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Chemical Thermodynamics: one-page cheat sheet

NEET Chemistry · Unit 4 · Physical Chemistry

ScoreChemby CSJ Sir
Weightage: 2–3 questions · 8–12 marks

1 · Systems & properties

Openexchanges matter and energy

Closedexchanges energy only

Isolatedexchanges neither

State function

p, V, T, U, H, S, G

(path independent)

Path function

q

and

w

Extensivemass,

V

U

H

S

G

C

Intensive

T

, density,

C_{m}

E^{\circ}

; a ratio of two extensive properties is intensive

q

and

w

are path functions, but

q + w = Δ U

is a state function.

2 · First law, work & processes

Δ U = q + w w = - p_{ext} (V_{f} - V_{i})

w_{rev} = - 2.303 n R T lo g \frac{V _{f}}{V _{i}} = - 2.303 n R T lo g \frac{p _{i}}{p _{f}}

Process	Condition	Result
Isothermal	$Δ T = 0$	$Δ U = 0$ , so $q = - w$
Isochoric	$Δ V = 0$	$w = 0$ , $Δ U = q_{v}$
Isobaric	$Δ p = 0$	$Δ H = q_{p}$
Adiabatic	$q = 0$	$Δ U = w$

Free expansion

p_{ext} = 0

, so

w = 0

Heat absorbed and work done on the system are positive (IUPAC). Reversible isothermal expansion gives maximum work.

3 · Enthalpy

Δ H = Δ U + Δ n_{g} R T

Δ n_{g} = 0

Δ H = Δ U

_{2}

+ I

_{2} \to

2HI)

Δ n_{g} > 0

Δ H > Δ U

(PCl

_{5} \to

PCl

_{3}

+ Cl

_{2}

)

Δ n_{g} < 0

Δ H < Δ U

_{2}

+ 3H

_{2} \to

2NH

_{3}

)

$Δ n_{g}$ counts gaseous products minus gaseous reactants only.

4 · Heat capacity

C_{p} - C_{v} = R γ = \frac{C _{p}}{C _{v}}

Gas	$C_{v}$	$C_{p}$	$γ$
Monoatomic	$\frac{3}{2} R$	$\frac{5}{2} R$	1.67
Diatomic	$\frac{5}{2} R$	$\frac{7}{2} R$	1.40
Non-linear polyatomic	$3 R$	$4 R$	1.33

Ideal gas

Δ U = n C_{v} Δ T

Δ H = n C_{p} Δ T

for any process

5 · Thermochemistry

Hess's law

Δ_{r} H = Δ H_{1} + Δ H_{2} + \dots

Formation

Δ_{r} H^{\circ} = \sum Δ_{f} H^{\circ} (products) - \sum Δ_{f} H^{\circ} (reactants)

Elements

Δ_{f} H^{\circ} = 0

in the standard state (graphite, O

_{2}

, S rhombic)

Bond enthalpy

Δ_{r} H^{\circ} = \sum BE(reactants) - \sum BE(products)

, gas phase only

Born–Haber

Δ_{f} H = Δ_{sub} H + Δ_{i} H + \frac{1}{2} Δ_{bond} H + Δ_{eg} H - Δ_{lattice} H

6 · Entropy

Δ S = \frac{q _{rev}}{T}, Δ S_{total} = Δ S_{sys} + Δ S_{surr} > 0

Surroundings

Δ S_{surr} = - Δ H / T

at constant

T, P

S

risessolid

\to

liquid

\to

gas; heating; more moles of gas; dissociation

S

fallscooling a crystal to 0 K; gas moles decrease

7 · Gibbs energy

Δ G = Δ H - T Δ S, Δ G^{\circ} = - 2.303 R T lo g K

$Δ H$	$Δ S$	Spontaneous?
$-$	$+$	at all $T$
$-$	$-$	low $T$ only
$+$	$+$	high $T$ only
$+$	$-$	never

Equilibrium

Δ G = 0

, so

T_{eq} = Δ H^{\circ} /Δ S^{\circ}

8 · Types of enthalpy change

Combustion1 mol burnt completely in O

_{2}

; always negative

Fusion, vaporisation

Δ_{sub} H = Δ_{fus} H + Δ_{vap} H

Neutralisationstrong acid + strong base: about

- 57.1 kJ mol^{- 1}

of water formed

Lattice

Δ_{lattice} H

is positive as defined (crystal

\to

gaseous ions)

Seen in NEET 2026

$2 A (g) + B (g) \to 2 D (g)$ ; $Δ U^{⊖} = - 10$ kJ, $Δ S^{⊖} = - 44$ J K $^{- 1}$ at 298 K. Find $Δ G^{⊖}$ .

Δ n_{g} = - 1 : Δ H^{⊖} = - 10000 - 8.314 \times 298 = - 12478 J

Δ G^{⊖} = - 12478 + 298 \times 44 \approx + 634 J \Rightarrow non-spontaneous

Six traps to remember

$q$ and $w$ depend on path; $Δ U$ , $Δ H$ , $Δ S$ do not.
Reversible isothermal expansion gives maximum work; free expansion gives zero.
$Δ H = Δ U + Δ n_{g} R T$ : count gas moles only.
Bond-enthalpy formula is reactants minus products; formation formula is the reverse.
$Δ S_{surr} = - Δ H / T$ ; check the total entropy.
$Δ G$ sign depends on $T$ only when $Δ H$ and $Δ S$ have the same sign.

Chemical Thermodynamics: free one-page cheat sheet