Thermodynamics

The chapter has a clean three-act structure this article follows. Act one sets up the vocabulary: system, surroundings, boundary, state functions versus path functions, and the first law, $\Delta U = q + w$, with the IUPAC sign convention — then $p$-$V$ work for reversible and irreversible processes, enthalpy $H = U + pV$ as the constant-pressure state function, and the ideal-gas relation $C_p - C_V = R$. Act two is applied enthalpy bookkeeping: measuring $\Delta U$ and $\Delta H$ by calorimetry; standard states and standard enthalpy of reaction; phase-transition enthalpies (fusion, vaporisation, sublimation); standard enthalpy of formation and thermochemical equations; Hess's law; and the family of named enthalpies — combustion, atomisation, bond enthalpy, lattice enthalpy (Born-Haber cycle), solution, and dilution. Act three asks what the first law cannot answer: why do processes run in one direction only? We examine why "enthalpy decrease" fails as a spontaneity criterion, introduce entropy $S$ and the second law ($\Delta S_{total} > 0$), note the third law, and combine enthalpy and entropy into Gibbs energy $G = H - TS$, ending at the bridge equation $\Delta_rG^\ominus = -RT\ln K$ linking thermodynamics to equilibrium.

System, surroundings, and the vocabulary of state

Every thermodynamic calculation begins by drawing a boundary. The system is the part of the universe under study; everything else is the surroundings; together they make the universe. The boundary may be a real physical wall (a beaker, a bomb calorimeter) or a purely imaginary surface enclosing a chosen region of space. Systems are classified by what can cross that boundary. An open system exchanges both matter and energy with its surroundings — an open beaker of reacting liquid, losing vapour and heat to the bench. A closed system exchanges energy but not matter — the same reaction sealed in a stoppered conducting flask. An isolated system exchanges neither — reactants inside a thermos flask. A wall permitting no heat transfer is an adiabatic wall (the process is adiabatic); one that permits heat transfer is thermally conducting (diathermic). The state of a system is fixed once enough macroscopic properties — $p$, $V$, $T$, amount $n$ — are specified; the minimum number needed depends on the system, but once fixed every other property has a definite value. Quantities like $p$, $V$, $T$, $U$, $H$, $S$, $G$ are state functions: value depends only on current state, never on path. Heat $q$ and work $w$ are path functions — the same net change can be reached with different splits between heat and work, so $q$ and $w$ individually depend on how the change was carried out, even though $q + w$ does not. Thermodynamic properties also split by how they scale with system size. An extensive property depends on the quantity of matter present — mass, volume, $U$, $H$, $S$, heat capacity; halve the gas and each halves too. An intensive property does not — temperature, pressure, density, and molar quantities ($V_m$, $C_m$) are intensive; splitting a gas sample with a partition leaves temperature and pressure of each half unchanged.

Internal energy, heat, work, and the first law

The internal energy $U$ of a system is the sum of all forms of energy stored inside it — translational, rotational, vibrational, electronic, and chemical bond energy — and it is a state function. There is no way to measure an absolute value of $U$, only changes $\Delta U$, just as one can measure a change in altitude without knowing "absolute" height above Earth's centre. Internal energy can change in three ways: heat can flow into or out of the system, work can be done on or by it, or matter can enter or leave (for closed and isolated systems, only the first two apply). Joule's classic churning experiments — doing the same amount of mechanical work on insulated water by paddle-wheel stirring or by an electrical immersion heater — showed that identical work always produces an identical temperature change, regardless of how the work was delivered. This established that adiabatic work $w_{ad}$ between two fixed states is path-independent, $\Delta U = U_2 - U_1 = w_{ad}$ — exactly why $U$ qualifies as a state function. By IUPAC convention used throughout this chapter, $w$ is positive when work is done on the system (its internal energy rises) and $q$ is positive when heat flows into the system. Combining both routes for changing internal energy gives the mathematical statement of the first law of thermodynamics:

where: $\Delta U$ is the change in internal energy of the system, $q$ is heat absorbed by the system (positive) or released by it (negative), and $w$ is work done on the system (positive) or by it (negative). The first law is simply the law of conservation of energy applied to a chemical system: energy is neither created nor destroyed, only exchanged as heat and work, so for a truly isolated system ($q = 0$, $w = 0$) $\Delta U = 0$ identically. Note that $q$ and $w$ individually depend on the path taken between the initial and final states — only their sum, $\Delta U$, is path-independent.

