NEET Chemistry · NCERT Class 11

Thermodynamics notes

Thermodynamic Terms & Processes, First Law of Thermodynamics, Calorimetry & Enthalpy, Hess Law & Reaction Enthalpies and Spontaneity & Gibbs Energy — explained topic by topic with key points, formulas, memory tricks and common mistakes.

  • 6 topics
  • 54 min read
  • 28 formulas
  • 45 PYQs
1

Chapter Overview

7 min

Thermodynamics studies heat, work, energy changes and the direction of chemical or physical processes. In chemistry, it helps predict whether a reaction releases heat, absorbs heat, or can occur spontaneously. The chapter begins with thermodynamic terms such as system, surroundings, state functions, path functions, heat, work and different processes. The first law connects heat, work and internal energy through conservation of energy. Enthalpy makes heat changes easier to study at constant pressure, especially using calorimetry. Hess Law allows calculation of reaction enthalpy through alternative paths, including bond enthalpy, lattice enthalpy and Born-Haber cycles. Finally, entropy and Gibbs energy decide spontaneity and equilibrium. For NEET, sign conventions, formula application and conceptual comparisons are extremely important.

Key points7

  • Energy is conserved but can be transferred as heat or work.
  • Internal energy and enthalpy are state functions; heat and work are path functions.
  • Expansion work is negative for a system because the system loses energy by doing work.
  • Exothermic processes have negative ΔH, while endothermic processes have positive ΔH.
  • Entropy generally increases during melting, vaporization, expansion and formation of more gaseous moles.
  • A process is spontaneous at constant temperature and pressure when ΔG is negative.
  • NEET frequently asks ΔH-ΔU relation, work sign, Hess Law calculations and ΔG sign conditions.

Memory tricks2

Thermodynamics Flow: Terms first, First Law next, heat measurement by calorimetry, path shortcut by Hess Law, and final decision by Gibbs energy.

Three Big Symbols: U means internal energy, H means heat content at constant pressure, G means go/no-go for spontaneity.

Examples2

Burning Fuel

Combustion of petrol releases heat, so ΔH is negative. Thermodynamics helps calculate heat released and predict feasibility.

Ice Melting

Ice melting absorbs heat, but entropy increases. At temperatures above 0°C, the TΔS term makes melting spontaneous.

Common mistakes2

Mixing Physics and Chemistry Work Signs

In chemistry, work done on the system is positive and expansion work done by the system is negative.

Using Celsius in Gibbs Equation

Always use kelvin in ΔG = ΔH - TΔS. Also convert entropy from J to kJ if enthalpy is in kJ.

Reference tables2

Quick revision8

Formulas4

First Law of Thermodynamics
ΔU=q+wΔU = q + w

Change in internal energy equals heat supplied to the system plus work done on the system.

ΔU
change in internal energy of the system
q
heat absorbed by the system
w
work done on the system
Enthalpy
H=U+pVH = U + pV

Enthalpy is a state function useful for heat changes at constant pressure.

H
enthalpy
U
internal energy
pV
pressure-volume term
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Diagrams3

2

Thermodynamic Terms & Processes

9 min

Thermodynamic terms define the language of the chapter. A system is the part of the universe chosen for study, while surroundings include everything outside it. Systems may be open, closed or isolated depending on whether matter and energy can be exchanged. Properties such as pressure, volume, temperature and internal energy describe the state of a system. State functions depend only on initial and final states, whereas path functions such as heat and work depend on how the change occurs. Internal energy is the total microscopic energy stored in a system. Heat is energy transfer due to temperature difference, while work is energy transfer by force-displacement or pressure-volume change. Processes may be isothermal, adiabatic, isochoric or isobaric.

Key points7

  • The universe in thermodynamics is system plus surroundings.
  • A boundary may be real or imaginary, fixed or movable, conducting or insulating.
  • State of a system is fixed when state variables such as p, V, T and composition are fixed.
  • Heat and work are not stored in a system; they are modes of energy transfer.
  • Internal energy changes due to heat and work exchange.
  • For ideal gas isothermal process, internal energy change is zero because internal energy depends only on temperature.
  • For adiabatic process, q = 0; for isochoric process, work due to volume change is zero.

