Chapter 5: Thermodynamics

Introduction

Overview

• Thermodynamics studies energy forms and transformations, mainly heat and work.

• Laws apply only when systems are at or move between equilibrium states.

• Chemical thermodynamics focuses on chemical changes and substances.

Solved Examples

• Ex 1: Identify the type of thermodynamics studying heat in chemical reactions.
  Answer: Chemical thermodynamics.

Practice Set

• Q1: Thermodynamics primarily studies:
  A) Electrical energy
  B) Energy transformations
  C) Nuclear reactions
  D) Mechanical forces

• Q2: Laws of thermodynamics apply when the system is:
  A) Non-equilibrium
  B) Equilibrium
  C) Isolated only
  D) Closed only

• Q3: Chemical thermodynamics deals with:
  A) Physical changes only
  B) Chemical changes only
  C) Mechanical work only
  D) Heat only

Answer Key

• Q1: B - Thermodynamics studies energy transformations, especially heat and work.

• Q2: B - Laws apply when systems are at or move between equilibrium states.

• Q3: B - Chemical thermodynamics focuses on chemical changes and substances.

Memory Aids

• Thermodynamics = Energy + Transformations.

• Laws valid only at equilibrium.

• Chemical thermodynamics = chemical changes only.

Thermodynamic Terms

Key Definitions

• System: Part of universe under study; rest is surroundings.

• Boundary: Real or imaginary surface separating system and surroundings.

• Types of Systems:

  • Open: Exchange matter and energy.

  • Closed: Exchange energy only.

  • Isolated: No exchange of matter or energy.

Solved Examples

• Ex 1: Classify a system that exchanges only energy.
  Answer: Closed system.

Practice Set

• Q1: A system exchanging both matter and energy is:
  A) Closed
  B) Open
  C) Isolated
  D) Adiabatic

• Q2: Which system exchanges neither matter nor energy?
  A) Open
  B) Closed
  C) Isolated
  D) Cyclic

• Q3: Boundary is:
  A) Part of surroundings
  B) Surface separating system and surroundings
  C) The system itself
  D) Energy inside system

Answer Key

• Q1: B - Open systems exchange both matter and energy.

• Q2: C - Isolated systems exchange neither matter nor energy.

• Q3: B - Boundary separates system from surroundings.

Memory Aids

• Open = Both matter & energy exchange.

• Closed = Only energy exchange.

• Isolated = No exchange.

The State of the System

State and Processes

• Thermodynamic state: Set of measurable properties (P, T, V, n) defining system.

• State variables/functions: Properties depending only on state, not path.

• Thermodynamic processes:

  • Isothermal: Constant temperature (ΔT=0)

  • Adiabatic: No heat exchange (q=0)

  • Isobaric: Constant pressure (ΔP=0)

  • Isochoric: Constant volume (ΔV=0)

  • Processes can be cyclic, reversible, or irreversible.

• Reversible: direction may be reversed at any stage

• Irreversible process : Proceeds only in one direction and cannot be reversed.

  Cyclic process : System returns to its original state after a number of steps. For such a process, ΔU =0 and ΔH =0.

• Exothermic process : Accompanied by the evolution of heat.

• Endothermic process : Accompanied by the absorption of heat.

Solved Examples

• Ex 1: Identify process with no heat exchange.
  Answer: Adiabatic process.

• Ex 2: Process at constant pressure is called?
  Answer: Isobaric process.

Practice Set

• Q1: Which process occurs at constant volume?
  A) Isobaric
  B) Isochoric
  C) Isothermal
  D) Adiabatic

• Q2: In which process is temperature constant?
  A) Isothermal
  B) Adiabatic
  C) Isobaric
  D) Irreversible

• Q3: A cyclic process:
  A) Changes state variables permanently
  B) Returns system to original state
  C) Is always irreversible
  D) Has ΔU ≠ 0

Answer Key

• Q1: B - Isochoric means constant volume.

• Q2: A - Isothermal means constant temperature.

• Q3: B - Cyclic process returns system to original state; ΔU=0.

Memory Aids

• Isothermal = T constant.

• Adiabatic = q=0, no heat exchange.

• Isobaric = P constant.

• Isochoric = V constant.

• Cyclic = returns to start state.

Internal Energy as a State Function

Concepts and Laws

• Internal energy (U): Total energy inside system (chemical, electrical, mechanical).

• U changes when:

  • Heat (q) enters or leaves system.

  • Work (w) is done on or by system.

  • Matter enters or leaves system.

• First law of thermodynamics: Energy of isolated system is constant (conservation of energy).

• Mathematical form: ΔU = q + w or q = ΔU − w

• q and w are path functions; ΔU is a state function.

• Sign conventions:

  • Heat absorbed by system: +ve

  • Heat evolved by system: −ve

  • Work done on system: +ve

  • Work done by system: −ve

• For adiabatic change (q=0): ΔU = wad

• At constant volume (no work): ΔU = qv

Solved Examples

• Ex 1: Calculate ΔU if system absorbs 100 J heat and does 40 J work.
  Solution: ΔU = q + w = 100 + (−40) = 60 J.

Practice Set

• Q1: If system absorbs 50 J heat and work done on system is 20 J, ΔU is:
  A) 30 J
  B) 70 J
  C) −30 J
  D) −70 J

• Q2: For adiabatic process, heat exchanged is:
  A) Positive
  B) Negative
  C) Zero
  D) Variable

• Q3: Work done by system is 80 J and heat evolved is 30 J, ΔU is:
  A) +110 J
  B) −110 J
  C) +50 J
  D) −50 J

Answer Key

• Q1: B - ΔU = q + w = 50 + 20 = 70 J.

• Q2: C - Adiabatic means no heat exchange (q=0).

• Q3: D - q = −30 J, w = −80 J (work done by system), ΔU = q + w = −110 J.

Memory Aids

• ΔU = q + w (First law).

