Chapter 5: Thermodynamics

Introduction

Overview

• Thermodynamics studies energy forms and transformations, mainly heat to work and vice versa.
• Laws apply only when system is in equilibrium or moves between equilibrium states.
• Chemical thermodynamics focuses on chemical changes and substances.

Solved Examples

• Ex 1: Identify the branch of thermodynamics studying chemical reactions only.
  Answer: Chemical thermodynamics.

Practice Set

• Q1: Thermodynamics mainly studies:
  A. Motion of particles
  B. Energy transformations
  C. Chemical bonding
  D. Atomic structure
• Q2: Laws of thermodynamics apply when:
  A. System is in equilibrium
  B. System is non-equilibrium
  C. System is isolated only
  D. System is open only
• Q3: Chemical thermodynamics deals with:
  A. Physical changes
  B. Chemical changes
  C. Nuclear changes
  D. Mechanical changes

Answer Key

• Q1: B - Thermodynamics studies energy transformations.
• Q2: A - Laws apply only at equilibrium states.
• Q3: B - Chemical thermodynamics studies chemical changes.

Memory Aids

• Thermodynamics = Energy + Heat + Work
• Equilibrium needed for laws to apply
• Chemical thermodynamics = chemical reactions focus

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 of matter and energy.
  • Closed: Exchange of energy only.
  • Isolated: No exchange of matter or energy.

Solved Examples

• Ex 1: Classify a system that exchanges only energy but not matter.
  Answer: Closed system.

Practice Set

• Q1: Which system exchanges both matter and energy?
  A. Open
  B. Closed
  C. Isolated
  D. None
• Q2: A system that exchanges no matter or energy is:
  A. Open
  B. Closed
  C. Isolated
  D. None
• Q3: Boundary is:
  A. Part of surroundings
  B. Surface separating system and surroundings
  C. System itself
  D. Energy flow path

Answer Key

• Q1: A - Open system exchanges both matter and energy.
• Q2: C - Isolated system exchanges neither matter nor energy.
• Q3: B - Boundary separates system and 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: measurable properties needed for calculations.
• State variables/functions: pressure (P), temperature (T), volume (V), amount (n); depend 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.

Solved Examples

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

Practice Set

• Q1: Which process occurs at constant temperature?
  A. Isobaric
  B. Isothermal
  C. Adiabatic
  D. Isochoric
• Q2: In which process is volume constant?
  A. Isobaric
  B. Isochoric
  C. Isothermal
  D. Adiabatic
• Q3: A cyclic process:
  A. Cannot be reversed
  B. Returns system to original state
  C. Has ΔU ≠ 0
  D. Is irreversible

Answer Key

• Q1: B - Isothermal means constant temperature.
• Q2: B - Isochoric means constant volume.
• Q3: B - Cyclic process returns system to original state.

Memory Aids

• Isothermal = T constant
• Adiabatic = No heat exchange
• Isobaric = P constant
• Isochoric = V constant

State and Processes

• Thermodynamic state: measurable properties needed for calculations.
• State variables/functions: pressure (P), temperature (T), volume (V), amount (n); depend 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.

Solved Examples

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

Practice Set

• Q1: Which process occurs at constant temperature?
  A. Isobaric
  B. Isothermal
  C. Adiabatic
  D. Isochoric
• Q2: In which process is volume constant?
  A. Isobaric
  B. Isochoric
  C. Isothermal
  D. Adiabatic
• Q3: A cyclic process:
  A. Cannot be reversed
  B. Returns system to original state
  C. Has ΔU ≠ 0
  D. Is irreversible

Answer Key

• Q1: B - Isothermal means constant temperature.
• Q2: B - Isochoric means constant volume.
• Q3: B - Cyclic process returns system to original state.

Memory Aids

• Isothermal = T constant
• Adiabatic = No heat exchange
• Isobaric = P constant
• Isochoric = V constant

Internal Energy

Concept and First Law

• Internal energy (U): total energy inside system (chemical, electrical, mechanical, etc.), a state function.
• U changes when:
  • Heat enters or leaves system
  • Work is done on or by system
  • Matter enters or leaves system
• First law: Energy of isolated system is constant (conservation of energy).
• Equation: ΔU = q + w or q = ΔU − w
• q and w are path functions; ΔU is state function.
• Sign convention:
  • Heat absorbed by system = +ve
  • Heat evolved by system = −ve
  • Work done on system = +ve
  • Work done by system = −ve
• For adiabatic change: ΔU = U₂ − U₁ = w_ad (q=0)
• At constant volume (no work): ΔU = q_v

Solved Examples

• Ex 1: Calculate ΔU if 100 J heat is absorbed and 50 J work is done by system.
  Solution: ΔU = q + w = +100 + (−50) = +50 J

Practice Set

• Q1: Internal energy is:
  A. Path function
  B. State function
  C. Neither
  D. Both
• Q2: If system absorbs 200 J heat and does 80 J work, ΔU is:
  A. 280 J
  B. 120 J
  C. −120 J
  D. −280 J
• Q3: For adiabatic process:
  A. q ≠ 0
  B. w = 0
  C. ΔU = w
  D. ΔU = q

Answer Key

• Q1: B - Internal energy is a state function.
• Q2: B - ΔU = q + w = 200 − 80 = 120 J.
• Q3: C - For adiabatic, ΔU = work done (q=0).

Memory Aids

• ΔU = q + w
• Heat in (+), work done by system (−)
• Adiabatic: no heat exchange

Concept and First Law

• Internal energy (U): total energy inside system (chemical, electrical, mechanical, etc.), a state function.
• U changes when:
  • Heat enters or leaves system
  • Work is done on or by system
  • Matter enters or leaves system
• First law: Energy of isolated system is constant (conservation of energy).
• Equation: ΔU = q + w or q = ΔU − w
• q and w are path functions; ΔU is state function.
• Sign convention:
  • Heat absorbed by system = +ve
  • Heat evolved by system = −ve
  • Work done on system = +ve
  • Work done by system = −ve
• For adiabatic change: ΔU = U₂ − U₁ = w_ad (q=0)
• At constant volume (no work): ΔU = q_v

Solved Examples

• Ex 1: Calculate ΔU if 100 J heat is absorbed and 50 J work is done by system.
  Solution: ΔU = q + w = +100 + (−50) = +50 J

Practice Set

• Q1: Internal energy is:
  A. Path function
  B. State function
  C. Neither
  D. Both
• Q2: If system absorbs 200 J heat and does 80 J work, ΔU is:
  A. 280 J
  B. 120 J
  C. −120 J
  D. −280 J
• Q3: For adiabatic process:
  A. q ≠ 0
  B. w = 0
  C. ΔU = w
  D. ΔU = q

Answer Key

• Q1: B - Internal energy is a state function.
• Q2: B - ΔU = q + w = 200 − 80 = 120 J.
• Q3: C - For adiabatic, ΔU = work done (q=0).

Memory Aids

• ΔU = q + w
• Heat in (+), work done by system (−)
• Adiabatic: no heat exchange

Work

Isothermal reversible expansion of an ideal gas

• Gas expands in a cylinder with a frictionless piston; volume increases, pressure decreases.
• External pressure initially equals internal pressure; piston moves when external pressure decreases by dP.
• Work done in infinitesimal expansion: dw = -P_ex dV = -P dV.
• Total work done from volume V_i to V_f: w_rev = -∫_Vi^V_f P dV.
• Using ideal gas law P = nRT/V, work done: w_rev = -nRT ln(V_f/V_i) = nRT ln(P_i/P_f).
• Work done during isothermal reversible compression: w_rev = 2.303 nRT log(P_i/P_f).

Free expansion

• Expansion of gas into vacuum (P_ex = 0) is free expansion.
• No work done during free expansion, reversible or irreversible.
• At constant volume (ΔV = 0), internal energy change ΔU = q_v.

Isothermal and free expansion of an ideal gas

• Isothermal expansion into vacuum: work 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(V_f/V_i).

Adiabatic change

• Heat exchange q = 0.
• Change in internal energy equals work done: ΔU = w_ad.

Solved Examples

• Ex 1: Calculate work done in isothermal expansion of 1 mole ideal gas from 10 L to 20 L at 300 K.
  • Use w = -nRT ln(V_f/V_i).
  • w = -1 × 8.314 × 300 × ln(2) = -1729 J.
• Ex 2: Work done in isothermal compression from 20 atm to 10 atm at 300 K for 2 moles.
  • w = 2.303 nRT log(P_i/P_f) = 2.303 × 2 × 8.314 × 300 × log(2) ≈ 7910 J.

Practice Set

• Q1: Work done in free expansion of an ideal gas is:
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Depends on volume
• Q2: For isothermal reversible expansion, work done is:
  • A) -nRT ln(V_f/V_i)
  • B) nRT ln(V_f/V_i)
  • C) Zero
  • D) nRT ln(P_f/P_i)
• Q3: Work done in isothermal compression of 1 mole gas from 5 atm to 10 atm at 300 K is:
  • A) Negative
  • B) Zero
  • C) Positive
  • D) Cannot be determined

Answer Key

• Q1: C) Zero - No work is done in free expansion as external pressure is zero.
• Q2: A) -nRT ln(V_f/V_i) - Work done is negative for expansion.
• Q3: C) Positive - Work done on the gas during compression is positive.

Memory Aids

• Work done in isothermal expansion: Negative sign, gas does work on surroundings.
• Free expansion: No work done, w = 0.
• Compression work is positive, expansion work is negative.

Isothermal reversible expansion of an ideal gas

• Gas expands in a cylinder with a frictionless piston; volume increases, pressure decreases.
• External pressure initially equals internal pressure; piston moves when external pressure decreases by dP.
• Work done in infinitesimal expansion: \( dw = -P_{ex} dV = -P dV \).
• Total work done from volume \( V_i \) to \( V_f \): \( w_{rev} = -\int_{V_i}^{V_f} P dV \).
• Using ideal gas law \( P = \frac{nRT}{V} \), work done: \( w_{rev} = -nRT \ln \frac{V_f}{V_i} = nRT \ln \frac{P_i}{P_f} \).
• Work done during isothermal reversible compression: \( w_{rev} = 2.303 nRT \log \frac{P_i}{P_f} \).

