.
Standard enthalpy of combustion (Δ<sub>c</sub>H°)The enthalpy change when 1 mole of a substance is completely burnt in excess oxygen; always negative.
Bond enthalpyThe energy required to break one mole of a specific covalent bond in a gaseous molecule to give gaseous atoms/fragments.
Hess's lawThe total enthalpy change for a reaction is the same whether it occurs in one step or several steps, because enthalpy is a state function.
Spontaneous processA process that proceeds on its own once initiated, without continuous external intervention; independent of the rate of the process.
Entropy (S)A state function measuring the degree of randomness or disorder of a system; entropy of the universe increases in every spontaneous process.
Gibbs energy (G)A state function, G = H − TS, whose decrease at constant T and p (ΔG < 0) is the criterion for a spontaneous process.
Third law of thermodynamicsThe entropy of a perfectly crystalline substance is zero at absolute zero (0 K).

Keep this on one pageFormula sheet

NameExpressionNotes
First law of thermodynamicsΔU = q + wq, w positive when the system gains heat/energy (IUPAC convention used in NCERT)
Work of expansionw = −pextΔVfor expansion/compression against a constant external pressure
Reversible isothermal workw = −2.303 nRT log(V2/V1)ideal gas, constant T; ΔU = 0, so q = −w
EnthalpyH = U + pVstate function; qp = ΔH at constant pressure
ΔH–ΔU relationΔH = ΔU + ΔngRTΔng = moles of gaseous products − gaseous reactants
Heat capacity relationCp − CV = Rfor one mole of an ideal gas
Enthalpy of reactionΔrH° = ΣΔfH°(products) − ΣΔfH°(reactants)each term multiplied by its stoichiometric coefficient
Bond enthalpy estimateΔrH° = Σ(bonds broken) − Σ(bonds formed)gives an approximate ΔH using mean bond enthalpies
Entropy change (reversible)ΔS = qrev/Tat constant T; ΔStotal ≥ 0 for spontaneous/equilibrium processes
Gibbs energy changeΔG = ΔH − TΔSat constant T, p; ΔG < 0 spontaneous, = 0 equilibrium, > 0 non-spontaneous
ΔG° and equilibrium constantΔG° = −2.303 RT log KR = 8.314 J K−1 mol−1

Practise with a pencilDiagrams that earn marks

System, boundary, and surroundings

Draw an outer rectangle labelled 'Surroundings' and an inner circle or flask labelled 'System', with the line between them labelled 'Boundary'. Draw arrows crossing the boundary labelled 'q' (heat) and 'w' (work) to show open/closed systems, and a dashed boundary with no arrows for an isolated system. Label all three system types beside their respective sketches.

p–V diagram for isothermal and adiabatic expansion

Plot pressure (y-axis) against volume (x-axis). Draw two curves starting from the same point: the isothermal curve (labelled T = constant, a shallower hyperbola-like curve) and the adiabatic curve (labelled q = 0, a steeper curve that falls faster). Shade the area under the isothermal curve to represent the larger work done in isothermal expansion; note that the adiabatic curve lies below the isothermal curve after the same volume change because the gas cools.

Enthalpy level diagram (exothermic vs endothermic)

Draw two vertical enthalpy-axis diagrams side by side. In the first, place 'Reactants' on a higher horizontal line and 'Products' on a lower line, with a downward arrow labelled ΔH < 0 (exothermic). In the second, reverse the positions with an upward arrow labelled ΔH > 0 (endothermic). Add the reaction equation beside each.

Hess's law cycle for formation of CO(g)

Draw C(graphite) + O2(g) at the top left and CO2(g) at the bottom, connected by a direct downward arrow labelled ΔH1 = −393.5 kJ mol−1. Draw a second path from C(graphite) + O2(g) through CO(g) + ½O2(g) (top right) down to CO2(g), with the first arrow labelled ΔH (unknown, target) and the second labelled ΔH2 = −283.0 kJ mol−1. Show that both paths end at the same point, illustrating that ΔH1 = ΔH + ΔH2.