• Heat in (+), heat out (−).

• Work on system (+), work by system (−).

• Adiabatic: q=0.

• Constant volume: ΔU = qv.

Work

Isothermal Reversible Expansion of an Ideal Gas

• Gas expands slowly in a cylinder with a frictionless piston; volume increases, pressure decreases.

• Work done in infinitesimal expansion: \( dw = -P_{ex} dV = -P dV \).

• Total work from initial volume \( V_i \) to final volume \( V_f \): \( w_{rev} = -nRT \ln \frac{V_f}{V_i} = nRT \ln \frac{P_i}{P_f} \).

• Work done during isothermal compression: \( w_{rev} = 2.303 nRT \log \frac{P_i}{P_f} \) (positive value).

Free Expansion

• Expansion into vacuum (\( P_{ex} = 0 \)) is free expansion.

• No work is done: \( w = 0 \), whether reversible or irreversible.

• Internal energy change: \( \Delta U = q - P_{ex} \Delta V \).

• At constant volume, \( \Delta U = q_v \).

Isothermal and Free Expansion of an Ideal Gas

• Isothermal expansion into vacuum: \( w = 0 \) since \( P_{ex} = 0 \).

• Isothermal irreversible change: \( q = -P_{ex} (V_f - V_i) \).

• Isothermal reversible change: \( q = w = 2.303 nRT \log \frac{V_f}{V_i} \).

• Adiabatic change: \( q = 0 \), \( \Delta U = w_{ad} \).

Solved Examples

Ex 1: Calculate work done when 1 mole of ideal gas expands isothermally and reversibly from 10 L to 20 L at 300 K.

• Use \( w = -nRT \ln \frac{V_f}{V_i} \).

• \( w = -1 \times 8.314 \times 300 \times \ln(2) = -1729 \text{ J} \).

Practice Set

• Q1: Work done during isothermal reversible expansion is

  • A) Positive

  • B) Negative

  • C) Zero

  • D) Depends on gas

• Q2: For free expansion of an ideal gas, work done is

  • A) Equal to \( nRT \)

  • B) Zero

  • C) Negative

  • D) Depends on volume

• Q3: Work done in isothermal reversible compression from pressure \( P_i \) to \( P_f \) is

  • A) \( 2.303 nRT \log \frac{P_f}{P_i} \)

  • B) \( -2.303 nRT \log \frac{P_i}{P_f} \)

  • C) \( 2.303 nRT \log \frac{P_i}{P_f} \)

  • D) Zero

Answer Key

• Q1: B - Work done by gas during expansion is negative (energy leaves system).

• Q2: B - No work done in free expansion as external pressure is zero.

• Q3: C - Work done in isothermal reversible compression is positive and given by \( 2.303 nRT \log \frac{P_i}{P_f} \).

Memory Aids

• Isothermal expansion: work negative, volume ↑, pressure ↓.

• Free expansion: no work done.

• Compression work positive, opposite sign to expansion.

Enthalpy

Definition and Properties

• Enthalpy (H) = Internal energy (U) + Pressure × Volume (PV).

• Extensive and state function; absolute value unknown.

• Change in enthalpy: \( \Delta H = \Delta U + P \Delta V \).

• Exothermic: \( \Delta H < 0 \); Endothermic: \( \Delta H > 0 \).

Relation with Heat and Work

• First law: \( q = \Delta U - w \), with \( w = -P \Delta V \).

• At constant volume: \( w = 0 \), so \( q_v = \Delta U \).

• At constant pressure: \( q_p = \Delta U + P \Delta V = \Delta H \).

• \( P \Delta V = \Delta n_g RT \), where \( \Delta n_g \) = moles gaseous products - reactants.

• Thus, \( \Delta H = \Delta U + \Delta n_g RT \).

• If \( \Delta n_g > 0 \), then \( \Delta H > \Delta U \); if \( \Delta n_g < 0 \), then \( \Delta H < \Delta U \); if zero, \( \Delta H = \Delta U \).

• For reactions in solids/liquids or closed vessels (\( \Delta V = 0 \)), \( \Delta H = \Delta U \).

Solved Examples

Ex 1: For a reaction with \( \Delta U = -100 \text{ kJ} \) and \( \Delta n_g = 2 \) at 298 K, calculate \( \Delta H \).

• \( \Delta H = \Delta U + \Delta n_g RT = -100 + 2 \times 8.314 \times 298 / 1000 = -95.05 \text{ kJ} \).

Practice Set

• Q1: Enthalpy is

  • A) Extensive and state function

  • B) Intensive and state function

  • C) Extensive and path function

  • D) Intensive and path function

• Q2: At constant pressure, heat absorbed equals

  • A) \( \Delta U \)

  • B) \( \Delta H \)

  • C) Work done

  • D) Zero

• Q3: For a reaction with no gas volume change, \( \Delta H \) is

  • A) Greater than \( \Delta U \)

  • B) Less than \( \Delta U \)

  • C) Equal to \( \Delta U \)

  • D) Zero

Answer Key

• Q1: A - Enthalpy depends on system size and is a state function.

• Q2: B - Heat at constant pressure equals enthalpy change.

• Q3: C - No gas volume change means \( \Delta H = \Delta U \).

Memory Aids

• \( H = U + PV \).

• At constant P, \( q_p = \Delta H \).

• Gas volume change affects \( \Delta H \) vs \( \Delta U \).

Heat Capacity

Definitions

• Heat capacity (C): heat needed to raise temperature by 1°C.

• Molar heat capacity (Cm): heat to raise 1 mole by 1°C, \( C_m = C/n \).

• Specific heat capacity (c): heat to raise 1 g by 1°C, \( q = mc \Delta T \).

• Water's specific heat: 1 cal g-1 K-1 = 4.18 J g-1 K-1.

Relation Between Cp and Cv

• \( C_p = \frac{\Delta H}{\Delta T} \), \( C_v = \frac{\Delta U}{\Delta T} \).

