Chemical Thermodynamics Formula Sheet — JEE Main Chemistry

Q: What are the most important Chemical Thermodynamics formulas for JEE Main?

This Chemical Thermodynamics formula sheet covers all the high-yield Chemistry formulas, definitions and theorems you need for JEE Main, across Concepts of System and types of systems, Surroundings, Work, heat, energy, Extensive and intensive properties, State functions — each shown with the key result and, where useful, a worked example.

Q: Is this Chemical Thermodynamics formula sheet free?

Yes — the full chapter formula sheet is free to read online, no login or payment required.

Q: How should I revise Chemical Thermodynamics formulas?

Blurt the Chemical Thermodynamics formulas from memory, then check against this sheet to find your gaps — and practise a few previous-year questions on the chapter to make sure you can apply them under time pressure.

Every key Chemical Thermodynamics formula, definition and theorem for JEE Main Chemistry in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNTA · 15 sub-topics

Concepts of System and types of systems
Surroundings
Work, heat, energy
Extensive and intensive properties
State functions
First law of thermodynamics - internal energy and enthalpy
Heat capacity and specific heat
Measurement of ΔU and ΔH
Hess's law of constant heat summation
Enthalpy of bond dissociation, combustion, formation, atomization, sublimation, phase transition, ionization, solution and dilution
Second law of thermodynamics (brief introduction)
Introduction of entropy as a state function
Gibbs energy change for spontaneous and non-spontaneous processes
Criteria for equilibrium
Third law of thermodynamics (brief introduction)

Basic Concepts & First Law

Defining the system

▸ $System$ = part of the universe under study; $surroundings$ = everything else; $universe = system + surroundings$ .
▸ $Open$ : exchanges matter $+$ energy (open beaker). $Closed$ : energy only (sealed flask). $Isolated$ : neither (thermos flask).
▸ $State functions$ (U, H, p, V, T, S, G) depend only on the state, not the path; q and w are $path$ functions.

Internal energy (

U

): The total energy stored in a system (kinetic + potential + bond, etc.). It is a state function; only changes

Δ U

can be measured, not absolute values.

Δ U

depends on initial and final states alone.

Heat (

q

) and work (

w

)

▸ $Heat$ q: energy exchanged due to a temperature difference (random molecular motion).
▸ $Work$ w: ordered energy transfer; for gases, mainly $pressure-volume$ work.
▸ $Adiabatic$ process: no heat exchange ( $q = 0$ ), so $Δ U = w_{a d}$ .

Process	Held constant	Ideal-gas result
Isothermal	$T$	$Δ U = Δ H = 0$ , $q = - w$
Adiabatic	$q = 0$	$Δ U = w = n C_{V} Δ T$
Isobaric	$p$	$q_{p} = Δ H = n C_{p} Δ T$
Isochoric	$V$	$w = 0$ , $q_{V} = Δ U = n C_{V} Δ T$
Cyclic	returns to start	$Δ U = Δ H = 0$

Extensive vs intensive

▸ $Extensive$ (depend on amount, additive): mass, V, U, H, S, G, heat capacity C.
▸ $Intensive$ (independent of amount): T, p, density, molar/specific properties ( $C_{m}$ , $Δ H$ per mole), refractive index. The ratio of two extensive properties is intensive.

Open, closed and isolated systems showing matter and energy exchange, with the first-law sign convention where heat into and work on the system are positive — Classify the system, then apply $Δ U = q + w$ with the IUPAC sign convention.

First law of thermodynamics

Δ U = q + w

Energy of an isolated system is constant (

q = w = 0 \Rightarrow Δ U = 0

). Sign convention: heat absorbed by the system and work done on the system are

positive

★ Remember · Sign convention (IUPAC)

q > 0

: heat flows into the system.

w > 0

: work done on the system (compression).

q < 0

w < 0

: system releases heat / does work on surroundings.

🎯 Exam · What thermodynamics does (and doesn't) tell you

It predicts

whether

a process is feasible and the energy involved, from initial and final states only — it says

nothing

about the

rate

(that is kinetics).