Memory tricks3

Open Closed Isolated: Open shares both matter and energy, Closed closes matter only, Isolated is alone and shares nothing.

Process Name Trick: Iso means same: isothermal same temperature, isochoric same volume, isobaric same pressure. Adiabatic means heat absent.

State Function Shortcut: State functions are like altitude: only start and end matter. Path functions are like distance walked: route matters.

Examples3

Open System

Hot tea in an open cup exchanges heat with air and also loses water vapour, so it is an open system.

Isochoric Process

Heating gas in a rigid sealed steel container increases pressure and temperature, but volume is constant, so PV work is zero.

Adiabatic Process

Rapid compression in a bicycle pump is approximately adiabatic; temperature rises because work is done on the gas.

Common mistakes3

Calling heat a state function

Heat is not contained in a system. It is energy in transit due to temperature difference, so q is a path function.

Confusing isolated and closed systems

A closed system can exchange energy but not matter. An isolated system exchanges neither matter nor energy.

Using wrong sign for expansion

For expansion, ΔV is positive, so w = -pextΔV is negative. The system does work on surroundings.

Reference tables4

Quick revision8

Formulas4

Universe Relation
Universe = System + Surroundings\text{Universe }=\text{ System }+\text{ Surroundings}

Thermodynamic analysis divides the universe into the part studied and everything outside it.

system
part selected for thermodynamic study
surroundings
everything outside the system
PV Work
w=−pextΔVw = -\text{pext}ΔV

Work done during expansion or compression against constant external pressure.

w
work done on the system
pext
external pressure
ΔV
change in volume of system
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Diagrams3

3

First Law of Thermodynamics

9 min

The First Law of Thermodynamics is the law of conservation of energy applied to thermodynamic systems. It states that energy can neither be created nor destroyed; it can only be converted from one form to another. For a chemical system, the change in internal energy equals heat supplied to the system plus work done on the system: ΔU = q + w. If heat is absorbed, q is positive; if heat is released, q is negative. If work is done on the system, w is positive; if the system expands and does work, w is negative. Enthalpy is defined as H = U + pV and becomes especially useful for constant-pressure reactions. For ideal gases, ΔH and ΔU differ by ΔngRT.

Key points6

  • Internal energy is a state function, but heat and work are path functions.
  • For cyclic processes, ΔU = 0 because final state equals initial state.
  • For isochoric processes, ΔV = 0, so PV work is zero and qv = ΔU.
  • For isobaric processes, heat at constant pressure equals enthalpy change.
  • Δng includes only gaseous moles: moles of gaseous products minus moles of gaseous reactants.
  • In numerical problems, always keep units consistent: L atm, J or kJ must be converted properly.

Memory tricks2

Chemistry Work Sign: Expansion: system spends energy, so w is negative. Compression: system receives work, so w is positive.

ΔH and ΔU Trick: ΔH differs from ΔU only when gaseous moles change significantly: remember the gas term ΔngRT.

Examples3

Numerical Example: Expansion Work

A gas expands from 2 L to 5 L against 1 atm. w = -pextΔV = -1 × 3 = -3 L atm = -303.9 J.

Numerical Example: First Law

If a system absorbs 500 J heat and does 200 J work, q = +500 J and w = -200 J, so ΔU = +300 J.

ΔH and ΔU Example

For N2(g) + 3H2(g) → 2NH3(g), Δng = 2 - 4 = -2, so ΔH = ΔU - 2RT.

Common mistakes3

Counting solids and liquids in Δng

In ΔH = ΔU + ΔngRT, count only gaseous reactants and gaseous products.

Using system pressure instead of external pressure

For irreversible expansion work against constant pressure, use external pressure: w = -pextΔV.

Forgetting unit conversion

1 L atm = 101.3 J. Convert work to J or kJ before adding to heat.

Reference tables3

Quick revision8

Formulas6

First Law
ΔU=q+wΔU = q + w

Relates internal energy change to heat and work using chemistry sign convention.

