Chemistry XI • Ch 5 Thermodynamics
Beginner - Story Mode
1. System & Surroundings - What You StudyPart 1
Imagine you are studying your tea cup. The tea inside the cup is the system — the part you focus on. The room air, table, everything outside the cup is the surroundings. The cup wall is the boundary that separates them. System + surroundings = universe.

Types of system:
Open system can exchange both matter and energy — an open tea cup, you can add sugar (matter) and heat escapes (energy).
Closed system can exchange only energy, not matter — a closed bottle with tight cap, you can't add sugar but heat can still go out.
Isolated system exchanges nothing — a sealed thermos flask, no heat, no matter goes out, tea stays hot for hours.

State vs Path functions:
State functions depend only on where you are, not how you came — like your height at a hilltop is same whether you came by road A or road B. Examples: temperature, pressure, internal energy U, enthalpy H, entropy S, Gibbs G.
Path functions depend on path taken — like distance traveled, work w, heat q.

Extensive vs Intensive:
Extensive depends on amount — mass, volume, U, H — more tea, more mass.
Intensive is independent of amount — temperature, pressure, density, boiling point — same for 1 spoon or 1 bucket.
Open house = people and food come and go, Closed house = only smell goes out, people can't, Isolated vault = nothing comes or goes. Height is state, distance is path.
Revision: System = focus (tea), surroundings = outside, boundary = wall, universe = system+surroundings. Open exchanges matter+energy (open cup), closed only energy (closed bottle), isolated nothing (thermos). State function depends on state not path — T,P,U,H,S,G like height. Path function depends on path — w,q like distance. Extensive depends on amount mass volume U H, intensive independent T P density b.p.
System

Quick Revision Box

  • System studied, surroundings rest, universe = system+surroundings
  • Open: matter+energy open cup, Closed: energy only closed bottle, Isolated: nothing thermos
  • State: T P U H S G, Path: q w
  • Extensive: mass vol U H amount dependent, Intensive: T P density independent
2. Internal Energy & First Law - Bank BalancePart 2
Internal energy U is total energy inside a system — kinetic energy of molecules moving, vibrating, rotating + potential energy of attraction between them. You can never know absolute U, only change ΔU.

Work w: When gas expands, it pushes piston and does work on surroundings — work is negative (system losing energy). When surroundings compress gas, work is positive (work done on system). Formula for expansion work against constant external pressure: w = -Pext × ΔV = -Pext(V2-V1). If expansion, V2>V1, ΔV positive, w negative.

Heat q: Heat absorbed by system is positive (feels hot), heat released is negative (feels cold).

First Law of Thermodynamics: ΔU = q + w — energy conservation. Energy cannot be created or destroyed, only converted. Think of bank balance: ΔU is balance change, q is heat deposit, w is work deposit. If you add heat and work is done on system, internal energy increases.
Revision: U = total energy inside kinetic+potential, only ΔU measurable. Work w=-PextΔV expansion negative (system does work), compression positive. Heat q positive absorbed, negative released. First law ΔU=q+w energy conservation bank balance ΔU=deposit q + deposit w.
First Law

Quick Revision Box

  • U = KE+PE inside, ΔU only measurable
  • w=-Pext ΔV expansion V2>V1 w negative, compression positive
  • q absorbed + released -
  • ΔU=q+w conservation like bank
3. Work - Slow vs Fast PushPart 3
Irreversible work against constant external pressure: w_irr = -Pext(V2-V1) — like pushing against a fixed weight suddenly.

Reversible work is infinitely slow, Pext is almost equal to internal pressure Pint at every step — maximum work obtained. Formula: w_rev = -nRT ln(V2/V1) = -2.303 nRT log(V2/V1). Reversible work is more than irreversible: |w_rev| > |w_irr| because slow push uses full internal pressure throughout.

On a P-V diagram, area under the curve is work. Irreversible is a rectangle (small area), reversible is a hyperbolic curve (large area).

Analogy: If you push a heavy box quickly, you get tired quickly and do less effective work. If you push slowly with full control balancing weight, you do maximum work. Reversible = slow = maximum work.
Revision: Irreversible w_irr=-Pext(V2-V1) constant Pext rectangle small area. Reversible infinitely slow Pext≈Pint max work w_rev=-nRT ln(V2/V1)=-2.303nRT log(V2/V1) hyperbolic large area |w_rev|>|w_irr|. P-V diagram area=work. Slow push = max work.
Work

Quick Revision Box

  • Irreversible w=-Pext(V2-V1)
  • Reversible w=-nRT ln(V2/V1) max work
  • P-V area = work, reversible area bigger
4. Enthalpy - Heat at Constant PressurePart 4
Enthalpy H = U + PV — it is heat content at constant pressure. Chemists love it because most reactions in lab happen at constant pressure (open in air).

At constant volume, heat change = ΔU. At constant pressure, heat change = ΔH.

Relation for gases: ΔH = ΔU + ΔngRT where Δng = moles of gaseous products - moles of gaseous reactants. Example: N2 + 3H2 → 2NH3, Δng = 2 - (1+3) = -2, so ΔH = ΔU -2RT.