• Difference: \( C_p - C_v = R \) (ideal gas).

• Ratio: \( \gamma = \frac{C_p}{C_v} \).

• Values: Monatomic gas \( \gamma = 1.66 \), Diatomic \( 1.40 \), Triatomic \( 1.30 \).

Solved Examples

Ex 1: Calculate \( C_p \) if \( C_v = 20.8 \text{ J/mol·K} \) for an ideal gas.

• \( C_p = C_v + R = 20.8 + 8.314 = 29.1 \text{ J/mol·K} \).

Practice Set

• Q1: Heat capacity is

  • A) Heat per mole per °C

  • B) Heat per gram per °C

  • C) Heat to raise temperature by 1°C

  • D) Heat per volume per °C

• Q2: For an ideal gas, \( C_p - C_v = \)

  • A) 0

  • B) R

  • C) \( \gamma \)

  • D) 1

• Q3: \( \gamma \) for diatomic gas is approximately

  • A) 1.66

  • B) 1.40

  • C) 1.30

  • D) 1.00

Answer Key

• Q1: C - Heat capacity is heat required to raise temperature by 1°C.

• Q2: B - Difference between \( C_p \) and \( C_v \) is gas constant R.

• Q3: B - Diatomic gases have \( \gamma \approx 1.40 \).

Memory Aids

• \( C_p - C_v = R \) for ideal gases.

• \( \gamma \) decreases with molecular complexity.

• Water has high specific heat capacity.

Efficiency of Heat Engine

Key Points

• Efficiency \( \eta = \frac{w}{q_2} = \frac{T_2 - T_1}{T_2} = \frac{q_2 - q_1}{q_2} \).

• \( q_2 \): heat absorbed at high temperature \( T_2 \).

• \( q_1 \): heat rejected at low temperature \( T_1 \).

• \( w \): net work done by engine.

• Unit conversions: 1 J = 0.2390 cal, 1 cal = 4.184 J, 1 litre-atm = 101.3 J.

Solved Examples

Ex 1: Calculate efficiency of engine operating between 500 K and 300 K.

• \( \eta = \frac{500 - 300}{500} = 0.4 = 40\% \).

Practice Set

• Q1: Efficiency of heat engine is

  • A) \( \frac{T_1}{T_2} \)

  • B) \( \frac{T_2 - T_1}{T_2} \)

  • C) \( \frac{T_2}{T_1} \)

  • D) \( \frac{T_1 - T_2}{T_1} \)

• Q2: Heat rejected by engine is

  • A) \( q_2 \)

  • B) \( q_1 \)

  • C) Work done

  • D) Zero

• Q3: If \( q_2 = 1000 \text{ J} \) and efficiency is 30%, work done is

  • A) 300 J

  • B) 700 J

  • C) 1300 J

  • D) 1000 J

Answer Key

• Q1: B - Efficiency formula is \( \frac{T_2 - T_1}{T_2} \).

• Q2: B - Heat rejected is \( q_1 \) at low temperature.

• Q3: A - Work done = efficiency × heat absorbed = 0.3 × 1000 = 300 J.

Memory Aids

• Efficiency depends on temperature difference.

• Heat rejected always less than heat absorbed.

Joule–Thomson Effect

Concept

• Temperature change when gas expands adiabatically from high to low pressure through a small opening.

• Joule-Thomson coefficient: \( \mu_{JT} = \left( \frac{\partial T}{\partial P} \right)_H \).

• For ideal gases, \( \mu_{JT} = 0 \); no temperature change on expansion.

Adiabatic Reversible Expansion

• Relations: \( PV^\gamma = \text{constant} \), \( TV^{\gamma -1} = \text{constant} \), \( T^\gamma P^{1-\gamma} = \text{constant} \).

• Work done: \( w = \frac{nR (T_1 - T_2)}{\gamma - 1} \).

Solved Examples

Ex 1: Calculate work done when 1 mole of gas cools from 400 K to 300 K adiabatically with \( \gamma = 1.4 \).

• \( w = \frac{1 \times 8.314 \times (400 - 300)}{1.4 - 1} = 2078.5 \text{ J} \).

Practice Set

• Q1: Joule-Thomson effect is zero for

  • A) Real gases

  • B) Ideal gases

  • C) Liquids

  • D) Solids

• Q2: In adiabatic reversible expansion, \( PV^\gamma = \) ?

  • A) Variable

  • B) Constant

  • C) Zero

  • D) Infinity

• Q3: Work done in adiabatic expansion depends on

  • A) Temperature difference and \( \gamma \)

  • B) Pressure only

  • C) Volume only

  • D) None

Answer Key

• Q1: B - Ideal gases show no Joule-Thomson effect.

• Q2: B - \( PV^\gamma \) remains constant in adiabatic reversible expansion.

• Q3: A - Work depends on temperature difference and heat capacity ratio \( \gamma \).

Memory Aids

• Joule-Thomson effect zero for ideal gases.

• Adiabatic expansion: \( PV^\gamma = \text{constant} \).

• Work formula involves \( (T_1 - T_2)/(\gamma -1) \).

Measurement of ΔU and ΔH: Calorimetry

Measurement of ΔU

• Performed using a bomb calorimeter, a sealed vessel where volume is constant.

• Formula: ΔU = C × Δt × (M / m)

• Where:

  • m = mass of substance

  • M = molar mass of substance

  • C = heat capacity of calorimeter

  • Δt = temperature change observed

• Measures internal energy change at constant volume.

Measurement of ΔH

• Reaction carried out in an insulated calorimeter vessel.

• Temperature rise or fall is recorded depending on exothermic or endothermic nature.

• Measures enthalpy change at constant pressure.

Solved Examples

• Ex 1: Calculate ΔU if 2 g of a substance (M = 50 g/mol) causes a 3°C rise in a calorimeter with heat capacity 100 J/°C.

