Thermodynamics Class 11 Chemistry Notes: Laws and Formulas

These thermodynamics class 11 chemistry notes revise the whole chapter in the order a teacher builds it: thermodynamic vocabulary, the first law, enthalpy and calorimetry, Hess’s law, then entropy and Gibbs energy. They follow the rationalised NCERT Class 11 Chemistry textbook used in the current CBSE session, so every definition, sign convention and formula here matches your syllabus.

Thermodynamics answers three questions the chapter opens with (NCERT, p. 137): how do we measure the energy change of a reaction, will the reaction occur, and what drives it? The subject compares the initial and final states of a macroscopic system, not how fast the change happens. Rates belong to chemical kinetics; direction and energy are this chapter’s job.

Use this page as a revision pass: read the chapter map, drill the formula sheet, copy the four worked examples, then check the exercise map to see what each NCERT question tests.

The Chapter Map: From Systems to Gibbs Energy

Hold the route in your head before touching formulas. NCERT builds the chapter in nine steps, and each step exists to answer one question:

• System vocabulary — what part of the universe are we studying, and can it exchange matter or energy with its surroundings?
• State functions and internal energy — which quantities describe a state, and what does a change in energy mean?
• First law (\( \Delta U = q + w \)) — how heat and work combine into the energy balance of a system.
• Enthalpy — the heat of a reaction measured at constant pressure, the condition of most open flasks.
• Calorimetry — how labs actually measure \( \Delta U \) and \( \Delta H \).
• Reaction enthalpies and Hess’s law — calculating heats of reaction from formation data and combining reactions.
• Bond, lattice and solution enthalpies — estimating enthalpies from bonds and ionic structures.
• Entropy and spontaneity — why some reactions go and others do not.
• Gibbs energy — the single criterion that settles spontaneity and links to the equilibrium constant.

The scope warning that keeps the chapter coherent: thermodynamics works on macroscopic systems in equilibrium and compares only initial and final states. It never predicts rate (NCERT, p. 137).

System and Surroundings: The First Distinction

Every thermodynamics problem starts by drawing a line. The system is the part of the universe under observation; the surroundings are everything outside it that can interact with it; system plus surroundings make the universe. The real or imaginary wall separating them is the boundary (NCERT, p. 138).

NCERT’s working example: a reaction mixture in a beaker is the system, and the room where the beaker stands is the surroundings (Fig 5.1). The boundary may be the beaker walls or simply an imaginary surface enclosing the reactants.

Systems are classified by what crosses that boundary:

Fig 5.2 shows the three types side by side. The classification decides which form of the first law you can use: an isolated system exchanges nothing, so \( q = 0 \), \( w = 0 \) and \( \Delta U = 0 \).

State Functions, Internal Energy, Work and Heat

A state function (or state variable) is a property whose value depends only on the present state of the system, not on how that state was reached (NCERT, p. 139). Pressure, volume, temperature and amount are state variables. Because the value is fixed by the state, the change in a state function is independent of path.

Fresh analogy — altitude difference, not route distance. The altitude difference between two hill stations is fixed: you measure the same value whether you climb directly or descend and climb back. That is a state function. The distance you actually walked depends on the route, exactly as work and heat depend on the path.

NCERT makes the same point with temperature: going from \( 25^\circ\text{C} \) to \( 35^\circ\text{C} \) is always a change of \( +10^\circ\text{C} \), even if you cool the system first and then heat it (NCERT, p. 140).

Internal energy \( U \) is the sum of all forms of energy in the system — chemical, electrical, mechanical, and any other kind (NCERT, p. 139). Only the change \( \Delta U \) can be measured; an absolute value cannot, because you cannot account for the motion of every particle inside (NCERT, p. 141).

Joule’s paddle-wheel experiments (1840–1850) are the evidence that \( U \) is a state function: 1 kJ of mechanical work from rotating paddles and 1 kJ of electrical work from an immersion rod produced the same temperature rise in the same water. Equal work, same final state, whatever the path (NCERT, p. 139–140).

Chemistry follows the IUPAC sign convention (NCERT, p. 140):

• Heat \( q \) is positive when it enters the system, negative when it leaves.
• Work \( w \) is positive when done on the system, negative when done by the system.

