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Class 11 Chemistry Chapter 5 Thermodynamics (Chemistry) — Formulas & Key Points

Q: What is the first law of thermodynamics formula in CBSE Class 11 Chemistry Chapter 5?

ΔU = q + w, where ΔU is the change in internal energy, q is heat transferred to the system (positive if absorbed), and w is work done on the system (positive if compressed, negative if expanded). This is the IUPAC sign convention used in NCERT.

Q: How do I calculate enthalpy change (ΔH) from standard enthalpies of formation?

Use ΔrH° = Σ ΔfH°(products) − Σ ΔfH°(reactants), multiplying each ΔfH° by its stoichiometric coefficient. Remember ΔfH° of any element in its standard state is zero. Ensure all values are at the same temperature (usually 298 K).

Q: What is the difference between ΔH and ΔU in thermodynamics?

ΔU is the change in internal energy; it equals heat at constant volume (qᵥ). ΔH = ΔU + Δ(PV) is the change in enthalpy; it equals heat at constant pressure (qₚ). For ideal gases, ΔH = ΔU + ΔnRT, where Δn is the change in moles of gas.

Q: How do I apply Hess's law to find an unknown enthalpy change?

Write the target equation. Manipulate given reactions (reverse to flip ΔH sign, multiply to scale ΔH) so they sum to the target. Add the ΔH values algebraically. Intermediate species must cancel. Hess's law works because enthalpy is a state function.

Q: What is the Gibbs free energy equation and when is a reaction spontaneous?

ΔG = ΔH − TΔS. A reaction is spontaneous at constant T and P if ΔG 0. Temperature can shift spontaneity depending on the signs of ΔH and ΔS.

Thermodynamics in CBSE Class 11 Chemistry Chapter 5 introduces energy changes in chemical reactions and the criteria for spontaneity. Mastering the formulas—first law, enthalpy, entropy, Gibbs free energy—is essential for numericals worth 6–8 marks in the board exam. This sheet organizes every NCERT equation, constant, and definition into tables, highlights common sign and unit pitfalls, and provides memory tricks and solved examples for rapid revision.

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Key takeaways

✓Internal energy change (ΔU) relates to heat (q) and work (w) via the first law: ΔU = q + w, with sign conventions crucial for CBSE exams.
✓Enthalpy change ΔH = ΔU + Δ(PV); at constant pressure, ΔH equals heat absorbed or released (qₚ).
✓Entropy (ΔS) measures disorder; spontaneous processes at constant T and P occur when ΔG = ΔH − TΔS < 0.
✓Standard enthalpy of formation (ΔfH°) of elements in their standard states is zero by definition.
✓Hess's law allows calculation of ΔrH° by summing enthalpy changes of intermediate steps, independent of pathway.
✓Bond enthalpy averages over many molecules; ΔrH = Σ(bonds broken) − Σ(bonds formed) estimates reaction enthalpy.
✓Sign conventions: heat absorbed and work done by system are positive in IUPAC; work done on system is negative.

Core Thermodynamic Formulas and Laws

Thermodynamics quantifies energy transformations. The first law of thermodynamics states that energy is conserved: the change in internal energy (ΔU) of a system equals the heat supplied (q) plus work done on the system (w). NCERT uses the IUPAC sign convention: heat absorbed by the system is positive, heat released is negative; work done by the system (expansion) is negative, work done on the system (compression) is positive. Enthalpy (H) is a state function defined as H = U + PV, particularly useful at constant pressure. The table below consolidates the essential laws and state functions with their mathematical forms and usage notes.

ΔU is a state function; it depends only on initial and final states, not the path taken.
Work in expansion/compression: w = −Pₑₓₜ ΔV (against constant external pressure).
For reversible isothermal expansion of an ideal gas: w = −nRT ln(V₂/V₁).
At constant volume, qᵥ = ΔU (no PV work).
At constant pressure, qₚ = ΔH.

Enthalpy Change Formulas and Standard States

Enthalpy change (ΔH) is the heat absorbed or released at constant pressure. Standard enthalpy changes (ΔH°) are measured at 298 K and 1 bar pressure with all substances in their standard states (pure, most stable form). The standard enthalpy of formation (ΔfH°) of any element in its reference form is defined as zero. Reaction enthalpy can be calculated from standard enthalpies of formation using ΔrH° = Σ ΔfH°(products) − Σ ΔfH°(reactants). NCERT Chapter 5 defines several types of standard molar enthalpy changes, each denoted by a subscript. Understanding these definitions is critical for conceptual MCQs and numericals in CBSE exams.