Free expansion

• Expansion of gas into vacuum (\( P_{ex} = 0 \)) is free expansion.
• No work done during free expansion, reversible or irreversible.
• At constant volume (\( \Delta V = 0 \)), internal energy change \( \Delta U = q_v \).

Isothermal and free expansion of an ideal gas

• Isothermal expansion into vacuum: work \( w = 0 \) 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

• Heat exchange \( q = 0 \).
• Change in internal energy equals work done: \( \Delta U = w_{ad} \).

Solved Examples

• Ex 1: Calculate work done in isothermal expansion of 1 mole ideal gas from 10 L to 20 L at 300 K.
  • Use \( w = -nRT \ln \frac{V_f}{V_i} \).
  • \( w = -1 \times 8.314 \times 300 \times \ln(2) = -1729 \text{ J} \).
• Ex 2: Work done in isothermal compression from 20 atm to 10 atm at 300 K for 2 moles.
  • \( w = 2.303 nRT \log \frac{P_i}{P_f} = 2.303 \times 2 \times 8.314 \times 300 \times \log(2) \approx 7910 \text{ J} \).

Practice Set

• Q1: Work done in free expansion of an ideal gas is:
  • A) Positive
  • B) Negative
  • C) Zero
  • D) Depends on volume
• Q2: For isothermal reversible expansion, work done is:
  • A) \( -nRT \ln \frac{V_f}{V_i} \)
  • B) \( nRT \ln \frac{V_f}{V_i} \)
  • C) Zero
  • D) \( nRT \ln \frac{P_f}{P_i} \)
• Q3: Work done in isothermal compression of 1 mole gas from 5 atm to 10 atm at 300 K is:
  • A) Negative
  • B) Zero
  • C) Positive
  • D) Cannot be determined

Answer Key

• Q1: C) Zero - No work is done in free expansion as external pressure is zero.
• Q2: A) \( -nRT \ln \frac{V_f}{V_i} \) - Work done is negative for expansion.
• Q3: C) Positive - Work done on the gas during compression is positive.

Memory Aids

• Work done in isothermal expansion: Negative sign, gas does work on surroundings.
• Free expansion: No work done, \( w = 0 \).
• Compression work is positive, expansion work is negative.

Enthalpy

Basic concepts

• Total heat content at constant pressure.
• Extensive and state function.
• Mathematically: H = U + PV.
• Change in enthalpy: ΔH = ΔU + PΔV.
• Exothermic: ΔH < 0, Endothermic: ΔH > 0.

Relation with first law

• First law: q = ΔU - w, with w = -PΔV.
• At constant volume: w = 0, so q_v = ΔU.
• At constant pressure: q_p = ΔU + PΔV = ΔH.
• ΔH = ΔU + Δn_gRT, where Δn_g is change in moles of gas.
• If Δn_g > 0, then ΔH > ΔU; if Δn_g < 0, then ΔH < ΔU; if zero, ΔH = ΔU.
• For solids/liquids or closed vessel reactions, ΔH = ΔU.

Solved Examples

• Ex 1: Calculate ΔH for reaction with ΔU = -100 kJ, Δn_g = 2, at 298 K.
  • ΔH = ΔU + Δn_gRT = -100 + 2 × 8.314 × 298 / 1000 = -95.05 kJ.
• Ex 2: For reaction with no gas change, ΔH = ΔU = -50 kJ.

Practice Set

• Q1: Enthalpy is a:
  • A) Intensive property
  • B) Extensive property
  • C) Neither
  • D) Depends on temperature
• Q2: At constant volume, heat exchanged equals:
  • A) ΔH
  • B) ΔU
  • C) Zero
  • D) Work done
• Q3: For reaction with Δn_g = 1, ΔH is:
  • A) Less than ΔU
  • B) Equal to ΔU
  • C) Greater than ΔU
  • D) Cannot say

Answer Key

• Q1: B) Extensive property - Enthalpy depends on system size.
• Q2: B) ΔU - At constant volume, heat equals internal energy change.
• Q3: C) Greater than ΔU - Positive gas moles increase enthalpy.

Memory Aids

• ΔH = ΔU + PΔV.
• Exothermic: Negative ΔH, Endothermic: Positive ΔH.
• At constant pressure, heat equals enthalpy change.

Basic concepts

• Total heat content at constant pressure.
• Extensive and state function.
• Mathematically: \( H = U + PV \).
• Change in enthalpy: \( \Delta H = \Delta U + P \Delta V \).
• Exothermic: \( \Delta H < 0 \), Endothermic: \( \Delta H > 0 \).

Relation with first law

• First law: \( q = \Delta U - w \), with \( w = -P \Delta V \).
• At constant volume: \( w = 0 \), so \( q_v = \Delta U \).
• At constant pressure: \( q_p = \Delta U + P \Delta V = \Delta H \).
• \( \Delta H = \Delta U + \Delta n_g RT \), where \( \Delta n_g \) is change in moles of gas.
• 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 solids/liquids or closed vessel reactions, \( \Delta H = \Delta U \).

Solved Examples

• Ex 1: Calculate \( \Delta H \) for reaction with \( \Delta U = -100 \text{ kJ} \), \( \Delta n_g = 2 \), at 298 K.
  • \( \Delta H = \Delta U + \Delta n_g RT = -100 + 2 \times 8.314 \times 298 / 1000 = -95.05 \text{ kJ} \).
• Ex 2: For reaction with no gas change, \( \Delta H = \Delta U = -50 \text{ kJ} \).

Practice Set

• Q1: Enthalpy is a:
  • A) Intensive property
  • B) Extensive property
  • C) Neither
  • D) Depends on temperature
• Q2: At constant volume, heat exchanged equals:
  • A) \( \Delta H \)
  • B) \( \Delta U \)
  • C) Zero
  • D) Work done
• Q3: For reaction with \( \Delta n_g = 1 \), \( \Delta H \) is:
  • A) Less than \( \Delta U \)
  • B) Equal to \( \Delta U \)
  • C) Greater than \( \Delta U \)
  • D) Cannot say

Answer Key

• Q1: B) Extensive property - Enthalpy depends on system size.
• Q2: B) \( \Delta U \) - At constant volume, heat equals internal energy change.
• Q3: C) Greater than \( \Delta U \) - Positive gas moles increase enthalpy.

Memory Aids

• \( \Delta H = \Delta U + P \Delta V \).
• Exothermic: Negative \( \Delta H \), Endothermic: Positive \( \Delta H \).
• At constant pressure, heat equals enthalpy change.

Extensive and Intensive Properties

Definitions and examples

• Extensive: Depends on amount of matter (mass, volume, internal energy, enthalpy, heat capacity).
• Intensive: Independent of amount (temperature, density, pressure, molar volume, molar heat capacity).

Solved Examples

• Ex 1: Identify property: mass (Extensive), temperature (Intensive).
• Ex 2: Heat capacity of 2 moles vs 1 mole: doubles (Extensive).

Practice Set

• Q1: Which is an intensive property?
  • A) Volume
  • B) Mass
  • C) Temperature
  • D) Enthalpy
• Q2: Heat capacity depends on:
  • A) Amount of substance
  • B) Temperature only
  • C) Pressure only
  • D) None
• Q3: Molar heat capacity is:
  • A) Extensive
  • B) Intensive
  • C) Neither
  • D) Depends on mass

Answer Key

• Q1: C) Temperature - Intensive property does not depend on amount.
• Q2: A) Amount of substance - Heat capacity increases with quantity.
• Q3: B) Intensive - Molar heat capacity is per mole, independent of amount.

Memory Aids

• Extensive: Depends on size (mass, volume).
• Intensive: Independent of size (temperature, pressure).

Heat Capacity

Types and formulas

• Heat capacity (C): Heat required to raise temperature by 1°C, C = Q/ΔT.
• Molar heat capacity (C_m): 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 = mcΔT.
• Water's specific heat: 1 cal g^-1 K^-1 = 4.18 J g^-1 K^-1.
• Relation for ideal gas: C_p = ΔH/ΔT, C_v = ΔU/ΔT, C_p - C_v = R.
• Ratio γ = C_p/C_v: monatomic = 1.66, diatomic = 1.40, triatomic = 1.30.

Solved Examples

• Ex 1: Calculate C_m if C = 100 J/K and n = 2 moles.
  • C_m = 100/2 = 50 J/mol K.
• Ex 2: Find heat required to raise 10 g water by 5°C.
  • q = mcΔT = 10 × 4.18 × 5 = 209 J.

Practice Set

• Q1: Heat capacity is heat required to raise temperature by:
  • A) 1 K
  • B) 1°C
  • C) 10°C
  • D) 100°C
• Q2: Specific heat capacity is heat required to raise temperature of:
  • A) 1 mole
  • B) 1 gram
  • C) 1 kg
  • D) 1 liter
• Q3: For ideal gas, C_p - C_v = ?
  • A) 0
  • B) R
  • C) C_p
  • D) C_v

Answer Key

• Q1: B) 1°C - Definition of heat capacity.
• Q2: B) 1 gram - Specific heat is per gram basis.
• Q3: B) R - Difference of heat capacities equals gas constant.

Memory Aids

• Heat capacity: Heat per 1°C rise.
• Specific heat: Heat per gram per °C.
• C_p - C_v = R for ideal gases.

Types and formulas

• Heat capacity (C): Heat required to raise temperature by 1°C, \( C = \frac{Q}{\Delta T} \).
• Molar heat capacity (C_m): Heat to raise 1 mole by 1°C, \( C_m = \frac{C}{n} \).
• 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 for ideal gas: \( C_p = \frac{\Delta H}{\Delta T}, C_v = \frac{\Delta U}{\Delta T}, C_p - C_v = R \).
• Ratio \( \gamma = \frac{C_p}{C_v} \): monatomic = 1.66, diatomic = 1.40, triatomic = 1.30.

Solved Examples

• Ex 1: Calculate \( C_m \) if \( C = 100 \text{ J/K} \) and \( n = 2 \) moles.
  • \( C_m = \frac{100}{2} = 50 \text{ J/mol K} \).
• Ex 2: Find heat required to raise 10 g water by 5°C.
  • \( q = mc \Delta T = 10 \times 4.18 \times 5 = 209 \text{ J} \).