Bank-account analogy for signs. Run the system like an account. Deposits into the account are positive, withdrawals are negative. Heat entering is a deposit (\( +q \)); work done by the system is a withdrawal and is already negative when it enters the sum.

The balance change \( \Delta U \) depends only on the opening and closing balance — not on the order of deposits and withdrawals — which is just path independence in another disguise.

Work and heat themselves are path functions: for a given change of state, \( q \) and \( w \) can vary with the route, but their sum \( q + w = \Delta U \) is fixed by the state (NCERT, p. 141).

First Law of Thermodynamics: The Energy Balance

The first law of thermodynamics is the law of conservation of energy written for a thermodynamic system:

\[ \Delta U = q + w \]

It states that the energy of an isolated system is constant (NCERT, p. 141). If \( q = 0 \) and \( w = 0 \), then \( \Delta U = 0 \). The equation also encodes a deeper fact: \( \Delta U \) depends only on initial and final state, even though \( q \) and \( w \) individually do not.

Pressure-volume work. For a change against a constant external pressure \( p_{ex} \):

\[ w = -p_{ex}\Delta V \]

The negative sign is forced by the convention, not chosen. Expansion makes \( \Delta V \) positive, so \( w \) is negative — work done by the system. Compression makes \( \Delta V \) negative, so \( w \) becomes positive — work done on the system (NCERT, p. 141–143).

Reversible isothermal expansion. When pressure changes in infinitesimal steps and the gas stays in near-equilibrium with its surroundings, the process is reversible. For \( n \) moles of an ideal gas at constant temperature (isothermal), integrating \( -p\,dV \) with \( pV = nRT \) gives:

\[ w_{rev} = -nRT \ln\frac{V_f}{V_i} = -2.303\,nRT \log\frac{V_f}{V_i} \]

(equation 5.5; both forms are examinable, NCERT, p. 143).

Free expansion is expansion into a vacuum: \( p_{ex} = 0 \), so \( w = 0 \). Joule found experimentally that \( q = 0 \) as well, so \( \Delta U = 0 \) for an ideal gas (NCERT, p. 143).

The special cases are worth tabulating:

The trap in the table: in the isothermal rows, \( q \) and \( w \) are each non-zero and opposite in sign, yet \( \Delta U \) is zero. For an ideal gas, isothermal means the internal energy does not change (NCERT, p. 143, 160).

Enthalpy: Heat at Constant Pressure

Most reactions are run in open flasks at constant atmospheric pressure, not at constant volume. The heat absorbed under that condition is not \( \Delta U \) but a new state function, enthalpy (NCERT, p. 144):

\[ H = U + pV \]

Why this definition works: at constant pressure, \( q_p = \Delta U + p\Delta V = (U_2 + pV_2) – (U_1 + pV_1) \). The combination \( U + pV \) appears in both states, so it earns a name. Hence:

\[ \Delta H = q_p \]

Because \( U \), \( p \) and \( V \) are state functions, \( H \) is too — \( \Delta H \) is path-independent even though \( q \) is not. The sign language is simple: \( \Delta H \) is negative for exothermic reactions (heat evolved) and positive for endothermic reactions (heat absorbed), NCERT, p. 144.

For solids and liquids, \( p\Delta V \) is negligible, so \( \Delta H \approx \Delta U \). For gases, the difference is significant and is fixed by the gas-mole change (NCERT, p. 145):

\[ \Delta H = \Delta U + \Delta n_g RT \]

Here \( \Delta n_g \) = moles of gaseous products − moles of gaseous reactants. Only gases count — solids and liquids contribute no significant volume term. The \( \Delta n_g RT \) term is the pressure-volume work exchanged with the atmosphere as the reaction changes volume.

Heat capacity links heat to the temperature rise it causes: \( q = C\Delta T \), where \( C \) is the heat capacity of the object. Molar heat capacity \( C_m = C/n \) is per mole; specific heat capacity \( c \) is per unit mass, so the working form (eq. 5.11) is \( q = c \cdot m \cdot \Delta T \) (NCERT, p. 145–146).

Water’s large heat capacity is why a swimming pool warms slowly.

For one mole of an ideal gas, \( C_p – C_v = R \) (eq. 5.13). The reason: heating at constant pressure must also do expansion work against the atmosphere, so raising the temperature by 1 K at constant pressure needs more heat than at constant volume — the extra is exactly \( R \) (NCERT, p. 146).

