Thermodynamics Notes for NEET: Class 11 NCERT Chemistry

Introduction

Thermodynamics studies the relationship between heat, work, and energy in chemical processes. It tells you whether a reaction will happen spontaneously (without external help) and how much energy is released or absorbed. For NEET, this chapter gives 2 to 3 questions per year.

The key areas for NEET are: enthalpy calculations (Hess's law, bond enthalpies), the Gibbs energy criterion for spontaneity (ΔG = ΔH − TΔS), and the sign conventions for ΔH, ΔS, and ΔG.

Basic Concepts: System, State, and Process

System and Surroundings

The system is the part of the universe you are studying (the reaction vessel, the gas in a cylinder). Everything else is the surroundings. Together they form the universe.

Type of system	Can exchange	Example
Open	Matter and energy	Open beaker of solution
Closed	Energy only (not matter)	Gas in sealed cylinder with piston
Isolated	Neither matter nor energy	Thermos flask (ideal); universe itself

State Functions and Path Functions

A state function depends only on the current state of the system (temperature, pressure, composition), not on how the system reached that state. Examples: internal energy (U), enthalpy (H), entropy (S), Gibbs energy (G), temperature, pressure, volume.

A path function depends on the path taken. Examples: heat (q) and work (w).

Types of Processes

Isothermal: constant temperature (ΔT = 0, so ΔU = 0 for ideal gas).
Adiabatic: no heat exchange with surroundings (q = 0).
Isobaric: constant pressure (q = ΔH).
Isochoric: constant volume (w = 0, q = ΔU).
Reversible: infinitely slow; system is always in equilibrium.
Irreversible: spontaneous, fast; system not in equilibrium.

First Law of Thermodynamics

Energy cannot be created or destroyed; it can only be converted from one form to another.

Δ U = q + w

where ΔU is the change in internal energy of the system, q is heat absorbed by the system (positive when heat flows in), and w is work done on the system (positive when surroundings do work on system).

For expansion against an external pressure P_ext:

w = - P_{e x t} Δ V

At constant volume (ΔV = 0): w = 0, so ΔU = q_v (heat at constant volume).

Enthalpy and Heat at Constant Pressure

Most laboratory reactions happen at constant atmospheric pressure. The heat exchanged at constant pressure is called enthalpy change (ΔH).

H = U + P V

Δ H = Δ U + Δ (P V) = Δ U + P Δ V (at const. P)

For reactions involving gases, using ideal gas law (PV = nRT):

Δ H = Δ U + Δ n_{g} R T

where Δn_g = (moles of gaseous products) − (moles of gaseous reactants).

Sign Convention for ΔH

Exothermic reaction: Heat is released to surroundings. ΔH is negative (system loses heat).
Endothermic reaction: Heat is absorbed from surroundings. ΔH is positive.

Heat Capacity

Heat capacity (C) is the heat required to raise the temperature of a substance by 1°C (or 1 K).

C_v: molar heat capacity at constant volume. ΔU = nC_vΔT.
C_p: molar heat capacity at constant pressure. ΔH = nC_pΔT.
For an ideal gas: C_p − C_v = R (R = 8.314 J mol⁻¹ K⁻¹).

Specific heat capacity (s) is per gram. Heat absorbed: q = msΔT.

Standard Enthalpies of Reactions

Standard state is defined as: pure substance at 1 bar pressure and 25°C (298 K). Standard enthalpy is denoted ΔH° (or ΔH⊖).

Standard Enthalpy of Formation (ΔH°_f)

The enthalpy change when one mole of a compound is formed from its constituent elements in their standard states. By convention, ΔH°_f = 0 for elements in their standard states (e.g., O₂(g), H₂(g), C(graphite), Na(s)).

Δ H_{r e a c t i o n}^{\circ} = \sum Δ H_{f}^{\circ} (products) - \sum Δ H_{f}^{\circ} (reactants)

Standard Enthalpy of Combustion (ΔH°_c)

Enthalpy change when one mole of a substance burns completely in oxygen under standard conditions. Always negative (exothermic). C + O₂ → CO₂; ΔH°_c(C, graphite) = −393.5 kJ/mol.

Standard Enthalpy of Neutralisation

For strong acid + strong base: always ≈ −57.1 kJ/mol (reaction is really H⁺ + OH⁻ → H₂O). For weak acid or weak base: less exothermic because some energy is used to ionise the weak electrolyte.