🚫 Examiner Trap · Examiner traps

(1) IUPAC sign:

q > 0

heat IN,

w > 0

work done ON the system

\Rightarrow Δ U = q + w

(some books use

Δ U = q - w

— know which). (2) U,H,S,G are STATE functions; q,w are PATH functions. (3) Adiabatic

q = 0 \Rightarrow Δ U = w

. (4) Thermodynamics gives feasibility, NOT rate (that's kinetics).

Work, Enthalpy & Heat Capacity

Irreversible (const.

p_{e x}

)

w = - p_{e x} Δ V

Reversible isothermal

w = - 2.303 nRT lo g \frac{V _{2}}{V _{1}}

Free expansion (

p_{e x} = 0

)

w = 0

Const. volume (

Δ V = 0

)

Δ U = q_{V}

Gas in a cylinder compressed by an external pressure, and the pressure-volume diagram where the shaded rectangular area equals the work done — Work = area under the $p$ – $V$ curve; reversible work is the maximum.

Enthalpy (

H

): The heat content at constant pressure:

H = U + p V

. Since most reactions occur in open vessels at constant atmospheric pressure,

Δ H = q_{p}

(heat exchanged at constant p). H is a state function.

Relating ΔH and ΔU

Δ H = Δ U + Δ n_{g} RT

Δ n_{g}

= (moles of gaseous products) − (moles of gaseous reactants). For reactions with no gas change or only solids/liquids,

Δ H \approx Δ U

★ Remember · Exo vs endo

Δ H < 0

: exothermic (heat released).

Δ H > 0

: endothermic (heat absorbed).

Heat absorbed

q = C Δ T = m c Δ T

At const.

V

/ const.

p

q_{V} = C_{V} Δ T = Δ U; q_{p} = C_{p} Δ T = Δ H

Ideal gas (Mayer)

C_{p} - C_{V} = R

🎯 Exam · Calorimetry

Bomb calorimeter

measures

q_{V} = Δ U

(constant volume, no work).

Coffee-cup

(constant p) measures

q_{p} = Δ H

. Heat lost by reaction

=

heat gained by calorimeter

= m c Δ T

🚫 Examiner Trap · Examiner traps

(1)

w = - p_{e x} Δ V

: on

expansion

(

Δ V > 0

) work is negative (system does work). (2) Reversible work is the MAXIMUM magnitude; free expansion (

p_{e x} = 0

) gives

w = 0

. (3)

Δ H = Δ U + Δ n_{gRT}

— use

Δ n_{g}

gas

moles only. (4) Bomb calorimeter measures

q_{V} = Δ U

; coffee-cup

q_{p} = Δ H

Thermochemistry — Enthalpy Types

Standard reaction enthalpy (

Δ_{r} H^{⊖}

): The enthalpy change of a reaction when all reactants and products are in their standard states (pure form at 1 bar, data usually at 298 K). The

^{⊖}

superscript denotes standard conditions.

Two enthalpy level diagrams: exothermic with products below reactants and delta H negative, endothermic with products above reactants and delta H positive — $Δ_{r} H = H_{products} - H_{reactants}$ — sign reveals exo/endothermic.

Thermochemical-equation rules

▸Coefficients = moles (not molecules); $Δ_{r} H$ has units kJ mo $l^{- 1}$ (per mole of reaction).
▸ $Reverse$ the reaction $\Rightarrow$ reverse the sign of $Δ H$ .
▸ $Multiply$ coefficients by n $\Rightarrow$ multiply $Δ H$ by n (enthalpy is extensive).

Enthalpy of	Symbol	Process
Formation	$Δ_{f} H^{⊖}$	1 mol compound from elements
Combustion	$Δ_{c} H^{⊖}$	1 mol burnt in excess $O_{2}$
Atomization	$Δ_{a} H^{⊖}$	1 mol → gaseous atoms
Fusion	$Δ_{f u s} H^{⊖}$	solid → liquid
Vaporization	$Δ_{v a p} H^{⊖}$	liquid → gas
Sublimation	$Δ_{s u b} H^{⊖}$	solid → gas
Solution	$Δ_{so l} H^{⊖}$	1 mol dissolved

★ Remember · Standard enthalpy of formation

Δ_{f} H^{⊖}

of an element in its

reference state

(e.g.