Pressure-volume work, reversible and irreversible

Chemical systems most commonly do pressure-volume ($p$-$V$) work — a gas expanding pushes back the atmosphere, or is compressed by an external pressure. Consider a gas at pressure $p$, volume $V_i$, in a cylinder with a frictionless piston of area $A$, and an external pressure $p_{ex} > p$ pushing the piston in by distance $l$ until pressures equalise. The volume change is $\Delta V = l\,A = V_f - V_i$, the force on the piston is $p_{ex}A$, and the work done on the gas is force times distance:

where: the minus sign is required by the IUPAC convention so compression ($\Delta V < 0$) gives positive $w$, correctly signalling work done on the gas. If $p_{ex}$ changes in finite steps, total work is $-\Sigma\,p\,\Delta V$, the sum of rectangular areas under a $p$-$V$ plot. In the limit where $p_{ex}$ differs from the internal pressure by only an infinitesimal $dp$ at every instant, the process becomes reversible — it can, at any moment, be turned back by an infinitesimal change, passing through an unbroken sequence of near-equilibrium states. All other processes are irreversible. For a reversible change, $p_{ex} \approx p_{in}$, so $w_{rev} = -\int_{V_i}^{V_f} p\,dV$, integrated below to the logarithmic form. A special case is free expansion into a vacuum ($p_{ex} = 0$), for which $w = 0$ regardless of reversibility, since there is no opposing force to push against.

For isothermal changes of an ideal gas, $\Delta U = 0$ always (since $U$ depends only on $T$), so the first law forces $q = -w$ in every case. Isothermal irreversible: $q = -w = p_{ex}(V_f - V_i)$. Isothermal reversible: $q = -w = nRT\ln(V_f/V_i)$ — derived in full below. Adiabatic ($q = 0$): all the internal-energy change appears as work, $\Delta U = w_{ad}$. And if volume is held fixed ($\Delta V = 0$), $\Delta U = q_V$ — exactly what makes bomb calorimetry a direct measurement of $\Delta U$.

Enthalpy — the state function for constant pressure

Most real chemistry happens in an open flask at roughly constant atmospheric pressure, not constant volume. At constant pressure, $\Delta U = q_p - p\Delta V$, so $q_p = (U_2 + pV_2) - (U_1 + pV_1)$, motivating a new state function, enthalpy:

where: $U$, $p$, and $V$ are all state functions, so $H$ is automatically a state function too, and $\Delta H$ is independent of path — which means $q_p$, unlike $q$ in general, is also independent of path. At constant pressure, $\Delta H = q_p$: enthalpy change is the heat absorbed (endothermic, $\Delta H > 0$) or released (exothermic, $\Delta H < 0$) by a process carried out at constant pressure. For finite changes at constant $p$, $\Delta H = \Delta U + p\Delta V$. For a reaction involving ideal gases, using $pV = nRT$ for both reactant and product gas totals at the same $T$ and $p$ gives $p\Delta V = \Delta n_gRT$, where $\Delta n_g$ is moles of gaseous products minus moles of gaseous reactants — so:

where: $\Delta n_g$ counts only gas-phase species (solids and liquids barely change volume, so contribute negligibly). When $\Delta n_g = 0$, $\Delta H = \Delta U$ exactly. This equation, derived in full below, is the everyday tool for converting bomb-calorimeter data ($\Delta U$) to tabulated thermochemical data ($\Delta H$).

