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State Board · Class 12 · Chemistry

Lesson 4: Chemical Thermodynamics

7 concepts · 56 practice questions with answers · Free to read

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Basic terms of thermodynamics

From daily life

A thermos (isolated), an open pot (open) and a closed pot (closed) are three kinds of systems.

Simple definition

A system is the part under study, the rest is the surroundings. A state function depends only on the initial and final states.

SystemsSystems and state functionsSystemMatterEnergyOpen✔✔Closed✘✔Isolated✘✘State functions: U, H, S, G, P, V, T (path-independent)Path functions: q, w (depend on the path)Extensive: m, V, U, H, S · Intensive: T, P, densityIsothermal: T const · adiabatic: q = 0 · isochoric: V const · isobaric: P const

In the figure : Systems and state functions.

Rule 1: Types of system — Open: exchanges matter and energy. Closed: energy only. Isolated: neither.

  • Easy : An open cup of tea is open.
  • Medium : A sealed tin is closed.
  • Exam-type : An ideal thermos is isolated.

Rule 2: State and path functions — State: U, H, S, G, P, V, T. Path: q, w. The change in a state function over a cycle is zero.

  • Easy : Distance is path-dependent; displacement is state-dependent.
  • Medium : ΔU is independent of path.
  • Exam-type : q and w depend on the path.

Rule 3: Properties and processes — Extensive: m, V, U, H. Intensive: T, P, density. Isothermal T const, adiabatic q = 0, isochoric V const, isobaric P const.

  • Easy : Halve the system: V halves, T stays.
  • Medium : A closed rigid vessel is isochoric.
  • Exam-type : An open beaker is isobaric.

Reasoning: why U is a state function but w is not

  • Step 1: Two paths between the same states.
  • Step 2: ΔU is the same on both (depends on states only).
  • Step 3: w differs, so q differs to keep q + w = ΔU.

CET level (MCQ)

  • CET 1) Not a state function? 1) w 2) U 3) H 4) S — Answer: 1).
  • CET 2) Exchanges both matter and energy? 1) open 2) closed 3) isolated 4) none — Answer: 1).

Real-life use: Thermos flasks, engines, refrigerators.

Remember

Open: matter + energy; closed: energy; isolated: none; q, w are path functions.

Common mistake

Treating q and w as state functions.

CET trap

ΔU, ΔH, ΔS, ΔG depend only on the end states.

Shortcut

Memory clue: "UHSG belong to state; q, w to path".

Spot the difference

SystemMatterEnergy
Open✔✔
Closed✘✔
Isolated✘✘

Practice questions

Easy
  1. 1System vs surroundings.
  2. 2What is a state function?
  3. 3Intensive vs extensive.
Medium
  1. 4Adiabatic process.
  2. 5Isothermal process.
  3. 6ΔU in a cycle?
Challenging
  1. 7Open vs closed system.
  2. 8Name two state functions.
Show answers (try them yourself first)
  1. 1The part studied vs the rest.
  2. 2Depends only on the end states.
  3. 3T, P vs m, V.
  4. 4q = 0.
  5. 5T constant.
  6. 6Zero.
  7. 7Matter exchange or not.
  8. 8U and H.

Work and heat

From daily life

A bicycle pump heats up when you compress air — work done on the gas.

Simple definition

Pressure–volume work: w = –P_ext ΔV; reversible isothermal: w = –2.303 nRT log(V₂/V₁); IUPAC sign: work done by the system ⇒ w negative.

WorkP–V graph: work = area under the curveVPV₁V₂Reversible (isothermal)w = –2.303 nRT log(V₂/V₁)1 mol, 298 K, 10 → 20 Lw = –1717 JConstant external pressurew = –Pext ΔV1 atm, ΔV = 2 Lw = –202.65 JFree expansion (vacuum): Pext = 0 ⇒ w = 0 · reversible work is the maximumIUPAC sign: work done by the system ⇒ w negative

In the figure : P–V graph.

