Subtopics - Thermodynamics and Thermochemistry (NEET)
Eight topic blocks: thermodynamic terms (systems, properties, processes), internal energy and first law, enthalpy with heat capacity, PV work for isothermal and adiabatic expansions with Joule-Thomson effect, thermochemistry with ten types of heats of reaction, Hess's law with Born-Haber cycle, second law with Carnot cycle and entropy, and Gibbs free energy with spontaneity criteria and the third law.
1) Thermodynamic Terms and System Types
Defines system, surroundings, and boundary. Classifies systems as isolated, closed, and open. Distinguishes extensive properties (volume, mass, entropy, enthalpy, heat capacity) from intensive properties (temperature, pressure, density, molar volume). Defines four process types: adiabatic, isothermal, isobaric, and isochoric.
2) Internal Energy, Zeroth Law, and First Law of Thermodynamics
Internal energy (E) is the sum of all forms of molecular energy. The zeroth law establishes thermal equilibrium and the concept of temperature. The first law (dq = dE + dw) is conservation of energy applied to thermodynamic systems. Special cases: adiabatic (dw = -dE), cyclic (dq = dw), and isochoric (dq = dE).
3) Enthalpy and Heat Capacity
Enthalpy H = E + PV is the heat content at constant pressure. Delta-H = delta-E + P delta-V at constant pressure. Specific heat capacity (per gram) and molar heat capacity (per mole) are defined at constant P and constant V. Cp - Cv = R. Gamma = Cp/Cv varies with atomicity: 1.66 (monoatomic), 1.40 (diatomic), 1.33 (triatomic).
4) Work, PV Work, and Joule-Thomson Effect
Defines heat and work as path functions. PV work at constant pressure: w = P delta-V. Isothermal reversible expansion: w(rev) = 2.303 nRT log(V2/V1). Adiabatic expansion follows PV^gamma = constant and TV^(gamma-1) = constant. Joule-Thomson effect: adiabatic expansion through a porous plug lowers temperature (isoenthalpic process). Most gases cool at room temperature except H2 and He.
5) Thermochemistry and Types of Heats of Reaction
Heat of reaction is the enthalpy change for a balanced equation at a given temperature. Sign convention: exothermic (delta-H negative), endothermic (delta-H positive). Ten types of heats: vaporisation, fusion, precipitation, sublimation, formation, standard heat of formation, combustion, neutralisation, solution, and dilution. Factors affecting heat of reaction: physical state, allotropic form, temperature (Kirchhoff's equation), and constant V vs constant P.
6) Hess's Law and Born-Haber Cycle
Hess's law: enthalpy change is path-independent. delta-H(reaction) = sum of delta-Hf(products) minus sum of delta-Hf(reactants). Applications: calculating heats of formation for compounds that cannot be synthesised directly, and determining lattice energy via the Born-Haber cycle. The Born-Haber cycle relates lattice energy to sublimation energy, bond dissociation energy, ionisation energy, electron affinity, and heat of formation.
7) Second Law of Thermodynamics, Carnot Cycle, and Entropy
The second law states that heat cannot be completely converted into work in a cyclic process. The Carnot cycle demonstrates maximum efficiency: eta = (T2 - T1)/T2 < 1. Entropy S measures disorder: dS = dq(rev)/T. For isothermal reversible expansion: delta-S = 2.303 nR log(V2/V1). Entropy of the universe always increases for spontaneous processes.
8) Gibbs Free Energy, Spontaneity, and Third Law
Gibbs free energy G = H - TS gives the direction of spontaneous change at constant T and P. delta-G = delta-H - T delta-S (Gibbs-Helmholtz equation). delta-G < 0: spontaneous; delta-G = 0: equilibrium; delta-G > 0: non-spontaneous. Relation to equilibrium constant: delta-G(standard) = -2.303 RT log K. The third law: entropy of a perfect crystal is zero at absolute zero.
Thermodynamics and Thermochemistry Download Notes & Weightage Plan
For each topic in the Thermodynamics and Thermochemistry chapter below, you get (2) the exact resources to download and how to use them, and (3) a simple importance & time plan so NEET students know what to do first and what to revise last.
Thermodynamic Terms and System Types
Definitions of system, surroundings, boundary, three system types, extensive vs intensive properties, and four process types.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: Identify: thermos flask = isolated, sealed test tube = closed. Know that density, temperature, and pressure are intensive. Heat capacity is extensive (depends on amount of substance).
- High-risk Area: Heat capacity is extensive but specific heat capacity is intensive. Students confuse the two. Total heat capacity doubles when the amount of substance doubles. Specific heat (per gram) does not change.
- Best Practice Style: For property classification: ask whether the value changes when the amount of substance changes. If yes, it is extensive. If no, it is intensive.