  • ΔU = 100 × 3 × (50 / 2) = 7500 J

• Ex 2: In a calorimeter, temperature falls by 2°C during a reaction. Is the reaction exothermic or endothermic?

  • Temperature fall indicates endothermic reaction.

Practice Set

• Q1: Which calorimeter is used to measure ΔU?

  • A. Coffee cup calorimeter

  • B. Bomb calorimeter

  • C. Differential scanning calorimeter

  • D. Adiabatic calorimeter

• Q2: In the formula ΔU = C × Δt × (M / m), what does C represent?

  • A. Molar mass of substance

  • B. Heat capacity of calorimeter

  • C. Mass of substance

  • D. Temperature change

• Q3: A reaction in a calorimeter causes temperature to rise by 5°C. The heat capacity of calorimeter is 200 J/°C. If 4 g of substance (M = 40 g/mol) reacted, calculate ΔU.

  • A. 10000 J

  • B. 1000 J

  • C. 5000 J

  • D. 2500 J

Answer Key

• Q1: B. Bomb calorimeter - It operates at constant volume to measure ΔU.

• Q2: B. Heat capacity of calorimeter - C is the calorimeter's heat capacity.

• Q3: A. 10000 J - ΔU = 200 × 5 × (40 / 4) = 10000 J.

Memory Aids

• Bomb calorimeter = constant volume → measures ΔU.

• Insulated vessel calorimeter = constant pressure → measures ΔH.

• ΔU formula: Heat capacity × temperature change × (molar mass / mass).

Enthalpy Change, Reaction Enthalpy

Definition and Calculation

• Reaction enthalpy (ΔrH): Enthalpy change during a chemical reaction.

• Calculated as: ΔrH = ΣaiHproducts − ΣbiHreactants

• a and b are stoichiometric coefficients of products and reactants.

• Example: CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

  • ΔrH = [Hm(CO2) + 2Hm(H2O)] − [Hm(CH4) + 2Hm(O2)]

• Hm = molar enthalpy.

• Standard enthalpy of reaction: Enthalpy change when all substances are in their standard states (pure form at 1 bar).

Enthalpy Changes During Phase Transformations

• Enthalpy of fusion (ΔfusH°): Solid → Liquid at melting point.

  • Example: H2O(s) → H2O(l) at 273 K; ΔfusH° = +6.0 kJ mol−1

• Enthalpy of vaporisation (ΔvapH°): Liquid → Vapour at boiling point.

  • Example: H2O(l) → H2O(g) at 373 K; ΔvapH° = +40.79 kJ mol−1

• Enthalpy of sublimation (ΔsubH°): Solid → Vapour directly.

  • Example: Iodine sublimation ΔsubH° = +62.4 kJ mol−1

  • Relation: ΔsubH° = ΔfusH° + ΔvapH°

• Enthalpy of formation (ΔfH°): Formation of 1 mole of substance from elements in standard states.

  • Example: Cgraphite(s) + 2H2(g) → CH4(g); ΔfH° = −74.8 kJ mol−1

Hess’s Law of Constant Heat Summation

• If a reaction occurs in multiple steps, total enthalpy change equals sum of enthalpy changes of steps at same temperature.

• Mathematically: ΔrH = ΔrH1 + ΔrH2 + ΔrH3

Solved Examples

• Ex 1: Calculate ΔrH for CH4 combustion if Hm(CO2) = −393.5 kJ/mol, Hm(H2O) = −285.8 kJ/mol, Hm(CH4) = −74.8 kJ/mol, Hm(O2) = 0 kJ/mol.

  • ΔrH = [−393.5 + 2(−285.8)] − [−74.8 + 2(0)] = −890.3 kJ/mol

• Ex 2: Using Hess’s law, find ΔrH for A → B given ΔrH1 (A → C) = +50 kJ, ΔrH2 (C → D) = −30 kJ, ΔrH3 (D → B) = +20 kJ.

  • ΔrH = 50 − 30 + 20 = +40 kJ

Practice Set

• Q1: What is the enthalpy change called when one mole of a solid converts to liquid at melting point?

  • A. Enthalpy of vaporisation

  • B. Enthalpy of fusion

  • C. Enthalpy of sublimation

  • D. Enthalpy of formation

• Q2: For the reaction Cgraphite(s) + O2(g) → CO2(g), the ΔfH° is −393.5 kJ/mol. What does this represent?

  • A. Enthalpy of vaporisation of CO2

  • B. Enthalpy of formation of CO2

  • C. Enthalpy of fusion of graphite

  • D. Enthalpy of sublimation of CO2

• Q3: Using Hess’s law, if ΔrH for A → C is +60 kJ, C → D is −25 kJ, and D → B is +15 kJ, find ΔrH for A → B.

  • A. +100 kJ

  • B. +50 kJ

  • C. +40 kJ

  • D. −20 kJ

Answer Key

• Q1: B. Enthalpy of fusion is the heat required to convert solid to liquid at melting point.

• Q2: B. Enthalpy of formation is heat change when 1 mole of compound forms from elements in standard states.

• Q3: C. ΔrH = 60 − 25 + 15 = +50 kJ (corrected: sum is +50 kJ, so option B is correct)

Memory Aids

• Enthalpy of fusion: solid to liquid (melting).

• Enthalpy of vaporisation: liquid to gas (boiling).

• Enthalpy of sublimation: solid to gas directly; equals fusion + vaporisation enthalpies.

• Hess’s law: total enthalpy change independent of path; sum of steps.

Enthalpy of Combustion

Definition and Key Points

• Change in enthalpy when 1 mole of a substance burns completely in oxygen.

• Example: C(s) + O2(g) → CO2(g); ΔcH° = –393.5 kJ mol–1

• ΔcH° is always negative (exothermic reaction).

Solved Examples

• Ex 1: Calculate heat released when 2 moles of carbon burn.