Born–Haber cycle for an ionic solid (e.g. NaCl)

Draw a stepped energy-level diagram starting from Na(s) + ½Cl2(g) at the bottom left. Show upward steps for sublimation of Na(s) → Na(g), ionization Na(g) → Na+(g), dissociation ½Cl2(g) → Cl(g), and electron gain Cl(g) → Cl−(g) → each labelled with its enthalpy. Then show a large downward arrow for lattice enthalpy from Na+(g) + Cl−(g) to NaCl(s), and a direct arrow from the starting materials straight to NaCl(s) labelled ΔfH°, closing the cycle.

TextbookNCERT questions the board keeps asking

5.1 Choose the correct answer: A thermodynamic state function is a quantity (i) used to determine heat changes (ii) whose value is independent of path (iii) used to determine pressure–volume work (iv) whose value depends on temperature only.1 m
(ii) whose value is independent of path. A state function depends only on the initial and final states of the system, not on how the change was carried out — that is exactly the property distinguishing it from a path function.
5.2 For the process to occur under adiabatic conditions, the correct condition is: (i) ΔT = 0 (ii) Δp = 0 (iii) q = 0 (iv) w = 0. State which is correct and explain briefly.1 m
(iii) q = 0. An adiabatic process is defined as one in which no heat is exchanged between the system and surroundings. Temperature is not held constant — in fact it usually changes, because ΔU = w when q = 0, so work done on/by the gas directly changes its internal energy and temperature.
5.3 The enthalpies of all elements in their standard states are: (i) unity (ii) zero (iii) < 0 (iv) different for each element. Explain your choice.1 m
(ii) zero. By convention, the standard enthalpy of formation of an element in its most stable standard state is taken as zero, since it is used as the reference point from which the formation enthalpies of all compounds are measured.
5.6 1 mol of an ideal gas is expanded isothermally and reversibly at 300 K from 5 L to 10 L. Calculate w for this process. (R = 8.314 J K−1 mol−1)2 m
For a reversible isothermal expansion, w = −2.303 nRT log(V2/V1). Here n = 1, R = 8.314 J K−1 mol−1, T = 300 K, V1 = 5 L, V2 = 10 L.
w = −2.303 × 1 × 8.314 × 300 × log(10/5) = −2.303 × 8.314 × 300 × log 2 = −2.303 × 8.314 × 300 × 0.301 ≈ −1729 J.
So w ≈ −1730 J (work is done by the gas on the surroundings, hence negative).
5.9 2 mol of an ideal gas expanded isothermally and reversibly from 1 L to 10 L at 300 K. What is the enthalpy change (ΔH) for this process? Calculate the entropy change also if qrev is used.3 m
For an ideal gas, both ΔU and ΔH depend only on temperature. Since the process is isothermal (ΔT = 0), ΔU = 0 and therefore ΔH = ΔU + Δ(pV) = ΔU + Δ(nRT) = 0 (as n, R, T are all constant). So ΔH = 0.
To find ΔS for the surroundings from the process, first find qrev = −w = 2.303 nRT log(V2/V1) = 2.303 × 2 × 8.314 × 300 × log(10/1) = 2.303 × 2 × 8.314 × 300 × 1 ≈ 11488 J.
ΔSsystem = qrev/T = 11488/300 ≈ 38.3 J K−1.
5.15 Calculate the standard enthalpy of formation of CH3OH(l) from the following data:
CH3OH(l) + 3/2 O2(g) → CO2(g) + 2H2O(l), ΔH° = −726 kJ mol−1
C(graphite) + O2(g) → CO2(g), ΔH° = −393 kJ mol−1
H2(g) + ½O2(g) → H2O(l), ΔH° = −286 kJ mol−1
3 m
Target: C(graphite) + 2H2(g) + ½O2(g) → CH3OH(l), ΔfH° = ?
By Hess's law, this equals: [equation 2] + 2×[equation 3] − [equation 1].
ΔfH° = (−393) + 2(−286) − (−726) = −393 − 572 + 726 = −239 kJ mol−1.
So the standard enthalpy of formation of methanol is −239 kJ mol−1.
5.17 The equilibrium constant for a reaction is 10. What will be the value of ΔG°? R = 8.314 J K−1 mol−1, T = 300 K.3 m
ΔG° = −2.303 RT log K = −2.303 × 8.314 × 300 × log 10 = −2.303 × 8.314 × 300 × 1 ≈ −5744 J mol−1 ≈ −5.74 kJ mol−1.
Since ΔG° is negative, the reaction is thermodynamically favourable (K > 1) at 300 K.
5.20 For the reaction, 2A(g) + B(g) → 2D(g), ΔU° = −10.5 kJ and ΔS° = −44.1 JK−1. Calculate ΔG° for the reaction, and predict whether the reaction may occur spontaneously at 298 K.5 m
First find ΔH° from ΔU°: Δng = 2 − (2+1) = −1. ΔH° = ΔU° + ΔngRT = −10.5 kJ + (−1)(8.314 × 10−3 kJ K−1mol−1)(298 K) = −10.5 − 2.478 = −12.978 kJ.
ΔG° = ΔH° − TΔS° = −12.978 kJ − (298 K)(−44.1 × 10−3 kJ K−1) = −12.978 + 13.14 = +0.163 kJ ≈ +163 J.
Since ΔG° is positive (though small), the reaction is non-spontaneous under standard conditions at 298 K.
5.21 Comment on the thermodynamic stability of NO(g), given: ½N2(g) + ½O2(g) → NO(g), ΔH° = 90 kJ mol−1, ΔS° = 12.5 JK−1mol−1; and NO(g) + ½O2(g) → NO2(g), ΔH° = −74 kJ mol−1, ΔS° = −121 JK−1mol−1. (T = 298 K)5 m
For NO formation: ΔG° = ΔH° − TΔS° = 90 − 298(12.5×10−3) = 90 − 3.725 = 86.3 kJ mol−1 (positive, so NO is thermodynamically unstable relative to N2 and O2, and formed only by supplying energy, e.g. lightning or high temperature).
For NO → NO2: ΔG° = −74 − 298(−121×10−3) = −74 + 36.06 = −37.9 kJ mol−1 (negative, so NO readily and spontaneously oxidises further to NO2 in air, confirming NO itself is unstable).
5.13 Given N2(g) + 3H2(g) → 2NH3(g), ΔH° = −92.4 kJ mol−1. What is the standard enthalpy of formation of NH3 gas?2 m
The given ΔH° is for formation of 2 mol NH3 from its elements in their standard states, so it is 2 × ΔfH°(NH3).
ΔfH°(NH3) = −92.4/2 = −46.2 kJ mol−1.
E1 In an adiabatic process, no transfer of heat takes place between system and surroundings. Can the temperature of the system change in an adiabatic process?2 m
Yes. In an adiabatic process q = 0, so by the first law ΔU = w. Any work done on or by the system directly changes the internal energy, and since internal energy of an ideal gas depends on temperature, the temperature does change — it rises during adiabatic compression (work done on the gas) and falls during adiabatic expansion (work done by the gas).
E2 State whether ΔH = ΔU for the reaction: PCl5(g) → PCl3(g) + Cl2(g). Justify your answer with a calculation of Δng.2 m
No, ΔH ≠ ΔU here. Δng = moles of gaseous products − moles of gaseous reactants = (1 + 1) − 1 = +1, which is not zero, so ΔH = ΔU + ΔngRT = ΔU + RT ≠ ΔU. ΔH equals ΔU only when Δng = 0.