Internal Energy, Zeroth Law, and First Law of Thermodynamics
Internal energy as a state function, the zeroth law (thermal equilibrium), and the first law (dq = dE + dw) with its three special cases.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: State functions: E, H, S, G, T, P, V. Path functions: q (heat) and w (work). The combination q + w = delta-E is a state function even though q and w individually are not.
- High-risk Area: Students sometimes think q + w (which equals delta-E) must also be a path function because q and w are path functions. But delta-E is a state function. The sum of two path functions can be a state function, which is counterintuitive. NEET tests this exact idea.
- Best Practice Style: If the question asks about state functions, q and w are always path functions. Everything else on the standard list (E, H, S, G, T, P, V) is a state function.
H = E + PV, delta-H at constant pressure, specific and molar heat capacities, Cp - Cv = R, and gamma values for monoatomic, diatomic, and triatomic gases.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: delta-n(g) = total moles of gaseous products minus total moles of gaseous reactants. Do NOT count solids or liquids. Use R = 8.314 J and T = 298 K unless specified otherwise.
- High-risk Area: Counting delta-n(g) incorrectly by including solid or liquid species. For N2(g) + 3H2(g) to 2NH3(g): delta-n(g) = 2 - 4 = -2, not -1 or 0. Students who rush may miscount the 3 moles of H2.
- Best Practice Style: Write out all gaseous species on each side. Count. Subtract. Then substitute into delta-H = delta-E + delta-n(g) RT.
Work, PV Work, and Joule-Thomson Effect
Defines PV work. Isothermal reversible expansion: w = 2.303 nRT log(V2/V1), maximum work. Adiabatic expansion: PV^gamma = constant. Free expansion: all thermodynamic quantities zero. Joule-Thomson effect and inversion temperature.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: For isothermal reversible work: w = 2.303 nRT log(V2/V1). Expansion (V2 > V1): w is positive (work done by system). For adiabatic: PV^gamma = constant, which means V increases and T decreases.
- High-risk Area: Confusing isothermal and adiabatic: in isothermal expansion delta-T = 0 but q is not zero (heat is absorbed). In adiabatic expansion q = 0 but delta-T is not zero (temperature falls). Students who mix these two get both wrong.
- Best Practice Style: Isothermal: T fixed, system exchanges heat. Adiabatic: q = 0, T changes. These are the two fundamental contrasts.
Thermochemistry and Types of Heats of Reaction
Defines ten types of heats. Key values: heat of neutralisation SA+SB = 13.7 kcal, heat of sublimation = fusion + vaporisation. Factors: physical state, allotropy, temperature (Kirchhoff), and constant P vs V.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: Standard enthalpy of formation of elements in most stable form = 0. Heat of neutralisation for SA+SB = 13.7 kcal; for WA+SB it is less because energy goes to dissociate the weak acid first. The difference gives the enthalpy of ionisation of the weak acid.
- High-risk Area: Forgetting that heat of combustion is ALWAYS exothermic (delta-H negative). Students sometimes write delta-H for combustion as positive. Also: standard heat of formation of O2(g), N2(g), C(graphite) etc. is zero, not undefined.
- Best Practice Style: For any thermochemistry question: first identify the type of heat (formation, combustion, neutralisation, etc.), then apply the definition precisely. Formation = one mole from elements. Combustion = one mole burns completely.
Hess's Law and Born-Haber Cycle
Hess's law: path independence of enthalpy. delta-H(rxn) = sum delta-Hf(products) - sum delta-Hf(reactants). Born-Haber cycle for ionic lattice energy.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: Hess's law is the most numerically tested topic in NEET thermodynamics. Know the formula: delta-H(rxn) = sum of delta-Hf(products) minus sum of delta-Hf(reactants). For Born-Haber: NEET asks to identify which step in the cycle represents a given thermochemical quantity.
- High-risk Area: Students reverse a thermochemical equation but forget to reverse the sign of delta-H. If A to B has delta-H = +100 kJ, then B to A has delta-H = -100 kJ. Also: multiplying the equation by 2 means delta-H is multiplied by 2. Students who forget this get answers off by a factor of 2.
- Best Practice Style: When combining equations using Hess's law: (1) write the target equation, (2) manipulate given equations one at a time to match the target, (3) when reversing, change the sign, (4) when multiplying, multiply delta-H, (5) add all delta-H values.
Second Law of Thermodynamics, Carnot Cycle, and Entropy
Second law: heat cannot fully convert to work. Carnot efficiency. Entropy as disorder: dS = dq(rev)/T. Entropy change for isothermal expansion.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: Entropy increases: solid to liquid to gas; fewer moles to more moles of gas; mixing; dissolution. Entropy decreases: gas to liquid to solid; more moles to fewer moles of gas; crystallisation.