ΔU
change in internal energy
q
heat absorbed by system
w
work done on system
Expansion or Compression Work
w=−pext(V2−V1)=−pextΔVw = -\text{pext}(V2 - V1) = -\text{pext}ΔV

Work done when a gas expands or compresses against constant external pressure.

pext
external pressure
V1
initial volume
V2
final volume
ΔV
change in volume
Enthalpy Definition
H=U+pVH = U + pV

Defines enthalpy as internal energy plus pressure-volume energy.

H
enthalpy
U
internal energy
pV
pressure-volume term
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Diagrams3

4

Calorimetry & Enthalpy

9 min

Calorimetry is the experimental measurement of heat exchanged during physical or chemical changes. The basic equation q = mcΔT connects heat with mass, specific heat capacity and temperature change. Heat capacity is the heat required to raise the temperature of a body by 1 K, while specific heat capacity is for unit mass. In chemistry, enthalpy change is the heat exchanged at constant pressure. Standard enthalpy changes are measured under standard conditions, usually 1 bar and specified temperature, commonly 298 K. Standard enthalpy of formation is the enthalpy change when one mole of a compound forms from elements in their standard states. Enthalpies of combustion, neutralization and solution are important reaction enthalpies frequently tested in NEET numericals.

Key points6

  • A calorimeter is designed to isolate heat exchange between reaction and measured surroundings.
  • In coffee-cup calorimetry, pressure is usually constant, so measured heat relates to ΔH.
  • In bomb calorimetry, volume is constant, so measured heat relates to ΔU.
  • The heat lost by hot body equals heat gained by cold body if no heat is lost to surroundings.
  • Standard enthalpy values depend on physical states of reactants and products.
  • Enthalpy of solution may be positive or negative depending on lattice breaking and hydration energy.

Memory tricks3

Calorimetry Formula: q = mcΔT: mass carries heat capacity through temperature change.

Formation Enthalpy Rule: Elements in their standard states have ΔfH° = 0. Oxygen gas, graphite carbon and hydrogen gas are common examples.

Combustion Sign: Combustion means burning and heat release, so enthalpy of combustion is usually negative.

Examples3

Solved Example: Heating Water

50 g water is heated by 10 K. Using c = 4.18 J g−1 K−1, q = 50 × 4.18 × 10 = 2090 J.

Solved Example: Neutralization

If neutralization releases 2.85 kJ for 0.05 mol water formed, ΔH = -2.85/0.05 = -57 kJ mol−1.

Standard Formation Example

Formation of CO2: C(graphite) + O2(g) → CO2(g). The enthalpy change is ΔfH° of CO2.

Common mistakes3

Forgetting the negative sign for reaction heat

If solution gains heat, reaction loses heat. Therefore qreaction = -qsolution.

Using mass in kg with c in J g−1 K−1

Keep units consistent. If c is in J g−1 K−1, mass must be in grams.

Not converting heat to per mole

Calorimeter gives total heat for the sample used. Enthalpy is usually reported in kJ mol−1.

Reference tables3

Quick revision8

Formulas5

Heat Capacity
C=q/ΔTC = q / ΔT

Heat capacity is heat required to raise temperature of a body by one kelvin.

C
heat capacity
q
heat supplied
ΔT
temperature change
Specific Heat Capacity
c=q/(mΔT)c = q / (mΔT)

Specific heat capacity is heat required to raise temperature of unit mass by one kelvin.

c
specific heat capacity
m
mass
ΔT
temperature change
Calorimetry Equation
q=mcΔTq = mcΔT

Used to calculate heat gained or lost by a substance from temperature change.

q
heat exchanged
m
mass of substance
c
specific heat capacity
ΔT
final temperature minus initial temperature
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Diagrams3

5

Hess Law & Reaction Enthalpies

10 min

Hess Law states that the enthalpy change of a reaction is the same whether the reaction occurs in one step or many steps, provided the initial and final states are the same. This works because enthalpy is a state function. Therefore, difficult reaction enthalpies can be calculated by adding, reversing or multiplying known thermochemical equations. Reaction enthalpy can also be calculated using standard enthalpies of formation or average bond enthalpies. Enthalpy of bond dissociation measures energy needed to break one mole of bonds in gaseous molecules. Enthalpy of atomization forms gaseous atoms from an element. Lattice enthalpy measures energy change during ionic crystal formation or separation. Born-Haber cycle applies Hess Law to ionic solids and connects sublimation, ionization, dissociation, electron gain and lattice enthalpy.