Previous years and frequently askedBoard questions with model answers

MCQ Which of the following is a state function?
  1. Heat
  2. Work
  3. Enthalpy
  4. Both heat and work
Frequently asked1 m
Answer: (c)
Enthalpy depends only on the initial and final states of the system, not on the path taken, so it is a state function; heat and work are path functions.
MCQ For an isothermal, reversible expansion of an ideal gas, which quantity is zero?
  1. q
  2. w
  3. ΔU
  4. ΔH is negative
Frequently asked1 m
Answer: (c)
For an ideal gas, U depends only on T. Since the process is isothermal, ΔT = 0, so ΔU = 0 (and ΔH = 0 too, since ΔH also depends only on T for an ideal gas).
MCQ The enthalpy of formation of an element in its standard state is:
  1. Always positive
  2. Always negative
  3. Zero
  4. Equal to its enthalpy of combustion
CBSE 20201 m
Answer: (c)
By convention, the reference point for enthalpy of formation is the element in its most stable standard state, so its ΔfH° is defined as zero.
MCQ For which of the following reactions will ΔH = ΔU?
  1. N2(g) + 3H2(g) → 2NH3(g)
  2. H2(g) + I2(g) → 2HI(g)
  3. 2SO2(g) + O2(g) → 2SO3(g)
  4. CaCO3(s) → CaO(s) + CO2(g)
Frequently asked1 m
Answer: (b)
ΔH = ΔU only when Δng = 0. For H2 + I2 → 2HI, moles of gas are 2 on both sides, so Δng = 0; the other options have a change in gaseous moles.
MCQ Which of these processes has a decrease in entropy?
  1. Melting of ice
  2. Sublimation of dry ice
  3. Dissolving common salt in water
  4. Freezing of water to ice
Frequently asked1 m
Answer: (d)
Freezing converts a liquid (more disordered) to a solid (highly ordered, fixed lattice), so entropy of the system decreases; the other three options all increase disorder.
MCQ For a reaction at constant temperature and pressure to be spontaneous, ΔG must be:
  1. Positive
  2. Zero
  3. Negative
  4. Equal to ΔH
CBSE 20231 m
Answer: (c)
A negative ΔG at constant T and p is the thermodynamic criterion for a spontaneous process; ΔG = 0 corresponds to equilibrium and ΔG > 0 to a non-spontaneous forward process.
MCQ The reaction that will always be spontaneous at all temperatures is one where:
  1. ΔH > 0, ΔS > 0
  2. ΔH < 0, ΔS < 0
  3. ΔH < 0, ΔS > 0
  4. ΔH > 0, ΔS < 0
Frequently asked1 m
Answer: (c)
ΔG = ΔH − TΔS. If ΔH is negative and ΔS is positive, then ΔG = (negative) − T(positive) is negative at every temperature T, so the reaction is spontaneous at all T.
MCQ What is the value of ΔStotal (system + surroundings) when a system reaches equilibrium?
  1. Greater than zero
  2. Less than zero
  3. Equal to zero
  4. Cannot be determined
Frequently asked1 m
Answer: (c)
At equilibrium, the total entropy of the universe has reached its maximum value for the process, so ΔStotal = 0; it is only positive for a spontaneous process still moving towards equilibrium.
Assertion–Reason Assertion (A): All exothermic reactions are spontaneous. Reason (R): A reaction with ΔH < 0 always has ΔG < 0.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is false but R is true
  4. Both A and R are false
Frequently asked1 m
Answer: (d)
Both statements are false. Spontaneity depends on ΔG = ΔH − TΔS, not on ΔH alone; an exothermic reaction with a sufficiently negative ΔS can still have ΔG > 0 at high temperature, so it need not be spontaneous.
Assertion–Reason Assertion (A): Enthalpy and internal energy are both state functions. Reason (R): Heat and work are also state functions.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. Both A and R are false
Frequently asked1 m
Answer: (c)
A is true — H and U depend only on the state. R is false — heat and work are path functions; their values depend on how the process is carried out.
Assertion–Reason Assertion (A): The molar heat capacity of a gas at constant pressure is greater than at constant volume. Reason (R): At constant pressure, part of the heat supplied is used to do expansion work in addition to raising the internal energy.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
Frequently asked1 m
Answer: (a)
Both are true and R correctly explains A: Cp − CV = R because heating at constant pressure must also supply energy for the gas to expand against the surroundings.
Very short answer Define enthalpy of combustion and enthalpy of atomization with one example each.CBSE 20192 m