Heat capacity and the Cp − Cv relation

The temperature rise produced by a given quantity of heat depends on the system's heat capacity $C$: $q = C\Delta T$. A large heat capacity (like that of water) means a lot of heat is absorbed for only a small temperature rise. The molar heat capacity is $C_m = C/n$, and the specific heat capacity $c$ is heat per unit mass per degree, so $q = c\,m\,\Delta T = C\Delta T$. Heat capacity depends on the conditions under which heat is added. At constant volume, $q_V = C_V\Delta T = \Delta U$; at constant pressure, $q_p = C_p\Delta T = \Delta H$. A gas heated at constant pressure must also expand and do work against the surroundings, so it needs more heat than at constant volume for the same temperature rise — the derivation below turns this into the compact result $C_p - C_V = R$ for one mole of an ideal gas.

Calorimetry — measuring ΔU and ΔH in the laboratory

Calorimetry measures the heat evolved or absorbed in a process from the temperature change of a calorimeter of known heat capacity. Two standard setups correspond to the two conditions above. A bomb calorimeter burns a combustible sample inside a sealed steel vessel (the "bomb") immersed in a water bath; because the bomb's volume cannot change, no $p$-$V$ work is possible even for gas-producing reactions, so the measured heat is exactly $q_V = \Delta U$. A calorimeter open to constant atmospheric pressure instead measures $q_p = \Delta H$ directly, and this is called the enthalpy of reaction, $\Delta_rH$.

Standard states, Hess's law, and enthalpies of reaction

Because enthalpy depends on conditions, chemists fix a standard state: the pure form of a substance at $1\ \text{bar}$ pressure (and a specified temperature, conventionally $298\ \text{K}$ unless stated otherwise). Quantities measured under these conditions carry the superscript $\ominus$, as in $\Delta_rH^\ominus$. The reaction enthalpy is the weighted difference of the standard molar enthalpies of products and reactants:

where: $a_i$ and $b_i$ are the stoichiometric coefficients of products and reactants in the balanced equation, and $H_m$ denotes molar enthalpy. Because these absolute molar enthalpies are not directly measurable, chemists instead tabulate standard enthalpy of formation, $\Delta_fH^\ominus$: the enthalpy change when one mole of a compound forms from its constituent elements, each in its most stable reference state at $298\ \text{K}$ and $1\ \text{bar}$ (for example, $\text{H}_2(g)$, $\text{O}_2(g)$, $\text{C}_{\text{graphite}}(s)$, $\text{S}_{\text{rhombic}}(s)$). By convention, $\Delta_fH^\ominus$ of an element in its reference state is exactly zero. Any reaction enthalpy can then be built from formation enthalpies:

This works because it is a special case of Hess's law of constant heat summation: since enthalpy is a state function, the enthalpy change of an overall reaction is the same whether it proceeds in one step or through any sequence of intermediate steps, so long as initial and final states match — formally, if $A \to B$ has enthalpy $\Delta_rH$ along a direct route and $\Delta_rH_1 + \Delta_rH_2 + \Delta_rH_3 + \dots$ along an alternative route through intermediates, then $\Delta_rH = \Delta_rH_1 + \Delta_rH_2 + \Delta_rH_3 + \dots$. Three bookkeeping conventions follow: coefficients refer to moles, never molecules; reversing an equation flips the sign of $\Delta_rH$; and multiplying an equation by a factor multiplies $\Delta_rH$ by the same factor, since enthalpy is extensive. Enthalpy changes accompany phase transitions too, always at constant temperature and pressure: standard enthalpy of fusion, $\Delta_{fus}H^\ominus$ (melting; always positive — for water, $+6.00\ \text{kJ mol}^{-1}$ at $273\ \text{K}$); standard enthalpy of vaporisation, $\Delta_{vap}H^\ominus$ (boiling; for water at its boiling point, $+40.79\ \text{kJ mol}^{-1}$; at the lower temperature of $298\ \text{K}$ it is a different, slightly larger $+44.01\ \text{kJ mol}^{-1}$ — always check which temperature a table quotes); and standard enthalpy of sublimation, $\Delta_{sub}H^\ominus$ (solid directly to vapour, as for dry ice at $195\ \text{K}$, $25.2\ \text{kJ mol}^{-1}$). By Hess's law, sublimation is simply the sum of fusion and vaporisation.