Rule 1: At constant pressure — w = –P_ext (V₂ – V₁); 1 atm = 101.325 kPa; 1 L atm = 101.325 J.

  • Easy : 1 atm, ΔV = 2 L ⇒ w = –202.65 J.
  • Medium : Compression ⇒ w positive.
  • Exam-type : Into vacuum w = 0.

Rule 2: Reversible isothermal — w = –nRT ln(V₂/V₁); this is the maximum work (area under the curve).

  • Easy : 1 mol, 298 K, 10 ⇒ 20 L: –1717 J.
  • Medium : Doubled volume ⇒ ln 2 = 0.693.
  • Exam-type : On compression w is positive.

Rule 3: Heat — q = m c ΔT = C ΔT; energy flows from hot to cold; q > 0 ⇒ the system absorbs heat.

  • Easy : Water c = 4.18 J g⁻¹ K⁻¹.
  • Medium : 100 g, 10 K ⇒ 4180 J.
  • Exam-type : Heat and temperature are different.

Calculation: 1 mol gas at 298 K, 10 ⇒ 20 L (reversible)

  • Step 1: w = –nRT ln(V₂/V₁).
  • Step 2: = –1 × 8.314 × 298 × 0.6931.
  • Step 3: = –1717 J (work done by the system).

CET level (MCQ)

  • CET 1) w in free expansion? 1) 0 2) positive 3) negative 4) infinite — Answer: 1).
  • CET 2) 1 L atm = ? 1) 101.3 J 2) 8.314 J 3) 1 J 4) 4.18 J — Answer: 1).

Real-life use: Engines, pumps, steam turbines.

Remember

w = –P_ext ΔV; reversible = maximum; expansion ⇒ w negative.

Common mistake

Writing w positive for expansion.

CET trap

IUPAC: work on the system is +, by the system is –.

Shortcut

Memory clue: "giving work = negative".

Spot the difference

Processw
Free expansion0
Constant P_ext–P_ext ΔV
Reversible–nRT ln(V₂/V₁)

Practice questions

Easy
  1. 1Explain w = –P_ext ΔV.
  2. 2Why is reversible work maximum?
  3. 31 atm, ΔV = 2 L ⇒ w?
Medium
  1. 4Heat vs temperature.
  2. 51 L atm = ? J
  3. 6Sign of w in compression.
Challenging
  1. 71 mol, 298 K, 10 ⇒ 20 L ⇒ w?
  2. 8What is free expansion?
Show answers (try them yourself first)
  1. 1System works against external pressure.
  2. 2P_ext ≈ P_int at every step.
  3. 3–202.65 J.
  4. 4Energy flow vs average KE.
  5. 5101.325 J.
  6. 6Positive.
  7. 7–1717 J.
  8. 8Expansion into vacuum.

Internal energy and the first law

From daily life

Burning petrol gives an engine heat and work — energy is never destroyed.

Simple definition

First law: ΔU = q + w; an isolated system has constant U; for an ideal gas isothermal ΔU = 0.

First lawFirst law of thermodynamics: ΔU = q + wSystemUq > 0heat inw < 0work outIsothermal (ideal gas)ΔU = 0 ⇒ q = –wAdiabaticq = 0 ⇒ ΔU = wIsochoric (ΔV = 0)w = 0 ⇒ ΔU = qVCyclicΔU = 0q = +800 J and 300 J work done by the system ⇒ ΔU = +500 Jq = +500 J and 200 J work done on the system ⇒ ΔU = +700 J

In the figure : First law.

Rule 1: Sign convention — q > 0: heat absorbed by the system; w > 0: work done on the system; ΔU = q + w.

  • Easy : q = +500, w = +200 ⇒ ΔU = +700 J.
  • Medium : q = +800, w = –300 ⇒ ΔU = +500 J.
  • Exam-type : q = –50, w = +50 ⇒ ΔU = 0.