Enthalpy of combustion is the enthalpy change when 1 mole of a substance is completely burnt in excess of oxygen, e.g. CH4(g) + 2O2(g) → CO2(g) + 2H2O(l), ΔcH° = −890.3 kJ mol−1.
Enthalpy of atomization is the enthalpy change on breaking 1 mole of a substance completely into gaseous atoms, e.g. CH4(g) → C(g) + 4H(g), ΔaH° = +1665 kJ mol−1.
Very short answer Distinguish between an isothermal process and an adiabatic process.Frequently asked2 m
An isothermal process occurs at constant temperature (dT = 0), so for an ideal gas ΔU = 0 and q = −w. An adiabatic process occurs with no heat exchange with the surroundings (q = 0), so ΔU = w and the temperature of the gas changes as work is done on or by it.
Very short answer What is Hess's law of constant heat summation? State one use of this law.CBSE 20222 m
Hess's law states that the total enthalpy change of a reaction is the same whether the reaction occurs in a single step or in a series of steps, because enthalpy is a state function and depends only on the initial and final states.
Use: it allows calculation of the enthalpy change for a reaction that cannot be measured directly (e.g. formation of CO(g)) by combining the enthalpies of reactions that can be measured experimentally.
Very short answer For the reaction 2A(g) + B(g) → 2C(g), predict the sign of ΔS. Justify.Frequently asked2 m
ΔS is negative. On the reactant side there are 3 moles of gas (2 + 1) and on the product side there are 2 moles of gas; a decrease in the number of gas moles means the system becomes more ordered, so entropy decreases.
Short answer Derive the relation ΔH = ΔU + ΔnRT for a reaction involving ideal gases at constant temperature and pressure.Frequently asked3 m
By the first law, at constant pressure, ΔU = qp + w, and w = −pΔV, so qp = ΔU + pΔV.
By definition, H = U + pV, so at constant pressure ΔH = ΔU + pΔV = qp.
For an ideal gas, pV = nRT, so at constant T and p, pΔV = ΔngRT, where Δng is the change in moles of gas.
Substituting: ΔH = ΔU + ΔngRT.
Short answer 1 mole of an ideal gas at 300 K is expanded isothermally from an initial volume of 1 L to 10 L against a constant external pressure of 1 atm. Calculate the work done, q, and ΔU for the process. (1 L atm = 101.3 J)CBSE 20213 m
This is an irreversible expansion against constant external pressure, so w = −pextΔV = −1 atm × (10 − 1) L = −9 L atm = −9 × 101.3 J = −911.7 J.
Since the process is isothermal for an ideal gas, ΔU = 0.
By the first law, ΔU = q + w, so q = ΔU − w = 0 − (−911.7) = +911.7 J.
Short answer Calculate the standard reaction enthalpy for C2H4(g) + H2(g) → C2H6(g), given ΔfH°[C2H4(g)] = +52.3 kJ mol−1, ΔfH°[C2H6(g)] = −84.7 kJ mol−1, ΔfH°[H2(g)] = 0.Frequently asked3 m
ΔrH° = ΔfH°(products) − ΔfH°(reactants) = ΔfH°[C2H6(g)] − {ΔfH°[C2H4(g)] + ΔfH°[H2(g)]}
= (−84.7) − (52.3 + 0) = −84.7 − 52.3 = −137.0 kJ mol−1.
Short answer Explain, with reasoning, why the melting of ice at room temperature (above 0°C) is a spontaneous process even though it is endothermic.Frequently asked3 m
For melting, ΔH is positive (heat is absorbed to break the ordered structure of ice) and ΔS is also positive (a solid changes into a more disordered liquid). Spontaneity depends on ΔG = ΔH − TΔS. At room temperature, T is large enough that the TΔS term outweighs the positive ΔH, making ΔG negative overall, so the process is spontaneous even though it absorbs heat.
Long answer (a) Derive the relation Cp − CV = R for one mole of an ideal gas. (b) State the second law of thermodynamics. (c) For a certain reaction, ΔH° = +30 kJ mol−1 and ΔS° = +100 J K−1mol−1. Above what temperature will the reaction become spontaneous?CBSE 20205 m
(a) At constant volume, qV = ΔU = CVΔT. At constant pressure, qp = ΔH = CpΔT. For one mole of ideal gas, ΔH = ΔU + Δ(pV) = ΔU + RΔT, so CpΔT = CVΔT + RΔT. Dividing by ΔT: Cp − CV = R.
(b) The total entropy of the universe (system + surroundings) always increases in a spontaneous process; ΔStotal > 0 for a spontaneous process and ΔStotal = 0 at equilibrium.
(c) For spontaneity, ΔG < 0, i.e. ΔH < TΔS, so T > ΔH/ΔS = 30000 J mol−1 / 100 J K−1mol−1 = 300 K. The reaction becomes spontaneous above 300 K.
Long answer (a) Using Hess's law, calculate the enthalpy of formation of CS2(l) from: C(graphite) + O2(g) → CO2(g), ΔH° = −393.3 kJ mol−1; S(rhombic) + O2(g) → SO2(g), ΔH° = −296.8 kJ mol−1; CS2(l) + 3O2(g) → CO2(g) + 2SO2(g), ΔH° = −1103.9 kJ mol−1. (b) Explain why bond enthalpies calculated from thermochemical data are only average/approximate values.Frequently asked5 m