O_{2} (g)

, C(graphite),

H_{2} (g)

) is taken as

zero

Δ_{s u b} H = Δ_{f u s} H + Δ_{v a p} H

(state-function additivity).

🎯 Exam · Solution = lattice + hydration

Δ_{so l} H^{⊖} = Δ_{l a tt i ce} H^{⊖} + Δ_{h y d} H^{⊖}

. Fusion, vaporization, sublimation are always endothermic (

+

More named enthalpies (syllabus list)

▸ $Bond dissociation$ : break 1 mol of a bond in the gas phase (always $+$ ).
▸ $Hydration$ : 1 mol of gaseous ions into water (always $-$ ).
▸ $Ionisation$ (of a weak electrolyte): energy to ionise it fully in solution. Neutralisation of a strong acid by a strong base is $- 57.1$ kJ/mol (just $H^{+} + O H^{-} \to H_{2} O$ ); with a WEAK acid or base it is less negative, the difference being the ionisation enthalpy.
▸ $Resonance energy$ $= Δ H_{o b ser v e d} - Δ H_{c a l c u l a t e d}$ (from bond enthalpies of one Lewis structure); e.g. benzene is more stable than the Kekulé estimate.

🚫 Examiner Trap · Examiner traps

(1)

Δ_{f} H^{⊖}

of an element in its reference state is

zero

(

O_{2}

, C-graphite,

H_{2}

). (2) Reverse a reaction

\Rightarrow

flip the sign; multiply coefficients

\Rightarrow

multiply

Δ H

(extensive). (3) Fusion/vaporisation/sublimation are always endothermic (

+

). (4)

Δ_{s u b} H = Δ_{f u s} H + Δ_{v a p} H

Hess's Law & Bond Enthalpy

Hess's law of constant heat summation: If a reaction occurs in several steps, its overall enthalpy change is the sum of the enthalpy changes of the steps — because enthalpy is a state function (path-independent).

Hess cycle with a direct path from A to B and an indirect path A to C to D to B, where the indirect enthalpy changes sum to the direct one — Direct $Δ_{r} H$ = sum along any indirect route ( $Δ H_{1} + Δ H_{2} + Δ H_{3}$ ).

From formation enthalpies

Δ_{r} H^{⊖} = \sum a_{i} Δ_{f} H^{⊖} (products) - \sum b_{i} Δ_{f} H^{⊖} (reactants)

a_{i}, b_{i}

= stoichiometric coefficients. 'Products minus reactants.'

From bond enthalpies (gas phase)

Δ_{r} H^{⊖} = \sum B.E.(bonds broken) - \sum B.E.(bonds formed)

'Reactant bonds − product bonds.' Valid only when all species are gaseous. For diatomics, bond enthalpy = bond dissociation enthalpy; for polyatomics use

mean

bond enthalpy (e.g. mean C–H in

C H_{4} = 416

kJ mo

l^{- 1}

🎯 Exam · Born–Haber cycle (lattice enthalpy)

Lattice enthalpy can't be measured directly, so it is found by a Hess cycle:

Δ_{f} H = Δ_{s u b} H + \frac{1}{2} Δ_{b o n d} H + Δ_{i} H + Δ_{e g} H - Δ_{l a tt i ce} H

(NaCl). The sum of all enthalpy changes around the closed cycle is zero.

✎ Example · Hess's law in action

Given

C (s) + O_{2} \to C O_{2}, Δ H = - 393.5

and

CO + \frac{1}{2} O_{2} \to C O_{2}, Δ H = - 283.0

kJ. Find

Δ H

for

C + \frac{1}{2} O_{2} \to CO

Target = (reaction 1) − (reaction 2).
$Δ H = - 393.5 - (- 283.0)$ .