Key points6

  • Thermochemical equations must be balanced exactly before using enthalpy values.
  • Physical states matter because enthalpy depends on state of substances.
  • Average bond enthalpy gives approximate values because bond strength depends on molecular environment.
  • Atomization enthalpy produces gaseous atoms and is always endothermic.
  • Lattice enthalpy increases with higher ionic charges and smaller ionic radii.
  • Born-Haber cycles are useful for calculating unknown lattice enthalpy or checking ionic solid stability.

Memory tricks3

Hess Law Shortcut: Reverse reaction, reverse sign. Multiply reaction, multiply heat. Add reactions, add heats.

Bond Enthalpy Trick: Broken bonds cost energy; formed bonds pay back energy. So ΔH = broken minus formed.

Lattice Enthalpy Trend: Small ions and high charges make a tight lattice, so lattice enthalpy magnitude becomes large.

Examples3

Formation Enthalpy Calculation

For CH4 + 2O2 → CO2 + 2H2O, ΔrH° = [ΔfH°CO2 + 2ΔfH°H2O] - [ΔfH°CH4 + 2ΔfH°O2]. Since O2 is an element in standard state, its ΔfH° is zero.

Bond Enthalpy Calculation

For H2 + Cl2 → 2HCl, ΔH ≈ BE(H-H) + BE(Cl-Cl) - 2BE(H-Cl).

Born-Haber Use

If all steps except lattice enthalpy are known for NaCl formation, Hess Law allows the unknown lattice enthalpy to be calculated.

Common mistakes3

Forgetting to change sign on reversing equation

If the thermochemical equation is reversed, ΔH must change sign.

Using bond enthalpy for solids or liquids directly

Bond enthalpies are usually gas-phase average values; they give approximate reaction enthalpies.

Ignoring physical states

H2O(l) and H2O(g) have different enthalpies. Always include physical states in Hess Law calculations.

Reference tables3

Quick revision8

Formulas4

Hess Law
ΔHtotal =ΔH1+ΔH2+ΔH3+...Δ\text{Htotal }= ΔH1 + ΔH2 + ΔH3 + ...

Overall reaction enthalpy equals the sum of enthalpy changes of individual steps.

ΔHtotal
enthalpy change of overall reaction
ΔH1, ΔH2
enthalpy changes of individual steps
Reaction Enthalpy from Formation Enthalpies
ΔrH°=ΣνΔfH°(products)−ΣνΔfH°(reactants)ΔrH° = ΣνΔfH°(\text{products}) - ΣνΔfH°(\text{reactants})

Calculates standard reaction enthalpy using standard enthalpies of formation.

ΔrH°
standard reaction enthalpy
ν
stoichiometric coefficient
ΔfH°
standard enthalpy of formation
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Diagrams3

6

Spontaneity & Gibbs Energy

10 min

Spontaneity tells whether a process can occur on its own under given conditions, but it does not tell how fast it occurs. A spontaneous process may be fast, like acid-base neutralization, or slow, like rusting. Entropy measures randomness or dispersal of energy; it generally increases during expansion, mixing, melting, vaporization and formation of more gaseous particles. Total entropy of the universe increases for a spontaneous process. For chemical reactions at constant temperature and pressure, Gibbs free energy is the most useful criterion: ΔG = ΔH - TΔS. If ΔG is negative, the process is spontaneous; if positive, non-spontaneous; if zero, equilibrium. Gibbs energy is also related to equilibrium by ΔG° = -RT ln K, linking thermodynamics with chemical equilibrium.