(a) Target: C(graphite) + 2S(rhombic) → CS2(l), ΔfH° = ?
By Hess's law: ΔfH°(CS2) = [ΔH°(C→CO2) + 2×ΔH°(S→SO2)] − ΔH°(combustion of CS2)
= [(−393.3) + 2(−296.8)] − (−1103.9) = [−393.3 − 593.6] + 1103.9 = −986.9 + 1103.9 = +117.0 kJ mol−1.
(b) In a polyatomic molecule with several bonds of the same type (e.g. the four C–H bonds in methane), each successive bond does not require exactly the same amount of energy to break, because the environment around the remaining bonds changes as atoms are removed. The bond enthalpy quoted is therefore the average of these different values, so calculations using it give only an approximate reaction enthalpy.
Long answer State the criteria for spontaneity of a process in terms of ΔG. Discuss all four possible combinations of the signs of ΔH and ΔS and their effect on spontaneity, with one example of each where possible.Frequently asked5 m
At constant temperature and pressure, ΔG = ΔH − TΔS decides spontaneity: ΔG < 0 spontaneous, ΔG = 0 equilibrium, ΔG > 0 non-spontaneous.
(i) ΔH < 0, ΔS > 0: ΔG is always negative — spontaneous at all temperatures, e.g. 2H2O2(l) → 2H2O(l) + O2(g).
(ii) ΔH > 0, ΔS < 0: ΔG is always positive — non-spontaneous at all temperatures.
(iii) ΔH < 0, ΔS < 0: ΔG is negative only at low T (below T = ΔH/ΔS) — spontaneous only at low temperature, e.g. freezing of water below 273 K.
(iv) ΔH > 0, ΔS > 0: ΔG is negative only at high T (above T = ΔH/ΔS) — spontaneous only at high temperature, e.g. melting of ice above 273 K, decomposition of CaCO3 at high temperature.
Case-based Read the passage and answer the questions that follow: A chemist places 2 mol of an ideal gas in a cylinder fitted with a frictionless piston. The gas is allowed to expand reversibly and isothermally at 298 K from a volume of 5 L to 20 L. In a separate experiment, the same gas sample is compressed adiabatically, causing its temperature to rise sharply.
(i) What is the value of ΔU for the isothermal expansion? Why?
(ii) Write the expression used to calculate the work done in the isothermal reversible expansion.
(iii) In the adiabatic compression, is q positive, negative, or zero? Explain.
(iv) Why does the temperature rise during the adiabatic compression?
Frequently asked4 m
(i) ΔU = 0, because internal energy of an ideal gas depends only on temperature, and the process is isothermal (temperature is constant).
(ii) w = −2.303 nRT log(V2/V1).
(iii) q = 0, because an adiabatic process is defined as one in which no heat is exchanged between the system and the surroundings (the boundary is thermally insulated).
(iv) Since q = 0, the first law gives ΔU = w. Work is done on the gas during compression (w is positive), so ΔU increases; since internal energy of an ideal gas is directly related to its temperature, the temperature rises.
Case-based Read the passage and answer the questions that follow: Dissolving ammonium chloride (NH4Cl) in water is a process that feels cold to the touch, yet it happens readily and completely on its own when the salt is added to water at room temperature. A student argues that since the process absorbs heat, it should not occur spontaneously.
(i) Is the student's reasoning correct? Why or why not?
(ii) What is the sign of ΔH for this dissolution? What is the sign of ΔS?
(iii) Using ΔG = ΔH − TΔS, explain how a process with positive ΔH can still be spontaneous.
(iv) Name the criterion (in terms of ΔG) that correctly decides spontaneity at constant temperature and pressure.
Frequently asked4 m
(i) No, the student's reasoning is incorrect. Enthalpy change alone does not determine spontaneity; spontaneity depends on the Gibbs energy change, ΔG, which also involves entropy and temperature.
(ii) ΔH is positive (the process is endothermic, it feels cold as heat is absorbed from the surroundings). ΔS is positive, since solid ions become free, hydrated, more disordered ions in solution.
(iii) In ΔG = ΔH − TΔS, if ΔS is sufficiently positive, the term −TΔS becomes large and negative at ordinary temperatures, and can outweigh the positive ΔH, making the overall ΔG negative. This is exactly what happens in NH4Cl dissolving in water.
(iv) At constant T and p, a process is spontaneous if ΔG < 0, at equilibrium if ΔG = 0, and non-spontaneous if ΔG > 0.

