Summary and Quick Revision

Card 1 — Thermodynamic Terms

The vocabulary the whole chapter runs on. Everything below is definition-level recall.

Term	Meaning
System	The part of the universe under observation
Surroundings	Everything else that can interact with the system
Boundary	The real or imaginary wall separating the two
Universe	System $+$ surroundings

Types of system

Type	Matter exchanged	Energy exchanged	Standard example
Open	yes	yes	Reactants in an open beaker
Closed	no	yes	Reactants in a sealed copper or steel vessel
Isolated	no	no	Hot tea in a stoppered thermos flask

Key Point: Every system exchanging matter also exchanges energy, so there is no fourth type. A sealed but conducting vessel is closed; a sealed and insulated vessel is isolated.

Walls

Wall	Heat passes	Consequence
Adiabatic	no	$q = 0$, so $\Delta U = w$
Diathermic	yes	Thermal equilibrium with surroundings; $T_{sys} = T_{surr}$

State of a system and state variables

The state of a system is fixed by its state variables: $p$, $V$, $T$ and $n$. For a fixed amount of an ideal gas, any two of $p$, $V$, $T$ fix the third through $pV = nRT$. Thermodynamics applies only to systems at equilibrium or moving between equilibrium states.

State functions against path functions

Key Point (Definition): A state function depends only on the present state of the system, not on how the system reached it. Its change is (final $-$ initial). A path function depends on the route taken and has no "change" of its own.

State functions	Path functions
$U$, $H$, $S$, $G$, $p$, $V$, $T$, $n$, density, $C_p$, $C_V$	$q$ (heat), $w$ (work)

• Written $\Delta U$, $\Delta H$, $\Delta S$, $\Delta G$ — never $\Delta q$ or $\Delta w$.
• $q$ and $w$ are individually path-dependent, yet the sum $q + w$ is not: $q + w = \Delta U$ is a state function. That is the content of the first law.
• Over any complete cycle, every state function returns to its starting value: $\Delta U = \Delta H = \Delta S = \Delta G = 0$ for a cyclic process, while $q$ and $w$ need not be zero.

Extensive against intensive

Key Point (Definition): An extensive property depends on the amount of matter present. An intensive property does not; it is the same for a part of the system as for the whole.

Extensive	Intensive
mass, volume, $n$, $U$, $H$, $S$, $G$, $C$ (heat capacity), total charge	temperature, pressure, density, viscosity, refractive index, molar mass, molarity, specific heat $c$, molar heat capacity $C_m$, $S^{\circ}$ per mole, surface tension

Test: halve the system. An extensive property halves; an intensive one does not change. Any extensive property divided by amount of substance becomes intensive — heat capacity $C$ is extensive, molar heat capacity $C_m$ is intensive.

Types of process

Process	Condition held fixed	Immediate consequence
Isothermal	$T$ constant	$\Delta U = 0$ and $\Delta H = 0$ for an ideal gas
Isobaric	$p$ constant	$q_p = \Delta H$
Isochoric	$V$ constant	$w = 0$, $q_V = \Delta U$
Adiabatic	no heat exchange	$q = 0$, $\Delta U = w$
Cyclic	returns to initial state	$\Delta U = 0$, so $q = -w$

[NEET] Statement-type questions almost always test one of three things: which properties are intensive, which quantities are state functions, and which system type a described container is. All three are in the tables above.

Card 2 — Internal Energy, Work, Heat and the First Law

Key Point (Definition): Internal energy $U$ is the total energy stored in a system — translational, rotational, vibrational, electronic, nuclear and the energy of interaction between particles. It is a state function and an extensive property. Only $\Delta U$ can be measured, never the absolute value of $U$.

Two and only two ways exist to change $U$ in a closed system: transfer of work and transfer of heat.

The IUPAC sign convention

Quantity	Positive when	Negative when
$q$	heat is absorbed by the system (endothermic)	heat is released by the system (exothermic)
$w$	work is done on the system (compression)	work is done by the system (expansion)
$\Delta U$	the internal energy of the system rises	the internal energy of the system falls

Key Point: Energy entering the system is positive; energy leaving the system is negative. That one sentence generates the whole table. Some physics texts define $w$ with the opposite sign and write $\Delta U = q - w$; every equation in this chapter uses the IUPAC form.

The first law

$$\Delta U = q + w$$

Key Point: The energy of an isolated system is constant. Energy can be converted from one form to another but can neither be created nor destroyed.

$\Delta U$ for a chemical change is $U_{products} - U_{reactants}$, and for a general change $U_2 - U_1$.