Rule 2: Special processes — Isothermal (ideal gas): q = –w. Adiabatic: ΔU = w. Isochoric: ΔU = q_V.

  • Easy : Adiabatic compression raises T.
  • Medium : Adiabatic expansion cools.
  • Exam-type : A bomb calorimeter measures q_V.

Rule 3: U of an ideal gas — U depends only on T; ΔU = n C_V ΔT; isothermal ⇒ ΔU = 0.

  • Easy : Monatomic C_V = 3R/2.
  • Medium : 1 mol, ΔT = 100 K ⇒ ΔU = 1247 J.
  • Exam-type : C_P – C_V = R.

Calculation: q = +800 J, work done by system 300 J

  • Step 1: w = –300 J (given out).
  • Step 2: ΔU = q + w = 800 + (–300).
  • Step 3: ΔU = +500 J.

CET level (MCQ)

  • CET 1) In an adiabatic process? 1) ΔU = w 2) ΔU = q 3) ΔU = 0 4) w = 0 — Answer: 1).
  • CET 2) Isothermal expansion of an ideal gas ⇒ ΔU? 1) 0 2) positive 3) negative 4) infinite — Answer: 1).

Real-life use: Engine efficiency, calorimetry.

Remember

ΔU = q + w; isothermal 0; adiabatic = w; isochoric = q.

Common mistake

Taking w positive when the system does work.

CET trap

Signs follow the system: in = +, out = –.

Shortcut

Memory clue: "whatever comes in is positive".

Spot the difference

ProcessConditionΔU
IsothermalΔT = 00
Adiabaticq = 0w
Isochoricw = 0q_V

Practice questions

Easy
  1. 1State the first law.
  2. 2What is adiabatic?
  3. 3q = 800, w by system 300 ⇒ ΔU?
Medium
  1. 4ΔU in isochoric process?
  2. 5What does U of an ideal gas depend on?
  3. 6ΔU in a cyclic process?
Challenging
  1. 7C_P – C_V?
  2. 8Adiabatic compression ⇒ T?
Show answers (try them yourself first)
  1. 1Conservation of energy.
  2. 2q = 0.
  3. 3+500 J.
  4. 4= q_V.
  5. 5Only T.
  6. 60.
  7. 7R.
  8. 8Increases.

Enthalpy and phase transitions

From daily life

Ice absorbs heat while melting yet stays at 0 °C — latent heat.

Simple definition

Enthalpy H = U + PV; at constant pressure ΔH = q_P; ΔH = ΔU + Δn_g RT.

Heating curveHeating curve of water (18 g)q (kJ)Tmelting 6.01vaporising 40.7water 7.52ice 0.38ΔfusH = 6.01 kJ/molΔvapH = 40.7 kJ/molq = m c ΔTc(water) = 4.18 J g⁻¹K⁻¹c(ice) = 2.09 J g⁻¹K⁻¹Total: ice at –10 °C ⇒ steam at 100 °C = 0.38 + 6.01 + 7.52 + 40.7 = 54.6 kJH = U + PV; ΔH = ΔU + Δng RTN₂ + 3H₂ → 2NH₃: ΔH = –92.4 ⇒ ΔU = –87.4 kJ

In the figure : Heating curve of water.

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Enthalpy and phase transitions

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Rule 1: ΔH and ΔU — ΔH = ΔU + Δn_g RT, with Δn_g = moles of gaseous products – reactants.

  • Easy : N₂ + 3H₂ → 2NH₃: Δn_g = –2.
  • Medium : ΔH = –92.4 ⇒ ΔU = –87.4 kJ.
  • Exam-type : Δn_g = 0 ⇒ ΔH = ΔU.

Rule 2: Phase changes — ΔfusH (ice) = 6.01 kJ/mol; ΔvapH (water, 373 K) = 40.7 kJ/mol; ΔsubH = ΔfusH + ΔvapH.