Extensive vs intensive properties. Extensive properties scale with the amount of matter (mass, volume, \( U \), \( H \), heat capacity); intensive properties do not (temperature, pressure, density). NCERT’s Fig 5.6 halves a gas’s volume with a partition: volume halves (extensive) but temperature stays the same (intensive) (NCERT, p. 145).

Calorimetry: Measuring Energy Changes in the Lab

Calorimetry is the experimental measurement of heat changes: a reaction runs in a calorimeter of known heat capacity, and the measured temperature rise is converted to heat with \( q = C\Delta T \). Two conditions give two different answers: \( q_v \) measures \( \Delta U \), \( q_p \) measures \( \Delta H \).

A bomb calorimeter (Fig 5.7) is a sealed steel vessel — the bomb — immersed in a water bath, with the sample burnt in pure oxygen. Because the bomb is sealed, its volume cannot change, so no pressure-volume work is done even when gases are involved. The temperature rise of the water bath therefore gives \( q_v = \Delta U \) (NCERT, p. 146).

The sign rule for the data: heat lost by the reaction equals heat gained by the calorimeter with the opposite sign, so \( q_{reaction} = -C_{cal}\Delta T \).

When a bomb-calorimeter reaction involves gases, the measured value is \( \Delta U \); convert to \( \Delta H \) with \( \Delta H = \Delta U + \Delta n_g RT \) before reporting. This is the correction exercised in NCERT’s own bomb experiments and tested in exercise 5.8.

A constant-pressure calorimeter (Fig 5.8) is open to the atmosphere, so the heat measured at constant pressure is \( q_p = \Delta H \). This quantity is also called the heat of reaction or enthalpy of reaction, \( \Delta_r H \). Exothermic reactions give negative \( q_p \); endothermic give positive (NCERT, p. 147).

Reaction Enthalpies, Standard States and Hess’s Law

The reaction enthalpy \( \Delta_r H \) is defined as products minus reactants (NCERT, p. 147):

\[ \Delta_r H = \sum a_i H_{\text{products}} – \sum b_i H_{\text{reactants}} \]

where \( a_i \) and \( b_i \) are the stoichiometric coefficients. Absolute enthalpies are unknowable, so in practice you compute \( \Delta_r H^\ominus \) from formation data.

Standard states and formation enthalpies. The standard state of a substance is its pure form at 1 bar, usually quoted at 298 K, shown by the superscript \( \ominus \) (NCERT, p. 148).

The standard enthalpy of formation \( \Delta_f H^\ominus \) is the enthalpy change when one mole of a compound forms from its elements in their most stable reference states — \( H_2(g) \), \( O_2(g) \), C(graphite), S(rhombic) (NCERT, p. 150). By convention, elements in their reference states have \( \Delta_f H^\ominus = 0 \).

Phase changes are enthalpy changes too, always endothermic (positive) at constant temperature (NCERT, p. 148):

Real-life application — why sweating cools the body. Vaporising water needs a large heat input because hydrogen bonds between water molecules must be broken before molecules escape the liquid (NCERT, p. 148). NCERT’s swimmer problem makes this concrete: a swimmer leaving a pool carries a film of about 18 g of water — exactly one mole.

Vaporising it at 298 K absorbs 44.01 kJ, most of it drawn from the body, which is why the skin cools (NCERT, p. 149). Sweating is the same process on a smaller scale: the enthalpy of vaporisation of water removes heat from the body surface, so evaporation cools you.

Once formation enthalpies are known, reaction enthalpies are arithmetic (eq. 5.15, NCERT, p. 151):

\[ \Delta_r H^\ominus = \sum a_i \Delta_f H^\ominus(\text{products}) – \sum b_i \Delta_f H^\ominus(\text{reactants}) \]

NCERT’s example is the decomposition of calcium carbonate, \( CaCO_3(s) \rightarrow CaO(s) + CO_2(g) \), whose formation enthalpies are \( -1206.9 \), \( -635.1 \) and \( -393.5\ \text{kJ mol}^{-1} \). The arithmetic gives \( \Delta_r H^\ominus = +178.3\ \text{kJ mol}^{-1} \) — endothermic, so lime-making needs continuous heat (NCERT, p. 151).