Bond enthalpy and lattice enthalpy

Reaction enthalpies can also be estimated from the energy stored in individual chemical bonds. The bond dissociation enthalpy is the enthalpy change when one mole of a specific covalent bond in a gaseous molecule is broken to give gaseous fragments — for a diatomic molecule this equals the enthalpy of atomisation, $\Delta_aH^\ominus$, such as $\text{H}_2(g) \to 2\text{H}(g)$, $\Delta_aH^\ominus = 435.0\ \text{kJ mol}^{-1}$. For a polyatomic molecule like methane, successive C–H bond dissociation enthalpies differ slightly at each step ($+427$, $+439$, $+452$, $+347\ \text{kJ mol}^{-1}$, summing to the full atomisation enthalpy $1665\ \text{kJ mol}^{-1}$), so chemists instead quote the mean bond enthalpy — for methane, $\Delta_{C-H}H^\ominus = \tfrac14(1665) = 416\ \text{kJ mol}^{-1}$. Using tabulated mean bond enthalpies, gas-phase reaction enthalpy is approximated as:

where: the sum runs over every bond broken in the reactants and every bond formed in the products, all species being gaseous — breaking bonds always costs energy (endothermic) and forming bonds always releases it (exothermic), so a reaction is exothermic overall whenever the bonds formed are collectively stronger than the bonds broken. For ionic solids, the relevant quantity is lattice enthalpy, $\Delta_{lattice}H^\ominus$: the enthalpy change when one mole of an ionic solid dissociates completely into its gaseous ions, always endothermic, for example $\text{NaCl}(s) \to \text{Na}^+(g) + \text{Cl}^-(g)$, $\Delta_{lattice}H^\ominus = +788\ \text{kJ mol}^{-1}$. Because this cannot be measured directly, it is obtained indirectly from an energy cycle called the Born-Haber cycle, which links it by Hess's law to enthalpies that can be measured: sublimation of the metal, ionisation enthalpy, bond dissociation of the non-metal, electron gain enthalpy, and the standard enthalpy of formation of the solid.

Once the lattice enthalpy is known, the enthalpy of solution, $\Delta_{sol}H^\ominus$ — the enthalpy change when one mole of an ionic solid dissolves in a specified amount of solvent — follows from a second Hess's-law cycle: dissolving first breaks the lattice apart (endothermic, $+\Delta_{lattice}H$) and then hydrates the separated ions (exothermic, $\Delta_{hyd}H$), so $\Delta_{sol}H^\ominus = \Delta_{lattice}H^\ominus + \Delta_{hyd}H^\ominus$. For $\text{NaCl}$, $+788 + (-784) = +4\ \text{kJ mol}^{-1}$, a small net endothermic value — which is why a very high lattice enthalpy that hydration cannot offset can leave a salt effectively insoluble. A related quantity, enthalpy of dilution, is the small heat released on adding more solvent to an already-dissolved solute, and it shrinks toward zero as the solution approaches infinite dilution.

Spontaneity: why enthalpy alone is not the answer

A spontaneous process is one with the intrinsic tendency to occur on its own, without continuous external intervention — instantly (an explosive reaction) or over years (rusting iron), but once started it will not reverse itself without outside help. Rate is irrelevant to spontaneity; only direction matters, and a spontaneous process is inherently an irreversible process in the thermodynamic sense. It is tempting to guess spontaneous reactions simply release energy, as falling stones lose potential energy on their own. Many familiar exothermic reactions support this: $\tfrac12\text{N}_2(g) + \tfrac32\text{H}_2(g) \to \text{NH}_3(g)$ ($\Delta_rH^\ominus = -46.1\ \text{kJ mol}^{-1}$) and $\text{H}_2(g) + \tfrac12\text{O}_2(g) \to \text{H}_2\text{O}(l)$ ($\Delta_rH^\ominus = -285.8\ \text{kJ mol}^{-1}$) are both spontaneous and exothermic. But the criterion breaks decisively: $\tfrac12\text{N}_2(g) + \text{O}_2(g) \to \text{NO}_2(g)$ ($\Delta_rH^\ominus = +33.2\ \text{kJ mol}^{-1}$) and $\text{C}_{\text{graphite}}(s) + 2\text{S}(l) \to \text{CS}_2(l)$ ($\Delta_rH^\ominus = +128.5\ \text{kJ mol}^{-1}$) are both spontaneous and endothermic. Enthalpy decrease is at best a contributing factor to spontaneity, never a universal criterion.