Every special case in one table

Condition	What it forces	First law becomes
Isolated system	$q = 0$ and $w = 0$	$\Delta U = 0$
Adiabatic	$q = 0$	$\Delta U = w$
Constant volume (rigid vessel)	$w = 0$	$\Delta U = q_V$
Constant pressure	$w = -p\Delta V$	$\Delta U = q_p - p\Delta V$, so $q_p = \Delta H$
Cyclic process	$\Delta U = 0$	$q = -w$
Isothermal, ideal gas	$\Delta U = 0$	$q = -w$
Free expansion into vacuum	$p_{ex} = 0$, insulated	$q = 0$, $w = 0$, $\Delta U = 0$

Sign readings worth memorising

Observation	$q$	$w$	Effect on $\Delta U$
Gas expands against the atmosphere	—	negative	lowers $U$
Gas is compressed by a piston	—	positive	raises $U$
Exothermic reaction	negative	—	lowers $U$
Endothermic reaction	positive	—	raises $U$
Adiabatic compression	$0$	positive	$U$ rises, so $T$ rises
Adiabatic expansion	$0$	negative	$U$ falls, so $T$ falls

Adiabatic compression heats a gas and adiabatic expansion cools it, with no heat crossing the boundary at all. Both follow from $\Delta U = w$.

Units

$$1\,\mathrm{cal} = 4.184\,\mathrm{J} \qquad 1\,\mathrm{L\,atm} = 101.3\,\mathrm{J} \qquad 1\,\mathrm{kJ} = 10^{3}\,\mathrm{J}$$

$$R = 8.314\,\mathrm{J\,K^{-1}\,mol^{-1}} = 0.08206\,\mathrm{L\,atm\,K^{-1}\,mol^{-1}} \approx 2\,\mathrm{cal\,K^{-1}\,mol^{-1}}$$

[JEE/NEET] When a question reports "work done by the gas $= 200\,\mathrm{J}$", the value to substitute is $w = -200\,\mathrm{J}$. Reading the direction from the wording before writing the sign is the single highest-yield habit in this chapter.

Card 3 — Work of Expansion

All of the work in this chapter is pressure-volume work. The pressure in every expression is the external pressure, not the pressure of the gas inside.

$$w = -p_{ex}\,\Delta V = -p_{ex}(V_f - V_i)$$

Process	Work expression	$p$-$V$ area
Free expansion into vacuum	$w = 0$ (since $p_{ex} = 0$)	no area at all
Single step against constant $p_{ex}$	$w = -p_{ex}\Delta V$	one low rectangle
Several steps of falling $p_{ex}$	$w = -\sum p_{ex}\Delta V$	a staircase, larger area
Reversible isothermal, ideal gas	$w = -2.303\,nRT\log\dfrac{V_f}{V_i}$	full area under the smooth isotherm
Constant volume (rigid vessel)	$w = 0$	a vertical line, zero area
General variable pressure	$w = -\displaystyle\int_{V_i}^{V_f} p_{ex}\,\mathrm{d}V$	area under the actual curve

Reversible isothermal expansion of an ideal gas

$$w_{rev} = -2.303\,nRT\log\frac{V_f}{V_i} = -2.303\,nRT\log\frac{p_i}{p_f}$$

Boyle's law at fixed $T$ gives $V_f/V_i = p_i/p_f$, so either ratio may be used. Since $\Delta U = 0$ for an isothermal ideal gas,

$$q_{rev} = -w_{rev} = +2.303\,nRT\log\frac{V_f}{V_i}$$

Key Point: A reversible process runs through a continuous sequence of equilibrium states, driven by an infinitesimal pressure difference. It delivers the maximum work an expansion between two given states can give, and requires the minimum work for a compression between the same two states.

$$\lvert w_{rev} \rvert > \lvert w_{irrev} \rvert \quad \text{for expansion between the same two states}$$

Signs, checked

Change	$\Delta V$	Sign of $w$	Meaning
Expansion	positive	$w < 0$	the system does work on the surroundings, losing energy
Compression	negative	$w > 0$	the surroundings do work on the system, adding energy
No volume change	zero	$w = 0$	nothing to push

Free expansion

$p_{ex} = 0$ into an evacuated space, so $w = 0$ whatever the volume change. If the container is also insulated, $q = 0$ and $\Delta U = 0$, and for an ideal gas the temperature does not change either.

Key Point: "Expands into vacuum" is an instruction to write $w = 0$ before reading any numbers. The volumes quoted in such a question do not enter the answer.

Unit conversion

$$1\,\mathrm{L\,atm} = 101.3\,\mathrm{J}$$

Pressure in atm times volume in litres gives L atm; multiply by $101.3$ to reach joules. A work value of a few L atm should land in the hundreds of joules.

Reading a $p$-$V$ diagram

• The magnitude of the work is the area under the curve between $V_i$ and $V_f$.
• Moving rightwards (expansion) makes $w$ negative; leftwards (compression) makes $w$ positive.
• A closed cycle traced clockwise gives $w_{net} < 0$; anticlockwise gives $w_{net} > 0$. For any cycle $\Delta U = 0$, so $q_{net} = -w_{net}$.

[JEE Main] Choose $R$ to match the units in the options: $8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}$ gives joules, $0.08206\,\mathrm{L\,atm\,K^{-1}\,mol^{-1}}$ gives L atm. $T$ is always in kelvin.