- High-risk Area: For the reaction 2H2(g) + O2(g) to 2H2O(l): entropy decreases (3 moles gas to 0 moles gas). Students who see an exothermic combustion may think entropy increases, but the physical state change (gas to liquid) dominates.
- Best Practice Style: Count moles of gas on each side. If gas moles increase, delta-S is positive. If gas moles decrease, delta-S is negative. This quick check works for most NEET questions.
Gibbs Free Energy, Spontaneity, and Third Law
G = H - TS. delta-G = delta-H - T delta-S. Criteria for spontaneity based on delta-G sign. Four cases from signs of delta-H and delta-S. Relation to equilibrium constant. Third law: S = 0 at 0 K for perfect crystals.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: The four-case table is the highest-yield single item in this chapter. If delta-H is negative and delta-S is positive, delta-G is always negative: spontaneous at all temperatures. If both have the same sign, temperature becomes the deciding factor.
- High-risk Area: When delta-H and delta-S are both negative: delta-G = (negative) - T(negative) = negative + T(positive). At low T, the delta-H term dominates and delta-G is negative (spontaneous). At high T, the T delta-S term dominates and delta-G becomes positive (non-spontaneous). Students often get confused about which term dominates at which temperature.
- Best Practice Style: Use the four-case table. For the borderline cases (both same sign): the reaction is spontaneous when the dominant term makes delta-G negative. If delta-H < 0 and delta-S < 0, the enthalpy driving force wins at low T. If delta-H > 0 and delta-S > 0, the entropy driving force wins at high T.
Thermodynamics and Thermochemistry Chapter NEET Traps & Common Mistakes (Topic-Wise)
Each subtopic below is of the Thermodynamics and Thermochemistry chapter and shows what NEET students usually do wrong in NEET examination, a short example of the mistake, and how NEET frames the question to trick you with close options are given below.
Mistake Snapshot (What Students Do Wrong)
- Wrong temperature dependence of spontaneity: When delta-H < 0 and delta-S < 0, the reaction is spontaneous at LOW temperature (enthalpy-driven), not high temperature. Students who think exothermic always means spontaneous at all temperatures miss the entropy penalty at high T.
- Thinking delta-G = 0 means no reaction occurs: delta-G = 0 means the system is at EQUILIBRIUM, not that the reaction is impossible. Both forward and reverse reactions occur at equal rates.
For the freezing of water at -5 degrees C: delta-H < 0 (exothermic), delta-S < 0 (liquid to solid, more ordered). At low T (below 0 degrees C), delta-G < 0: spontaneous. At high T (above 0 degrees C), delta-G > 0: non-spontaneous (ice melts instead). At exactly 0 degrees C, delta-G = 0: equilibrium.
How NEET Frames The Trap
NEET gives delta-H and delta-S signs and asks at what temperature the reaction is spontaneous.
Q. A reaction has delta-H = -50 kJ/mol and delta-S = -100 J/mol K. The reaction is spontaneous at
A. T < 500 K B. T > 500 K C. All temperatures D. No temperature
Trick: delta-G = delta-H - T delta-S = -50000 - T(-100) = -50000 + 100T. For spontaneity: delta-G < 0, so -50000 + 100T < 0, giving T < 500 K. Option A is correct. Option B reverses the inequality. Option C ignores the entropy term.
Mistake Snapshot (What Students Do Wrong)
- Forgetting to reverse delta-H when reversing an equation: If A to B has delta-H = +100 kJ, then B to A has delta-H = -100 kJ. Students who add equations without reversing the sign get answers that are off by twice the delta-H of the reversed step.
- Not multiplying delta-H when multiplying the equation coefficients: If C + O2 to CO2 has delta-H = -94 kcal, then 2C + 2O2 to 2CO2 has delta-H = -188 kcal. Forgetting to multiply gives half the correct answer.
Given: C + O2 to CO2, delta-H = -94 kcal; CO + 1/2 O2 to CO2, delta-H = -67.5 kcal. Find delta-H for C + 1/2 O2 to CO. Reverse the second equation: CO2 to CO + 1/2 O2, delta-H = +67.5 kcal. Add to first: C + O2 + CO2 to CO2 + CO + 1/2 O2. Cancel CO2: C + 1/2 O2 to CO, delta-H = -94 + 67.5 = -26.5 kcal. Without reversing the sign: delta-H = -94 + (-67.5) = -161.5 kcal, which is wrong.
How NEET Frames The Trap
NEET gives two thermochemical equations and asks for the delta-H of a third equation derived by combining them.
Q. Given: H2(g) + 1/2 O2(g) to H2O(l), delta-H = -286 kJ and H2(g) + 1/2 O2(g) to H2O(g), delta-H = -242 kJ. The enthalpy of vaporisation of water is
A. +44 kJ B. -44 kJ C. +528 kJ D. -528 kJ
Trick: H2O(l) to H2O(g): reverse first equation and add to second. delta-H = +286 + (-242) = +44 kJ (Option A). Option B forgets to reverse the sign. Options C and D multiply incorrectly.