Standard Enthalpy of Atomisation

Enthalpy to convert 1 mol of a substance (element or compound) into gaseous atoms.

H₂(g) → 2H(g): ΔH° = +436 kJ/mol (= 2 × bond dissociation enthalpy of H-H).
For monoatomic elements like Fe(s), it equals the enthalpy of sublimation.

Lattice Enthalpy

The enthalpy change when 1 mole of an ionic compound is formed from its gaseous ions. It is always highly negative (exothermic). Cannot be measured directly but is calculated using the Born-Haber cycle (application of Hess's law).

Hess's Law of Constant Heat Summation

Hess's Law: The total enthalpy change is the same regardless of the route taken. Add or reverse intermediate reactions to reach the target.

Example 1

Example 2

Example 3

Target: Formation of CO₂ from C and O₂

C(s) + O₂(g) → CO₂(g)

ΔH = -393 kJ/mol (find by combining the steps below)

Given reactions

(1) C(s) + ½O₂(g) → CO(g) ΔH = -111 kJ/mol

(2) CO(g) + ½O₂(g) → CO₂(g) ΔH = -283 kJ/mol

Solution steps

The total enthalpy change for a reaction is the same regardless of whether it takes place in one step or several steps. This is a consequence of enthalpy being a state function.

Mathematically: if a reaction can be written as the sum of other reactions, its ΔH is the sum of their ΔH values.

If you reverse a reaction: ΔH changes sign. If you multiply a reaction by n: ΔH is multiplied by n.

Example: Find ΔH for C(graphite) + ½O₂(g) → CO(g) from:

Reaction 1: C(graphite) + O₂(g) → CO₂(g); ΔH₁ = −393.5 kJ/mol
Reaction 2: CO(g) + ½O₂(g) → CO₂(g); ΔH₂ = −283 kJ/mol

Desired = Reaction 1 − Reaction 2: ΔH = −393.5 − (−283) = −110.5 kJ/mol.

Bond Enthalpy

The bond dissociation enthalpy is the energy required to break one mole of a specific bond in gaseous molecules. For polyatomic molecules, average bond enthalpies are used.

Δ H_{r e a c t i o n} = \sum Bond enthalpies of bonds broken - \sum Bond enthalpies of bonds formed

Bonds broken: always endothermic (+ sign). Bonds formed: always exothermic (− sign).

Important values: H-H = 436, C-H = 414, C-C = 346, C=C = 614, C≡C = 839, N≡N = 946, O=O = 498, H-O = 460, H-Cl = 431 kJ/mol.

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Second Law of Thermodynamics and Entropy

The second law states: all spontaneous processes in an isolated system proceed in the direction of increasing entropy. Entropy (S) is a measure of the disorder or randomness of a system.

Δ S = \frac{q _{r e v}}{T}

For a spontaneous process in an isolated system: ΔS_universe> 0.

Entropy Changes in Processes

Solid → liquid → gas: entropy increases (ΔS > 0).
Dissolving a solid in liquid: usually ΔS > 0.
Reaction increasing moles of gas: ΔS > 0 (more disorder).
Cooling, condensation, freezing: ΔS < 0 (more order).

Third Law of Thermodynamics

The entropy of a pure, perfect crystalline substance at absolute zero (0 K) is zero. This provides an absolute reference point for measuring entropy.

Gibbs Free Energy

Adjust ΔH, ΔS, and temperature T to calculate Gibbs free energy ΔG = ΔH − TΔS. See if the reaction is spontaneous.

Combustion (ΔH<0, ΔS>0)

Ice melting (ΔH>0, ΔS>0)

N₂+3H₂→2NH₃ (ΔH<0, ΔS<0)

CaCO₃ decomp (ΔH>0, ΔS>0)

ΔH (enthalpy)

-890 kJ/mol

Exothermic (heat released)

ΔS (entropy)

+243 J/mol·K

Disorder increases

Temperature (T)

298 K (25°C)

ΔG = ΔH − TΔS

= -890 − 298 × (243/1000) = -890 − (72.41) = -962.41 kJ/mol

Spontaneous

ΔG < 0: Reaction proceeds forward spontaneously.