Card 4 — Enthalpy

Most reactions are run in open vessels at atmospheric pressure, where the system may change volume and do work. Enthalpy is the state function built to handle that case.

$$H = U + pV$$

$H$ is a state function because $U$, $p$ and $V$ all are, and it is extensive. Absolute $H$ cannot be measured; only $\Delta H$ can.

$$\Delta H = \Delta U + p\Delta V \quad \text{(constant } p\text{)} \qquad \Longrightarrow \qquad \boxed{\Delta H = q_p}$$

Key Point: $\Delta H$ is the heat exchanged at constant pressure and $\Delta U$ the heat exchanged at constant volume. Both are state functions; the heats they equal are path functions that happen to be fixed once the path condition is stated.

Reaction type	Sign of $\Delta H$	Heat	Surroundings
Exothermic	$\Delta H < 0$	released by the system	get warmer
Endothermic	$\Delta H > 0$	absorbed by the system	get cooler

$\Delta H = H_{products} - H_{reactants}$, so a negative value means the products lie lower in enthalpy than the reactants.

Relating the two

For a reaction involving gases treated as ideal, $p\Delta V = \Delta n_g RT$, giving

$$\Delta H = \Delta U + \Delta n_g RT$$

Key Point (Definition): $\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$, counted from the balanced equation. Solids, liquids and species in solution are not counted.

At $298\,\mathrm{K}$, $RT = 8.314 \times 298 = 2477.6\,\mathrm{J\,mol^{-1}} = 2.478\,\mathrm{kJ\,mol^{-1}}$, so each unit of $\Delta n_g$ shifts $\Delta H$ from $\Delta U$ by about $2.5\,\mathrm{kJ\,mol^{-1}}$.

Worked $\Delta n_g$ values

Reaction	$\Delta n_g$	Relation
$\mathrm{CaCO_3(s)} \rightarrow \mathrm{CaO(s)} + \mathrm{CO_2(g)}$	$1 - 0 = +1$	$\Delta H > \Delta U$
$\mathrm{N_2(g)} + 3\,\mathrm{H_2(g)} \rightarrow 2\,\mathrm{NH_3(g)}$	$2 - 4 = -2$	$\Delta H < \Delta U$
$\mathrm{H_2(g)} + \mathrm{Cl_2(g)} \rightarrow 2\,\mathrm{HCl(g)}$	$2 - 2 = 0$	$\Delta H = \Delta U$
$\mathrm{CH_4(g)} + 2\,\mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)} + 2\,\mathrm{H_2O(l)}$	$1 - 3 = -2$	water is liquid, not counted
$\mathrm{C_6H_{12}O_6(s)} + 6\,\mathrm{O_2(g)} \rightarrow 6\,\mathrm{CO_2(g)} + 6\,\mathrm{H_2O(l)}$	$6 - 6 = 0$	$\Delta H = \Delta U$
$\mathrm{C(graphite,s)} + \mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)}$	$1 - 1 = 0$	graphite not counted
$\mathrm{C(graphite,s)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{CO(g)}$	$1 - \tfrac{1}{2} = +\tfrac{1}{2}$	fractions are allowed
$2\,\mathrm{H_2(g)} + \mathrm{O_2(g)} \rightarrow 2\,\mathrm{H_2O(l)}$	$0 - 3 = -3$	large negative correction
$\mathrm{H_2O(l)} \rightarrow \mathrm{H_2O(g)}$	$1 - 0 = +1$	phase changes have $\Delta n_g$ too
$\mathrm{NH_4NO_3(s)} \rightarrow \mathrm{N_2O(g)} + 2\,\mathrm{H_2O(g)}$	$3 - 0 = +3$

When the two are equal

$\Delta H = \Delta U$ exactly when $\Delta n_g = 0$. Two situations produce it:

1. Equal moles of gas on both sides, as in $\mathrm{H_2(g)} + \mathrm{Cl_2(g)} \rightarrow 2\,\mathrm{HCl(g)}$.
2. No gases at all — reactions entirely among solids, liquids or solutions, where the volume change is negligible.

Note what does not qualify: a sealed rigid vessel. Fixing the volume makes $q_V$ equal $\Delta U$, but $\Delta H$ still differs from $\Delta U$ by $\Delta n_g RT$, because at constant volume $\Delta H - \Delta U = V\Delta p$, which for a reaction quoted at a fixed temperature is the same $\Delta n_g RT$ as always.

Key Point: The two differ by at most a few kilojoules per mole while reaction enthalpies run to hundreds. The correction is small but it is not optional, and its sign follows the sign of $\Delta n_g$.

[Board] Convert $RT$ to kilojoules before adding it to a $\Delta U$ quoted in $\mathrm{kJ\,mol^{-1}}$. Mixing $2477.6\,\mathrm{J}$ into a kilojoule equation is the commonest arithmetic slip here.