ΔfH° (formation): enthalpy change when 1 mole of compound forms from elements in standard states.
ΔcH° (combustion): enthalpy change when 1 mole of substance burns completely in O₂.
ΔatH° (atomisation): enthalpy change to dissociate 1 mole of substance into gaseous atoms.
Bond enthalpy: energy required to break 1 mole of a specific bond in gaseous molecules (average value).
ΔsubH° (sublimation): solid → gas. ΔvapH° (vaporisation): liquid → gas. ΔfusH° (fusion): solid → liquid.

Hess's Law and Calculation Strategy

Hess's law states that the total enthalpy change for a reaction is the same regardless of the number of steps, because enthalpy is a state function. Mathematically, if a reaction can be expressed as the sum of two or more reactions, ΔrH° (net) = ΔrH°₁ + ΔrH°₂ + … This principle allows calculation of unknown enthalpy changes from known values. NCERT Class 11 Chemistry Chapter 5 illustrates Hess's law with formation and combustion data. When manipulating equations, remember: reversing a reaction changes the sign of ΔH; multiplying coefficients multiplies ΔH by the same factor. This is a favourite topic for 3–4 mark numericals in CBSE boards.

Write the target reaction clearly.
Identify given reactions and their ΔH values.
Reverse or multiply reactions as needed so that they sum to the target.
Add the enthalpy changes algebraically, applying sign and factor changes.
Check that intermediate species cancel out.

Entropy and the Second Law of Thermodynamics

Entropy (S) is a state function that measures the randomness or disorder of a system. The second law of thermodynamics states that the entropy of an isolated system increases in any spontaneous process (ΔSₜₒₜₐₗ > 0). For a reversible process, ΔS = qᵣₑᵥ/T. The standard molar entropy S° is always positive and increases with temperature, molecular complexity, and phase (solid < liquid < gas). Entropy change for a reaction is ΔrS° = Σ S°(products) − Σ S°(reactants). Units: J K⁻¹ mol⁻¹. NCERT emphasizes that while ΔSsystem may decrease, ΔSuniverse = ΔSsystem + ΔSsurroundings must be positive for spontaneity.

Entropy is an extensive property; it scales with amount of substance.
At absolute zero (third law), the entropy of a perfect crystal is zero.
Phase transitions: ΔS = ΔH(transition)/T at equilibrium temperature.
Mixing of gases or dissolving a solute increases entropy.
For exothermic reactions, ΔSsurroundings = −ΔHsystem/T (surroundings gain heat).

Gibbs Free Energy and Spontaneity Criteria

Gibbs free energy (G) combines enthalpy and entropy to predict spontaneity at constant temperature and pressure: G = H − TS. The change in Gibbs energy is ΔG = ΔH − TΔS. A process is spontaneous if ΔG < 0, at equilibrium if ΔG = 0, and non-spontaneous if ΔG > 0. Standard Gibbs energy of reaction ΔrG° = Σ ΔfG°(products) − Σ ΔfG°(reactants). The relationship ΔG° = −RT ln K links thermodynamics to equilibrium constants (covered in later chapters). NCERT Chapter 5 explains that even if ΔH is positive (endothermic), a reaction can be spontaneous if TΔS is large and positive. Conversely, exothermic reactions with negative ΔS may be non-spontaneous at high T.

ΔG < 0: spontaneous (product-favoured).
ΔG = 0: system at equilibrium.
ΔG > 0: non-spontaneous (reactant-favoured).
Temperature dependence: ΔG = ΔH − TΔS shows how spontaneity can reverse with T.
Standard free energy of formation ΔfG° of elements in standard state is zero.

Key Terms and Definitions (NCERT Verbatim)

Precise definitions are tested in 1-mark MCQs and 2-mark short answers. NCERT Class 11 Chemistry Chapter 5 defines system (part of universe under study), surroundings (rest of universe), boundary (interface separating system and surroundings). Systems are classified as open (exchange matter and energy), closed (exchange energy only), or isolated (no exchange). A state function depends only on the state, not the path (e.g. U, H, S, G). A path function depends on the route taken (e.g. q, w). An isothermal process occurs at constant temperature, adiabatic at zero heat transfer (q = 0), isobaric at constant pressure, isochoric at constant volume. Extensive properties (U, H, S, V) scale with amount; intensive properties (T, P, density) do not.

System: portion of matter under thermodynamic investigation.
Surroundings: everything outside the system that can exchange energy or matter.
State function: property determined by current state (independent of history).
Path function: depends on the specific process path (not a state function).
Reversible process: proceeds infinitely slowly through equilibrium states; maximum work.
Irreversible process: real processes; ΔSuniverse > 0.
Internal energy (U): total energy contained in a system.
Enthalpy (H): U + PV; heat content at constant pressure.
Entropy (S): measure of molecular disorder or randomness.
Gibbs free energy (G): H − TS; criterion for spontaneity at constant T, P.