Thermochemical equations follow three conventions (NCERT, p. 151–152):

• Coefficients mean moles, never molecules.
• The quoted \( \Delta_r H \) is per mole of reaction as written — halving the equation halves the enthalpy change (enthalpy is extensive).
• Reversing the equation reverses the sign: \( N_2 + 3H_2 \rightarrow 2NH_3 \) has \( \Delta_r H^\ominus = -91.8\ \text{kJ mol}^{-1} \), and the reverse decomposition has \( +91.8\ \text{kJ mol}^{-1} \).

Hess’s law of constant heat summation 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, at the same temperature (NCERT, p. 152). It holds because enthalpy is a state function — the route from reactants to products is irrelevant.

Hess’s law is not a formality; some reactions cannot be measured directly. \( C(\text{graphite}) + \frac{1}{2}O_2(g) \rightarrow CO(g) \) always produces some \( CO_2 \), so NCERT combines two measurable reactions: \( C + O_2 \rightarrow CO_2 \) (\( -393.5\ \text{kJ mol}^{-1} \)) and \( CO + \frac{1}{2}O_2 \rightarrow CO_2 \) (\( -283.0\ \text{kJ mol}^{-1} \)).

Reverse the second (\( +283.0 \)), add: \( \Delta_r H^\ominus = -110.5\ \text{kJ mol}^{-1} \) (NCERT, p. 152–153).

Bond Enthalpy, Lattice Enthalpy and Enthalpy of Solution

Bond dissociation enthalpy is the enthalpy change when one mole of a particular covalent bond is broken in a gaseous molecule. For a diatomic molecule it is the molecule’s own bond value: \( H_2(g) \rightarrow 2H(g) \) has \( \Delta_{H-H}H^\ominus = 435.0\ \text{kJ mol}^{-1} \) (NCERT, p. 154).

In a polyatomic molecule the same bond can break at different energies. The four C–H bonds of methane are identical in length but break at successively different energies — 427, 439, 452 and 347 \( \text{kJ mol}^{-1} \).

Chemists therefore use the mean bond enthalpy: the atomization enthalpy of \( CH_4 \) is 1665 \( \text{kJ mol}^{-1} \), and one quarter of it — \( 416\ \text{kJ mol}^{-1} \) — is the mean C–H bond enthalpy (NCERT, p. 154–155).

For gas-phase reactions, the reaction enthalpy can be estimated from bonds (eq. 5.17):

\[ \Delta_r H^\ominus \approx \sum \text{bond enthalpies of reactants} – \sum \text{bond enthalpies of products} \]

Breaking reactant bonds costs energy; forming product bonds releases it. The rule works well only when all species are gaseous (NCERT, p. 155).

Lattice enthalpy is the enthalpy change when one mole of an ionic compound dissociates into gaseous ions:

\[ NaCl(s) \rightarrow Na^+(g) + Cl^-(g), \qquad \Delta_{lattice}H^\ominus = +788\ \text{kJ mol}^{-1} \]

It cannot be measured directly, so chemists use a Born-Haber cycle (Fig 5.9): a closed loop of steps whose enthalpy changes sum to zero by Hess’s law (NCERT, p. 156). For NaCl the cycle runs: sublimation of sodium (+108.4), ionisation of sodium (+496), dissociation of half a mole of \( Cl_2 \) (+121), electron gain by chlorine (−348.6), then the lattice step itself.

Summing gives \( 411.2 + 108.4 + 121 + 496 – 348.6 = +788\ \text{kJ mol}^{-1} \). The ionisation and electron-gain enthalpies come from Unit 3 — revise them in the chemical bonding notes.

Enthalpy of solution combines the lattice term and the hydration term:

\[ \Delta_{sol}H^\ominus = \Delta_{lattice}H^\ominus + \Delta_{hyd}H^\ominus \]

For NaCl: \( +788 + (-784) = +4\ \text{kJ mol}^{-1} \) — almost no heat change, which is why dissolving salt in water feels thermally neutral (NCERT, p. 157). If the lattice enthalpy is very high, dissolution may not occur at all; this is one reason fluorides are often less soluble than the corresponding chlorides.