Card 5 — Heat Capacity and Calorimetry

$$q = C\,\Delta T$$

Symbol	Name	Definition	Unit
$C$	Heat capacity	heat needed to raise the temperature of the whole sample by $1\,\mathrm{K}$	$\mathrm{J\,K^{-1}}$
$c$	Specific heat capacity	heat needed per gram per kelvin	$\mathrm{J\,g^{-1}\,K^{-1}}$
$C_m$	Molar heat capacity	heat needed per mole per kelvin	$\mathrm{J\,K^{-1}\,mol^{-1}}$

$$q = m\,c\,\Delta T = n\,C_m\,\Delta T$$

$C$ is extensive; $c$ and $C_m$ are intensive. For water, $c = 4.18\,\mathrm{J\,g^{-1}\,K^{-1}}$ and $C_m = 75.3\,\mathrm{J\,K^{-1}\,mol^{-1}}$.

Two heat capacities, two conditions

At constant volume	At constant pressure
$w = 0$, so all the heat raises $U$	some heat is spent doing expansion work
$q_V = \Delta U = n\,C_V\,\Delta T$	$q_p = \Delta H = n\,C_p\,\Delta T$
$C_V = \left(\dfrac{\partial U}{\partial T}\right)_V$	$C_p = \left(\dfrac{\partial H}{\partial T}\right)_p$

Heating at constant pressure requires more heat for the same temperature rise, because part of the energy leaves as work pushing the surroundings back. So $C_p > C_V$ always.

Key Point: For one mole of an ideal gas, $C_p - C_V = R$. The proof is two lines: $H = U + pV = U + nRT$, so for one mole $\Delta H = \Delta U + R\Delta T$, and dividing by $\Delta T$ gives $C_p = C_V + R$.

Gas type	$C_V$	$C_p$	$\gamma = C_p/C_V$
Monatomic	$\tfrac{3}{2}R = 12.47$	$\tfrac{5}{2}R = 20.79$	$1.67$
Diatomic	$\tfrac{5}{2}R = 20.79$	$\tfrac{7}{2}R = 29.10$	$1.40$

(Values in $\mathrm{J\,K^{-1}\,mol^{-1}}$.)

Calorimetry: the instrument decides the quantity

	Bomb calorimeter	Coffee-cup calorimeter
Vessel	thick sealed steel bomb in a water bath	insulated polystyrene cup, open to air
Condition	constant volume	constant pressure
Work done	$w = 0$	expansion work possible
Measures	$\Delta U$ directly	$\Delta H$ directly
Heat counted as	$q = -C_{cal}\Delta T$	$q = -m\,c\,\Delta T$
Typical use	enthalpies of combustion	neutralisation, dissolution, dilution

$C_{cal}$ is the calorimeter constant, the combined heat capacity of bomb, water and fittings, in $\mathrm{J\,K^{-1}}$. Working rule for both instruments: heat gained by the calorimeter equals heat lost by the reaction, so the reaction quantity carries the opposite sign to the temperature rise.

$$\Delta U = \frac{q}{n} \quad\text{(bomb)} \qquad\qquad \Delta H = \frac{q}{n} \quad\text{(cup)}$$

with $n$ the moles of the limiting substance, so that the answer comes out per mole.

Converting a bomb result

$$\Delta H = \Delta U + \Delta n_g RT$$

A bomb hands you $\Delta U$; tables quote $\Delta H$. The correction is a couple of kilojoules per mole of gas, and its sign follows $\Delta n_g$.

[NEET] Match instrument to quantity before starting any arithmetic. "Bomb calorimeter" plus "find $\Delta H$" means a $\Delta n_g RT$ correction is coming; "polystyrene cup" plus "find $\Delta H$" means none is needed.

Card 6 — Reaction Enthalpy, Standard States and Hess's Law

Key Point (Definition): The reaction enthalpy $\Delta_r H$ is the enthalpy change for the reaction exactly as written, with the stoichiometric coefficients read as moles. Its unit is $\mathrm{kJ\,mol^{-1}}$, meaning per mole of reaction as the equation stands.

$$\Delta_r H = \sum a_i H_{products} - \sum b_i H_{reactants}$$

Standard state

Key Point (Definition): The standard state of a substance at a specified temperature is its pure form at $1\,\mathrm{bar}$. The superscript circle marks it: $\Delta_r H^{\circ}$. Data are almost always quoted at $298\,\mathrm{K}$, but the standard state fixes pressure, not temperature.

At $298\,\mathrm{K}$ and $1\,\mathrm{bar}$ water is a liquid, so standard equations end in $\mathrm{H_2O(l)}$.

Thermochemical equation rules

Operation on the equation	Effect on $\Delta_r H$
Reverse the equation	reverse the sign, same magnitude
Multiply every coefficient by $n$	multiply $\Delta_r H$ by $n$
Divide every coefficient by $n$	divide $\Delta_r H$ by $n$
Add two equations	add the two $\Delta_r H$ values
Change the physical state of any species	$\Delta_r H$ changes; the equation is a different one

Physical state labels are part of the equation:

$$\mathrm{H_2(g)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{H_2O(l)}; \qquad \Delta_r H^{\circ} = -285.8\,\mathrm{kJ\,mol^{-1}}$$

$$\mathrm{H_2(g)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{H_2O(g)}; \qquad \Delta_r H^{\circ} = -241.8\,\mathrm{kJ\,mol^{-1}}$$

The gap of $44.0\,\mathrm{kJ\,mol^{-1}}$ is exactly the enthalpy of vaporisation of water at $298\,\mathrm{K}$.