∴

Δ H = - 110.5 kJ mol^{- 1}

🚫 Examiner Trap · Examiner traps

(1) From formation:

products - reactants

. From bond enthalpy:

bonds broken - bonds formed

(reactants

-

products) — OPPOSITE order, easy to flip. (2) Bond-enthalpy method needs ALL species gaseous. (3) Use

mean

bond enthalpy for polyatomics. (4) Born–Haber: lattice enthalpy from a closed Hess cycle (can't measure directly).

Spontaneity, Entropy & the Second Law

Spontaneous process: A process with the potential to proceed on its own without external help. It is irreversible and can only be reversed by an external agency. Spontaneity says nothing about the rate (e.g.

H_{2} + O_{2}

is spontaneous but slow).

⚠️ Watch out · Enthalpy decrease is NOT the sole criterion

Many

endothermic

reactions are spontaneous (e.g.

N_{2} + O_{2} \to 2NO, Δ H > 0

). So

Δ H < 0

alone cannot decide spontaneity — entropy must be included.

Entropy (

S

): A state function measuring the degree of randomness or disorder of a system. For a reversible change at temperature T,

Δ S = q_{re v} / T

(units J

K^{- 1}

l^{- 1}

Second law of thermodynamics

Δ S_{t o t a l} = Δ S_{sys} + Δ S_{s u rr} > 0 (spontaneous)

Δ S_{t o t a l} > 0

for any spontaneous (irreversible) change;

Δ S_{t o t a l} = 0

at equilibrium. The entropy of the universe always increases.

Predicting the sign of

Δ S

▸Increases ( $+$ ): solid → liquid → gas, dissolving, more gas moles in products, mixing.
▸Decreases ( $-$ ): gas → liquid/solid, fewer gas moles, association/ordering.
▸ $Δ S_{s u rr} = - \frac{Δ H _{sys}}{T}$ — an exothermic reaction raises the surroundings' entropy.

Entropy changes you can calculate

Δ S_{t r an s} = \frac{Δ H _{t r an s}}{T _{t r an s}}; Δ S_{i so t h er ma l} = n R ln \frac{V _{2}}{V _{1}} = 2.303 n R lo g \frac{p _{1}}{p _{2}}

phase change at its transition temperature (melting, boiling); ideal gas at constant T. Heating at constant p:

Δ S = n C_{p} ln \frac{T _{2}}{T _{1}}

. Kirchhoff (not named in the syllabus):

Δ H_{T_{2}} = Δ H_{T_{1}} + Δ C_{p} (T_{2} - T_{1})

🚫 Examiner Trap · Examiner traps

(1)

Δ H < 0

alone does NOT decide spontaneity — endothermic spontaneous reactions exist (entropy-driven). (2)

S_{g a s} > S_{l i q u i d} > S_{so l i d}

; more gas moles

\Rightarrow Δ S > 0

. (3)

Δ S_{s u rr} = - Δ H_{sys} / T

. (4) Second law:

Δ S_{t o t a l} > 0

for spontaneous,

= 0

at equilibrium — use the TOTAL, not just the system.

Gibbs Energy & Spontaneity

Gibbs energy (

G

G = H - TS

, an extensive state function.

Δ G

is the maximum useful (non-expansion) work a process can deliver at constant T and p — the 'free energy' of the reaction.

Gibbs–Helmholtz equation

Δ G = Δ H - T Δ S

Combines the enthalpy (

Δ H

) and entropy (

T Δ S

) drives into one spontaneity criterion at constant T,p.

★ Remember · The single criterion for spontaneity

Δ G < 0

: spontaneous (exergonic).

Δ G = 0

: equilibrium.

Δ G > 0

: non-spontaneous (the reverse is spontaneous).

Plot of delta G versus temperature as four straight lines for the four sign combinations of delta H and delta S, showing crossover where delta G equals zero — Slope $= - Δ S$ ; the line crosses $Δ G = 0$ at $T = Δ H /Δ S$ .