  • ΔcH° for 1 mole C = –393.5 kJ

  • For 2 moles: Heat = 2 × (–393.5) = –787 kJ

Practice Set

• Q1: What is the sign of enthalpy of combustion?

  • A) Positive

  • B) Negative

  • C) Zero

  • D) Depends on substance

• Q2: Combustion of 1 mole of methane releases –890 kJ. What is ΔcH°?

  • A) +890 kJ mol–1

  • B) –890 kJ mol–1

  • C) 0 kJ mol–1

  • D) +393.5 kJ mol–1

• Q3: Which of the following has the most negative enthalpy of combustion?

  • A) C2H6

  • B) C3H8

  • C) C4H10

  • D) CH4

Answer Key

• Q1: B) Negative – Combustion releases heat, so enthalpy change is negative.

• Q2: B) –890 kJ mol–1 – Given heat released equals enthalpy of combustion.

• Q3: C) C4H10 – Larger hydrocarbons release more heat on combustion.

Memory Aids

• Combustion = burning → always exothermic → ΔH negative.

• More carbon atoms → more negative ΔcH°.

Enthalpy of Atomization

Definition and Key Points

• Energy change when 1 mole of bonds breaks to form gaseous atoms.

• Example: CH4(g) → C(g) + 4H(g); ΔaH = 1665 kJ mol–1

• Bond dissociation enthalpy is related; it is energy to break 1 mole of covalent bonds in gas phase.

Solved Examples

• Ex 1: Calculate bond dissociation enthalpy of H–H if H2 → 2H requires 436 kJ.

  • Bond dissociation enthalpy = 436 kJ mol–1

Practice Set

• Q1: Enthalpy of atomization is always:

  • A) Negative

  • B) Positive

  • C) Zero

  • D) Depends on compound

• Q2: Which process represents atomization?

  • A) H2O(l) → H2O(g)

  • B) CH4(g) → C(g) + 4H(g)

  • C) NaCl(s) → Na+(g) + Cl–(g)

  • D) C(graphite) → C(diamond)

• Q3: Bond dissociation enthalpy is related to:

  • A) Energy to form bonds

  • B) Energy to break bonds

  • C) Energy of phase change

  • D) Energy of ionization

Answer Key

• Q1: B) Positive – Energy is required to break bonds.

• Q2: B) CH4(g) → C(g) + 4H(g) – Breaking bonds to atoms.

• Q3: B) Energy to break bonds – Bond dissociation enthalpy is bond breaking energy.

Memory Aids

• Atomization = bond breaking → always endothermic (+ΔH).

• Bond dissociation enthalpy = energy to break 1 mole of bonds.

Bond Enthalpy

Definition and Key Points

• Standard enthalpy change related to bond enthalpies of reactants and products.

• Formula: ΔrH° = Σ bond enthalpiesreactants – Σ bond enthalpiesproducts

• Used to estimate enthalpy changes in gas phase reactions.

Solved Examples

• Ex 1: Calculate ΔrH° for H2 + Cl2 → 2HCl given:

  • H–H = 436 kJ/mol, Cl–Cl = 243 kJ/mol, H–Cl = 431 kJ/mol

  • ΔrH° = (436 + 243) – 2×431 = 679 – 862 = –183 kJ/mol

Practice Set

• Q1: Bond enthalpy is the energy required to:

  • A) Form bonds

  • B) Break bonds

  • C) Ionize atoms

  • D) Change phase

• Q2: For reaction A–B + C–D → A–D + C–B, ΔrH° is:

  • A) Sum bonds broken – sum bonds formed

  • B) Sum bonds formed – sum bonds broken

  • C) Sum bonds broken + sum bonds formed

  • D) Zero

• Q3: Calculate ΔrH° for H2 + F2 → 2HF given H–H=436, F–F=158, H–F=565 kJ/mol.

  • A) –542 kJ/mol

  • B) +542 kJ/mol

  • C) –158 kJ/mol

  • D) +158 kJ/mol

Answer Key

• Q1: B) Break bonds – Bond enthalpy is bond breaking energy.

• Q2: A) Sum bonds broken – sum bonds formed – standard formula.

• Q3: A) –542 kJ/mol – (436+158) – 2×565 = 594 – 1130 = –542 kJ/mol.

Memory Aids

• ΔH = bonds broken – bonds formed.

• Breaking bonds requires energy (+), forming bonds releases energy (–).

Lattice Enthalpy

Definition and Key Points

• Enthalpy change when 1 mole of ionic solid dissociates into gaseous ions.

• Calculated using Born–Haber cycle.

• Example for NaCl:

  • ΔfH°(NaCl) = ΔsubH°(Na) + ½ ΔdissH°(Cl2) + ΔionH°(Na) + ΔegH°(Cl) + ΔlatticeH°(NaCl)

Solved Examples

• Ex 1: Use Born–Haber cycle to find lattice enthalpy if other values are known.

  • Rearrange formula to find ΔlatticeH°.

Practice Set

• Q1: Lattice enthalpy is energy:

  • A) Released when ions form solid

  • B) Required to dissociate solid into ions

  • C) Required to vaporize solid

  • D) Released when atoms ionize

• Q2: Born–Haber cycle helps to calculate:

  • A) Ionization energy

  • B) Lattice enthalpy

  • C) Bond enthalpy

  • D) Enthalpy of combustion

• Q3: Which step is NOT part of Born–Haber cycle for NaCl?

  • A) Sublimation of Na

  • B) Dissociation of Cl2

  • C) Ionization of Cl

  • D) Electron gain by Cl

Answer Key

• Q1: B) Required to dissociate solid into ions – lattice enthalpy is dissociation energy.

• Q2: B) Lattice enthalpy – Born–Haber cycle calculates lattice enthalpy.

• Q3: C) Ionization of Cl – Cl gains electron, does not lose (ionization is for Na).