Hess's law

Key Point: Hess's law of constant heat summation — the enthalpy change of a reaction is the same whether it takes place in one step or in several, provided the initial and final states are the same. It is a direct consequence of $H$ being a state function.

Practical form: if a target equation can be built by reversing, scaling and adding known equations, then combining their $\Delta H$ values in exactly the same way gives the target $\Delta H$.

A worked chain in three lines. Given

$$\mathrm{C(graphite,s)} + \mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)}; \qquad \Delta H_1 = -393.5\,\mathrm{kJ\,mol^{-1}}$$

$$\mathrm{CO(g)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{CO_2(g)}; \qquad \Delta H_2 = -283.0\,\mathrm{kJ\,mol^{-1}}$$

Reverse the second and add: $\Delta_f H^{\circ}[\mathrm{CO(g)}] = \Delta H_1 - \Delta H_2 = -393.5 + 283.0 = -110.5\,\mathrm{kJ\,mol^{-1}}$.

The formation-enthalpy formula

$$\Delta_r H^{\circ} = \sum a_i\,\Delta_f H^{\circ}(\text{products}) - \sum b_i\,\Delta_f H^{\circ}(\text{reactants})$$

with $a_i$ and $b_i$ the coefficients in the balanced equation.

Key Point: $\Delta_f H^{\circ}$ of any element in its reference state is exactly zero: $\mathrm{O_2(g)}$, $\mathrm{H_2(g)}$, $\mathrm{N_2(g)}$, $\mathrm{C(graphite,s)}$, $\mathrm{S(rhombic,s)}$, $\mathrm{Br_2(l)}$, $\mathrm{Hg(l)}$, $\mathrm{Na(s)}$. Non-reference forms are not zero: $\Delta_f H^{\circ}[\mathrm{C(diamond,s)}] = +1.90$ and $\Delta_f H^{\circ}[\mathrm{O_3(g)}] = +142.7\,\mathrm{kJ\,mol^{-1}}$.

Formation data worth carrying in

Substance	$\Delta_f H^{\circ}$ / kJ mol$^{-1}$	Substance	$\Delta_f H^{\circ}$ / kJ mol$^{-1}$
$\mathrm{H_2O(l)}$	$-285.8$	$\mathrm{CaO(s)}$	$-635.1$
$\mathrm{H_2O(g)}$	$-241.8$	$\mathrm{CaCO_3(s)}$	$-1206.9$
$\mathrm{CO_2(g)}$	$-393.5$	$\mathrm{NaCl(s)}$	$-411.2$
$\mathrm{CO(g)}$	$-110.5$	$\mathrm{Fe_2O_3(s)}$	$-824.2$
$\mathrm{CH_4(g)}$	$-74.8$	$\mathrm{HCl(g)}$	$-92.3$
$\mathrm{NH_3(g)}$	$-46.2$	$\mathrm{C_2H_5OH(l)}$	$-277.7$

Sample use: for $\mathrm{CaCO_3(s)} \rightarrow \mathrm{CaO(s)} + \mathrm{CO_2(g)}$,

$$\Delta_r H^{\circ} = (-635.1) + (-393.5) - (-1206.9) = +178.3\,\mathrm{kJ\,mol^{-1}}$$

[Board] Write "products minus reactants" at the top of the page before substituting. Reversing that order is the single commonest source of a sign error in a Hess's law question.

Card 7 — The Named Enthalpies

Every enthalpy the chapter names, with its symbol, its definition, its invariable sign and a value to anchor it.