Exothermic ΔH negative — heat released, surroundings feel hot — burning wood, energy of products lower than reactants.
Endothermic ΔH positive — heat absorbed, surroundings feel cold — ice melting needs heat, products have higher energy than reactants.
Open pan cooking heat = enthalpy (constant pressure), sealed pressure cooker heat = internal energy (constant volume). Exothermic = giving away heat you feel warm, Endothermic = taking heat you feel cold.
Revision: Enthalpy H=U+PV heat at constant pressure. Constant V heat=ΔU, constant P heat=ΔH. Gases ΔH=ΔU+ΔngRT Δng=gaseous products-reactants N2+3H2→2NH3 Δng=-2. Exothermic ΔH -ve heat released burning products lower, Endothermic ΔH +ve heat absorbed ice melting products higher.
Enthalpy

Quick Revision Box

  • H=U+PV, qp=ΔH qv=ΔU
  • ΔH=ΔU+ΔngRT Δng=gas products-reactants
  • Exothermic ΔH -ve burning, Endothermic +ve ice melting
5. Types of Enthalpy ChangesPart 5
We measure enthalpy changes at standard state: 298K (25°C), 1 bar pressure, pure form.

Enthalpy of Formation ΔfH°: Formation of 1 mole compound from its elements in their standard states. Elements in standard state have ΔfH° = 0, like O2 gas, H2 gas, C graphite = 0.

Enthalpy of Combustion ΔcH°: Burning 1 mole substance in oxygen — always exothermic negative, like CH4 + 2O2 → CO2 + 2H2O ΔH = -890 kJ/mol.

Enthalpy of Neutralization: Acid + Base → salt + water, H+ + OH- → H2O, always -57.1 kJ/mol for strong acid-strong base.

Enthalpy of Atomization ΔaH°: Breaking 1 mole gaseous molecule into atoms, like H2 → 2H, CH4 → C + 4H.

Phase Transition Enthalpies: Fusion ΔfusH Solid→Liquid melting, Vaporization ΔvapH Liquid→Gas boiling, Sublimation ΔsubH Solid→Gas like camphor disappearing.
Revision: Standard 298K 1 bar. Formation ΔfH° 1 mole from elements elements 0 O2 H2 C graphite. Combustion ΔcH° burning 1 mole O2 always -ve CH4 -890kJ. Neutralization H+OH-→H2O -57.1 kJ strong acid base. Atomization ΔaH° gas→atoms H2→2H. Phase: fusion Solid→Liq melting, vaporization Liq→Gas boiling, sublimation Solid→Gas camphor.
Enthalpy Types

Quick Revision Box

  • Standard 298K 1 bar
  • ΔfH° formation 1 mol from elements elements 0
  • ΔcH° combustion burning exothermic
  • Neutralization -57.1 kJ strong acid base
  • Atomization gas→atoms, fusion solid→liquid, vaporization liquid→gas, sublimation solid→gas
6. Hess's Law - Path Doesn't MatterPart 6
Hess's Law says enthalpy change is same regardless of path because enthalpy is a state function — like height of a hill. Whether you go directly from ground floor to 3rd floor by lift, or via 1st and 2nd floor by stairs, total height change is same.

This is super useful to calculate ΔH for reactions that cannot be measured directly.

Example: C + ½ O2 → CO cannot be measured directly because some CO2 also forms when carbon burns. But you can measure:
(1) C + O2 → CO2 ΔH = -393 kJ
(2) CO + ½ O2 → CO2 ΔH = -283 kJ
Subtract (2) from (1): C + ½ O2 → CO ΔH = -393 - (-283) = -110 kJ. So Hess's law lets you find hidden path enthalpy by combining known paths.
Revision: Hess's Law ΔH same regardless path state function like height ground to 3rd floor direct lift or via stairs same. Useful for indirect reactions. Ex: C+O2→CO2 -393 CO+½O2→CO2 -283 subtract → C+½O2→CO -110 kJ not directly measurable.
Hess

Quick Revision Box

  • H state function like height path independent
  • Total ΔH sum of steps
  • Example C+½O2→CO cannot direct but via CO2 cycle -110 kJ
  • Reverse reaction sign flips
7. Bond Enthalpy - Breaking vs MakingPart 7
Bond enthalpy is energy needed to break 1 mole bonds in gaseous phase.

For diatomic molecules, it's called bond dissociation enthalpy — like H-H is 435 kJ/mol.
For polyatomic molecules like CH4, we take average of 4 C-H bonds — average bond enthalpy 413 kJ/mol.

Reaction enthalpy from bond enthalpies:
Breaking bonds needs energy (positive), forming bonds releases energy (negative). So ΔrH = Σ Bond Enthalpy of reactants broken - Σ Bond Enthalpy of products formed.