Entropy and the second law

The missing driving force appears clearly in a process with no enthalpy change at all: two ideal gases diffusing into each other after a partition is removed inside an isolated container. There is no heat exchange and no work, yet the mixing happens spontaneously and never reverses. What changes is predictability — before mixing, a molecule sampled from the left chamber is certainly gas A; after mixing, it could be either. The system has become more disordered, and it never spontaneously un-mixes. This disorder is quantified by entropy, $S$, a state function measuring the degree of randomness in a system: the more disordered the arrangement, the higher the entropy. Crystalline solids have the lowest entropy of the three states of matter for a given substance; gases, free to roam the whole container, have the highest. Reactions that increase gas-phase particles or otherwise scramble structure tend to have positive $\Delta S$; the reverse — gas condensing, ions locking into a crystal — gives negative $\Delta S$. For a reversible process:

where: $q_{rev}$ is heat absorbed reversibly at absolute temperature $T$; dividing by $T$ reflects that the same heat adds more disorder to a cold, already-ordered system than to a hot, already-chaotic one. The full criterion for spontaneity considers both system and surroundings:

This is the second law of thermodynamics: in any spontaneous process, the total entropy of the universe increases; at equilibrium it is maximum and stationary. It explains why exothermic reactions are so commonly spontaneous even without the system's own entropy rising — heat released to the surroundings raises $\Delta S_{surr} \approx -\Delta H_{sys}/T$, and if this outweighs any decrease in $\Delta S_{sys}$, the total stays positive. This is why iron oxidation, $4\text{Fe}(s) + 3\text{O}_2(g) \to 2\text{Fe}_2\text{O}_3(s)$, is spontaneous despite $\Delta S_{sys} = -549.4\ \text{J K}^{-1}\text{mol}^{-1}$ (gas is consumed, a compact solid forms): the reaction is strongly exothermic ($\Delta_rH^\ominus = -1648\ \text{kJ mol}^{-1}$), giving $\Delta S_{surr} = +5530\ \text{J K}^{-1}\text{mol}^{-1}$, so $\Delta S_{total} = +4980.6\ \text{J K}^{-1}\text{mol}^{-1}$ — positive, hence spontaneous. The third law of thermodynamics fixes the reference point for entropy: a perfectly ordered, pure crystalline substance has entropy approaching zero as $T \to 0\ \text{K}$, since only one microscopic arrangement remains. This is not true of glasses, solutions, or supercooled liquids, which retain residual disorder even at $0\ \text{K}$. Practically, the third law allows absolute molar entropies to be tabulated from thermal data by summing $q_{rev}/T$ increments from $0\ \text{K}$ upward, and standard entropy changes can then be calculated by a Hess's-law-type sum exactly analogous to equation (6).

Gibbs energy and equilibrium

Tracking two separate entropy changes for every reaction is cumbersome. J. Willard Gibbs combined them into a single system-only state function, Gibbs energy (or Gibbs free energy):

where: $G$ is extensive and a state function, exactly like $H$ and $S$. At constant temperature, $\Delta G = \Delta H - T\Delta S$ (the Gibbs equation), and — as derived rigorously below from the second law — this single quantity gives a complete, self-contained spontaneity test at constant temperature and pressure:

Because $\Delta G$ combines an enthalpy term and a $T\Delta S$ term with opposite possible signs, temperature itself can flip the spontaneity of a reaction. The four sign combinations of $\Delta H$ and $\Delta S$ give the following behaviour.