Memory Aids

• Born–Haber cycle = sum of steps to form ionic solid.

• Lattice enthalpy = energy to break ionic solid into gaseous ions.

Enthalpy of Solution

Definition and Key Points

• Enthalpy change when 1 mole of substance dissolves in solvent.

• Formula: ΔsolH° = ΔlatticeH° + ΔhydH°

• Usually positive (endothermic) for ionic compounds.

• Solubility increases with temperature if ΔsolH° is positive.

• High lattice enthalpy may prevent dissociation.

Solved Examples

• Ex 1: Calculate ΔsolH° if lattice enthalpy = +780 kJ/mol and hydration enthalpy = –720 kJ/mol.

  • ΔsolH° = 780 + (–720) = +60 kJ/mol (endothermic)

Practice Set

• Q1: Enthalpy of solution is:

  • A) Heat change on dissolving 1 mole solute

  • B) Heat change on melting solute

  • C) Heat change on vaporizing solute

  • D) Heat change on ionization

• Q2: If lattice enthalpy > hydration enthalpy, solution process is:

  • A) Exothermic

  • B) Endothermic

  • C) No change

  • D) Spontaneous

• Q3: Solubility of salts generally increases with temperature because:

  • A) ΔsolH° is negative

  • B) ΔsolH° is positive

  • C) Lattice enthalpy decreases

  • D) Hydration enthalpy increases

Answer Key

• Q1: A) Heat change on dissolving 1 mole solute – definition of enthalpy of solution.

• Q2: B) Endothermic – lattice enthalpy > hydration enthalpy means heat absorbed.

• Q3: B) ΔsolH° is positive – endothermic dissolution increases with temperature.

Memory Aids

• ΔsolH° = lattice + hydration enthalpy.

• Positive ΔsolH° → solubility ↑ with temperature.

Enthalpy of Dilution

Definition and Key Points

• Heat change when additional solvent is added to a solution at constant T and P.

• Heat withdrawn from surroundings (usually exothermic).

• Depends on original concentration of solution.

Solved Examples

• Ex 1: Adding water to concentrated acid releases heat (exothermic dilution).

Practice Set

• Q1: Enthalpy of dilution is heat change when:

  • A) Solute dissolves

  • B) Solvent evaporates

  • C) More solvent added to solution

  • D) Solution freezes

• Q2: Enthalpy of dilution depends on:

  • A) Temperature only

  • B) Original concentration

  • C) Volume only

  • D) Pressure only

• Q3: Dilution of concentrated acid is usually:

  • A) Endothermic

  • B) Exothermic

  • C) No heat change

  • D) Spontaneous

Answer Key

• Q1: C) More solvent added to solution – definition of enthalpy of dilution.

• Q2: B) Original concentration – affects heat change on dilution.

• Q3: B) Exothermic – heat released when acid diluted.

Memory Aids

• Dilution = adding solvent → heat change depends on concentration.

• Concentrated acids release heat on dilution.

Enthalpy of Neutralisation

Definition and Key Points

• Heat change when 1 gram equivalent of acid neutralizes 1 gram equivalent of base in dilute solution.

• Example: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l); ΔneuH = –57.1 kJ mol–1

• Usually exothermic (heat released).

Solved Examples

• Ex 1: Calculate heat released when 0.5 mole HCl neutralizes NaOH.

  • Heat = 0.5 × (–57.1) = –28.55 kJ

Practice Set

• Q1: Enthalpy of neutralisation is heat change when:

  • A) Acid reacts with base

  • B) Acid dissolves in water

  • C) Base melts

  • D) Salt forms solid

• Q2: Neutralisation of strong acid and strong base releases:

  • A) Heat

  • B) No heat

  • C) Heat absorbed

  • D) Gas

• Q3: ΔneuH for HCl + NaOH is approximately:

  • A) –57 kJ/mol

  • B) +57 kJ/mol

  • C) 0 kJ/mol

  • D) –393 kJ/mol

Answer Key

• Q1: A) Acid reacts with base – definition of enthalpy of neutralisation.

• Q2: A) Heat – neutralisation is exothermic.

• Q3: A) –57 kJ/mol – standard enthalpy of neutralisation for strong acid-base.

Memory Aids

• Neutralisation = acid + base → salt + water + heat.

• Strong acid + strong base → ΔH ≈ –57 kJ/mol.

Enthalpy of Transition

Definition and Key Points

• Enthalpy change when 1 mole of substance changes from one allotrope to another.

• Example: C(graphite) → C(diamond); ΔtransH = 1.9 kJ mol–1

Solved Examples

• Ex 1: Calculate heat required to convert 2 moles graphite to diamond.

  • Heat = 2 × 1.9 = 3.8 kJ

Practice Set

• Q1: Enthalpy of transition involves change in:

  • A) Phase

  • B) Allotropic form

  • C) Ionization state

  • D) Chemical composition

• Q2: Transition enthalpy for graphite to diamond is:

  • A) Negative

  • B) Positive

  • C) Zero

  • D) Very large

• Q3: Which is more stable allotrope of carbon at room temperature?

  • A) Diamond

  • B) Graphite

  • C) Both equal

  • D) Depends on pressure

Answer Key

• Q1: B) Allotropic form – transition between allotropes.

• Q2: B) Positive – energy required to convert graphite to diamond.

• Q3: B) Graphite – more stable at room temperature.

Memory Aids

• Transition = allotrope change.

• Graphite more stable than diamond at normal conditions.

Enthalpy of Hydration

Definition and Key Points

• Heat change when 1 mole of anhydrous substance combines with water to form hydrated compound.

• Example: CuSO4(s) + 5H2O → CuSO4·5H2O(s); ΔhydH = –78.2 kJ mol–1

• Usually exothermic (heat released).

Solved Examples

• Ex 1: Calculate heat released when 3 moles CuSO4 hydrate.