Example: H2 + Cl2 → 2HCl
Broken: H-H 435 + Cl-Cl 242 = 677 kJ
Formed: 2 × H-Cl = 2 × 431 = 862 kJ
ΔrH = 677 - 862 = -185 kJ — exothermic, because forming stronger bonds releases more energy than breaking.
Revision: Bond enthalpy energy to break 1 mole gas bonds. Diatomic dissociation H-H 435. Polyatomic average CH4 C-H 413 average 4. ΔrH = Σ bonds broken - Σ bonds formed breaking +ve forming -ve. Ex: H2+Cl2→2HCl broken 435+242=677 formed 2×431=862 ΔrH=677-862=-185 kJ exothermic.
Bond

Quick Revision Box

  • Bond enthalpy break 1 mol gas
  • H-H 435, C-H avg 413
  • ΔrH = Σ broken - Σ formed
  • H2+Cl2→2HCl 435+242-862=-185 kJ
8. Entropy - Disorder MeasurePart 8
Entropy S is measure of disorder or randomness. More disorder means more entropy.

Solid is ordered — low entropy, liquid is medium, gas is highly disordered — high entropy. When you melt ice, solid→liquid, entropy increases. When water vapour condenses to liquid, entropy decreases.

Think of your room: Clean, well-arranged room low entropy, messy room with clothes everywhere high entropy. Nature prefers messy (high entropy).

Second Law of Thermodynamics: Entropy of universe always increases for a spontaneous process. ΔSuniv = ΔSsys + ΔSsurr > 0 for spontaneous. For system, ΔS = q_rev / T — reversible heat divided by temperature.
Revision: Entropy S = disorder randomness Solid low ordered, Liquid medium, Gas high disordered. Melting solid→liquid S increases. Messy room high S clean low. Second Law ΔSuniv=ΔSsys+ΔSsurr>0 spontaneous ΔS=q_rev/T.
Entropy

Quick Revision Box

  • S disorder randomness solid
  • Second Law ΔSuniv=ΔSsys+ΔSsurr>0 spontaneous
  • ΔS=q_rev/T
  • Melting ↑ entropy, crystallization ↓
9. Gibbs Free Energy - Profit CheckPart 9
Gibbs Free Energy G = H - T×S combines enthalpy and entropy — it is the ultimate profit check for a reaction.

ΔG = ΔH - TΔS

Criteria:
ΔG negative → spontaneous like profit — reaction goes forward on its own.
ΔG positive → non-spontaneous like loss — needs external energy.
ΔG zero → equilibrium — balanced, no net change.

Four cases:
1) ΔH negative (exothermic) and ΔS positive (more disorder) → always spontaneous — both favor.
2) ΔH positive and ΔS negative → never spontaneous — both oppose.
3) ΔH negative and ΔS negative → spontaneous at low temperature (enthalpy dominates).
4) ΔH positive and ΔS positive → spontaneous at high temperature (entropy dominates).

Important relations: ΔG = -nFE°cell and ΔG° = -RT ln K = -2.303 RT log K — links free energy with equilibrium constant and cell potential.
Like business profit = income (H) - temperature × expenses (S). If profit ΔG negative you earn, reaction forward spontaneously.
Revision: Gibbs G=H-TS profit check ΔG=ΔH-TΔS. ΔG -ve spontaneous profit, +ve non-spontaneous loss, 0 equilibrium balanced. Cases: ΔH -ve ΔS +ve always spontaneous, ΔH +ve ΔS -ve never, ΔH -ve ΔS -ve spontaneous low T, ΔH +ve ΔS +ve spontaneous high T. Relations ΔG=-nFEcell ΔG°=-RT lnK=-2.303RT logK.
Gibbs

Quick Revision Box

  • G=H-TS ΔG=ΔH-TΔS
  • ΔG -ve spontaneous, +ve non, 0 equilibrium
  • -H +S always spontaneous, +H -S never, -H -S low T, +H +S high T
  • ΔG=-nFEcell ΔG°=-RT lnK
10. Third Law & Equilibrium LinkPart 10
Third Law of Thermodynamics: Entropy of a perfect crystal at absolute zero (0K) is zero — S=0. At 0K, there is perfect order, no disorder.

Standard Gibbs and Equilibrium:
ΔG° = ΔH° - TΔS°
Equilibrium constant K = e^(-ΔG°/RT) = 10^(-ΔG°/2.303RT).

If ΔG° negative, K > 1 — products favored. If ΔG° positive, K < 1 — reactants favored. If ΔG° = 0, K = 1 — balanced.

Summary: Thermodynamics tells you if a reaction is possible (ΔG) and how much heat involved (ΔH), but it does NOT tell you how fast it will happen — that's kinetics (speed of reaction).
Revision: Third Law perfect crystal at 0K S=0 perfect order. Standard ΔG°=ΔH°-TΔS° K=e^(-ΔG°/RT)=10^(-ΔG°/2.303RT). ΔG° -ve K>1 products favored, ΔG° +ve K<1 reactants, ΔG°=0 K=1 balanced. Thermodynamics tells if possible ΔG not how fast kinetics.
Third Law

Quick Revision Box

  • Third Law S=0 at 0K perfect crystal perfect order
  • ΔG°=ΔH°-TΔS°, K=e^(-ΔG°/RT)
  • ΔG° -ve K>1 products favored, +ve K<1 reactants
  • Thermo if possible not how fast