At constant $T$ and $p$, chemical equilibrium is the state at which $G$ is at a minimum, so the criterion for equilibrium is exactly $\Delta_rG = 0$. For a reaction run with all species in their standard states, this leads to the central bridge equation of the chapter, connecting thermodynamics directly to the equilibrium constant $K$:

where: $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$, $T$ is the absolute temperature, and $K$ is the equilibrium constant of the reaction under standard-state conventions. A large negative $\Delta_rG^\ominus$ forces $K \gg 1$ — the reaction goes essentially to completion; a large positive $\Delta_rG^\ominus$ forces $K \ll 1$ — negligible product forms; and $\Delta_rG^\ominus \approx 0$ gives $K \approx 1$, comparable amounts of reactants and products at equilibrium. This is the pivot on which the whole of the next chapter, chemical equilibrium, turns.

Board level - Bomb calorimeter combustion of graphite (NCERT Problem 5.6 pattern)

One gram of graphite is burnt in a bomb calorimeter in excess oxygen at $298\ \text{K}$ and $1\ \text{atm}$ according to $\text{C(graphite)} + \text{O}_2(g) \to \text{CO}_2(g)$. During the reaction the temperature rises from $298\ \text{K}$ to $299\ \text{K}$. The heat capacity of the calorimeter is $20.7\ \text{kJ/K}$. Find the enthalpy change for the combustion of one mole of graphite.

Step 1. The heat absorbed by the calorimeter is $q_{cal} = C_V \times \Delta T = 20.7\ \text{kJ/K} \times 1\ \text{K} = 20.7\ \text{kJ}$. Heat lost by the system equals heat gained by the calorimeter, so $q = -20.7\ \text{kJ}$ for combustion of $1\ \text{g}$ of graphite — negative, confirming an exothermic reaction.

Step 2. Because a bomb calorimeter holds volume fixed, no $p$-$V$ work occurs, so this heat is directly the internal-energy change: $\Delta U$ (per gram) $= -20.7\ \text{kJ/g}$. Scaling to one mole using the molar mass of carbon, $12.0\ \text{g mol}^{-1}$: $\Delta U = 12.0\ \text{g mol}^{-1} \times (-20.7\ \text{kJ/g}) = -2.48\times10^2\ \text{kJ mol}^{-1}$.

Step 3. Convert to $\Delta H$ using $\Delta H = \Delta U + \Delta n_gRT$. The reaction consumes one mole of gaseous $\text{O}_2$ and produces one mole of gaseous $\text{CO}_2$, so $\Delta n_g = 1 - 1 = 0$. Therefore $\Delta H = \Delta U$ exactly. Answer: $\Delta U = \Delta_cH^\ominus = -2.48\times10^{2}\ \text{kJ mol}^{-1}$ for the combustion of one mole of graphite.

Board level - Cooling and freezing water in two enthalpy steps (NCERT Problem 5.8 pattern)

Assuming water vapour behaves as a perfect gas, calculate the enthalpy change when $1\ \text{mol}$ of liquid water at $100\,^\circ\text{C}$ and $1\ \text{bar}$ is cooled and frozen into ice at $0\,^\circ\text{C}$. Given: $\Delta_{fus}H^\ominus = 6.00\ \text{kJ mol}^{-1}$, and heat capacity of liquid water $= 4.2\ \text{J g}^{-1\,\circ}\text{C}^{-1}$.

Step 1. Break the change into two Hess's-law steps: Step A cools liquid water from $100\,^\circ\text{C}$ to $0\,^\circ\text{C}$ (no phase change), and Step B freezes it to ice at $0\,^\circ\text{C}$.

Step 2. Compute Step A using $q = c\,m\,\Delta T$ with $m = 18\ \text{g}$ (one mole of water) and $\Delta T = 0 - 100 = -100\,^\circ\text{C}$: $\Delta H_1 = 18\ \text{g} \times 4.2\ \text{J g}^{-1\,\circ}\text{C}^{-1} \times (-100\,^\circ\text{C}) = -7560\ \text{J mol}^{-1} = -7.56\ \text{kJ mol}^{-1}$.