All four cases (NEET favorites)

ΔH −, ΔS +

Always spontaneous at all T

Current

ΔH +, ΔS −

Never spontaneous at any T

ΔH −, ΔS −

Spontaneous at low T (ΔH dominates)

ΔH +, ΔS +

Spontaneous at high T (TΔS dominates)

J.W. Gibbs combined the first and second laws to define a new state function, Gibbs energy (G):

G = H - T S

At constant temperature and pressure:

Δ G = Δ H - T Δ S

Gibbs energy change combines enthalpy (tendency toward lower energy) and entropy (tendency toward higher disorder). It tells you whether a process is spontaneous at constant T and P.

Spontaneity Criteria

ΔH	ΔS	ΔG = ΔH − TΔS	Spontaneity
Negative (exothermic)	Positive (increase)	Always negative	Spontaneous at all temperatures
Positive (endothermic)	Negative (decrease)	Always positive	Non-spontaneous at all temperatures
Negative	Negative	Negative at low T; positive at high T	Spontaneous at low T only
Positive	Positive	Positive at low T; negative at high T	Spontaneous at high T only

ΔG and Equilibrium

At equilibrium: ΔG = 0. The relation between standard Gibbs energy and the equilibrium constant K:

Δ G^{\circ} = - R T ln K

If ΔG° < 0: K > 1 (products favoured at equilibrium). If ΔG° > 0: K < 1 (reactants favoured). If ΔG° = 0: K = 1.

The temperature at which a reaction switches from non-spontaneous to spontaneous (ΔG = 0):

T_{t r an s i t i o n} = \frac{Δ H}{Δ S}

Worked NEET Problems

NEET-style problem · Hess's Law

Question

Given: (1) C + O₂ → CO₂; ΔH = −393.5 kJ/mol. (2) H₂ + ½O₂ → H₂O; ΔH = −285.8 kJ/mol. (3) CH₄ + 2O₂ → CO₂ + 2H₂O; ΔH = −890 kJ/mol. Find the standard enthalpy of formation of CH₄.

Solution

We want: C + 2H₂ → CH₄; ΔH = ?

Using Hess's law: (1) + 2×(2) − (3).

ΔH_f(CH₄) = ΔH₁ + 2ΔH₂ − ΔH₃ = −393.5 + 2(−285.8) − (−890).

= −393.5 − 571.6 + 890 = −75.1 kJ/mol.

NEET-style problem · ΔH and ΔU

Question

For the reaction N₂(g) + 3H₂(g) → 2NH₃(g), ΔH = −92 kJ/mol at 300 K. Calculate ΔU. (R = 8.314 J mol⁻¹ K⁻¹)

Solution

Δn_g = moles of gaseous products − moles of gaseous reactants = 2 − (1 + 3) = 2 − 4 = −2.

ΔH = ΔU + Δn_gRT

−92 × 10³ = ΔU + (−2)(8.314)(300)

ΔU = −92000 + 4988.4 = −87011.6 J/mol = −87.0 kJ/mol.

NEET-style problem · Gibbs Energy and Spontaneity

Question

For a reaction, ΔH = +50 kJ/mol and ΔS = +200 J mol⁻¹ K⁻¹. At what temperature does the reaction become spontaneous?

Solution

ΔG = ΔH − TΔS. For spontaneity, ΔG < 0.

ΔH − TΔS < 0. Since ΔH > 0 and ΔS > 0, ΔG is negative only when TΔS > ΔH.

T > ΔH/ΔS = 50000 J/mol ÷ 200 J mol⁻¹ K⁻¹ = 250 K.

The reaction becomes spontaneous above 250 K (about −23°C). At room temperature (298 K > 250 K), it is spontaneous.

Summary Cheat Sheet

Concept	Key Formula / Fact
First law	ΔU = q + w
Work at const. pressure	w = −PΔV
Enthalpy	ΔH = ΔU + Δn_g RT (for gases)
Exothermic	ΔH < 0; heat released
Endothermic	ΔH > 0; heat absorbed
Hess's law	ΔH is a state function; can add reactions
ΔH from bond enthalpies	Σ(bonds broken) − Σ(bonds formed)
ΔH from formation enthalpies	Σ ΔH°f(products) − Σ ΔH°f(reactants)
Entropy: system + surroundings	ΔS_universe > 0 for spontaneous process
Gibbs energy	ΔG = ΔH − TΔS
Spontaneous	ΔG < 0
Equilibrium	ΔG = 0
Non-spontaneous	ΔG > 0
ΔG° and K	ΔG° = −RT ln K
Transition temperature	T = ΔH / ΔS (when ΔG = 0)

Frequently asked questions

What is the difference between ΔH and ΔU in a reaction?