The closely related enthalpy of dilution is the heat change when extra solvent is added to a solution. NCERT’s HCl data show the enthalpy of solution drifting from \( -69.01 \) to \( -74.85\ \text{kJ mol}^{-1} \) as the solvent amount grows toward infinite dilution (NCERT, p. 157).

Spontaneity and Entropy: The Direction of Change

A spontaneous process has the potential to proceed without external help, and it cannot reverse itself without an external agency (NCERT, p. 158). Rate is not part of the definition: hydrogen and oxygen can sit mixed for years with no visible change, yet their combination is spontaneous.

Misconception autopsy — “spontaneous” does not mean fast, and “endothermic” does not mean impossible. Both errors come from reading the word casually; the chapter breaks both:

• Spontaneity is about potential, not speed. A lit match is an external trigger for \( H_2 + O_2 \); the reaction itself needs no outside help to be thermodynamically favoured.
• Endothermic reactions can be spontaneous. \( \frac{1}{2}N_2(g) + O_2(g) \rightarrow NO_2(g) \) has \( \Delta_r H^\ominus = +33.2\ \text{kJ mol}^{-1} \), and \( C(\text{graphite}) + 2S(l) \rightarrow CS_2(l) \) has \( +128.5\ \text{kJ mol}^{-1} \); both proceed without external help (NCERT, p. 158–159). A decrease in enthalpy is a contributory factor, not the criterion.

Entropy \( S \) is a measure of randomness or disorder in the system (NCERT, p. 159).

NCERT’s mental picture is gas diffusion (Fig 5.11): with a partition in place, any molecule picked from the left chamber is certainly gas A; once the partition is removed and the gases mix, a picked molecule could be either — the system is less predictable, more disordered, and its entropy has increased.

Heat randomises molecules, but the same heat causes more disorder at low temperature than at high temperature. That is why the entropy change is (NCERT, p. 160):

\[ \Delta S = \frac{q_{rev}}{T} \]

Like \( U \) and \( H \), \( S \) is a state function, so \( \Delta S \) is path-independent. Qualitative predictions from the chapter (see NCERT Problem 5.10):

• Solid → liquid → gas: entropy increases (solid is most ordered, gas most disordered).
• More gas molecules in the products → entropy increases.
• Dissolving a solid usually increases entropy.
• Cooling decreases entropy; crystallising a liquid lowers it.
• More particles, more disorder: \( H_2(g) \rightarrow 2H(g) \) increases entropy.

Second law of thermodynamics: for a spontaneous process the total entropy change of system plus surroundings is positive (NCERT, p. 160):

\[ \Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} \gt 0 \]

At equilibrium the entropy of the system is at its maximum, and \( \Delta S = 0 \).

The second law explains why a negative \( \Delta S_{sys} \) does not stop a reaction. Iron rusting, \( 4Fe(s) + 3O_2(g) \rightarrow 2Fe_2O_3(s) \), has \( \Delta S_{sys} = -549.4\ \text{J K}^{-1}\text{mol}^{-1} \) yet rusts anyway.

The reaction releases \( 1648 \times 10^3\ \text{J mol}^{-1} \) of heat; the surroundings absorb it and gain \( \Delta S_{surr} = -\Delta H/T = 5530\ \text{J K}^{-1}\text{mol}^{-1} \). Total entropy change is \( +4980.6\ \text{J K}^{-1}\text{mol}^{-1} \), so the process is spontaneous (NCERT, p. 161).

Third law of thermodynamics: the entropy of a perfect crystalline substance approaches zero as the temperature approaches absolute zero (NCERT, p. 162). It matters because it allows absolute entropies: summing \( q_{rev}/T \) increments from 0 K to 298 K gives the standard entropy of a pure substance, and those values feed Hess-law-type calculations.

Gibbs Energy: The Single Criterion That Settles It

Neither \( \Delta H \) nor \( \Delta S \) alone decides spontaneity in closed systems; the combination does. Define the Gibbs energy \( G = H – TS \). At constant temperature and pressure, the Gibbs equation gives (NCERT, p. 161):

\[ \Delta G = \Delta H – T\Delta S \]

Where it comes from: \( \Delta S_{surr} = -\Delta H_{sys}/T \), so \( T\Delta S_{total} = T\Delta S_{sys} – \Delta H_{sys} = -\Delta G \). Since spontaneity requires \( \Delta S_{total} \gt 0 \), it is exactly equivalent to \( \Delta G \lt 0 \).