Key points7

  • Spontaneity depends on both enthalpy and entropy, not enthalpy alone.
  • Exothermic reactions are often spontaneous but not always.
  • Endothermic reactions can be spontaneous if entropy increase is large and temperature is high.
  • Temperature must be in kelvin in Gibbs equation.
  • Entropy unit is commonly J mol−1 K−1, while enthalpy is often kJ mol−1, so unit conversion is essential.
  • At equilibrium, ΔG = 0 but ΔG° may not be zero unless K = 1.
  • A catalyst cannot change ΔG; it only changes the rate of reaching equilibrium.

Memory tricks3

Gibbs Spontaneity: Negative G means Go. Positive G means No. Zero G means equilibrium.

Temperature Cases: H negative, S positive: always yes. H positive, S negative: always no. Same signs depend on temperature.

Entropy Direction: Entropy likes gases, mixing, spreading, heating and more particles.

Examples3

Melting of Ice

Melting is endothermic, so ΔH is positive, but entropy increases. Above 273 K, TΔS becomes large enough to make ΔG negative.

Combustion

Combustion often has negative ΔH and may have favorable ΔG, but it still needs ignition because spontaneity does not remove activation energy.

Equilibrium Example

If K is greater than 1, ΔG° = -RT ln K is negative, meaning products are favored under standard-state comparison.

Common mistakes4

Equating spontaneity with speed

Spontaneous does not mean fast. Diamond converting to graphite is thermodynamically favored but extremely slow.

Forgetting unit conversion in ΔG

If ΔH is in kJ mol−1 and ΔS is in J mol−1 K−1, convert ΔS to kJ mol−1 K−1 before using ΔG = ΔH - TΔS.

Thinking catalyst changes ΔG

A catalyst lowers activation energy and changes rate, but it does not change ΔG, ΔH, ΔS or equilibrium constant.

Confusing ΔG and ΔG°

ΔG depends on actual conditions through Q, while ΔG° refers to standard conditions and relates to K.

Reference tables4

Quick revision8

Formulas5

Gibbs Free Energy Equation
ΔG=ΔH−TΔSΔG = ΔH - TΔS

Main equation for deciding spontaneity at constant temperature and pressure.

ΔG
Gibbs energy change
ΔH
enthalpy change
T
absolute temperature in kelvin
ΔS
entropy change
Entropy of Universe Criterion
ΔSuniverse =ΔSsystem +ΔSsurroundingsΔ\text{Suniverse }= Δ\text{Ssystem }+ Δ\text{Ssurroundings}

A process is spontaneous when total entropy of the universe increases.

ΔSuniverse
total entropy change of system plus surroundings
ΔSsystem
entropy change of system
ΔSsurroundings
entropy change of surroundings
Equilibrium Relation
ΔG°=−RT ln KΔG° = -\text{RT ln }K

Relates standard Gibbs energy change to equilibrium constant.

ΔG°
standard Gibbs energy change
R
gas constant
T
absolute temperature
K
equilibrium constant
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Diagrams4

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NEET PYQs from Thermodynamics

The latest 3 of 45 previous-year questions from this chapter.

  1. NEET 2026Set 11EasyQ1

    At a certain temperature T(K), during a process, 500 J is absorbed by the system and work of 200 J is done by the system. Then change in internal energy of the system is:

    • A400 J
    • B300 J
    • C700 J
    • D500 J
  2. NEET 2026Set 11EasyQ2

    An electric heater supplies heat to a system at a rate of 100 W. If the system performs work at a rate of 75 J/s, then the rate at which internal energy increases will be:

    • A75 W
    • B100 W
    • C125 W
    • D25 W
  3. NEET 2026Set 11MediumQ3

    Consider the following reaction: 2A(g) + B(g) → 2D(g) ΔU° = −10 kJ mol⁻¹ and ΔS° = −44 J K⁻¹ at 298 K Identify the correct option with ΔG° for the reaction and spontaneity of the reaction at 298 K. (Given: R = 8.31 J mol⁻¹ K⁻¹)

    • A−1.635 kJ mol⁻¹, spontaneous
    • B−0.63568 kJ mol⁻¹, spontaneous
    • C+0.63568 kJ mol⁻¹, non-spontaneous
    • D+1.635 kJ mol⁻¹, non-spontaneous

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