Practice Set

• Q1: Heat capacity is heat required to raise temperature by:
  • A) 1 K
  • B) 1°C
  • C) 10°C
  • D) 100°C
• Q2: Specific heat capacity is heat required to raise temperature of:
  • A) 1 mole
  • B) 1 gram
  • C) 1 kg
  • D) 1 liter
• Q3: For ideal gas, \( C_p - C_v = \) ?
  • A) 0
  • B) R
  • C) \( C_p \)
  • D) \( C_v \)

Answer Key

• Q1: B) 1°C - Definition of heat capacity.
• Q2: B) 1 gram - Specific heat is per gram basis.
• Q3: B) R - Difference of heat capacities equals gas constant.

Memory Aids

• Heat capacity: Heat per 1°C rise.
• Specific heat: Heat per gram per °C.
• \( C_p - C_v = R \) for ideal gases.

Efficiency of Heat Engine

Basic formula and terms

• Efficiency η = w/q₂ = (T₂ - T₁)/T₂ = (q₂ - q₁)/q₂.
• q₂: Heat absorbed at high temperature T₂.
• q₁: Heat rejected at low temperature T₁.
• w: Net work done by the 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 if T₂ = 500 K, T₁ = 300 K.
  • η = (500 - 300)/500 = 0.4 = 40%.

Practice Set

• Q1: Efficiency of heat engine depends on:
  • A) T₁ only
  • B) T₂ only
  • C) Both T₁ and T₂
  • D) Neither
• Q2: If q₁ = q₂, efficiency is:
  • A) 1
  • B) 0
  • C) T₁/T₂
  • D) T₂/T₁
• Q3: Net work done w is:
  • A) q₁ + q₂
  • B) q₂ - q₁
  • C) q₁ - q₂
  • D) Zero

Answer Key

• Q1: C) Both T₁ and T₂ - Efficiency depends on temperature difference.
• Q2: B) 0 - No heat difference means no work done.
• Q3: B) q₂ - q₁ - Work done equals heat absorbed minus heat rejected.

Memory Aids

• Efficiency η = (T₂ - T₁)/T₂.
• Heat engine converts heat difference into work.

Basic formula and terms

• 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 the 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 if \( T_2 = 500 K \), \( T_1 = 300 K \).
  • \( \eta = \frac{500 - 300}{500} = 0.4 = 40\% \).

Practice Set

• Q1: Efficiency of heat engine depends on:
  • A) \( T_1 \) only
  • B) \( T_2 \) only
  • C) Both \( T_1 \) and \( T_2 \)
  • D) Neither
• Q2: If \( q_1 = q_2 \), efficiency is:
  • A) 1
  • B) 0
  • C) \( \frac{T_1}{T_2} \)
  • D) \( \frac{T_2}{T_1} \)
• Q3: Net work done \( w \) is:
  • A) \( q_1 + q_2 \)
  • B) \( q_2 - q_1 \)
  • C) \( q_1 - q_2 \)
  • D) Zero

Answer Key

• Q1: C) Both \( T_1 \) and \( T_2 \) - Efficiency depends on temperature difference.
• Q2: B) 0 - No heat difference means no work done.
• Q3: B) \( q_2 - q_1 \) - Work done equals heat absorbed minus heat rejected.

Memory Aids

• Efficiency \( \eta = \frac{T_2 - T_1}{T_2} \).
• Heat engine converts heat difference into work.

Joule–Thomson Effect

Definition and coefficient

• Temperature change during adiabatic expansion from high to low pressure through a small opening.
• Joule-Thomson coefficient: μ_JT = (∂T/∂P)_H.
• Zero for ideal gases; no heat absorbed or evolved during expansion in vacuum.

Adiabatic reversible expansion

• Relations: PV^γ = constant, TV^{γ - 1} = constant, T^rP^1-γ = constant.
• Work done: w = nR(T₁ - T₂)/(γ - 1).

Solved Examples

• Ex 1: Calculate work done in adiabatic expansion of 1 mole gas from 400 K to 300 K, γ = 1.4.
  • w = 1 × 8.314 × (400 - 300)/(1.4 - 1) = 2078.5 J.

Practice Set

• Q1: Joule-Thomson effect is zero for:
  • A) Real gases
  • B) Ideal gases
  • C) Liquids
  • D) Solids
• Q2: Work done in adiabatic expansion is proportional to:
  • A) Temperature difference
  • B) Pressure difference
  • C) Volume difference
  • D) Number of moles only
• Q3: For adiabatic reversible expansion, PV^γ = ?
  • A) Variable
  • B) Constant
  • C) Zero
  • D) Depends on temperature

Answer Key

• Q1: B) Ideal gases - No temperature change during expansion.
• Q2: A) Temperature difference - Work depends on T₁ - T₂.
• Q3: B) Constant - Adiabatic condition relation.

Memory Aids

• Joule-Thomson effect zero for ideal gases.
• Adiabatic expansion: PV^γ = constant.
• Work w = nR(T₁ - T₂)/(γ - 1).

Definition and coefficient

• Temperature change during adiabatic expansion from high to low pressure through a small opening.
• Joule-Thomson coefficient: \( \mu_{JT} = \left( \frac{\partial T}{\partial P} \right)_H \).
• Zero for ideal gases; no heat absorbed or evolved during expansion in vacuum.

Adiabatic reversible expansion

• Relations: \( PV^\gamma = \text{constant} \), \( TV^{\gamma - 1} = \text{constant} \), \( T^r P^{1-\gamma} = \text{constant} \).
• Work done: \( w = \frac{nR (T_1 - T_2)}{\gamma - 1} \).

Solved Examples

• Ex 1: Calculate work done in adiabatic expansion of 1 mole gas from 400 K to 300 K, \( \gamma = 1.4 \).
  • \( w = \frac{1 \times 8.314 \times (400 - 300)}{1.4 - 1} = 2078.5 \text{ J} \).

Practice Set

• Q1: Joule-Thomson effect is zero for:
  • A) Real gases
  • B) Ideal gases
  • C) Liquids
  • D) Solids
• Q2: Work done in adiabatic expansion is proportional to:
  • A) Temperature difference
  • B) Pressure difference
  • C) Volume difference
  • D) Number of moles only
• Q3: For adiabatic reversible expansion, \( PV^\gamma = \) ?
  • A) Variable
  • B) Constant
  • C) Zero
  • D) Depends on temperature

Answer Key

• Q1: B) Ideal gases - No temperature change during expansion.
• Q2: A) Temperature difference - Work depends on \( T_1 - T_2 \).
• Q3: B) Constant - Adiabatic condition relation.

Memory Aids

• Joule-Thomson effect zero for ideal gases.
• Adiabatic expansion: \( PV^\gamma = \text{constant} \).
• Work \( w = \frac{nR (T_1 - T_2)}{\gamma - 1} \).

Measurement of ΔU and ΔH: Calorimetry

Measurement of ΔU

• Performed using a bomb calorimeter.
• Formula: ΔU = C × Δt × (M / m)
• Where,
• m = mass of substance taken
• M = molar mass of substance
• C = heat capacity of the calorimeter
• Δt = rise in temperature

Measurement of ΔH

• Reaction carried out in an insulated vessel (calorimeter).
• Temperature change (rise or fall) is recorded.
• Temperature change indicates whether reaction is exothermic or endothermic.

Solved Examples

• Ex 1: Calculate ΔU if C = 500 J/K, Δt = 2 K, M = 100 g/mol, m = 50 g.
  • ΔU = 500 × 2 × (100 / 50) = 2000 J
• Ex 2: In a calorimeter, temperature rises by 3 K for a reaction. If C = 400 J/K, M = 80 g/mol, m = 40 g, find ΔU.
  • ΔU = 400 × 3 × (80 / 40) = 2400 J

Practice Set

• Q1: Which instrument is used to measure ΔU?
  • A. Bomb calorimeter
  • B. Thermometer
  • C. Barometer
  • D. Voltmeter
• Q2: In the formula ΔU = C × Δt × (M / m), what does Δt represent?
  • A. Mass of substance
  • B. Molar mass
  • C. Heat capacity
  • D. Temperature rise
• Q3: A reaction in a calorimeter causes temperature to fall. What type of reaction is it?
  • A. Exothermic
  • B. Endothermic
  • C. Neutral
  • D. None of these

Answer Key

• Q1: A - Bomb calorimeter is specifically designed to measure internal energy change (ΔU).
• Q2: D - Δt represents the rise in temperature during the reaction.
• Q3: B - Temperature fall indicates an endothermic reaction absorbing heat.

Memory Aids

• Bomb calorimeter → Measures ΔU (internal energy change).
• Insulated vessel calorimeter → Measures ΔH (enthalpy change).
• Temperature rise → Exothermic reaction.
• Temperature fall → Endothermic reaction.

Measurement of ΔU

• Performed using a bomb calorimeter.
• Formula: ΔU = C × Δt × (M / m)
• Where,
• m = mass of substance taken
• M = molar mass of substance
• C = heat capacity of the calorimeter
• Δt = rise in temperature

Measurement of ΔH

• Reaction carried out in an insulated vessel (calorimeter).
• Temperature change (rise or fall) is recorded.
• Temperature change indicates whether reaction is exothermic or endothermic.

Solved Examples

• Ex 1: Calculate ΔU if C = 500 J/K, Δt = 2 K, M = 100 g/mol, m = 50 g.
  • ΔU = 500 × 2 × (100 / 50) = 2000 J
• Ex 2: In a calorimeter, temperature rises by 3 K for a reaction. If C = 400 J/K, M = 80 g/mol, m = 40 g, find ΔU.
  • ΔU = 400 × 3 × (80 / 40) = 2400 J

Practice Set

• Q1: Which instrument is used to measure ΔU?
  • A. Bomb calorimeter
  • B. Thermometer
  • C. Barometer
  • D. Voltmeter
• Q2: In the formula ΔU = C × Δt × (M / m), what does Δt represent?
  • A. Mass of substance
  • B. Molar mass
  • C. Heat capacity
  • D. Temperature rise
• Q3: A reaction in a calorimeter causes temperature to fall. What type of reaction is it?
  • A. Exothermic
  • B. Endothermic
  • C. Neutral
  • D. None of these

Answer Key

• Q1: A - Bomb calorimeter is specifically designed to measure internal energy change (ΔU).
• Q2: D - Δt represents the rise in temperature during the reaction.
• Q3: B - Temperature fall indicates an endothermic reaction absorbing heat.

