Thermodynamics — NEET Notes

Thermodynamic terms — system, surroundings, boundary

Thermodynamics begins by carving up the universe. The piece you are studying is the system; everything else is the surroundings; the imaginary or physical wall that separates them is the boundary. A beaker holding a reaction mixture is a system; the room is its surroundings; the glass wall is the boundary. NCERT is emphatic that "the universe = the system + the surroundings," and that for practical purposes the surroundings need only be the region close enough to the system to interact with it.

Systems are classified by what crosses the boundary — matter, energy, or neither. Three types of system exhaust the possibilities, and NEET tests the distinction directly through diagram-based questions and "which of the following is an open system" stems.

The state of a system is fully specified by its measurable macroscopic properties — pressure (p), volume (V), temperature (T), composition. Once a small set of these is fixed, all other state variables follow. Thermodynamics does not need positions and velocities of individual molecules; it only needs averaged, bulk descriptors.

State functions vs path functions

This distinction underlies every calculation in the chapter. A state function depends only on the current state of the system — the path used to reach that state is irrelevant. Internal energy U, enthalpy H, entropy S, Gibbs energy G, plus the obvious physical variables p, V, T are all state functions. A path function, by contrast, depends on how the change was carried out. Heat q and work w are the two famous path functions of thermodynamics — yet remarkably, their sum (q + w = ΔU) is path-independent.

Work, heat, and internal energy

The internal energy U of a system is the sum of all kinetic and potential energies stored in its molecules — translational, rotational, vibrational, electronic, intermolecular. We cannot measure its absolute value; we can only measure changes, ΔU. NCERT introduces U through Joule's classic experiment: 1 kJ of mechanical work done by paddles on water in an adiabatic vessel raised the temperature by the same amount as 1 kJ of electrical work via an immersion heater. The path differed; ΔU did not. So U is a state function, and for an adiabatic process, ΔU = w_ad.

Heat, q, is energy transferred across the boundary because of a temperature difference. Work, w, is every other kind of energy transfer — most often, in chemistry, pressure–volume work as gases expand or compress. For a system contained in a piston against external pressure p_ex:

w = − p_ex ΔV

Pressure–volume work, NCERT sign convention

For a reversible isothermal expansion of n moles of an ideal gas — a process where every infinitesimal step is in equilibrium with the surroundings — the work is w_rev = −nRT ln(V_f/V_i) = −2.303 nRT log(V_f/V_i). For a free expansion into vacuum, p_ex = 0, so w = 0; experimentally q = 0 too for an ideal gas, so ΔU = 0. This was tested directly in NEET 2020 (Q.176).

The first law of thermodynamics

Combine the two ways of changing internal energy — heat and work — and the first law of thermodynamics writes itself:

The energy of an isolated system is constant. ΔU = q + w

First law of thermodynamics — Equation 5.1, NCERT

This is the law of conservation of energy in chemical clothing. Energy is neither created nor destroyed — it only changes form. For an isolated system (q = 0, w = 0), ΔU = 0. The first law makes no statement about direction — that comes later, from the second law. It tells us only that the books must balance.

NEET 2017 (Q.33) is a perfect application. A gas in a well-insulated container expands against p_ex = 2.5 atm from 2.50 L to 4.50 L. Adiabatic, so q = 0. Then ΔU = w = −p_exΔV = −2.5 × 2 = −5 L·atm = −505 J. The negative sign tells you the gas lost internal energy — which is exactly what an expansion against pressure does.

Enthalpy and heat capacity

Bomb calorimeters work at constant volume, where ΔU = q_V. But most laboratory and biological reactions happen in open vessels at constant atmospheric pressure. NCERT introduces a new state function for these conditions — enthalpy:

H = U + pV  ⇒  ΔH = q_p  (at constant p)

Enthalpy — "heat content" — Equation 5.7, NCERT

Because U, p, and V are state functions, H is also a state function. The change ΔH at constant pressure equals the heat absorbed or released — making enthalpy the natural quantity to tabulate for chemical reactions. ΔH is negative for exothermic reactions (heat lost by the system) and positive for endothermic ones.