  • Heat = 3 × (–78.2) = –234.6 kJ

Practice Set

• Q1: Enthalpy of hydration is heat change when:

  • A) Hydrated compound loses water

  • B) Anhydrous compound gains water

  • C) Compound melts

  • D) Compound vaporizes

• Q2: Hydration enthalpy is usually:

  • A) Positive

  • B) Negative

  • C) Zero

  • D) Variable

• Q3: Which compound forms pentahydrate on hydration?

  • A) CuSO4

  • B) NaCl

  • C) MgSO4

  • D) KCl

Answer Key

• Q1: B) Anhydrous compound gains water – definition of enthalpy of hydration.

• Q2: B) Negative – hydration releases heat.

• Q3: A) CuSO4 – forms CuSO4·5H2O.

Memory Aids

• Hydration = water addition → heat released.

• CuSO4 commonly forms pentahydrate.

Spontaneity

Definition and Characteristics

• Spontaneous process: occurs without external influence, irreversible.

• Reversal requires external work.

Entropy (S)

• Measure of disorder/randomness in a system.

• For reversible process: \( \Delta S = \frac{q_{rev}}{T} \).

• Total entropy change for spontaneous process: \( \Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings} > 0 \).

• At equilibrium, entropy is maximum and \( \Delta S = 0 \).

• Entropy change during phase changes:

  • Fusion: \( \Delta S = \frac{\Delta H_{fus}}{T_m} \)

  • Vaporization: \( \Delta S = \frac{\Delta H_{vap}}{T_b} \)

  • Sublimation: \( \Delta S = \frac{\Delta H_{sub}}{T} \)

• \( T_m \): melting point, \( T_b \): boiling point, \( T \): sublimation temperature.

• Entropy change of reaction: \( \Delta_r S^\circ = \sum S^\circ_{products} - \sum S^\circ_{reactants} \).

Solved Examples

• Ex 1: Calculate entropy change when 1 mole of ice melts at 273 K, \( \Delta H_{fus} = 6.01 \text{ kJ/mol} \).

  • \( \Delta S = \frac{\Delta H_{fus}}{T_m} = \frac{6010}{273} = 22.02 \text{ J/K} \)

Practice Set

• Q1: Entropy change for vaporization is given by?

  • A. \( \frac{\Delta H_{vap}}{T_b} \)

  • B. \( \frac{\Delta H_{fus}}{T_m} \)

  • C. \( \frac{\Delta H_{sub}}{T} \)

  • D. \( \Delta H_{vap} \times T_b \)

• Q2: For a spontaneous process, total entropy change is?

  • A. Zero

  • B. Negative

  • C. Positive

  • D. Undefined

• Q3: At equilibrium, entropy change is?

  • A. Positive

  • B. Negative

  • C. Zero

  • D. Infinite

Answer Key

• Q1: A - Entropy change for vaporization is heat absorbed divided by boiling point.

• Q2: C - Spontaneous processes have positive total entropy change.

• Q3: C - At equilibrium, entropy change is zero.

Memory Aids

• Entropy increases with phase change: solid < liquid < gas.

• Spontaneous process: total entropy change > 0.

• At equilibrium: entropy change = 0.

Gibbs Free Energy

Definition and Equations

• Gibbs free energy (G): energy available to do useful work.

• Formula: \( G = H - TS \).

• Standard free energy change: \( \Delta_r G^\circ = \sum \Delta_f G^\circ_{products} - \sum \Delta_f G^\circ_{reactants} \).

• Gibbs–Helmholtz equation: \( \Delta G = \Delta H - T \Delta S \).

• Relation with emf: \( \Delta G^\circ = -nFE^\circ \), where \( E^\circ \) is standard emf.

Gibbs Energy and Spontaneity

• \( \Delta G < 0 \): spontaneous process.

• \( \Delta G > 0 \): non-spontaneous process.

• \( \Delta G = 0 \): equilibrium.

Effect of Temperature on Spontaneity

• \( \Delta H < 0, \Delta S > 0 \): spontaneous at all temperatures.

• \( \Delta H < 0, \Delta S < 0 \): spontaneous at low temperatures.

• \( \Delta H > 0, \Delta S > 0 \): spontaneous at high temperatures.

• \( \Delta H > 0, \Delta S < 0 \): never spontaneous.

Solved Examples

• Ex 1: Calculate \( \Delta G \) at 298 K for a reaction with \( \Delta H = -40 \text{ kJ} \) and \( \Delta S = -100 \text{ J/K} \).

  • Convert \( \Delta S \) to kJ/K: \( -0.1 \text{ kJ/K} \).

  • \( \Delta G = -40 - 298 \times (-0.1) = -40 + 29.8 = -10.2 \text{ kJ} \) (spontaneous).

Practice Set

• Q1: For \( \Delta G < 0 \), the process is?

  • A. Non-spontaneous

  • B. Spontaneous

  • C. At equilibrium

  • D. Irreversible

• Q2: Gibbs free energy equation is?

  • A. \( G = H + TS \)

  • B. \( G = H - TS \)

  • C. \( G = TS - H \)

  • D. \( G = H / TS \)

• Q3: If \( \Delta H > 0 \) and \( \Delta S < 0 \), reaction is?

  • A. Spontaneous at high T

  • B. Spontaneous at low T

  • C. Never spontaneous

  • D. Always spontaneous

Answer Key

• Q1: B - Negative \( \Delta G \) means spontaneous.

• Q2: B - Gibbs free energy formula is \( G = H - TS \).

• Q3: C - Reaction is never spontaneous if \( \Delta H > 0 \) and \( \Delta S < 0 \).

Memory Aids

• "G = H minus TS" for free energy.

• Spontaneity depends on signs of \( \Delta H \) and \( \Delta S \) and temperature.

• Remember: Negative \( \Delta G \) = spontaneous.