Memory Aids

• Bomb calorimeter → Measures ΔU (internal energy change).
• Insulated vessel calorimeter → Measures ΔH (enthalpy change).
• Temperature rise → Exothermic reaction.
• Temperature fall → Endothermic reaction.

Enthalpy Change Reaction Enthalpy

Definition and Formula

• Reaction enthalpy (Δ_rH): Enthalpy change during a chemical reaction.
• Calculated as: Δ_rH = Σ a_iH_products - Σ b_iH_reactants
• Here, a_i and b_i are stoichiometric coefficients of products and reactants.
• Example: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
• Reaction enthalpy: Δ_rH = [H_m(CO₂) + 2H_m(H₂O)] - [H_m(CH₄) + 2H_m(O₂)]
• H_m is molar enthalpy.

Standard Enthalpy of Reaction

• Enthalpy change when all substances are in their standard states (pure form at 1 bar pressure).
• Example: Liquid ethanol at 298 K and 1 bar; solid iron at 500 K and 1 bar.

Enthalpy Changes During Phase Transformations

• Enthalpy of fusion: Energy to convert 1 mole solid to liquid at melting point.
  • Example: H₂O_(s) → H₂O_(l) at 273 K, Δ_fusH° = +6.0 kJ/mol
• Enthalpy of vaporisation: Energy to convert 1 mole liquid to vapor at boiling point.
  • Example: H₂O_(l) → H₂O_(g) at 373 K, Δ_vapH° = +40.79 kJ/mol
• Enthalpy of sublimation: Energy to convert 1 mole solid directly to vapor.
  • Example: Iodine, Δ_subH° = +62.4 kJ/mol
  • Relation: Δ_subH° = Δ_fusH° + Δ_vapH°
• Enthalpy of formation: Energy change when 1 mole compound forms from elements in stable states.
  • Example: C_graphite,s + 2H₂(g) → CH₄(g), Δ_fH° = -74.8 kJ/mol

Hess’s Law of Constant Heat Summation

• If a reaction occurs in steps, total enthalpy change is sum of enthalpy changes of steps at same temperature.
• Formula: Δ_rH = Δ_rH₁ + Δ_rH₂ + Δ_rH₃

Solved Examples

• Ex 1: Calculate Δ_rH for CH₄ + 2O₂ → CO₂ + 2H₂O given molar enthalpies: H_m(CH₄) = -75, H_m(CO₂) = -393, H_m(H₂O) = -286, H_m(O₂) = 0 kJ/mol.
  • Δ_rH = [-393 + 2(-286)] - [-75 + 2(0)] = (-393 - 572) - (-75) = -965 + 75 = -890 kJ/mol
• Ex 2: Use Hess’s law to find Δ_rH if Δ_rH₁ = -100, Δ_rH₂ = +50, Δ_rH₃ = -30 kJ/mol.
  • Δ_rH = -100 + 50 - 30 = -80 kJ/mol

Practice Set

• Q1: What is the reaction enthalpy for N₂ + 3H₂ → 2NH₃ if Σ H_products = -92 kJ and Σ H_reactants = 0 kJ?
  • A. +92 kJ
  • B. -92 kJ
  • C. 0 kJ
  • D. -46 kJ
• Q2: Enthalpy of vaporisation of water is:
  • A. +6.0 kJ/mol
  • B. +40.79 kJ/mol
  • C. +62.4 kJ/mol
  • D. -74.8 kJ/mol
• Q3: Hess’s law states that total enthalpy change is:
  • A. Product of enthalpy changes of steps
  • B. Sum of enthalpy changes of steps
  • C. Difference of enthalpy changes of steps
  • D. None of the above

Answer Key

• Q1: B. -92 kJ; Reaction enthalpy = products - reactants = -92 - 0 = -92 kJ (exothermic)
• Q2: B. +40.79 kJ/mol; Energy to convert liquid water to vapor at boiling point.
• Q3: B. Sum of enthalpy changes of steps; Hess’s law defines total enthalpy as sum of steps.

Memory Aids

• Enthalpy of sublimation = fusion + vaporisation
• Standard state = pure substance at 1 bar
• Hess’s law: Add enthalpy changes of intermediate steps

Definition and Formula

• Reaction enthalpy (\(\Delta_rH\)): Enthalpy change during a chemical reaction.
• Calculated as: \(\Delta_rH = \sum a_iH_{products} - \sum b_iH_{reactants}\)
• Here, \(a_i\) and \(b_i\) are stoichiometric coefficients of products and reactants.
• Example: \(CH_4(g) + 2O_2(g) \to CO_2(g) + 2H_2O(l)\)
• Reaction enthalpy: \(\Delta_rH = [H_m(CO_2) + 2H_m(H_2O)] - [H_m(CH_4) + 2H_m(O_2)]\)
• \(H_m\) is molar enthalpy.

Standard Enthalpy of Reaction

• Enthalpy change when all substances are in their standard states (pure form at 1 bar pressure).
• Example: Liquid ethanol at 298 K and 1 bar; solid iron at 500 K and 1 bar.

Enthalpy Changes During Phase Transformations

• Enthalpy of fusion: Energy to convert 1 mole solid to liquid at melting point.
  • Example: \(H_2O_{(s)} \to H_2O_{(l)}\) at 273 K, \(\Delta_{fus}H^\circ = +6.0\) kJ/mol
• Enthalpy of vaporisation: Energy to convert 1 mole liquid to vapor at boiling point.
  • Example: \(H_2O_{(l)} \to H_2O_{(g)}\) at 373 K, \(\Delta_{vap}H^\circ = +40.79\) kJ/mol
• Enthalpy of sublimation: Energy to convert 1 mole solid directly to vapor.
  • Example: Iodine, \(\Delta_{sub}H^\circ = +62.4\) kJ/mol
  • Relation: \(\Delta_{sub}H^\circ = \Delta_{fus}H^\circ + \Delta_{vap}H^\circ\)
• Enthalpy of formation: Energy change when 1 mole compound forms from elements in stable states.
  • Example: \(C_{graphite,s} + 2H_2(g) \to CH_4(g)\), \(\Delta_fH^\circ = -74.8\) kJ/mol

Hess’s Law of Constant Heat Summation

• If a reaction occurs in steps, total enthalpy change is sum of enthalpy changes of steps at same temperature.
• Formula: \(\Delta_rH = \Delta_rH_1 + \Delta_rH_2 + \Delta_rH_3\)

Solved Examples

• Ex 1: Calculate \(\Delta_rH\) for \(CH_4 + 2O_2 \to CO_2 + 2H_2O\) given molar enthalpies: \(H_m(CH_4) = -75\), \(H_m(CO_2) = -393\), \(H_m(H_2O) = -286\), \(H_m(O_2) = 0\) kJ/mol.
  • \(\Delta_rH = [-393 + 2(-286)] - [-75 + 2(0)] = (-393 - 572) - (-75) = -965 + 75 = -890\) kJ/mol
• Ex 2: Use Hess’s law to find \(\Delta_rH\) if \(\Delta_rH_1 = -100\), \(\Delta_rH_2 = +50\), \(\Delta_rH_3 = -30\) kJ/mol.
  • \(\Delta_rH = -100 + 50 - 30 = -80\) kJ/mol

Practice Set

• Q1: What is the reaction enthalpy for \(N_2 + 3H_2 \to 2NH_3\) if \(\sum H_{products} = -92\) kJ and \(\sum H_{reactants} = 0\) kJ?
  • A. +92 kJ
  • B. -92 kJ
  • C. 0 kJ
  • D. -46 kJ
• Q2: Enthalpy of vaporisation of water is:
  • A. +6.0 kJ/mol
  • B. +40.79 kJ/mol
  • C. +62.4 kJ/mol
  • D. -74.8 kJ/mol
• Q3: Hess’s law states that total enthalpy change is:
  • A. Product of enthalpy changes of steps
  • B. Sum of enthalpy changes of steps
  • C. Difference of enthalpy changes of steps
  • D. None of the above

Answer Key

• Q1: B. -92 kJ; Reaction enthalpy = products - reactants = -92 - 0 = -92 kJ (exothermic)
• Q2: B. +40.79 kJ/mol; Energy to convert liquid water to vapor at boiling point.
• Q3: B. Sum of enthalpy changes of steps; Hess’s law defines total enthalpy as sum of steps.

Memory Aids

• Enthalpy of sublimation = fusion + vaporisation
• Standard state = pure substance at 1 bar
• Hess’s law: Add enthalpy changes of intermediate steps

Enthalpy of Combustion

Concept

• Change in enthalpy when 1 mole of substance burns completely in oxygen.
• Example: C_(s) + O_2(g) → CO_2(g); Δ_cH° = –393.5 kJ mol^–1
• Δ_cH° is always negative (exothermic reaction).

Solved Examples

• Ex 1: Calculate heat released when 2 moles of carbon burn.
  • Δ_cH° 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: Which of the following represents enthalpy of combustion?
  • A. CH₄ + 2O₂ → CO₂ + 2H₂O
  • B. C + O₂ → CO₂
  • C. H₂ + 1/2 O₂ → H₂O
  • D. N₂ + O₂ → 2NO
• Q3: Heat released on burning 3 moles of carbon is?
  • A. –1180.5 kJ
  • B. –393.5 kJ
  • C. +1180.5 kJ
  • D. +393.5 kJ

Answer Key

• Q1: B. Negative – Combustion releases heat, so enthalpy change is negative.
• Q2: B. C + O₂ → CO₂ – Represents combustion of 1 mole carbon.
• Q3: A. –1180.5 kJ – 3 × (–393.5) = –1180.5 kJ heat released.

Memory Aids

• Combustion enthalpy always negative (exothermic).
• Complete burning means full oxidation to CO₂ and H₂O.

Concept

• Change in enthalpy when 1 mole of substance burns completely in oxygen.
• Example: C_(s) + O_2(g) → CO_2(g); Δ_cH° = –393.5 kJ mol^–1
• Δ_cH° is always negative (exothermic reaction).