  • Easy : Melting 36 g ice = 2 × 6.01 = 12.02 kJ.
  • Medium : Vaporising 18 g water = 40.7 kJ.
  • Exam-type : Temperature stays constant.

Rule 3: Total heat — q = m c ΔT + n ΔH(phase). 18 g from ice at –10 °C to steam at 100 °C: 0.38 + 6.01 + 7.52 + 40.7 = 54.6 kJ.

  • Easy : Ice c = 2.09 J g⁻¹K⁻¹.
  • Medium : Water c = 4.18.
  • Exam-type : Most heat goes into vaporising.

Calculation: N₂ + 3H₂ → 2NH₃, ΔH = –92.4 kJ ⇒ ΔU at 298 K

  • Step 1: Δn_g = 2 – 4 = –2.
  • Step 2: ΔU = ΔH – Δn_g RT = –92.4 + 2 × 8.314 × 298 ÷ 1000.
  • Step 3: ΔU = –92.4 + 4.96 = –87.4 kJ.

CET level (MCQ)

  • CET 1) ΔH = ΔU when? 1) Δn_g = 0 2) Δn_g > 0 3) Δn_g < 0 4) always — Answer: 1).
  • CET 2) Heat to melt 36 g ice? 1) 12.02 kJ 2) 6.01 kJ 3) 40.7 kJ 4) 81.4 kJ — Answer: 1).

Real-life use: Refrigeration, steam engines, weather.

Remember

ΔH = ΔU + Δn_g RT; fusion 6.01; vaporisation 40.7; q_P = ΔH.

Common mistake

Counting liquids/solids in Δn_g.

CET trap

Δn_g counts gas moles only.

Shortcut

Memory clue: "melt 6, boil 40.7".

Spot the difference

ChangeΔH (kJ/mol)
Fusion6.01
Vaporisation40.7
Sublimation46.7

Practice questions

Easy
  1. 1What is enthalpy?
  2. 2Relation of ΔH and ΔU.
  3. 3What is latent heat?
Medium
  1. 4Melting 36 g ice.
  2. 5Why flat parts in the curve?
  3. 6Find ΔsubH.
Challenging
  1. 7Example of Δn_g.
  2. 8N₂ + 3H₂ ⇒ ΔU?
Show answers (try them yourself first)
  1. 1H = U + PV.
  2. 2ΔH = ΔU + Δn_g RT.
  3. 3Heat of a phase change.
  4. 412.02 kJ.
  5. 5Heat goes into phase change.
  6. 66.01 + 40.7 = 46.7 kJ/mol.
  7. 7–2 (N₂ + 3H₂ → 2NH₃).
  8. 8–87.4 kJ.

Thermochemistry and Hess's law

From daily life

How much heat an LPG cylinder gives comes from the enthalpy of combustion.

Simple definition

Hess's law: ΔH of a reaction is independent of the path; ΔrH° = ΣΔfH°(products) – ΣΔfH°(reactants).

HessHess's law: enthalpy-level diagramC(s) + O₂(g)ΔH = 0CO(g) + ½O₂(g)–110.5CO₂(g)–393.5–393.5–110.5–283.0–393.5 = (–110.5) + (–283.0): any route, same ΔHΔrH° = ΣΔfH°(products) – ΣΔfH°(reactants)CH₄ + 2O₂ → CO₂ + 2H₂O(l): –393.5 + 2(–285.8) – (–74.8) = –890.3 kJ

In the figure : Hess cycle.

Rule 1: Special enthalpies — ΔfH° = formation of 1 mol from elements; ΔfH° of an element = 0; ΔcH° = combustion of 1 mol; also ΔsolH, ΔdissH.

  • Easy : ΔfH°(CO₂) = –393.5.
  • Medium : ΔfH°(H₂O, l) = –285.8.
  • Exam-type : ΔfH°(CH₄) = –74.8.