Enthalpy	Symbol	Defined as the enthalpy change when	Sign	Anchor value
Combustion	$\Delta_c H^{\circ}$	one mole of a substance burns completely in excess oxygen	always $-$	$\mathrm{CH_4}$: $-890.3\,\mathrm{kJ\,mol^{-1}}$
Formation	$\Delta_f H^{\circ}$	one mole of a compound forms from its elements in their reference states	either	$\mathrm{H_2O(l)}$: $-285.8\,\mathrm{kJ\,mol^{-1}}$
Fusion	$\Delta_{fus}H^{\circ}$	one mole of solid melts at its melting point	always $+$	ice: $+6.00\,\mathrm{kJ\,mol^{-1}}$
Vaporisation	$\Delta_{vap}H^{\circ}$	one mole of liquid becomes vapour at its boiling point	always $+$	water at $373\,\mathrm{K}$: $+40.79\,\mathrm{kJ\,mol^{-1}}$
Sublimation	$\Delta_{sub}H^{\circ}$	one mole of solid becomes vapour directly	always $+$	$\mathrm{CO_2}$ at $195\,\mathrm{K}$: $+25.2\,\mathrm{kJ\,mol^{-1}}$
Atomisation	$\Delta_a H^{\circ}$	one mole of a substance is broken into free gaseous atoms	always $+$	$\mathrm{CH_4}$: $+1665\,\mathrm{kJ\,mol^{-1}}$
Bond dissociation	$\Delta_{bond}H^{\circ}$	one mole of a named bond breaks in the gas phase	always $+$	$\mathrm{H-H}$: $+435.8\,\mathrm{kJ\,mol^{-1}}$
Ionization	$\Delta_i H^{\circ}$	one mole of gaseous atoms loses one electron each	always $+$	$\mathrm{Na(g)}$: $+496\,\mathrm{kJ\,mol^{-1}}$
Electron gain	$\Delta_{eg}H^{\circ}$	one mole of gaseous atoms gains one electron each	usually $-$	$\mathrm{Cl(g)}$: $-348.6\,\mathrm{kJ\,mol^{-1}}$
Solution	$\Delta_{sol}H^{\circ}$	one mole of solute dissolves in a large excess of solvent	either	$\mathrm{NH_4Cl}$: about $+15\,\mathrm{kJ\,mol^{-1}}$
Dilution	$\Delta_{dil}H^{\circ}$	a solution is diluted further, per mole of solute	either	small
Lattice	$\Delta_{lattice}H^{\circ}$	one mole of an ionic solid separates into gaseous ions	always $+$	$\mathrm{NaCl}$: $+788\,\mathrm{kJ\,mol^{-1}}$
Hydration	$\Delta_{hyd}H^{\circ}$	one mole of gaseous ions is hydrated by water	always $-$	large negative

Phase-transition relation

$$\Delta_{sub}H^{\circ} = \Delta_{fus}H^{\circ} + \Delta_{vap}H^{\circ} \qquad \text{(same temperature)}$$

For every substance $\Delta_{vap}H^{\circ} \gg \Delta_{fus}H^{\circ}$: melting only loosens the arrangement, vaporising removes the attractions entirely. Reverse changes — freezing, condensation, deposition — carry the same magnitude with the opposite sign.

Bond dissociation against mean bond enthalpy

Bond dissociation enthalpy	Mean bond enthalpy
A specific bond in a specific molecule	The average over all bonds of that type in a molecule, or across many molecules
$\mathrm{H_2O} \rightarrow \mathrm{H} + \mathrm{OH}$: $502\,\mathrm{kJ\,mol^{-1}}$	$\mathrm{O-H}$ in water: $(502 + 427)/2 = 464.5\,\mathrm{kJ\,mol^{-1}}$
$\mathrm{OH} \rightarrow \mathrm{H} + \mathrm{O}$: $427\,\mathrm{kJ\,mol^{-1}}$	Tabulated values are means
Exact, but only for that one step	Approximate, so bond-enthalpy estimates of $\Delta_r H$ are approximate

For a diatomic molecule the two are the same number, and both equal the enthalpy of atomisation.

Estimating a reaction enthalpy from bond enthalpies

$$\Delta_r H^{\circ} = \sum \Delta_{bond}H^{\circ}(\text{bonds broken, reactants}) - \sum \Delta_{bond}H^{\circ}(\text{bonds formed, products})$$

Key Point: This is the one formula in the chapter that reads reactants minus products, because bond breaking absorbs energy and bond forming releases it. It is valid only when every species is gaseous.

Useful mean values in $\mathrm{kJ\,mol^{-1}}$: $\mathrm{H-H}$ $435.8$, $\mathrm{Cl-Cl}$ $242$, $\mathrm{H-Cl}$ $431$, $\mathrm{O=O}$ $498$, $\mathrm{O-H}$ $464$, $\mathrm{C-H}$ $414$, $\mathrm{C-C}$ $347$, $\mathrm{C=C}$ $611$, $\mathrm{N} \equiv \mathrm{N}$ $946$.

Enthalpy of solution and the Born-Haber cycle

$$\Delta_{sol}H^{\circ} = \Delta_{lattice}H^{\circ} + \Delta_{hyd}H^{\circ}$$

The lattice enthalpy is positive and the hydration enthalpy negative; whichever is larger in magnitude decides whether dissolving warms or cools the solution. A salt whose lattice enthalpy far exceeds its hydration enthalpy is sparingly soluble.

Lattice enthalpy cannot be measured directly. It comes from a Born-Haber cycle, Hess's law applied to the formation of an ionic solid:

$$\Delta_f H^{\circ}[\mathrm{NaCl}] = \Delta_{sub}H^{\circ}[\mathrm{Na}] + \Delta_i H^{\circ}[\mathrm{Na}] + \tfrac{1}{2}\Delta_{bond}H^{\circ}[\mathrm{Cl_2}] + \Delta_{eg}H^{\circ}[\mathrm{Cl}] - \Delta_{lattice}H^{\circ}$$

$$-411.2 = 108.4 + 496 + 121 - 348.6 - 788$$

[JEE Main] In a Born-Haber question, check first whether the given lattice enthalpy refers to the solid breaking into gaseous ions (positive) or to ions coming together to form the solid (negative). The two differ only in sign, and the whole answer turns on it.