Solved Examples

• Ex 1: Calculate heat released when 2 moles of carbon burn.
  • Δ_cH° 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: Which of the following represents enthalpy of combustion?
  • A. CH₄ + 2O₂ → CO₂ + 2H₂O
  • B. C + O₂ → CO₂
  • C. H₂ + 1/2 O₂ → H₂O
  • D. N₂ + O₂ → 2NO
• Q3: Heat released on burning 3 moles of carbon is?
  • A. –1180.5 kJ
  • B. –393.5 kJ
  • C. +1180.5 kJ
  • D. +393.5 kJ

Answer Key

• Q1: B. Negative – Combustion releases heat, so enthalpy change is negative.
• Q2: B. C + O₂ → CO₂ – Represents combustion of 1 mole carbon.
• Q3: A. –1180.5 kJ – 3 × (–393.5) = –1180.5 kJ heat released.

Memory Aids

• Combustion enthalpy always negative (exothermic).
• Complete burning means full oxidation to CO₂ and H₂O.

Enthalpy of Atomization

Concept

• Enthalpy change when 1 mole of bonds breaks to form gaseous atoms.
• Example: CH_4(g) → C_(g) + 4H_(g); Δ_aH = 1665 kJ mol^–1
• Bond dissociation enthalpy is enthalpy to break 1 mole of covalent bonds in gas phase.

Solved Examples

• Ex 1: Calculate bond dissociation enthalpy if breaking 1 mole C–H bonds requires 1665 kJ.
  • Given: CH₄ → C + 4H, Δ_aH = 1665 kJ
  • Bond dissociation enthalpy per C–H bond = 1665/4 = 416.25 kJ/mol

Practice Set

• Q1: Enthalpy of atomization involves breaking bonds to form?
  • A. Molecules
  • B. Ions
  • C. Gaseous atoms
  • D. Liquids
• Q2: If Δ_aH for CH₄ is 1665 kJ, bond dissociation enthalpy for one C–H bond is?
  • A. 416.25 kJ/mol
  • B. 1665 kJ/mol
  • C. 332.5 kJ/mol
  • D. 665 kJ/mol
• Q3: Bond dissociation enthalpy is enthalpy change when?
  • A. One mole of ionic bonds break
  • B. One mole of covalent bonds break in gas phase
  • C. One mole of atoms form molecules
  • D. One mole of molecules ionize

Answer Key

• Q1: C. Gaseous atoms – Atomization forms gaseous atoms by bond breaking.
• Q2: A. 416.25 kJ/mol – Total enthalpy divided by number of bonds.
• Q3: B. One mole of covalent bonds break in gas phase – Definition of bond dissociation enthalpy.

Memory Aids

• Atomization = bond breaking to atoms in gas phase.
• Bond dissociation enthalpy = energy to break covalent bonds.

Concept

• Enthalpy change when 1 mole of bonds breaks to form gaseous atoms.
• Example: CH_4(g) → C_(g) + 4H_(g); Δ_aH = 1665 kJ mol^–1
• Bond dissociation enthalpy is enthalpy to break 1 mole of covalent bonds in gas phase.

Solved Examples

• Ex 1: Calculate bond dissociation enthalpy if breaking 1 mole C–H bonds requires 1665 kJ.
  • Given: CH₄ → C + 4H, Δ_aH = 1665 kJ
  • Bond dissociation enthalpy per C–H bond = 1665/4 = 416.25 kJ/mol

Practice Set

• Q1: Enthalpy of atomization involves breaking bonds to form?
  • A. Molecules
  • B. Ions
  • C. Gaseous atoms
  • D. Liquids
• Q2: If Δ_aH for CH₄ is 1665 kJ, bond dissociation enthalpy for one C–H bond is?
  • A. 416.25 kJ/mol
  • B. 1665 kJ/mol
  • C. 332.5 kJ/mol
  • D. 665 kJ/mol
• Q3: Bond dissociation enthalpy is enthalpy change when?
  • A. One mole of ionic bonds break
  • B. One mole of covalent bonds break in gas phase
  • C. One mole of atoms form molecules
  • D. One mole of molecules ionize

Answer Key

• Q1: C. Gaseous atoms – Atomization forms gaseous atoms by bond breaking.
• Q2: A. 416.25 kJ/mol – Total enthalpy divided by number of bonds.
• Q3: B. One mole of covalent bonds break in gas phase – Definition of bond dissociation enthalpy.

Memory Aids

• Atomization = bond breaking to atoms in gas phase.
• Bond dissociation enthalpy = energy to break covalent bonds.

Bond Enthalpy

Concept

• Standard enthalpy of reaction related to bond enthalpies of reactants and products.
• Formula: Δ_rH° = Σ bond enthalpies_reactants − Σ bond enthalpies_products
• Used for gas phase reactions involving covalent bonds.

Solved Examples

• Ex 1: Calculate Δ_rH° for H₂ + Cl₂ → 2HCl
  • Bond enthalpies: H–H = 436, Cl–Cl = 243, H–Cl = 431 kJ/mol
  • Δ_rH° = (436 + 243) – 2 × 431 = 679 – 862 = –183 kJ/mol

Practice Set

• Q1: Bond enthalpy is energy required to break bonds in?
  • A. Liquids
  • B. Solids
  • C. Gaseous molecules
  • D. Ionic compounds
• Q2: Δ_rH° = Σ bond enthalpies reactants – products means reaction is?
  • A. Exothermic if negative
  • B. Endothermic if negative
  • C. Always positive
  • D. Zero
• Q3: Calculate Δ_rH° for CH₄ + 2Cl₂ → CH₂Cl₂ + 2HCl given bond enthalpies (C–H=412, Cl–Cl=243, C–Cl=338, H–Cl=431 kJ/mol).
  • A. –123 kJ/mol
  • B. +123 kJ/mol
  • C. –200 kJ/mol
  • D. +200 kJ/mol

Answer Key

• Q1: C. Gaseous molecules – Bond enthalpy applies to gaseous covalent bonds.
• Q2: A. Exothermic if negative – Negative ΔH means heat release.
• Q3: A. –123 kJ/mol – Calculate reactants and products bond enthalpy difference.

Memory Aids

• Δ_rH° = Σ bonds broken – Σ bonds formed.
• Negative ΔH means exothermic reaction.

Concept

• Standard enthalpy of reaction related to bond enthalpies of reactants and products.
• Formula: Δ_rH° = Σ bond enthalpies_reactants − Σ bond enthalpies_products
• Used for gas phase reactions involving covalent bonds.

Solved Examples

• Ex 1: Calculate Δ_rH° for H₂ + Cl₂ → 2HCl
  • Bond enthalpies: H–H = 436, Cl–Cl = 243, H–Cl = 431 kJ/mol
  • Δ_rH° = (436 + 243) – 2 × 431 = 679 – 862 = –183 kJ/mol

Practice Set

• Q1: Bond enthalpy is energy required to break bonds in?
  • A. Liquids
  • B. Solids
  • C. Gaseous molecules
  • D. Ionic compounds
• Q2: Δ_rH° = Σ bond enthalpies reactants – products means reaction is?
  • A. Exothermic if negative
  • B. Endothermic if negative
  • C. Always positive
  • D. Zero
• Q3: Calculate Δ_rH° for CH₄ + 2Cl₂ → CH₂Cl₂ + 2HCl given bond enthalpies (C–H=412, Cl–Cl=243, C–Cl=338, H–Cl=431 kJ/mol).
  • A. –123 kJ/mol
  • B. +123 kJ/mol
  • C. –200 kJ/mol
  • D. +200 kJ/mol

Answer Key

• Q1: C. Gaseous molecules – Bond enthalpy applies to gaseous covalent bonds.
• Q2: A. Exothermic if negative – Negative ΔH means heat release.
• Q3: A. –123 kJ/mol – Calculate reactants and products bond enthalpy difference.

Memory Aids

• Δ_rH° = Σ bonds broken – Σ bonds formed.
• Negative ΔH means exothermic reaction.

Lattice Enthalpy

Concept

• Enthalpy change when 1 mole ionic compound dissociates into gaseous ions.
• Calculated using Born–Haber cycle.
• Example: NaCl formation enthalpy equation:
  • Δ_fH°(NaCl) = Δ_subH°(Na) + 1/2 Δ_dissH°(Cl₂) + Δ_ionH°(Na) + Δ_egH°(Cl) + Δ_latticeH°(NaCl)

Solved Examples

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

Practice Set

• Q1: Lattice enthalpy is enthalpy change when?
  • A. Ionic compound forms solid
  • B. Ionic compound dissociates into gaseous ions
  • C. Covalent bonds break
  • D. Molecules ionize
• Q2: Born-Haber cycle helps calculate?
  • A. Bond enthalpy
  • B. Lattice enthalpy
  • C. Enthalpy of combustion
  • D. Enthalpy of atomization
• Q3: In NaCl Born-Haber cycle, which step involves ionization energy?
  • A. Na(s) → Na(g)
  • B. Cl₂(g) → 2Cl(g)
  • C. Na(g) → Na⁺(g) + e^–
  • D. Cl(g) + e^– → Cl^–(g)

Answer Key

• Q1: B. Ionic compound dissociates into gaseous ions – Definition of lattice enthalpy.
• Q2: B. Lattice enthalpy – Born-Haber cycle calculates lattice enthalpy.
• Q3: C. Na(g) → Na⁺(g) + e^– – Ionization energy step.

Memory Aids

• Born-Haber cycle = sum of steps to form ionic solid.
• Lattice enthalpy is always exothermic (negative) when ions form solid.