ΔU is the change in internal energy at constant volume. ΔH is the change in enthalpy at constant pressure (which is the heat exchanged in most lab reactions). They are related by ΔH = ΔU + Δn_g RT, where Δn_g is the change in moles of gas. For reactions with no gaseous species or no change in moles of gas, ΔH ≈ ΔU.

How do you use Hess's law to find an unknown enthalpy?

Write the target reaction at the top. Look at the given reactions and adjust them (reverse, multiply, or divide) so that when you add them together, you get the target. For each operation: reversing changes the sign of ΔH; multiplying by n multiplies ΔH by n. The ΔH of the target reaction is the algebraic sum of the ΔH values of the adjusted reactions.

When is a reaction spontaneous? I keep confusing the conditions.

A reaction is spontaneous when ΔG < 0 (where ΔG = ΔH − TΔS). If ΔH < 0 and ΔS > 0: always spontaneous. If ΔH > 0 and ΔS < 0: never spontaneous. If ΔH < 0 and ΔS < 0: spontaneous at low temperature (ΔH term dominates). If ΔH > 0 and ΔS > 0: spontaneous at high temperature (TΔS term dominates). The crossover temperature is T = ΔH/ΔS.

Why is the enthalpy of neutralisation of a weak acid less than −57 kJ/mol?

For a strong acid + strong base, the net ionic reaction is just H⁺(aq) + OH⁻(aq) → H₂O(l), releasing about −57.1 kJ/mol. When a weak acid is used (e.g., acetic acid), some energy must first be used to ionise the weak acid (ionisation is endothermic). This reduces the net heat released, making ΔH_neutralisation less negative (say −55.2 kJ/mol for acetic acid + NaOH).

What is the relation between ΔG° and the equilibrium constant K?

ΔG° = −RT ln K. If ΔG° is negative (large negative value), K is greater than 1, meaning products are favoured at equilibrium. If ΔG° is positive, K is less than 1, reactants are favoured. At equilibrium, ΔG = 0 (not ΔG°). ΔG° is the standard Gibbs energy change (all species at 1 bar, 298 K); ΔG = 0 describes the equilibrium condition under actual concentrations.

What does it mean for a reaction to have a positive entropy change (ΔS > 0)?

A positive ΔS means the products are more disordered (have more randomness) than the reactants. This happens when: the number of moles of gas increases in the reaction; a solid dissolves into ions; a liquid evaporates; a solid melts. ΔS > 0 makes ΔG more negative, favouring spontaneity, especially at high temperatures.

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ThermodynamicsNEET Chemistry · Class 11 · NCERT Chapter 5

Introduction

Basic Concepts: System, State, and Process

System and Surroundings

State Functions and Path Functions

Types of Processes

First Law of Thermodynamics

Enthalpy and Heat at Constant Pressure

Sign Convention for ΔH

Heat Capacity

Standard Enthalpies of Reactions

Standard Enthalpy of Formation (ΔH°f)

Standard Enthalpy of Combustion (ΔH°c)

Standard Enthalpy of Neutralisation

Standard Enthalpy of Atomisation

Lattice Enthalpy

Hess's Law of Constant Heat Summation

Bond Enthalpy

Second Law of Thermodynamics and Entropy

Entropy Changes in Processes

Third Law of Thermodynamics

Gibbs Free Energy

Spontaneity Criteria

ΔG and Equilibrium

Worked NEET Problems

Summary Cheat Sheet

Frequently asked questions

What is the difference between ΔH and ΔU in a reaction?

How do you use Hess's law to find an unknown enthalpy?

When is a reaction spontaneous? I keep confusing the conditions.

Why is the enthalpy of neutralisation of a weak acid less than −57 kJ/mol?

What is the relation between ΔG° and the equilibrium constant K?

What does it mean for a reaction to have a positive entropy change (ΔS > 0)?

Continue with the next chapter notes

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Standard Enthalpy of Formation (ΔH°_f)

Standard Enthalpy of Combustion (ΔH°_c)