• \( \Delta G \lt 0 \): the process is spontaneous.
• \( \Delta G \gt 0 \): the process is non-spontaneous.
• \( \Delta G = 0 \): the system is at equilibrium, and the free energy of the system is at its minimum (NCERT, p. 161–163).

Memory device — negative G means the reaction goes. If \( \Delta G \) is negative, the reaction is spontaneous; positive means it needs external help; zero means equilibrium. One sign check replaces three separate arguments.

Temperature changes which term wins. Table 5.4 in NCERT summarises the four cases (NCERT, p. 163); the idea is that the \( T\Delta S \) term grows with temperature, so at high enough \( T \) a positive entropy change can outweigh a positive enthalpy change:

This is why many industrial reactions with positive entropy change are run at high temperature: the entropy term eventually dominates the enthalpy term.

Gibbs energy also answers “how far?”. At equilibrium \( \Delta_r G = 0 \), which gives the link to the equilibrium constant (NCERT, p. 163):

\[ \Delta_r G^\ominus = -RT \ln K = -2.303\,RT \log K \]

Strongly exothermic reactions tend to have large negative \( \Delta_r G^\ominus \) and therefore large \( K \); strongly endothermic reactions have \( K \) much smaller than 1. The equilibrium chapter uses this relation directly — see the equilibrium notes for how \( K \) is used in calculations.

NCERT’s ozone problem is a clean example: for \( \frac{3}{2}O_2(g) \rightarrow O_3(g) \), \( K_p = 2.47 \times 10^{-29} \), so \( \Delta_r G^\ominus = +163\ \text{kJ mol}^{-1} \) at 298 K — positive, meaning ozone does not form spontaneously from oxygen under standard conditions (NCERT, p. 164).

Thermodynamics Class 11 Chemistry Notes: Definitions Table

One pass through the chapter’s vocabulary, in student language with a concrete example for each row.

Formula Sheet: Equations with Symbols and Units

Every examinable equation in one table, with the condition that makes it valid. The gas constant is \( R = 8.314\ \text{J K}^{-1}\text{mol}^{-1} \) for all calculations.

Constants to memorise: \( R = 8.314\ \text{J K}^{-1}\text{mol}^{-1} \), and at 298 K, \( RT \approx 2.48\ \text{kJ mol}^{-1} \) — the shortcut for \( \Delta n_g RT \) corrections at room temperature. Molar enthalpies are quoted in \( \text{kJ mol}^{-1} \); molar entropies in \( \text{J K}^{-1}\text{mol}^{-1} \), so convert before using \( T\Delta S \).

Worked Examples: Four Problems You Should Be Able to Do

Four solved problems, each demonstrating a method you will reuse. Copy the layout: name the method, fix the signs, convert the units, calculate, then state the physical conclusion.

Example 1 — Applying the First Law with Signs

Problem. A gas absorbs 250 J of heat and does 90 J of work on its surroundings. Calculate the change in internal energy of the gas.

Example 2 — Converting ΔU to ΔH with Δn_g

Problem. For \( N_2(g) + 3H_2(g) \rightarrow 2NH_3(g) \), \( \Delta U^\ominus = -88.4\ \text{kJ mol}^{-1} \) at 298 K. Calculate \( \Delta H^\ominus \).

Example 3 — Hess’s Law from Formation Enthalpies

Problem. Calculate \( \Delta_r H^\ominus \) for \( 2Fe_2O_3(s) + 3C(\text{graphite}) \rightarrow 4Fe(s) + 3CO_2(g) \). Data: \( \Delta_f H^\ominus(Fe_2O_3,\ s) = -824.2\ \text{kJ mol}^{-1} \), \( \Delta_f H^\ominus(CO_2,\ g) = -393.51\ \text{kJ mol}^{-1} \); Fe(s) and C(graphite) are elements in their reference states.

Example 4 — At What Temperature Does a Reaction Become Spontaneous?

Problem. A reaction has \( \Delta H = +120\ \text{kJ mol}^{-1} \) and \( \Delta S = +150\ \text{J K}^{-1}\text{mol}^{-1} \). Is it spontaneous at 400 K? At what temperature does it become spontaneous?

Common Mistakes: Error and Correction Pairs

Six traps specific to this chapter. Each row gives the wrong move, the correct rule, and a quick check that catches the error.