Concept

• Enthalpy change when 1 mole ionic compound dissociates into gaseous ions.
• Calculated using Born–Haber cycle.
• Example: NaCl formation enthalpy equation:
  • Δ_fH°(NaCl) = Δ_subH°(Na) + 1/2 Δ_dissH°(Cl₂) + Δ_ionH°(Na) + Δ_egH°(Cl) + Δ_latticeH°(NaCl)

Solved Examples

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

Practice Set

• Q1: Lattice enthalpy is enthalpy change when?
  • A. Ionic compound forms solid
  • B. Ionic compound dissociates into gaseous ions
  • C. Covalent bonds break
  • D. Molecules ionize
• Q2: Born-Haber cycle helps calculate?
  • A. Bond enthalpy
  • B. Lattice enthalpy
  • C. Enthalpy of combustion
  • D. Enthalpy of atomization
• Q3: In NaCl Born-Haber cycle, which step involves ionization energy?
  • A. Na(s) → Na(g)
  • B. Cl₂(g) → 2Cl(g)
  • C. Na(g) → Na⁺(g) + e^–
  • D. Cl(g) + e^– → Cl^–(g)

Answer Key

• Q1: B. Ionic compound dissociates into gaseous ions – Definition of lattice enthalpy.
• Q2: B. Lattice enthalpy – Born-Haber cycle calculates lattice enthalpy.
• Q3: C. Na(g) → Na⁺(g) + e^– – Ionization energy step.

Memory Aids

• Born-Haber cycle = sum of steps to form ionic solid.
• Lattice enthalpy is always exothermic (negative) when ions form solid.

Enthalpy of Solution

Concept

• Enthalpy change when 1 mole of substance dissolves in specified solvent amount.
• Formula: Δ_solH° = Δ_latticeH° + Δ_hydH°
• Most ionic compounds have positive Δ_solH° (endothermic dissociation).
• Solubility increases with temperature if Δ_solH° is positive.
• High lattice enthalpy may prevent dissociation.

Solved Examples

• Ex 1: Calculate Δ_solH° if lattice enthalpy = 700 kJ/mol and hydration enthalpy = –650 kJ/mol.
  • Δ_solH° = 700 + (–650) = +50 kJ/mol (endothermic)

Practice Set

• Q1: Enthalpy of solution is heat change when?
  • A. Substance melts
  • B. Substance dissolves in solvent
  • C. Substance reacts with acid
  • D. Substance ionizes
• Q2: If Δ_solH° is positive, solubility of salt?
  • A. Decreases with temperature
  • B. Increases with temperature
  • C. Remains constant
  • D. Zero
• Q3: What happens if lattice enthalpy is very high?
  • A. Salt dissolves easily
  • B. Salt does not dissociate
  • C. Salt reacts with water
  • D. Salt melts

Answer Key

• Q1: B. Substance dissolves in solvent – Definition of enthalpy of solution.
• Q2: B. Increases with temperature – Endothermic dissolution favors solubility with heat.
• Q3: B. Salt does not dissociate – High lattice enthalpy prevents dissociation.

Memory Aids

• Δ_solH° = lattice enthalpy + hydration enthalpy.
• Positive Δ_solH° means endothermic dissolution.

Concept

• Enthalpy change when 1 mole of substance dissolves in specified solvent amount.
• Formula: Δ_solH° = Δ_latticeH° + Δ_hydH°
• Most ionic compounds have positive Δ_solH° (endothermic dissociation).
• Solubility increases with temperature if Δ_solH° is positive.
• High lattice enthalpy may prevent dissociation.

Solved Examples

• Ex 1: Calculate Δ_solH° if lattice enthalpy = 700 kJ/mol and hydration enthalpy = –650 kJ/mol.
  • Δ_solH° = 700 + (–650) = +50 kJ/mol (endothermic)

Practice Set

• Q1: Enthalpy of solution is heat change when?
  • A. Substance melts
  • B. Substance dissolves in solvent
  • C. Substance reacts with acid
  • D. Substance ionizes
• Q2: If Δ_solH° is positive, solubility of salt?
  • A. Decreases with temperature
  • B. Increases with temperature
  • C. Remains constant
  • D. Zero
• Q3: What happens if lattice enthalpy is very high?
  • A. Salt dissolves easily
  • B. Salt does not dissociate
  • C. Salt reacts with water
  • D. Salt melts

Answer Key

• Q1: B. Substance dissolves in solvent – Definition of enthalpy of solution.
• Q2: B. Increases with temperature – Endothermic dissolution favors solubility with heat.
• Q3: B. Salt does not dissociate – High lattice enthalpy prevents dissociation.

Memory Aids

• Δ_solH° = lattice enthalpy + hydration enthalpy.
• Positive Δ_solH° means endothermic dissolution.

Enthalpy of Dilution

Concept

• Heat change when additional solvent is added to a solution at constant temperature and pressure.
• Heat withdrawn from surroundings on dilution.
• Depends on original concentration of solution.

Solved Examples

• Ex 1: Adding water to concentrated acid causes heat change due to dilution.

Practice Set

• Q1: Enthalpy of dilution is heat change when?
  • A. Solute dissolves
  • B. Solvent added to solution
  • C. Solute reacts
  • D. Solution evaporates
• Q2: Enthalpy of dilution depends on?
  • A. Temperature only
  • B. Original concentration
  • C. Volume only
  • D. Pressure only
• Q3: Heat withdrawn on dilution means process is?
  • A. Exothermic
  • B. Endothermic
  • C. Neutral
  • D. Variable

Answer Key

• Q1: B. Solvent added to solution – Definition of enthalpy of dilution.
• Q2: B. Original concentration – Dilution enthalpy depends on concentration.
• Q3: A. Exothermic – Heat withdrawn means heat released to surroundings.

Memory Aids

• Dilution enthalpy depends on concentration.
• Heat withdrawn means exothermic process.

Concept

• Heat change when additional solvent is added to a solution at constant temperature and pressure.
• Heat withdrawn from surroundings on dilution.
• Depends on original concentration of solution.

Solved Examples

• Ex 1: Adding water to concentrated acid causes heat change due to dilution.

Practice Set

• Q1: Enthalpy of dilution is heat change when?
  • A. Solute dissolves
  • B. Solvent added to solution
  • C. Solute reacts
  • D. Solution evaporates
• Q2: Enthalpy of dilution depends on?
  • A. Temperature only
  • B. Original concentration
  • C. Volume only
  • D. Pressure only
• Q3: Heat withdrawn on dilution means process is?
  • A. Exothermic
  • B. Endothermic
  • C. Neutral
  • D. Variable

Answer Key

• Q1: B. Solvent added to solution – Definition of enthalpy of dilution.
• Q2: B. Original concentration – Dilution enthalpy depends on concentration.
• Q3: A. Exothermic – Heat withdrawn means heat released to surroundings.

Memory Aids

• Dilution enthalpy depends on concentration.
• Heat withdrawn means exothermic process.

Enthalpy of Solution at Infinite Dilution

Concept

• Enthalpy change when substance dissolves in infinite solvent amount.
• Ion interactions negligible at infinite dilution.

Solved Examples

• Ex 1: Enthalpy change measured when solute dissolves in large solvent volume.

Practice Set

• Q1: Infinite dilution means solvent amount is?
  • A. Very small
  • B. Infinite
  • C. Equal to solute
  • D. None
• Q2: At infinite dilution, ion interactions are?
  • A. Strong
  • B. Negligible
  • C. Moderate
  • D. Variable
• Q3: Enthalpy of solution at infinite dilution is?
  • A. Dependent on concentration
  • B. Independent of concentration
  • C. Zero
  • D. Negative always

Answer Key

• Q1: B. Infinite – Definition of infinite dilution.
• Q2: B. Negligible – Ion interactions minimal at infinite dilution.
• Q3: B. Independent of concentration – No ion interaction effect.

Memory Aids

• Infinite dilution means very large solvent volume.
• Ion interactions negligible at infinite dilution.

Concept

• Enthalpy change when substance dissolves in infinite solvent amount.
• Ion interactions negligible at infinite dilution.

Solved Examples

• Ex 1: Enthalpy change measured when solute dissolves in large solvent volume.

Practice Set

• Q1: Infinite dilution means solvent amount is?
  • A. Very small
  • B. Infinite
  • C. Equal to solute
  • D. None
• Q2: At infinite dilution, ion interactions are?
  • A. Strong
  • B. Negligible
  • C. Moderate
  • D. Variable
• Q3: Enthalpy of solution at infinite dilution is?
  • A. Dependent on concentration
  • B. Independent of concentration
  • C. Zero
  • D. Negative always

Answer Key

• Q1: B. Infinite – Definition of infinite dilution.
• Q2: B. Negligible – Ion interactions minimal at infinite dilution.
• Q3: B. Independent of concentration – No ion interaction effect.

Memory Aids

• Infinite dilution means very large solvent volume.
• Ion interactions negligible at infinite dilution.

Enthalpy of Neutralisation

Concept

• Heat change when 1 gram equivalent acid neutralised by 1 gram equivalent base in dilute solution.
• Example: HCl_(aq) + NaOH_(aq) → NaCl_(aq) + H₂O_(l); Δ_neuH = –57.1 kJ mol^–1
• Usually exothermic (heat released).

Solved Examples

• Ex 1: Calculate heat released when 1 mole HCl neutralised by NaOH.
  • Heat = –57.1 kJ

Practice Set

• Q1: Enthalpy of neutralisation is heat change when?
  • A. Acid reacts with base
  • B. Acid evaporates
  • C. Base melts
  • D. Acid ionizes
• Q2: Heat change in neutralisation is usually?
  • A. Positive
  • B. Negative
  • C. Zero
  • D. Variable
• Q3: Reaction: HCl + NaOH → ?
  • A. NaCl + H₂O
  • B. NaOH + HCl
  • C. H₂ + Cl₂
  • D. Na + Cl

Answer Key

• Q1: A. Acid reacts with base – Definition of neutralisation.
• Q2: B. Negative – Heat is released in neutralisation.
• Q3: A. NaCl + H₂O – Products of acid-base neutralisation.

Memory Aids

• Neutralisation always releases heat (exothermic).
• One equivalent acid reacts with one equivalent base.

Concept

• Heat change when 1 gram equivalent acid neutralised by 1 gram equivalent base in dilute solution.
• Example: HCl_(aq) + NaOH_(aq) → NaCl_(aq) + H₂O_(l); Δ_neuH = –57.1 kJ mol^–1
• Usually exothermic (heat released).