Important Constants, Units, and Sign Conventions

Consistent units and correct signs are crucial for scoring full marks in numericals. The gas constant R appears in multiple formulas: R = 8.314 J K⁻¹ mol⁻¹ = 0.0821 L atm K⁻¹ mol⁻¹. Use J K⁻¹ mol⁻¹ for entropy and Gibbs energy calculations, L atm for PV work. Standard conditions: P° = 1 bar (10⁵ Pa), T = 298 K (25°C) unless stated otherwise. IUPAC sign convention (adopted by NCERT): heat absorbed by system q > 0, heat released q < 0; work done on system w > 0, work done by system (expansion) w < 0. Always write units in final answers: ΔH in kJ mol⁻¹, ΔS in J K⁻¹ mol⁻¹, ΔG in kJ mol⁻¹.

R = 8.314 J K⁻¹ mol⁻¹ (use for ΔG, ΔS, w calculations).
1 L atm = 101.3 J (conversion for PV work).
1 bar = 10⁵ Pa; standard pressure in NCERT is 1 bar.
Temperature in Kelvin: T(K) = T(°C) + 273.15.
ΔH, ΔG: kJ mol⁻¹; ΔS: J K⁻¹ mol⁻¹ (note kJ vs J).
Exothermic: ΔH < 0 (heat released). Endothermic: ΔH > 0 (heat absorbed).
Work: expansion (ΔV > 0) → w < 0; compression (ΔV < 0) → w > 0.

Memory Tricks, Mnemonics, and Common Pitfalls

Remembering sign conventions: 'Heat IN is positive, Work ON is positive' (IUPAC). For Gibbs spontaneity, use the mnemonic 'Negative G means GO!' (ΔG < 0 = spontaneous). To recall ΔG = ΔH − TΔS, think 'G goes down when H drops or S rises'. Hess's law: 'Enthalpy is path-independent; add steps like algebraic equations'. Common mistakes in CBSE exams include mixing up ΔU and ΔH (use ΔU = qᵥ at const. V, ΔH = qₚ at const. P), forgetting to convert ΔS from J to kJ when combining with ΔH in ΔG, using wrong sign for work (expansion should give negative w), and assuming ΔfH° for O₂ or H₂ is non-zero. Always double-check stoichiometric coefficients when summing ΔfH° or S° values.

Mnemonic for state functions: 'U H S G are states; q and w are paths'.
Sign rule: 'System GAINS heat (+q), LOSES work (−w when expanding)'.
ΔG and T: if ΔH and ΔS have same sign, spontaneity flips at some temperature.
Hess's law: reverse reaction flips sign of ΔH; multiply by n scales ΔH by n.
Entropy units: J K⁻¹ mol⁻¹ (not kJ); convert before using in ΔG = ΔH − TΔS.
Standard state of element: ΔfH° = 0, but S° ≠ 0 (third law exception at 0 K only).
Work in litre-atmosphere: convert to joules using 1 L atm = 101.3 J.

Three Solved Mini-Examples Applying Key Formulas

Worked examples cement formula application and reveal step-by-step logic for CBSE-style numericals. Each example below targets a different formula cluster: first law and work, Hess's law with formation enthalpies, and Gibbs free energy spontaneity. Practice these patterns for 3-mark and 5-mark questions. Always write given data, formula, substitution, and final answer with units. Show sign reasoning explicitly. These examples mirror the difficulty and format of NCERT in-text and end-of-chapter problems in Class 11 Chemistry Chapter 5, ensuring alignment with board exam standards and typical CBSE marking schemes that award method marks even if the arithmetic slips.

One-Glance Last-Minute Revision Box

This condensed checklist captures the absolute essentials for quick recap 24 hours before the CBSE Chemistry board exam or a school test on Chapter 5. Pin it above your desk or screenshot it on your phone. Revise sign conventions first—they account for half the silly mistakes. Then run through the four big formulas: first law, ΔH from ΔfH°, ΔS calculation, and ΔG spontaneity. Remember the Gibbs table (four cases of ΔH and ΔS). Know standard state definitions by heart (ΔfH° of elements = 0). Practice one Hess's law sum and one ΔG numerical the night before. Check that you can interconvert kJ and J for entropy. Finally, skim NCERT in-text examples 5.1 through 5.7 and solved problems at chapter end—CBSE often adapts these directly. With this sheet and 30 minutes of focussed revision, you will enter the exam confident and formula-ready.