Step 3. Compute Step B. Freezing is the reverse of fusion, so it releases exactly what fusion would absorb: $\Delta H_2 = -\Delta_{fus}H^\ominus = -6.00\ \text{kJ mol}^{-1}$. Adding the two steps by Hess's law: $\Delta H = \Delta H_1 + \Delta H_2 = -7.56 + (-6.00) = -13.56\ \text{kJ mol}^{-1}$. Since the volume change between liquid and solid water is negligible, $\Delta n_gRT \approx 0$, so $\Delta U \approx \Delta H$ as well. Answer: $\Delta H = \Delta U = -13.56\ \text{kJ mol}^{-1}$.

Competitive level (JEE / NEET) - Temperature at which a reaction turns spontaneous

For the reaction $2\text{A}(g) + \text{B}(g) \to \text{C}(g)$, $\Delta_rH^\ominus = +400\ \text{kJ mol}^{-1}$ and $\Delta_rS^\ominus = +0.2\ \text{kJ K}^{-1}\text{mol}^{-1}$ at $298\ \text{K}$. Assuming both are constant with temperature, find the minimum temperature at which the reaction becomes spontaneous.

Step 1. Both $\Delta_rH^\ominus$ and $\Delta_rS^\ominus$ are positive, so the reaction is non-spontaneous at low $T$ (unfavourable $+\Delta H$ dominates) but turns spontaneous once $T$ is large enough for the favourable $T\Delta S$ term to win.

Step 2. The crossover is where $\Delta_rG^\ominus$ changes sign: $0 = \Delta_rH^\ominus - T\Delta_rS^\ominus$, so $T = \Delta_rH^\ominus / \Delta_rS^\ominus$.

Step 3. Substitute, keeping units consistent: $T = \dfrac{400\ \text{kJ mol}^{-1}}{0.2\ \text{kJ K}^{-1}\text{mol}^{-1}} = 2000\ \text{K}$. Above this, $\Delta_rG^\ominus$ is negative and the reaction is spontaneous; below it, non-spontaneous. Answer: the reaction becomes spontaneous only above $T = 2000\ \text{K}$. Timing note: under two minutes. This is one division once you recognise "temperature at which a reaction becomes spontaneous" means set $\Delta G^\ominus = 0$ and solve $T = \Delta H^\ominus/\Delta S^\ominus$ — do not waste time computing $\Delta G^\ominus$ at $298\ \text{K}$ first. Watch units: mixing kJ for $\Delta H$ with J for $\Delta S$ is the commonest slip.

Competitive level (JEE / NEET) - Mean C–Cl bond enthalpy in CCl4 via a Hess's law cycle

Given: $\Delta_fH^\ominus(\text{CCl}_4, g) = -105.0\ \text{kJ mol}^{-1}$; $\Delta_aH^\ominus(\text{C, graphite}) = 715.0\ \text{kJ mol}^{-1}$; $\Delta_aH^\ominus(\text{Cl}_2) = 242\ \text{kJ mol}^{-1}$ (full $\text{Cl}_2 \to 2\text{Cl}$ atomisation). Find the enthalpy change for $\text{CCl}_4(g) \to \text{C}(g) + 4\text{Cl}(g)$ and hence the mean C–Cl bond enthalpy.

Step 1. Build a Hess's law cycle linking $\text{CCl}_4(g)$ to its gas-phase atoms via the elements: first decompose $\text{CCl}_4(g)$ back to graphite and chlorine gas (reverse of formation, enthalpy $-\Delta_fH^\ominus$), then atomise both elements (carbon once, chlorine doubled, since two moles of $\text{Cl}_2$ give the four moles of $\text{Cl}$ atoms needed).

Step 2. Reverse the formation step: $\text{CCl}_4(g) \to \text{C(graphite)} + 2\text{Cl}_2(g)$, enthalpy $= +105.0\ \text{kJ mol}^{-1}$.

Step 3. Atomise the elements: $\text{C(graphite)} \to \text{C}(g)$ gives $+715.0\ \text{kJ mol}^{-1}$; $2\text{Cl}_2(g) \to 4\text{Cl}(g)$ gives $2 \times 242 = +484\ \text{kJ mol}^{-1}$.