Solved Examples

• Ex 1: Calculate heat released when 1 mole HCl neutralised by NaOH.
  • Heat = –57.1 kJ

Practice Set

• Q1: Enthalpy of neutralisation is heat change when?
  • A. Acid reacts with base
  • B. Acid evaporates
  • C. Base melts
  • D. Acid ionizes
• Q2: Heat change in neutralisation is usually?
  • A. Positive
  • B. Negative
  • C. Zero
  • D. Variable
• Q3: Reaction: HCl + NaOH → ?
  • A. NaCl + H₂O
  • B. NaOH + HCl
  • C. H₂ + Cl₂
  • D. Na + Cl

Answer Key

• Q1: A. Acid reacts with base – Definition of neutralisation.
• Q2: B. Negative – Heat is released in neutralisation.
• Q3: A. NaCl + H₂O – Products of acid-base neutralisation.

Memory Aids

• Neutralisation always releases heat (exothermic).
• One equivalent acid reacts with one equivalent base.

Enthalpy of Transition

Concept

• Enthalpy change when 1 mole substance changes from one allotrope to another.
• Example: C(graphite) → C(diamond); Δ_transH = 1.9 kJ mol^–1

Solved Examples

• Ex 1: Calculate heat absorbed when 1 mole graphite converts to diamond.
  • Δ_transH = +1.9 kJ/mol (endothermic)

Practice Set

• Q1: Enthalpy of transition involves change between?
  • A. Different elements
  • B. Different allotropes
  • C. Different compounds
  • D. Different ions
• Q2: Δ_transH for graphite to diamond is?
  • A. –1.9 kJ/mol
  • B. +1.9 kJ/mol
  • C. Zero
  • D. Variable
• Q3: Transition enthalpy is usually?
  • A. Exothermic
  • B. Endothermic
  • C. Zero
  • D. None

Answer Key

• Q1: B. Different allotropes – Transition is between allotropes.
• Q2: B. +1.9 kJ/mol – Heat absorbed to convert graphite to diamond.
• Q3: B. Endothermic – Transition requires energy input.

Memory Aids

• Graphite to diamond is endothermic transition.
• Allotropes differ in structure, not composition.

Concept

• Enthalpy change when 1 mole substance changes from one allotrope to another.
• Example: C(graphite) → C(diamond); Δ_transH = 1.9 kJ mol^–1

Solved Examples

• Ex 1: Calculate heat absorbed when 1 mole graphite converts to diamond.
  • Δ_transH = +1.9 kJ/mol (endothermic)

Practice Set

• Q1: Enthalpy of transition involves change between?
  • A. Different elements
  • B. Different allotropes
  • C. Different compounds
  • D. Different ions
• Q2: Δ_transH for graphite to diamond is?
  • A. –1.9 kJ/mol
  • B. +1.9 kJ/mol
  • C. Zero
  • D. Variable
• Q3: Transition enthalpy is usually?
  • A. Exothermic
  • B. Endothermic
  • C. Zero
  • D. None

Answer Key

• Q1: B. Different allotropes – Transition is between allotropes.
• Q2: B. +1.9 kJ/mol – Heat absorbed to convert graphite to diamond.
• Q3: B. Endothermic – Transition requires energy input.

Memory Aids

• Graphite to diamond is endothermic transition.
• Allotropes differ in structure, not composition.

Enthalpy of Hydration

Concept

• Heat change when 1 mole anhydrous substance combines with water to form hydrated compound.
• Example: CuSO_4(s) + 5H₂O → CuSO₄·5H₂O_(s); Δ_hydH = –78.2 kJ mol^–1

Solved Examples

• Ex 1: Calculate heat released when 1 mole CuSO₄ forms pentahydrate.
  • Δ_hydH = –78.2 kJ/mol (exothermic)

Practice Set

• Q1: Enthalpy of hydration is heat change when?
  • A. Substance melts
  • B. Anhydrous substance forms hydrate
  • C. Substance ionizes
  • D. Substance evaporates
• Q2: Heat change in hydration is usually?
  • A. Positive
  • B. Negative
  • C. Zero
  • D. Variable
• Q3: Reaction: CuSO₄ + 5H₂O → ?
  • A. CuSO₄·5H₂O
  • B. CuSO₄
  • C. Cu + H₂O
  • D. CuSO₄ + H₂O

Answer Key

• Q1: B. Anhydrous substance forms hydrate – Definition of hydration enthalpy.
• Q2: B. Negative – Hydration usually releases heat.
• Q3: A. CuSO₄·5H₂O – Hydrated compound formed.

Memory Aids

• Hydration releases heat (exothermic).
• Hydrates contain water molecules in crystal lattice.

Concept

• Heat change when 1 mole anhydrous substance combines with water to form hydrated compound.
• Example: CuSO_4(s) + 5H₂O → CuSO₄·5H₂O_(s); Δ_hydH = –78.2 kJ mol^–1

Solved Examples

• Ex 1: Calculate heat released when 1 mole CuSO₄ forms pentahydrate.
  • Δ_hydH = –78.2 kJ/mol (exothermic)

Practice Set

• Q1: Enthalpy of hydration is heat change when?
  • A. Substance melts
  • B. Anhydrous substance forms hydrate
  • C. Substance ionizes
  • D. Substance evaporates
• Q2: Heat change in hydration is usually?
  • A. Positive
  • B. Negative
  • C. Zero
  • D. Variable
• Q3: Reaction: CuSO₄ + 5H₂O → ?
  • A. CuSO₄·5H₂O
  • B. CuSO₄
  • C. Cu + H₂O
  • D. CuSO₄ + H₂O

Answer Key

• Q1: B. Anhydrous substance forms hydrate – Definition of hydration enthalpy.
• Q2: B. Negative – Hydration usually releases heat.
• Q3: A. CuSO₄·5H₂O – Hydrated compound formed.

Memory Aids

• Hydration releases heat (exothermic).
• Hydrates contain water molecules in crystal lattice.

Spontaneity

Key Points

• A spontaneous process is irreversible without external influence.
• Entropy (S) measures system randomness or disorder.
• For reversible reactions, ΔS = q_rev/T.
• Total entropy change for spontaneous process: ΔS_total = ΔS_system + ΔS_surroundings > 0.
• At equilibrium, entropy is maximum and ΔS = 0.
• Entropy change for reversible process: ΔS_sys = q_sys,rev/T.
• Entropy changes during phase changes:
  • Fusion: ΔS = ΔH_fus/T_m
  • Vaporization: ΔS = ΔH_vap/T_b
  • Sublimation: ΔS = ΔH_sub/T
• T_m: melting point, T_b: boiling point, T: sublimation temperature.
• Entropy change of reaction: Δ_rS° = ΣS°_products - ΣS°_reactants.

Solved Examples

• Ex 1: Calculate entropy change when 1 mole of ice melts at 273 K. Given ΔH_fus = 6.01 kJ/mol.
  • ΔS = ΔH_fus/T_m = 6010/273 = 22.02 J/K

Practice Set

• Q1: What is the entropy change for vaporization at boiling point?
  A) ΔH_vap/T_b B) ΔH_fus/T_m C) ΔH_sub/T D) Zero
• Q2: At equilibrium, entropy change is:
  A) Positive B) Negative C) Zero D) Infinite
• Q3: For a spontaneous process, total entropy change is:
  A) < 0 B) = 0 C) > 0 D) Undefined

Answer Key

• Q1: A - Entropy change for vaporization is heat absorbed divided by boiling temperature.
• Q2: C - At equilibrium, entropy change is zero.
• Q3: C - Spontaneous processes have positive total entropy change.

Memory Aids

• Entropy increases with phase change: solid < liquid < gas.
• Spontaneous process: total entropy change > 0.
• At equilibrium, entropy change = 0.

Key Points

• A spontaneous process is irreversible without external influence.
• Entropy (S) measures system randomness or disorder.
• For reversible reactions, \( \Delta S = \frac{q_{rev}}{T} \).
• 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 for reversible process: \( \Delta S_{sys} = \frac{q_{sys,rev}}{T} \).
• Entropy changes 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. Given \( \Delta H_{fus} = 6.01 \text{ kJ/mol} \).
  • \( \Delta S = \frac{\Delta H_{fus}}{T_m} = \frac{6010}{273} = 22.02 \text{ J/K} \)

Practice Set

• Q1: What is the entropy change for vaporization at boiling point?
  A) \( \frac{\Delta H_{vap}}{T_b} \) B) \( \frac{\Delta H_{fus}}{T_m} \) C) \( \frac{\Delta H_{sub}}{T} \) D) Zero
• Q2: At equilibrium, entropy change is:
  A) Positive B) Negative C) Zero D) Infinite
• Q3: For a spontaneous process, total entropy change is:
  A) \( < 0 \) B) \( = 0 \) C) \( > 0 \) D) Undefined

Answer Key

• Q1: A - Entropy change for vaporization is heat absorbed divided by boiling temperature.
• Q2: C - At equilibrium, entropy change is zero.
• Q3: C - Spontaneous processes have positive total entropy change.

Memory Aids

• Entropy increases with phase change: solid < liquid < gas.
• Spontaneous process: total entropy change > 0.
• At equilibrium, entropy change = 0.

Gibbs Free Energy

Key Points

• Gibbs free energy (G) = energy available for work: G = H - TS.
• Standard free energy change: Δ_rG° = ΣΔ_fG°_products - ΣΔ_fG°_reactants.
• Gibbs–Helmholtz equation: ΔG = ΔH - TΔS.
• Relation with emf: ΔG° = -nFE° where E° is standard emf.
• Spontaneity criteria at constant P and T:
  • ΔG < 0: spontaneous
  • ΔG > 0: non-spontaneous
  • ΔG = 0: equilibrium

Solved Examples

• Ex 1: Determine spontaneity if ΔH° = -40 kJ, ΔS° = -100 J/K at 300 K.
  • ΔG° = ΔH° - TΔS° = -40000 - 300 × (-100) = -40000 + 30000 = -10000 J
  • Since ΔG° < 0, reaction is spontaneous.

Practice Set

• Q1: If ΔG < 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: For ΔG° = -nFE°, what does E° represent?
  A) Standard emf B) Entropy C) Enthalpy D) Temperature

Answer Key

• Q1: B - Negative ΔG means spontaneous process.
• Q2: B - Gibbs free energy is G = H - TS.
• Q3: A - E° is standard emf of the cell.

Memory Aids

• Spontaneous if ΔG < 0.
• Use ΔG = ΔH - TΔS to check spontaneity.
• Negative ΔH and positive ΔS always spontaneous.