**First Law:** ΔU = q + w. Signs: heat in (+), work on system (+).
**Enthalpy at const. P:** ΔH = qₚ. ΔrH° = Σ ΔfH°(prod) − Σ ΔfH°(react).
**Work (const. Pₑₓₜ):** w = −Pₑₓₜ ΔV. Reversible isothermal: w = −2.303 nRT log(V₂/V₁).
**Entropy:** ΔS = qᵣₑᵥ/T. ΔrS° = Σ S°(prod) − Σ S°(react). Units: J K⁻¹ mol⁻¹.
**Gibbs Free Energy:** ΔG = ΔH − TΔS. Spontaneous: ΔG < 0. ΔrG° = Σ ΔfG°(prod) − Σ ΔfG°(react).
**Hess's Law:** ΔH is path-independent; reverse flips sign, multiply scales ΔH.
**Standard states:** ΔfH°(element) = 0; ΔfG°(element) = 0. S°(element) ≠ 0 (except at 0 K).
**Gibbs 4 cases:** (−ΔH, +ΔS) always spontaneous; (+ΔH, −ΔS) never; others T-dependent.
**Constants:** R = 8.314 J K⁻¹ mol⁻¹; 1 L bar ≈ 100 J; T(K) = °C + 273.
**Unit check:** ΔH, ΔG in kJ mol⁻¹; ΔS in J K⁻¹ mol⁻¹. Convert before substituting in ΔG = ΔH − TΔS.

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Frequently asked questions

What is the first law of thermodynamics formula in CBSE Class 11 Chemistry Chapter 5?+

ΔU = q + w, where ΔU is the change in internal energy, q is heat transferred to the system (positive if absorbed), and w is work done on the system (positive if compressed, negative if expanded). This is the IUPAC sign convention used in NCERT.

How do I calculate enthalpy change (ΔH) from standard enthalpies of formation?+

Use ΔrH° = Σ ΔfH°(products) − Σ ΔfH°(reactants), multiplying each ΔfH° by its stoichiometric coefficient. Remember ΔfH° of any element in its standard state is zero. Ensure all values are at the same temperature (usually 298 K).

What is the difference between ΔH and ΔU in thermodynamics?+

ΔU is the change in internal energy; it equals heat at constant volume (qᵥ). ΔH = ΔU + Δ(PV) is the change in enthalpy; it equals heat at constant pressure (qₚ). For ideal gases, ΔH = ΔU + ΔnRT, where Δn is the change in moles of gas.

How do I apply Hess's law to find an unknown enthalpy change?+

Write the target equation. Manipulate given reactions (reverse to flip ΔH sign, multiply to scale ΔH) so they sum to the target. Add the ΔH values algebraically. Intermediate species must cancel. Hess's law works because enthalpy is a state function.

What is the Gibbs free energy equation and when is a reaction spontaneous?+

ΔG = ΔH − TΔS. A reaction is spontaneous at constant T and P if ΔG < 0, at equilibrium if ΔG = 0, and non-spontaneous if ΔG > 0. Temperature can shift spontaneity depending on the signs of ΔH and ΔS.

Why is entropy measured in J K⁻¹ mol⁻¹ but enthalpy in kJ mol⁻¹?+

Entropy changes are typically smaller in magnitude, so J K⁻¹ mol⁻¹ is convenient. Enthalpy changes are larger, hence kJ mol⁻¹. Always convert ΔS to kJ K⁻¹ mol⁻¹ (divide by 1000) before substituting into ΔG = ΔH − TΔS to match units.

What are the four cases of ΔH and ΔS for predicting spontaneity?+

(1) ΔH<0, ΔS>0: always spontaneous. (2) ΔH>0, ΔS<0: never spontaneous. (3) ΔH<0, ΔS<0: spontaneous at low T. (4) ΔH>0, ΔS>0: spontaneous at high T. Use ΔG = ΔH − TΔS to find the crossover temperature.

How do I remember the sign convention for heat and work?+

IUPAC (NCERT) convention: heat absorbed by the system is positive (+q), heat released is negative (−q); work done on the system (compression) is positive (+w), work done by the system (expansion) is negative (−w). Mnemonic: 'Heat IN and Work ON are positive.'

What is the standard enthalpy of formation, and why is it zero for elements?+

ΔfH° is the enthalpy change when 1 mole of a compound forms from its elements in their standard states at 1 bar and 298 K. By definition, ΔfH° of an element in its most stable form (e.g. O₂(g), C(graphite)) is zero, as there is no formation from simpler substances.

Can CBSETUTOR.ai help solve thermodynamics numericals step-by-step?+

Yes. Upload a photo of any Chapter 5 problem—Hess's law, ΔG calculation, or work-heat-energy sums—and receive instant, detailed solutions with formula breakdowns. The AI tutor explains sign conventions, unit conversions, and method marks. Try the 3-day free trial at ₹999/month for all subjects and classes.

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