Step 4. Sum by Hess's law: $\Delta_aH^\ominus(\text{CCl}_4) = 105.0 + 715.0 + 484 = 1304\ \text{kJ mol}^{-1}$. This breaks four equivalent C–Cl bonds, so the mean bond enthalpy is one quarter of the total: $1304/4 = 326\ \text{kJ mol}^{-1}$. Answer: $\Delta H = 1304\ \text{kJ mol}^{-1}$, mean C–Cl bond enthalpy $= 326\ \text{kJ mol}^{-1}$. Timing note: about three minutes. The trap is the chlorine atomisation term — $\Delta_aH^\ominus(\text{Cl}_2)$ as tabulated already breaks one whole mole of $\text{Cl}_2$ into two moles of $\text{Cl}$ atoms, so getting four moles of $\text{Cl}$ atoms requires doubling it. Draw the cycle before adding numbers; arithmetic slips are more common than conceptual ones here.

The single most persistent error is applying the older sign convention (work done by the system is positive) instead of the IUPAC convention used throughout this chapter ($\Delta U = q + w$, both positive when added to the system) — always check which convention a problem uses, and default to IUPAC unless told otherwise. A close second is forgetting that $\Delta U = \Delta H$ only when $\Delta n_g = 0$; students routinely equate bomb-calorimeter $\Delta U$ with tabulated $\Delta H$ without the $\Delta n_gRT$ correction. Extensive and intensive properties are frequently confused — dividing any extensive property by moles converts it into an intensive molar property, which is why $C_m$ and $V_m$ do not depend on sample size even though $C$ and $V$ do. On the spontaneity side, students often treat "$\Delta H < 0$" as sufficient for a reaction to proceed, ignoring the entropy and temperature dependence baked into $\Delta G = \Delta H - T\Delta S$. Equally common is confusing $\Delta S_{sys}$ with $\Delta S_{total}$: only the total entropy must increase — the system's own entropy can decrease (freezing, iron rusting) provided the surroundings gain more than enough to compensate. Do not apply the third law's "zero entropy at $0\ \text{K}$" to anything but a perfect crystalline solid; glasses and solutions retain residual entropy at absolute zero. Finally, a negative $\Delta G$ only guarantees a reaction is thermodynamically favourable, not fast — rate and mechanism belong to kinetics, an entirely separate branch of chemistry.

• Classify a given process as involving an open, closed, or isolated system, and identify which of matter, heat, and work can cross its boundary.
• State the first law of thermodynamics, apply the IUPAC sign convention correctly, and compute $\Delta U$ from given $q$ and $w$ (or vice versa).
• Calculate pressure-volume work for a single-step irreversible process and for an isothermal reversible process, and explain why free expansion into a vacuum does zero work.
• Define enthalpy, explain why it is the natural state function for constant-pressure processes, and convert between $\Delta U$ and $\Delta H$ using $\Delta H = \Delta U + \Delta n_gRT$.
• Derive and apply $C_p - C_V = R$ for an ideal gas, and distinguish molar heat capacity from specific heat capacity.
• Use Hess's law and standard enthalpies of formation to calculate an unknown reaction enthalpy, and correctly apply the sign and coefficient conventions for reversing or scaling thermochemical equations.
• Distinguish standard enthalpies of combustion, atomisation, bond enthalpy, and lattice enthalpy, and construct a Born-Haber cycle to obtain a lattice enthalpy indirectly.
• Explain why decrease in enthalpy alone cannot predict spontaneity, and state the second law's total-entropy criterion, $\Delta S_{total} > 0$.
• Compute $\Delta G$ from $\Delta H$ and $\Delta S$, use the sign of $\Delta G$ to judge spontaneity at a given temperature, and find the crossover temperature at which a reaction's spontaneity changes.
• Apply $\Delta_rG^\ominus = -RT\ln K$ to move between a standard Gibbs energy change and an equilibrium constant in either direction.

Ask the tutor: Want to work through a mixed problem that starts from a bomb-calorimeter reading, converts it to ΔH with the Δng correction, then uses that ΔH along with a given ΔS to decide the temperature at which the same reaction turns spontaneous?