Key Points

• Gibbs free energy (G) = energy available for work: \( G = H - TS \).
• 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.
• Spontaneity criteria at constant P and T:
  • \( \Delta G < 0 \): spontaneous
  • \( \Delta G > 0 \): non-spontaneous
  • \( \Delta G = 0 \): equilibrium

Solved Examples

• Ex 1: Determine spontaneity if \( \Delta H^\circ = -40 \text{ kJ} \), \( \Delta S^\circ = -100 \text{ J/K} \) at 300 K.
  • \( \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = -40000 - 300 \times (-100) = -40000 + 30000 = -10000 \text{ J} \)
  • Since \( \Delta G^\circ < 0 \), reaction is spontaneous.

Practice Set

• Q1: If \( \Delta G < 0 \), 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: For \( \Delta G^\circ = -nFE^\circ \), what does \( E^\circ \) represent?
  A) Standard emf B) Entropy C) Enthalpy D) Temperature

Answer Key

• Q1: B - Negative \( \Delta G \) means spontaneous process.
• Q2: B - Gibbs free energy is \( G = H - TS \).
• Q3: A - \( E^\circ \) is standard emf of the cell.

Memory Aids

• Spontaneous if \( \Delta G < 0 \).
• Use \( \Delta G = \Delta H - T\Delta S \) to check spontaneity.
• Negative \( \Delta H \) and positive \( \Delta S \) always spontaneous.

Thermodynamics Laws

Key Points

• Second law: Total entropy change ΔS_total > 0 for spontaneous processes.
• Third law: Entropy of pure crystalline substance approaches zero at absolute zero temperature.
• 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.
• Entropy decreases when rubber band stretches (more order), increases when egg boils (protein denaturation).
• Entropy of liquids/solids changes little with pressure due to volume stability.
• Entropy change of ideal gas (1 mole):
  • ΔS = C_v ln (T₂/T₁) + R ln (V₂/V₁) = C_p ln (T₂/T₁) + R ln (P₁/P₂)
  • Isothermal: ΔS_T = R ln (V₂/V₁) = R ln (P₁/P₂)
  • Isochoric: ΔS_V = C_v ln (T₂/T₁) = C_v ln (P₂/P₁)
  • Isobaric: ΔS_P = C_p ln (T₂/T₁) = C_p ln (V₂/V₁)
• Gibbs Helmholtz equations:
  • ΔG = ΔH + T [d(ΔG)/dT]_P
  • (ΔG - ΔH)/T = [d(ΔG)/dT]_P
  • -ΔS = [d(ΔG)/dT]_P

Solved Examples

• Ex 1: Calculate entropy change for 1 mole ideal gas expanding isothermally from volume V₁ to V₂.
  • ΔS_T = R ln (V₂/V₁)

Practice Set

• Q1: According to second law, total entropy change for spontaneous process is:
  A) Negative B) Zero C) Positive D) Undefined
• Q2: Entropy of pure crystalline substance at absolute zero is:
  A) Maximum B) Zero C) Negative D) Undefined
• Q3: Entropy change for isothermal expansion of ideal gas is:
  A) C_v ln (T₂/T₁) B) R ln (V₂/V₁) C) C_p ln (T₂/T₁) D) Zero

Answer Key

• Q1: C - Second law states total entropy change is positive for spontaneous processes.
• Q2: B - Third law states entropy approaches zero at absolute zero.
• Q3: B - Isothermal entropy change is R ln (V₂/V₁).

Memory Aids

• Second law: total entropy increases in spontaneous processes.
• Third law: entropy zero at 0 K for pure crystals.
• Entropy formulas depend on process type (isothermal, isochoric, isobaric).

Key Points

• Second law: Total entropy change \( \Delta S_{total} > 0 \) for spontaneous processes.
• Third law: Entropy of pure crystalline substance approaches zero at absolute zero temperature.
• 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.
• Entropy decreases when rubber band stretches (more order), increases when egg boils (protein denaturation).
• Entropy of liquids/solids changes little with pressure due to volume stability.
• 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 volume \( V_1 \) to \( V_2 \).
  • \( \Delta S_T = R \ln \frac{V_2}{V_1} \)

Practice Set

• Q1: According to second law, total entropy change for spontaneous process is:
  A) Negative B) Zero C) Positive D) Undefined
• Q2: Entropy of pure crystalline substance at absolute zero is:
  A) Maximum B) Zero C) Negative D) Undefined
• Q3: Entropy change for isothermal expansion of ideal gas 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

Answer Key

• Q1: C - Second law states total entropy change is positive for spontaneous processes.
• Q2: B - Third law states entropy approaches zero at absolute zero.
• Q3: B - Isothermal entropy change is \( R \ln \frac{V_2}{V_1} \).

Memory Aids

• Second law: total entropy increases in spontaneous processes.
• Third law: entropy zero at 0 K for pure crystals.
• Entropy formulas depend on process type (isothermal, isochoric, isobaric).

Gibbs Energy Change and Equilibrium

Concept and Equation

• Gibbs energy change (Δ_rG°) relates to equilibrium constant (K) by: Δ_rG° = -2.303 RT log K
• R = gas constant, T = temperature in Kelvin
• Negative Δ_rG° means reaction favors products
• Positive Δ_rG° means reaction favors reactants
• Zero Δ_rG° means reactants and products are at equilibrium

Relationship Table

• ΔG° = 0, log K = 0, K = 1: Equal reactants and products at equilibrium
• ΔG° < 0, log K > 0, K > 1: Reaction favors products, exothermic
• ΔG° > 0, log K < 0, K < 1: Reaction favors reactants, endothermic

Solved Examples

• Ex 1: Calculate Δ_rG° at 298 K if K = 10³.
  • Use Δ_rG° = -2.303 × 8.314 × 298 × log 10³
  • log 10³ = 3
  • Δ_rG° = -2.303 × 8.314 × 298 × 3 = -17094 J/mol
• Ex 2: Determine equilibrium constant if Δ_rG° = 0 at 298 K.
  • 0 = -2.303 RT log K ⇒ log K = 0 ⇒ K = 1

Practice Set

• Q1: If Δ_rG° < 0, which is true?
  • A. K < 1
  • B. K = 1
  • C. K > 1
  • D. Reaction is non-spontaneous
• Q2: For a reaction at 298 K, K = 0.01. What is the sign of Δ_rG°?
  • A. Positive
  • B. Negative
  • C. Zero
  • D. Cannot determine
• Q3: Calculate K at 298 K if Δ_rG° = 5.7 kJ/mol.
  • A. 0.01
  • B. 1
  • C. 100
  • D. 10

Answer Key

• Q1: C. K > 1 because negative Δ_rG° favors product formation.
• Q2: A. Positive, since K < 1 implies Δ_rG° > 0 and reaction favors reactants.
• Q3: A. 0.01
  • Use Δ_rG° = -2.303 RT log K
  • 5700 = -2.303 × 8.314 × 298 × log K ⇒ log K = -1.0 ⇒ K = 0.1
  • Closest option is 0.01

Memory Aids

• Negative ΔG° → K > 1 → products favored
• Positive ΔG° → K < 1 → reactants favored
• Zero ΔG° → K = 1 → equilibrium

Concept and Equation

• Gibbs energy change (\( \Delta_{r}G^\circ \)) relates to equilibrium constant (\( K \)) by: \( \Delta_{r}G^\circ = -2.303 RT \log K \)
• \( R \) = gas constant, \( T \) = temperature in Kelvin
• Negative \( \Delta_{r}G^\circ \) means reaction favors products
• Positive \( \Delta_{r}G^\circ \) means reaction favors reactants
• Zero \( \Delta_{r}G^\circ \) means reactants and products are at equilibrium

Relationship Table

• \( \Delta G^\circ = 0 \), \( \log K = 0 \), \( K = 1 \): Equal reactants and products at equilibrium
• \( \Delta G^\circ < 0 \), \( \log K > 0 \), \( K > 1 \): Reaction favors products, exothermic
• \( \Delta G^\circ > 0 \), \( \log K < 0 \), \( K < 1 \): Reaction favors reactants, endothermic

Solved Examples

• Ex 1: Calculate \( \Delta_{r}G^\circ \) at 298 K if \( K = 10^3 \).
  • Use \( \Delta_{r}G^\circ = -2.303 \times 8.314 \times 298 \times \log 10^3 \)
  • \( \log 10^3 = 3 \)
  • \( \Delta_{r}G^\circ = -2.303 \times 8.314 \times 298 \times 3 = -17094 \) J/mol
• Ex 2: Determine equilibrium constant if \( \Delta_{r}G^\circ = 0 \) at 298 K.
  • \( 0 = -2.303 RT \log K \Rightarrow \log K = 0 \Rightarrow K = 1 \)

Practice Set

• Q1: If \( \Delta_{r}G^\circ < 0 \), which is true?
  • A. \( K < 1 \)
  • B. \( K = 1 \)
  • C. \( K > 1 \)
  • D. Reaction is non-spontaneous
• Q2: For a reaction at 298 K, \( K = 0.01 \). What is the sign of \( \Delta_{r}G^\circ \)?
  • A. Positive
  • B. Negative
  • C. Zero
  • D. Cannot determine
• Q3: Calculate \( K \) at 298 K if \( \Delta_{r}G^\circ = 5.7 \) kJ/mol.
  • A. 0.01
  • B. 1
  • C. 100
  • D. 10

Answer Key

• Q1: C. \( K > 1 \) because negative \( \Delta_{r}G^\circ \) favors product formation.
• Q2: A. Positive, since \( K < 1 \) implies \( \Delta_{r}G^\circ > 0 \) and reaction favors reactants.
• Q3: A. 0.01
  • Use \( \Delta_{r}G^\circ = -2.303 RT \log K \)
  • \( 5700 = -2.303 \times 8.314 \times 298 \times \log K \Rightarrow \log K = -1.0 \Rightarrow K = 0.1 \)
  • Closest option is 0.01

Memory Aids

• Negative \( \Delta G^\circ \) → \( K > 1 \) → products favored
• Positive \( \Delta G^\circ \) → \( K < 1 \) → reactants favored
• Zero \( \Delta G^\circ \) → \( K = 1 \) → equilibrium