Rule 2: Using Hess's law — C + O₂ ⇒ CO₂ (–393.5) and CO + ½O₂ ⇒ CO₂ (–283.0) ⇒ C + ½O₂ ⇒ CO = –110.5 kJ.

  • Easy : Reverse the equation ⇒ flip the sign.
  • Medium : Multiply ⇒ multiply ΔH.
  • Exam-type : Add ⇒ add ΔH.

Rule 3: Enthalpy of combustion — CH₄ + 2O₂ → CO₂ + 2H₂O(l): ΔcH° = –393.5 + 2(–285.8) – (–74.8) = –890.3 kJ/mol.

  • Easy : 1.6 g CH₄ (0.1 mol) ⇒ 89.03 kJ.
  • Medium : 1 kg CH₄ ≈ 55 600 kJ.
  • Exam-type : Combustion is always exothermic.

Calculation: heat from burning 1.6 g CH₄

  • Step 1: ΔrH° = –393.5 + 2(–285.8) – (–74.8) = –890.3 kJ/mol.
  • Step 2: n = 1.6 ÷ 16 = 0.1 mol.
  • Step 3: Heat = 0.1 × 890.3 = 89.03 kJ.

CET level (MCQ)

  • CET 1) ΔfH° of an element in its standard state? 1) 0 2) –1 3) +1 4) 100 — Answer: 1).
  • CET 2) Hess's law rests on? 1) H a state function 2) q a path 3) S increase 4) G — Answer: 1).

Real-life use: Fuel value, food calories.

Remember

ΔrH = Σ products – Σ reactants; reverse flips the sign.

Common mistake

Subtracting products – reactants the wrong way round.

CET trap

Multiply ΔfH° by the stoichiometric coefficient.

Shortcut

Memory clue: "end minus start".

Spot the difference

SubstanceΔfH° (kJ/mol)
CO₂(g)–393.5
H₂O(l)–285.8
CH₄(g)–74.8

Practice questions

Easy
  1. 1State Hess's law.
  2. 2Find ΔH for C ⇒ CO.
  3. 3Enthalpy of combustion of CH₄.
Medium
  1. 4What is ΔfH°?
  2. 5Heat from 1.6 g CH₄?
  3. 6If the equation is reversed?
Challenging
  1. 7Why is ΔfH° of an element 0?
  2. 8Why is combustion exothermic?
Show answers (try them yourself first)
  1. 1ΔH is path-independent.
  2. 2–110.5 kJ.
  3. 3–890.3 kJ/mol.
  4. 41 mol from elements.
  5. 589.03 kJ.
  6. 6The sign flips.
  7. 7By definition it is the reference.
  8. 8Products have lower enthalpy.

Entropy and the second law

From daily life

Incense spreads through a room and never collects again — disorder grows.

Simple definition

Entropy S measures disorder; ΔS = q_rev/T. Second law: ΔS_universe > 0 for a spontaneous process.

EntropyEntropy: disorder and the second lawSolidLiquidGasS(solid) < S(liquid) < S(gas)ΔS = qrev / T ; S = k ln WMelting ice: 6010 / 273 = 22.0 J K⁻¹ mol⁻¹Boiling water: 40700 / 373 = 109.1 J K⁻¹ mol⁻¹1 mol gas, volume doubles: ΔS = R ln 2 = 5.76 J K⁻¹Second law: ΔSuniv = ΔSsys + ΔSsurr > 0 (spontaneous)Third law: S = 0 for a perfect crystal at 0 K · ΔSsurr = –ΔH/T

In the figure : Entropy.

Rule 1: Entropy change — S(solid) < S(liquid) < S(gas); S rises with T and V; doubling a gas volume gives ΔS = R ln 2 = 5.76 J K⁻¹ mol⁻¹.

  • Easy : CaCO₃ → CaO + CO₂: ΔS is +.
  • Medium : N₂ + 3H₂ → 2NH₃: ΔS is –.
  • Exam-type : Dissolving salt: ΔS is +.