The relationship between ΔH and ΔU is one of NEET's favourite quick-fire topics:

The amount of heat needed to raise the temperature of a sample by ΔT is q = C·ΔT, where C is the heat capacity. Specific heat capacity is per unit mass; molar heat capacity C_m = C/n is per mole. For an ideal gas, two heat capacities matter: C_V (constant volume) and C_p (constant pressure). They are linked by the Mayer relation:

C_p − C_V = R   (per mole, ideal gas)

Mayer's relation — derived from ΔH = ΔU + Δ(pV) and pV = RT

NEET 2021 (Q.55) asked exactly this and gave the four options C_P − C_V = R among them — the answer. The derivation: at constant volume q_V = C_VΔT = ΔU; at constant pressure q_p = C_pΔT = ΔH. For one mole of an ideal gas ΔH = ΔU + Δ(pV) = ΔU + RΔT. Substituting gives C_pΔT = C_VΔT + RΔT, so C_p − C_V = R.

Measurement of ΔU and ΔH — calorimetry

Calorimetry is the experimental technique used to measure energy changes. The reaction is run inside an insulated vessel — the calorimeter — and the heat is calculated from the temperature change of a surrounding liquid (or the calorimeter itself) of known heat capacity.

• Bomb calorimeter (constant volume): a sealed steel vessel ("the bomb") is immersed in a water bath. A combustible sample is ignited in pure O₂. Because the volume is fixed, no pV-work is done and the heat released equals ΔU. NCERT works through this for graphite: 1 g of graphite burnt at 298 K raises the temperature of a 20.7 kJ/K calorimeter by 1 K, giving q = −20.7 kJ and ΔU = −2.48 × 10² kJ mol⁻¹ for the combustion of carbon to CO₂.
• Coffee-cup (constant pressure) calorimeter: the reaction is run in an insulated open cup at atmospheric pressure. The heat exchanged is q_p = ΔH directly.

Enthalpies for different types of reactions

NCERT catalogues the standard enthalpies you must know by name. The standard state of a substance is its pure form at 1 bar; standard quantities carry the superscript ⊖ and are usually quoted at 298 K. By convention, the standard enthalpy of formation of any element in its reference state (most stable allotrope) is zero.

Bond enthalpy — NEET's reliable workhorse

Bond enthalpy lets you estimate Δ_rH for any gas-phase reaction using the simple bookkeeping rule:

Δ_rH = Σ (bond enthalpies of reactants) − Σ (bond enthalpies of products)

Bond-enthalpy method for gas-phase reactions

NEET 2018 (Q.80) used this idea cleanly: bond enthalpies of X₂, Y₂, XY are in the ratio 1 : 0.5 : 1, and ΔH for the formation of XY is −200 kJ mol⁻¹. Set X₂ = a, Y₂ = 0.5a, XY = a. Then for ½X₂ + ½Y₂ → XY, ΔH = (a/2 + 0.5a/2) − a = −0.25a = −200, giving a = 800 kJ mol⁻¹ as the bond dissociation energy of X₂.

Hess's law of constant heat summation

Because enthalpy is a state function, the path doesn't matter. Hess's law states that if a reaction takes place in several steps, its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions — provided every step is at the same temperature. This is the single most powerful calculation device in the chapter. Reactions whose enthalpy cannot be measured directly (for instance, the partial oxidation of graphite to CO without forming any CO₂) become tractable by combining other reactions that can be measured.

The same machinery handles formation enthalpies, lattice enthalpies (via the Born–Haber cycle), and the calculation of unmeasurable reactions from tabulated formation values. NEET regularly serves up Hess-style multi-equation problems and the only trap is sign-tracking — write each equation with its ΔH, reverse what you must (flipping the sign), multiply where needed (scaling ΔH), and add.

Spontaneity — what drives a reaction?