Thermodynamics Laws

Second Law

• For spontaneous processes, total entropy change \( \Delta S_{total} > 0 \).

Third Law

• Entropy of pure crystalline substance approaches zero as temperature approaches absolute zero (0 K).

Additional Notes on Entropy

• More symmetrical molecules have lower entropy than less symmetrical ones with similar mass and heat capacity.

• Simple structured solids have lower entropy than complex structured solids.

• Stretching rubber band decreases entropy (more ordered macromolecules).

• Boiling egg increases entropy (protein denaturation to random coils).

• Entropy of liquids/solids not much affected by pressure changes.

Entropy Change of Ideal Gas (1 mole)

• \( \Delta S = C_V \ln \frac{T_2}{T_1} + R \ln \frac{V_2}{V_1} = C_P \ln \frac{T_2}{T_1} + R \ln \frac{P_1}{P_2} \)

• Isothermal: \( \Delta S_T = R \ln \frac{V_2}{V_1} = R \ln \frac{P_1}{P_2} \)

• Isochoric: \( \Delta S_V = C_V \ln \frac{T_2}{T_1} = C_V \ln \frac{P_2}{P_1} \)

• Isobaric: \( \Delta S_P = C_P \ln \frac{T_2}{T_1} = C_P \ln \frac{V_2}{V_1} \)

Gibbs Helmholtz Equations

• \( \Delta G = \Delta H + T \left[ \frac{d(\Delta G)}{dT} \right]_P \)

• \( \frac{\Delta G - \Delta H}{T} = \left[ \frac{d(\Delta G)}{dT} \right]_P \)

• \( -\Delta S = \left[ \frac{d(\Delta G)}{dT} \right]_P \)

Solved Examples

• Ex 1: Calculate entropy change for 1 mole ideal gas expanding isothermally from 2 L to 4 L.

  • \( \Delta S = R \ln \frac{V_2}{V_1} = 8.314 \times \ln 2 = 5.76 \text{ J/K} \)

Practice Set

• Q1: Entropy change for isothermal expansion is?

  • A. \( C_V \ln \frac{T_2}{T_1} \)

  • B. \( R \ln \frac{V_2}{V_1} \)

  • C. \( C_P \ln \frac{T_2}{T_1} \)

  • D. Zero

• Q2: According to third law, entropy at 0 K is?

  • A. Infinite

  • B. Zero for pure crystalline substance

  • C. Negative

  • D. Undefined

• Q3: Gibbs Helmholtz equation relates \( \Delta G \) with?

  • A. \( \Delta H \) and temperature derivative of \( \Delta G \)

  • B. Pressure only

  • C. Volume only

  • D. Entropy only

Answer Key

• Q1: B - Isothermal entropy change depends on volume change.

• Q2: B - Entropy of pure crystalline substance approaches zero at 0 K.

• Q3: A - Gibbs Helmholtz relates \( \Delta G \), \( \Delta H \), and temperature derivative.

Memory Aids

• Entropy change formulas depend on process type (isothermal, isochoric, isobaric).

• Third law: entropy zero at absolute zero for perfect crystals.

• Gibbs Helmholtz links free energy change with enthalpy and temperature.

Gibbs Energy Change and Equilibrium

Relation between Gibbs Energy and Equilibrium Constant

• Gibbs energy change for a reaction, ΔrG°, is linked to equilibrium constant K by:
  ΔrG° = –2.303 RT log K

• R = gas constant, T = temperature in Kelvin

• Negative ΔrG° means reaction favors products (K > 1)

• Positive ΔrG° means reaction favors reactants (K < 1)

• ΔrG° = 0 means equilibrium with equal reactants and products (K = 1)

Gibbs Energy and Equilibrium States

• ΔG° = 0, log K = 0, K = 1: Reactants and products are in equal amounts at equilibrium

• ΔG° < 0, log K > 0, K > 1: Reaction favors products; exothermic reaction

• ΔG° > 0, log K < 0, K < 1: Reaction favors reactants; endothermic reaction

Solved Examples

• Ex 1: Calculate ΔrG° at 298 K if K = 103.
  Step 1: Use ΔrG° = –2.303 RT log K
  Step 2: R = 8.314 J/mol·K, T = 298 K, log 103 = 3
  Step 3: ΔrG° = –2.303 × 8.314 × 298 × 3 = –17,100 J/mol (–17.1 kJ/mol)

• Ex 2: For a reaction with ΔrG° = +5 kJ/mol at 298 K, find K.
  Step 1: Rearrange: log K = –ΔrG° / (2.303 RT)
  Step 2: log K = –5000 / (2.303 × 8.314 × 298) = –0.875
  Step 3: K = 10–0.875 ≈ 0.133 (favors reactants)

Practice Set

• Q1: If ΔrG° = 0, what is the value of K?
  A. 0
  B. 1
  C. >1
  D. <1

• Q2: For an exothermic reaction with K = 100 at 298 K, what is the sign of ΔrG°?
  A. Positive
  B. Negative
  C. Zero
  D. Cannot determine

• Q3: Calculate ΔrG° at 300 K if K = 0.01. (R = 8.314 J/mol·K)
  A. +22.8 kJ/mol
  B. –22.8 kJ/mol
  C. +11.4 kJ/mol
  D. –11.4 kJ/mol

Answer Key

• Q1: B. 1 — ΔrG° = 0 means log K = 0, so K = 1.

• Q2: B. Negative — K > 1 implies ΔrG° < 0, favoring products.

• Q3: A. +22.8 kJ/mol — ΔrG° = –2.303 × 8.314 × 300 × log 0.01 = +22,800 J/mol.

Memory Aids

• ΔrG° = –2.303 RT log K: Negative ΔG means product-favored (K > 1)

• K = 1 when ΔG = 0: equilibrium with equal reactants and products

• Exothermic reactions usually have negative ΔG and K > 1

• Endothermic reactions usually have positive ΔG and K < 1