Rule 2: ΔS in phase changes — ΔS = ΔH/T. Melting ice: 6010/273 = 22.0; boiling water: 40700/373 = 109.1 J K⁻¹ mol⁻¹.

  • Easy : ΔS is larger for boiling.
  • Medium : Liquid → gas adds much disorder.
  • Exam-type : Surroundings: ΔS_surr = –ΔH/T.

Rule 3: Second and third laws — ΔS_univ = ΔS_sys + ΔS_surr > 0. Third law: a perfect crystal at 0 K has S = 0, giving absolute entropy.

  • Easy : Cold-to-hot heat flow is not spontaneous.
  • Medium : S° (H₂O, l) = 69.9; (g) = 188.8 J K⁻¹ mol⁻¹.
  • Exam-type : S = k ln W.

Reasoning: why is ΔS positive on melting?

  • Step 1: A solid is well ordered.
  • Step 2: Molecules move more freely in the liquid.
  • Step 3: ΔS = 6010/273 = +22.0 J K⁻¹ mol⁻¹.

CET level (MCQ)

  • CET 1) Highest entropy? 1) gas 2) liquid 3) solid 4) equal — Answer: 1).
  • CET 2) A spontaneous process has? 1) ΔS_univ > 0 2) < 0 3) = 0 4) ΔH > 0 — Answer: 1).

Real-life use: Ice making, engines, reaction direction.

Remember

S gas > liquid > solid; ΔS = ΔH/T; spontaneous ⇒ ΔS_univ > 0.

Common mistake

Thinking spontaneous means ΔS_system > 0.

CET trap

ΔS_univ must be positive; ΔS_sys may be negative.

Shortcut

Memory clue: "more mess = more S".

Spot the difference

ChangeΔS
Melting+22.0
Boiling+109.1
Gas doubles V+5.76

Practice questions

Easy
  1. 1What is entropy?
  2. 2State the second law.
  3. 3State the third law.
Medium
  1. 4ΔS of melting ice?
  2. 5ΔS of boiling water?
  3. 6ΔS for doubling V of 1 mol gas?
Challenging
  1. 7Formula of ΔS_surr.
  2. 8Sign of ΔS for N₂ + 3H₂ ⇒ 2NH₃.
Show answers (try them yourself first)
  1. 1Measure of disorder.
  2. 2ΔS_univ > 0.
  3. 3S = 0 at 0 K.
  4. 422.0 J K⁻¹ mol⁻¹.
  5. 5109.1 J K⁻¹ mol⁻¹.
  6. 65.76 J K⁻¹.
  7. 7–ΔH/T.
  8. 8Negative (fewer gas moles).

Gibbs energy and spontaneity

From daily life

A ball rolls downhill by itself; reactions also move toward lower ΔG.

Simple definition

ΔG = ΔH – TΔS. ΔG < 0 spontaneous; ΔG = 0 equilibrium; ΔG > 0 non-spontaneous. ΔG° = –RT ln K = –nFE°.

GibbsGibbs energy: ΔG = ΔH – TΔST (K)ΔG0465 Kspontaneous (ΔG < 0)noN₂ + 3H₂ → 2NH₃ΔH = –92.4 kJΔS = –198.7 J/K298 K : ΔG = –33.2 kJΔH –, ΔS +always spontaneousΔH +, ΔS –never spontaneousΔH –, ΔS –spontaneous at low TΔH +, ΔS +spontaneous at high TΔG° = –RT ln K = –2.303 RT log K = –nFE°ΔG° = –5.7 kJ/mol at 298 K ⇒ log K = 1.0 ⇒ K ≈ 10

In the figure : ΔG versus T.

Live figure

Gibbs energy and spontaneity

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Rule 1: Four cases — ΔH –, ΔS + ⇒ always; ΔH +, ΔS – ⇒ never; ΔH –, ΔS – ⇒ low T; ΔH +, ΔS + ⇒ high T.