Card 8 — Entropy and the Second Law

Key Point (Definition): Entropy $S$ is a measure of the degree of randomness or disorder of a system, counting both the arrangement of the particles and the spread of energy among them. It is a state function and an extensive property, quoted in $\mathrm{J\,K^{-1}\,mol^{-1}}$.

$$S_{gas} \gg S_{liquid} > S_{solid}$$

Reading the sign of $\Delta S$ without data

Change	$\Delta S$
Solid $\rightarrow$ liquid $\rightarrow$ gas	positive
Gas $\rightarrow$ liquid $\rightarrow$ solid	negative
$\Delta n_g > 0$ (more moles of gas produced)	positive
$\Delta n_g < 0$ (moles of gas consumed)	negative
$\Delta n_g = 0$	small; decided by the non-gaseous species
Dissolving a solid in water	usually positive
Mixing two gases	positive
Raising the temperature	positive
Expanding a gas into a larger volume	positive
A perfect crystal at $0\,\mathrm{K}$	$S = 0$

Gases dominate the entropy account, so counting $\Delta n_g$ settles most sign questions in one step.

Reaction	$\Delta n_g$	Sign of $\Delta_r S$
$\mathrm{CaCO_3(s)} \rightarrow \mathrm{CaO(s)} + \mathrm{CO_2(g)}$	$+1$	positive
$\mathrm{N_2(g)} + 3\,\mathrm{H_2(g)} \rightarrow 2\,\mathrm{NH_3(g)}$	$-2$	negative
$4\,\mathrm{Fe(s)} + 3\,\mathrm{O_2(g)} \rightarrow 2\,\mathrm{Fe_2O_3(s)}$	$-3$	negative
$\mathrm{H_2O(l)} \rightarrow \mathrm{H_2O(g)}$	$+1$	positive
$\mathrm{H_2(g)} + \mathrm{I_2(g)} \rightarrow 2\,\mathrm{HI(g)}$	$0$	near zero

The quantitative definition

$$\Delta S = \frac{q_{rev}}{T}$$

$q_{rev}$ is the heat absorbed along a reversible path at temperature $T$. Since $S$ is a state function, $\Delta S$ between two states is the same for any path; the reversible path is simply the one that lets you compute it. A phase change at its own transition temperature is reversible as it stands:

$$\Delta_{fus}S = \frac{\Delta_{fus}H}{T_f} \qquad \Delta_{vap}S = \frac{\Delta_{vap}H}{T_b}$$

Ice at $273\,\mathrm{K}$: $\Delta_{fus}S = 6000/273 = 22.0\,\mathrm{J\,K^{-1}\,mol^{-1}}$. Water at $373\,\mathrm{K}$: $\Delta_{vap}S = 40790/373 = 109.4\,\mathrm{J\,K^{-1}\,mol^{-1}}$.

The surroundings

$$\Delta S_{surr} = -\frac{\Delta H_{sys}}{T}$$

An exothermic reaction has $\Delta H_{sys} < 0$, which makes $\Delta S_{surr}$ positive. The heat dumped into the surroundings raises their entropy, and that is how an exothermic reaction drives itself.

The second law

$$\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr}$$

$\Delta S_{total}$	Verdict
$> 0$	spontaneous
$= 0$	equilibrium
$< 0$	non-spontaneous; the reverse change is the spontaneous one

Key Point: For a spontaneous process the total entropy of system plus surroundings increases. The entropy of the system alone may fall — water freezes, plants build sugar, iron rusts — provided the surroundings gain more than the system loses.

For an isolated system there are no surroundings to exchange heat with, so $\Delta S_{surr} = 0$ and the criterion reduces to $\Delta S_{sys} > 0$.

Standard entropy change of a reaction

$$\Delta_r S^{\circ} = \sum \nu\,S^{\circ}(\text{products}) - \sum \nu\,S^{\circ}(\text{reactants})$$

Unlike formation enthalpies, standard entropies of elements are not zero. Every pure substance has a positive $S^{\circ}$ at $298\,\mathrm{K}$.

Substance	$S^{\circ}$ / J K$^{-1}$ mol$^{-1}$	Substance	$S^{\circ}$ / J K$^{-1}$ mol$^{-1}$
$\mathrm{C(graphite,s)}$	$5.7$	$\mathrm{H_2(g)}$	$130.7$
$\mathrm{CaO(s)}$	$39.7$	$\mathrm{H_2O(g)}$	$188.8$
$\mathrm{H_2O(l)}$	$70.0$	$\mathrm{N_2(g)}$	$191.6$
$\mathrm{CaCO_3(s)}$	$92.9$	$\mathrm{O_2(g)}$	$205.0$
$\mathrm{NH_3(g)}$	$192.5$	$\mathrm{CO_2(g)}$	$213.8$

[JEE/NEET] Convert $\Delta H$ from $\mathrm{kJ\,mol^{-1}}$ to $\mathrm{J\,mol^{-1}}$ before dividing by $T$ in $\Delta S_{surr}$, and keep the sign of $\Delta H$ inside the expression.