$Δ H$	$Δ S$	Spontaneous?
$-$	$+$	Yes, at all $T$
$-$	$-$	Yes, at low $T$
$+$	$+$	Yes, at high $T$
$+$	$-$	No, at any $T$

🎯 Exam · Crossover temperature

When

Δ H

and

Δ S

have the

same

sign, spontaneity flips at

T = \frac{Δ H}{Δ S}

(where

Δ G = 0

). This is why some reactions need heating (or cooling) to 'switch on'.

🚫 Examiner Trap · Examiner traps

(1)

Δ G = Δ H - T Δ S

is the single criterion:

< 0

spontaneous,

= 0

equilibrium,

> 0

non-spontaneous. (2) Same-sign

Δ H, Δ S

\Rightarrow

spontaneity flips at

T = Δ H /Δ S

. (3) Watch UNITS —

Δ S

in J/K but

Δ H

in kJ (convert!). (4)

Δ G

is the max NON-expansion (useful) work.

Gibbs Energy & Equilibrium

Gibbs energy and equilibrium constant

Δ_{r} G^{⊖} = - RT ln K = - 2.303 RT lo g K

At equilibrium

Δ_{r} G = 0

. This links thermodynamics to the position of equilibrium K.

Reading

Δ_{r} G^{⊖}

and K

▸ $Δ_{r} G^{⊖} < 0 \Rightarrow K > 1$ : products favoured (strongly exothermic reactions tend to have large K).
▸ $Δ_{r} G^{⊖} = 0 \Rightarrow K = 1$ : reactants and products balanced.
▸ $Δ_{r} G^{⊖} > 0 \Rightarrow K < 1$ : reactants favoured (strongly endothermic reactions form little product).

Combining both standard relations

Δ_{r} G^{⊖} = Δ_{r} H^{⊖} - T Δ_{r} S^{⊖} = - 2.303 RT lo g K

Lets you find K from calorimetric

Δ H^{⊖}

and

Δ S^{⊖}

— entropy can make K large even when

Δ H > 0

at high T.

🎯 Exam · Master checklist

Spontaneity

\Leftrightarrow Δ G < 0 \Leftrightarrow Δ S_{t o t a l} > 0 \Leftrightarrow K > 1

(for

Δ_{r} G^{⊖}

). Use

Δ H

(energy) and

T Δ S

(disorder)

together

— never

Δ H

alone.

✎ Example · Is the reaction spontaneous?

For a reaction

Δ H = + 30

kJ mo

l^{- 1}

Δ S = + 100

K^{- 1}

l^{- 1}

. Find the temperature above which it becomes spontaneous.

Spontaneous when $Δ G = Δ H - T Δ S < 0$ .
Crossover at $T = \frac{Δ H}{Δ S} = \frac{30000}{100}$ .

∴

T > 300 K

(above 300 K,

Δ G < 0

🚫 Examiner Trap · Examiner traps

(1)

Δ_{r} G^{⊖} = - RT ln K

Δ_{r} G^{⊖} < 0 \Rightarrow K > 1

(products). (2)

Δ_{r} G^{⊖}

(standard) decides K;

Δ_{r} G

(actual) decides direction NOW — don't confuse. (3) Even

Δ H > 0

can give large K at high T if

Δ S > 0

. (4) At equilibrium

Δ_{r} G = 0

, NOT

Δ_{r} G^{⊖} = 0

More JEE Main Chemistry formula sheets

Frequently Asked Questions

What are the most important Chemical Thermodynamics formulas for JEE Main?

This Chemical Thermodynamics formula sheet covers all the high-yield Chemistry formulas, definitions and theorems you need for JEE Main, across Concepts of System and types of systems, Surroundings, Work, heat, energy, Extensive and intensive properties, State functions — each shown with the key result and, where useful, a worked example.

Is this Chemical Thermodynamics formula sheet free?

Yes — the full chapter formula sheet is free to read online, no login or payment required.

How should I revise Chemical Thermodynamics formulas?

Blurt the Chemical Thermodynamics formulas from memory, then check against this sheet to find your gaps — and practise a few previous-year questions on the chapter to make sure you can apply them under time pressure.

Also useful: all formula sheetsJEE Main previous-year papersmost important chapters.