Exam Notes: What the NCERT Exercises Test

NCERT’s end-of-chapter exercises (5.1 to 5.20) cluster around a small set of skills. Use the map to find your weak spots and practise deliberately. Write the working yourself — the map tells you what a full-mark answer must include, not the answer itself.

Beyond the map, exercises 5.21 and 5.22 test thermodynamic stability of a compound and the entropy change of the surroundings, \( \Delta S_{surr} = -\Delta H/T \).

Mark-earning habits for every numerical:

• Write the sign before every \( q \), \( w \) and \( \Delta H \) value — the sign carries the mark.
• Write units at every step: \( \text{kJ mol}^{-1} \) for enthalpies, \( \text{J K}^{-1}\text{mol}^{-1} \) for entropies.
• Show \( \Delta n_g \) explicitly before using \( \Delta H = \Delta U + \Delta n_g RT \).
• State that a computed \( \Delta_r H \) is per mole of reaction as written.
• End spontaneity questions with a verdict: “\( \Delta G \lt 0 \), so the reaction is spontaneous.”

When an exercise feels uncertain, the official NCERT chapter PDF (kech105.pdf from ncert.nic.in) carries the fully solved in-chapter problems — use it to match your method against the textbook’s.

One-Page Revision Summary: Laws, Signs and Key Numbers

Five minutes before the exam — three tables and nothing else.

Table 1: The three laws.

Table 2: Sign conventions.

Table 3: Key numbers from the chapter.

Final line to carry in: negative G means the reaction goes. For more support, browse the Class 11 Chemistry notes hub, the Class 11 notes index, or start from the main CBSE notes page.

Reference: NCERT Class 11 Chemistry textbook, chapter Thermodynamics.

Frequently Asked Questions

Why is the first law written as q + w and not q – w?

Because of the IUPAC sign convention NCERT follows. Work is already signed: \( w \) is positive when work is done on the system and negative when the system does work, so adding \( q \) and \( w \) gives the correct balance in every case.

The \( q – w \) form belongs to the older physics convention where work done by the system was called positive (NCERT, p. 141).

When is the gas-mole correction between enthalpy and internal energy needed?

Whenever a reaction involves gases and you know one of \( \Delta U \) or \( \Delta H \) and need the other — bomb calorimetry gives \( \Delta U \), so convert before quoting \( \Delta H \). Count only gaseous species in \( \Delta n_g \). If \( \Delta n_g = 0 \), the equation collapses to \( \Delta H = \Delta U \), as in the combustion of graphite to \( CO_2 \) (NCERT, p. 145, 147).

Why can an endothermic reaction be spontaneous?

Spontaneity is decided by the total entropy change of the universe, not by the sign of \( \Delta H \). The Gibbs equation \( \Delta G = \Delta H – T\Delta S \) shows that a positive \( \Delta H \) can be outweighed by a large positive \( T\Delta S \). NCERT’s examples are \( NO_2 \) and \( CS_2 \) formation — both endothermic yet spontaneous (NCERT, p. 158–159).

What is the difference between bond dissociation enthalpy and mean bond enthalpy?

Bond dissociation enthalpy is the energy to break one mole of a specific bond in a gaseous substance — for a diatomic like \( H_2 \), it is the unique H–H value, \( 435.0\ \text{kJ mol}^{-1} \).

Mean bond enthalpy is an average for a bond type in a polyatomic molecule: the four C–H bonds of methane break at different energies, so the mean is the total atomization enthalpy divided by 4, \( 1665/4 = 416\ \text{kJ mol}^{-1} \) (NCERT, p. 154–155).

How do I use the Gibbs energy change to decide if a reaction is spontaneous?

Calculate \( \Delta G = \Delta H – T\Delta S \) at the temperature of interest. \( \Delta G \lt 0 \) means spontaneous, \( \Delta G \gt 0 \) means non-spontaneous, and \( \Delta G = 0 \) means equilibrium. Keep units consistent — if \( \Delta H \) is in kJ, convert \( \Delta S \) from J to kJ. When \( K \) is given instead, use \( \Delta_r G^\ominus = -2.303\,RT \log K \) (NCERT, p. 161–163).

Official source: download the NCERT textbook free from ncert.nic.in.