Mistake Snapshot (What Students Do Wrong)
- Including solids and liquids in delta-n(g): delta-n(g) counts ONLY gaseous species. For CaCO3(s) to CaO(s) + CO2(g): delta-n(g) = 1 - 0 = 1, not 2 - 1 = 1 (which coincidentally gives the right answer here but would fail for other reactions).
- Wrong sign or unit for R: When using delta-H = delta-E + delta-n(g)RT, R must be in the same unit system as delta-H. If delta-H is in kJ, use R = 8.314 x 10^-3 kJ/(mol K). Using R = 8.314 J gives an answer off by factor of 1000.
For N2(g) + 3H2(g) to 2NH3(g), delta-H = -92 kJ at 298 K. delta-n(g) = 2 - (1+3) = -2. delta-H = delta-E + delta-n(g)RT. -92 = delta-E + (-2)(8.314 x 10^-3)(298). -92 = delta-E - 4.95. delta-E = -92 + 4.95 = -87.05 kJ. If student counts delta-n(g) = -1 (forgetting 3H2 is 3 moles): delta-E = -89.5 kJ, which is wrong.
How NEET Frames The Trap
NEET gives delta-H and asks for delta-E, or vice versa. The distractors use wrong delta-n(g) counts.
Q. For the reaction C(s) + H2O(g) to CO(g) + H2(g), delta-H = 131.3 kJ. Calculate delta-E at 298 K.
A. 128.8 kJ B. 131.3 kJ C. 133.8 kJ D. 126.3 kJ
Trick: delta-n(g) = (1+1) - (1) = 1 (C is solid, not counted). delta-H = delta-E + delta-n(g)RT. 131.3 = delta-E + (1)(8.314 x 10^-3)(298) = delta-E + 2.48. delta-E = 131.3 - 2.48 = 128.8 kJ (Option A). Option B ignores the correction. Option C adds instead of subtracting.
Mistake Snapshot (What Students Do Wrong)
- Thinking isothermal means no heat exchange: Isothermal means constant temperature. Heat IS exchanged with surroundings to maintain T. In contrast, adiabatic means q = 0 (no heat exchange) and T changes.
- Applying isothermal work formula to adiabatic process: w = 2.303 nRT log(V2/V1) is valid only for isothermal reversible expansion. For adiabatic, the work depends on Cv(T1-T2) and the PV^gamma relations.
1 mole of ideal gas expands from 10 L to 20 L. Isothermal at 300K: w = 2.303(1)(8.314)(300) log(2) = 1729 J, and q = 1729 J (heat absorbed), delta-T = 0. Adiabatic: q = 0, T drops, w = Cv(T1-T2) = (3/2)(8.314)(T1-T2) for monoatomic gas.
How NEET Frames The Trap
NEET asks for work or delta-T under isothermal or adiabatic conditions. Applying the wrong formula gives the wrong answer.
Q. For an ideal gas undergoing adiabatic expansion, which is true?
A. Temperature decreases B. Temperature remains constant C. delta-E = 0 D. q = w
Trick: Adiabatic: q = 0; gas does work at expense of internal energy, so E decreases and T decreases. Option A is correct. Option B is isothermal. Option C is isothermal. Option D is cyclic process.
Mistake Snapshot (What Students Do Wrong)
- Assuming all neutralisations release 13.7 kcal: 13.7 kcal is for strong acid + strong base only. Weak acid + strong base releases less because part of the energy goes to dissociate the weak acid. The difference gives the enthalpy of ionisation of the weak acid.
- Forgetting that the ionic reaction H+ + OH- to H2O gives 13.7 kcal: The actual neutralisation step is always H+ + OH- to H2O with delta-H = -13.7 kcal. For weak acids, the observed heat is less because dissociation of the weak acid consumes some energy.
Heat of neutralisation of CH3COOH with NaOH = -11.7 kcal. Enthalpy of ionisation of CH3COOH = 13.7 - 11.7 = 2.0 kcal. The 2.0 kcal is consumed in dissociating acetic acid before neutralisation can occur.
How NEET Frames The Trap
NEET gives heat of neutralisation of a weak acid and asks for the enthalpy of ionisation, or vice versa.
Q. The enthalpy of neutralisation of HCN with NaOH is -12.1 kcal. The enthalpy of ionisation of HCN is
A. 1.6 kcal B. -1.6 kcal C. 25.8 kcal D. -25.8 kcal
Trick: Ionisation energy = 13.7 - 12.1 = 1.6 kcal (Option A). Ionisation is endothermic (positive). Option B makes it exothermic. Options C and D add instead of subtract.