The first law tells you energy is conserved; it stays mute on direction. Heat flows from hot to cold but not the reverse; gases diffuse to fill the container but never spontaneously unmix; iron rusts but never spontaneously un-rusts. A spontaneous process is one that occurs without the help of an external agency. (Spontaneous doesn't mean fast — H₂ and O₂ at room temperature will technically react, just imperceptibly slowly.) An irreversible process is spontaneous; a spontaneous process can only be reversed by external work.

Could decrease in enthalpy be the criterion? Many spontaneous reactions are exothermic — combustion, neutralisation, the formation of HCl from H₂ and Cl₂. But not all: dissolving NH₄Cl in water cools the solution, and the diffusion of two ideal gases has ΔH = 0 yet proceeds spontaneously. Enthalpy alone is not enough.

Entropy — the missing ingredient

The driver of those endothermic and zero-enthalpy spontaneous processes is entropy, S — a measure of the disorder or randomness of the system. The greater the disorder, the higher the entropy. Gases have higher entropy than liquids, which have higher entropy than solids. Heating any substance increases its entropy. For a chemical reaction, entropy goes up when the number of moles of gas increases.

Entropy is a state function. For a reversible heat transfer at temperature T:

ΔS = q_rev / T    (units: J K⁻¹ mol⁻¹)

Clausius definition of entropy change

The same amount of heat creates more disorder when added at low T (where molecules are already calm) than at high T (where they are already chaotic) — hence the inverse temperature dependence. The second law of thermodynamics, expressed in NCERT's terms, is that for any spontaneous process the total entropy of the system plus surroundings must increase:

ΔS_total = ΔS_system + ΔS_surroundings > 0  (spontaneous)

Second law of thermodynamics — entropy criterion

At equilibrium, ΔS_total = 0. The third law of thermodynamics states that the entropy of a pure, perfectly crystalline solid is zero at absolute zero — there is only one microstate, complete order. This lets us assign absolute entropies (not just changes) to substances.

Gibbs free energy — combining ΔH and ΔS

Tracking ΔS_total means tracking both the system and its surroundings — tedious. NCERT introduces a state function that bundles the entropy criterion into a system-only quantity. The Gibbs free energy is:

ΔG is also called the "free" energy because it represents the energy available to do useful work (other than pV-work). The TΔS term is the energy unavailable for work — it gets dispersed into the surroundings as entropy.

The signs of ΔH and ΔS produce four scenarios, each of which is a NEET PYQ in waiting:

The crossover temperature, where ΔG flips sign, is T = ΔH/ΔS (set ΔG = 0). NEET 2017 (Q.36): ΔH = 35.5 kJ mol⁻¹, ΔS = 83.6 J K⁻¹ mol⁻¹, both positive. T_eq = 35500/83.6 ≈ 425 K. Above this T, TΔS overtakes ΔH and ΔG goes negative. Answer: T > 425 K.

Gibbs energy and equilibrium

At chemical equilibrium, the forward and reverse reactions proceed at equal rates, and the free energy of the system is at a minimum. The fundamental link between thermodynamics and the equilibrium constant K is:

ΔG⊖ = −RT ln K = −2.303 RT log K

Equation 5.23, NCERT — the bridge between thermodynamics and equilibrium

The implications are sharp. A strongly exothermic reaction with large negative ΔG⊖ has K ≫ 1 and goes nearly to completion. A strongly endothermic reaction has a small K and stays mostly reactant. The reactions that fascinate chemists sit in the middle — where small temperature changes shift K by orders of magnitude.

NEET 2023 (Q.88) worked this exactly. Equilibrium concentrations 2, 3, 10, 6 mol L⁻¹ for A + B ⇌ C + D give K = (10 × 6)/(2 × 3) = 10. With R = 2 cal mol⁻¹ K⁻¹ and T = 300 K: ΔG⊖ = −2.303 × 2 × 300 × log 10 = −1381.8 cal. The answer was (4).

NEET PYQ Snapshot