Card 9 — Gibbs Energy

The second law needs the surroundings. Gibbs energy removes them, leaving a criterion written entirely in system properties.

$$G = H - TS \qquad\qquad \Delta G = \Delta H - T\Delta S \quad (\text{constant } T)$$

$G$ is a state function and an extensive property, with units of energy. Substituting $\Delta S_{surr} = -\Delta H/T$ into the second law gives the identity

$$\Delta G = -T\,\Delta S_{total}$$

so a negative $\Delta G$ and a positive $\Delta S_{total}$ always say the same thing.

$\Delta G$	Verdict	What it means physically
$\Delta G < 0$	spontaneous	the forward change proceeds on its own
$\Delta G = 0$	equilibrium	forward and reverse balance; $G$ is at its minimum
$\Delta G > 0$	non-spontaneous	the reverse change is spontaneous; the forward one needs work from outside

Key Point: The criterion $\Delta G < 0$ holds at constant temperature and constant pressure. A positive $\Delta G$ never means "impossible" — electrolysis of water has $\Delta G > 0$ and runs perfectly well while the power supply is on.

The four cases

$\Delta_r H$	$\Delta_r S$	$\Delta_r G = \Delta H - T\Delta S$	Spontaneity	Example
$-$	$+$	negative at every $T$	spontaneous at all temperatures	$2\,\mathrm{H_2O_2(l)} \rightarrow 2\,\mathrm{H_2O(l)} + \mathrm{O_2(g)}$
$+$	$-$	positive at every $T$	non-spontaneous at all temperatures	$3\,\mathrm{O_2(g)} \rightarrow 2\,\mathrm{O_3(g)}$
$-$	$-$	negative at low $T$, positive at high $T$	spontaneous only below the crossover	$\mathrm{N_2(g)} + 3\,\mathrm{H_2(g)} \rightarrow 2\,\mathrm{NH_3(g)}$
$+$	$+$	positive at low $T$, negative at high $T$	spontaneous only above the crossover	$\mathrm{CaCO_3(s)} \rightarrow \mathrm{CaO(s)} + \mathrm{CO_2(g)}$

$T$ multiplies $\Delta S$ only, so raising $T$ strengthens the entropy term and leaves the enthalpy term where it is. Both terms favourable, nothing can spoil it; both unfavourable, nothing can rescue it. When $\Delta H$ is favourable and $\Delta S$ is not, cooling helps because $T\lvert\Delta S\rvert$ shrinks; when $\Delta S$ is favourable and $\Delta H$ is not, heating helps because $T\Delta S$ grows.

The crossover temperature

Setting $\Delta G = 0$,

$$T = \frac{\Delta H}{\Delta S}$$

Key Point: $T = \Delta H/\Delta S$ is the temperature at which $\Delta G$ changes sign. It is meaningful only when $\Delta H$ and $\Delta S$ share a sign; opposite signs give a negative kelvin value, which is the arithmetic reporting that the sign never changes.

Limestone: $\Delta H^{\circ} = +178.3\,\mathrm{kJ\,mol^{-1}}$, $\Delta S^{\circ} = +160.6\,\mathrm{J\,K^{-1}\,mol^{-1}}$, so

$$T = \frac{178300}{160.6} = 1110\,\mathrm{K}$$

Above $1110\,\mathrm{K}$ the decomposition is spontaneous, which is why lime kilns run near $1200\,\mathrm{K}$.

Key Point: $\Delta H$ is tabulated in $\mathrm{kJ\,mol^{-1}}$ and $\Delta S$ in $\mathrm{J\,K^{-1}\,mol^{-1}}$. Convert before subtracting or dividing. Dividing $178.3$ by $160.6$ gives $1.11\,\mathrm{K}$ — wrong by a factor of a thousand and absurd on its face.

$\Delta G$ as maximum useful work

$$\Delta H = \Delta G + T\Delta S$$

$T\Delta S$ is the part dispersed as heat to satisfy the entropy requirement; $\Delta G$ is the part that can be harvested.

Key Point: At constant $T$ and $p$, $-\Delta G$ is the maximum non-expansion (useful) work obtainable from a change, which is why $G$ was long called the free energy. Expansion work against the atmosphere is not included.

For $\mathrm{H_2(g)} + \tfrac{1}{2}\mathrm{O_2(g)} \rightarrow \mathrm{H_2O(l)}$, $\Delta G^{\circ} = -237.1\,\mathrm{kJ\,mol^{-1}}$: one mole of hydrogen in a fuel cell can deliver at most $237.1\,\mathrm{kJ}$ of electrical work, and any real cell delivers less.

[Board] Work in joules throughout and convert to kilojoules only at the end. One conversion is easier to remember than two.