  • Easy : Melting ice: ΔH +, ΔS + ⇒ high T.
  • Medium : Combustion: ΔH –, ΔS + ⇒ always.
  • Exam-type : NH₃: ΔH –, ΔS – ⇒ low T.

Rule 2: Crossover temperature — At ΔG = 0, T = ΔH/ΔS. N₂ + 3H₂ → 2NH₃: 92.4 ÷ 0.1987 = 465 K.

  • Easy : At 298 K: ΔG = –33.2 kJ.
  • Medium : Not spontaneous above 465 K.
  • Exam-type : Yet slow, so a catalyst is used.

Rule 3: ΔG° and K — ΔG° = –2.303 RT log K; ΔG° = –5.7 kJ/mol at 298 K ⇒ log K = 1.0 ⇒ K ≈ 10; ΔG° = –nFE°.

  • Easy : Negative ΔG° ⇒ K > 1.
  • Medium : ΔG° = 0 ⇒ K = 1.
  • Exam-type : Positive ΔG° ⇒ K < 1.

Calculation: ΔH = –92.4 kJ, ΔS = –198.7 J/K, T = 298 K

  • Step 1: TΔS = 298 × (–0.1987) = –59.2 kJ.
  • Step 2: ΔG = –92.4 – (–59.2) = –33.2 kJ.
  • Step 3: Negative ⇒ spontaneous; ΔG = 0 at T = 465 K.

CET level (MCQ)

  • CET 1) Spontaneous requires? 1) ΔG < 0 2) ΔG > 0 3) ΔG = 0 4) ΔS < 0 — Answer: 1).
  • CET 2) ΔH +, ΔS + is spontaneous at? 1) high T 2) low T 3) always 4) never — Answer: 1).

Real-life use: Chemical industry, batteries, metabolism.

Remember

ΔG = ΔH – TΔS; ΔG° = –RT ln K; T = ΔH/ΔS.

Common mistake

Mixing ΔS in J with ΔH in kJ.

CET trap

Divide ΔS by 1000 to convert to kJ.

Shortcut

Memory clue: "Gibbs = H minus TS".

Spot the difference

ΔHΔSSpontaneous
–+always
+–never
––low T
++high T

Practice questions

Easy
  1. 1Explain ΔG = ΔH – TΔS.
  2. 2Write the four cases.
  3. 3Find the crossover temperature.
Medium
  1. 4Relation of ΔG° and K.
  2. 5How is ΔG = –33.2?
  3. 6ΔG° = –5.7 kJ ⇒ K?
Challenging
  1. 7What does ΔG = 0 mean?
  2. 8Where is ΔG° = –nFE° used?
Show answers (try them yourself first)
  1. 1Balance of ΔH and TΔS.
  2. 2(–,+), (+,–), (–,–), (+,+).
  3. 3ΔH/ΔS.
  4. 4ΔG° = –2.303 RT log K.
  5. 5–92.4 + 59.2.
  6. 6≈ 10.
  7. 7Equilibrium.
  8. 8In electrochemical cells.

Quick revision of the whole lesson

ConceptKey formula / fact
First lawΔU = q + w
Workw = –P_ext ΔV
EnthalpyΔH = ΔU + Δn_g RT
HessΣ products – Σ reactants
EntropyΔS = q_rev/T
Second lawΔS_univ > 0
GibbsΔG = ΔH – TΔS
ΔG° and KΔG° = –2.303 RT log K
Live figure

🚀 Quick revision of the whole lesson

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🎯 Most important points for the exam

  • Systems, state functions.
  • Work, heat.
  • First law.
  • Enthalpy, phases.
  • Hess's law.
  • Entropy.
  • Gibbs energy.
Study mantra

understand the idea → remember the formula → solve the example → solve problems yourself.