Test yourselfPractice quiz

Marks lost every yearCommon mistakes

  • Assuming all exothermic reactions (ΔH < 0) are spontaneous — spontaneity depends on ΔG = ΔH − TΔS, not on ΔH alone.
  • Confusing adiabatic (q = 0) with isothermal (ΔT = 0) — these are opposite conditions and are frequently swapped by mistake.
  • Forgetting to convert ΔS from J K−1mol−1 to kJ K−1mol−1 before combining it with ΔH (in kJ) in ΔG = ΔH − TΔS, leading to a wrong numerical answer.
  • Getting the sign wrong on reversing an equation in a Hess's law calculation — reversing a reaction must flip the sign of its ΔH.
  • Writing q and w with the wrong sign convention, especially forgetting that work done BY the system on the surroundings is negative in the IUPAC/NCERT convention.
  • Treating heat (q) and work (w) as state functions — they are path functions and depend on how the process is carried out, unlike ΔU or ΔH for the same overall change.
  • Using bond enthalpy values as if they give an exact ΔH — they give only an approximate value because average bond enthalpies are used for polyatomic molecules.
  • Forgetting that Δng must be computed using only gaseous species — solids and liquids are excluded when finding the change in moles of gas for ΔH = ΔU + ΔngRT.

How to prepareStrategy for this chapter

This chapter rewards precision more than memorisation. Start by locking down the sign convention (ΔU = q + w, with q and w positive when the system gains energy) and the classification of state vs path functions and extensive vs intensive properties — these show up as easy one-mark and VSA questions almost every year.

Next, master three derivations word for word: ΔH = ΔU + ΔngRT, Cp − CV = R, and the ΔG < 0 spontaneity criterion from ΔStotal ≥ 0. These appear as 3–5 mark questions repeatedly, and partial credit depends on showing every algebraic step, not just the final formula.

For numericals, practise Hess's law problems until combining and reversing equations becomes automatic, and always double-check units (kJ vs J) before substituting into ΔG = ΔH − TΔS. Learn the four ΔH/ΔS sign combinations and what each implies for spontaneity at different temperatures — this table is asked in some form virtually every year.

Finally, be ready to link this chapter forward: ΔG° = −2.303 RT log K connects directly to equilibrium (Class 11, next chapter) and to ΔG° = −nFE°cell in Class 12 electrochemistry, so JEE/NEET often combine thermodynamics with these topics in a single question.

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