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WBB • Class XI • Chemistry • Ch 6
Estimated Time: 90 minutes
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Thermodynamics

Chemical thermodynamics is the quantitative science governing energy transformations, heat exchanges, work, and spontaneity across physical and chemical processes. Chapter 6 explores the fundamental laws and mathematical state functions of thermodynamics as mandated by the WBCHSE Class 11 Chemistry curriculum. Beginning with system classifications (open, closed, isolated), state variables, and thermodynamic processes (isothermal, adiabatic, isobaric, isochoric, and cyclic), the chapter develops the First Law of Thermodynamics, establishing internal energy U as a state function and formulating expansion work under reversible and irreversible regimes. The concept of enthalpy H is introduced to quantify heat exchange at constant pressure, deriving the vital relationship connecting delta H and delta U in gaseous reactions along with Mayer relation for molar heat capacities (Cp minus Cv equals R). Thermochemistry applies these foundations to chemical reactions, employing standard enthalpies of formation, combustion, atomization, and bond dissociation alongside Hess law of constant heat summation. The chapter then addresses the directionality of natural phenomena through the Second Law of Thermodynamics, defining entropy S as a measure of molecular disorder and thermal dispersal. Finally, Josiah Willard Gibbs free energy G provides the universal criterion for chemical spontaneity and thermodynamic equilibrium at constant temperature and pressure, culminating in the Nernst Heat Theorem and the Third Law of Thermodynamics.

Why This Chapter Matters

Thermodynamics is the master framework that determines whether a chemical reaction can occur, how much useful work it can generate, and where its chemical equilibrium will lie. In industrial chemical synthesis, such as the production of methanol, sulfuric acid by the contact process, or ammonia by the Haber process, calculating enthalpy changes dictates cooling jacket requirements, while Gibbs free energy optimization determines operating temperatures and equilibrium yields. In biochemistry and molecular biology, the hydrolysis of adenosine triphosphate (ATP) couples an exergonic reaction (negative delta G) to power non-spontaneous cellular work, muscle contraction, and active transport across cell membranes. In metallurgical extraction, Ellingham diagrams plot standard free energies of oxide formation against temperature to select the most cost-effective reducing agent for smelting iron, zinc, or titanium. In sustainable energy and electrochemistry, free energy changes directly dictate the maximum electrical voltage delivered by fuel cells and lithium-ion batteries. For WBCHSE Class 11, WBJEE, JEE Main, and NEET aspirants, this chapter represents the supreme conceptual bridge connecting classical physics with reaction energetics and chemical equilibrium.

Before You Begin (Prerequisites)

  • Basic concepts of heat, temperature, kinetic energy, work, and conservation of mechanical energy from physics.
  • Stoichiometry, balanced chemical equations, and the mole concept.
  • Ideal gas equation (PV = nRT) and combined gas laws from the states of matter unit.
  • Basic logarithmic calculations (natural logarithms and base-10 logarithms).

Chapter Roadmap & Progression

1 Module 1: Basic Thermodynamic Conce...
2 Module 2: First Law of Thermodynami...
3 Module 3: Enthalpy ($H$), Heat Capa...
4 Module 4: Thermochemistry, Standard...
5 Module 5: Second Law of Thermodynam...
6 Module 6: Gibbs Free Energy ($G$),...

Complete Concept Guide (100% Curriculum Coverage)

Module 1: Basic Thermodynamic Concepts, Systems & Processes

1.1 System, Surroundings and Boundaries

In thermodynamics, the universe is partitioned into two distinct conceptual domains:

  • System: The specific portion of the physical universe selected for thermodynamic study or observation (e.g., a reaction mixture inside a beaker or a gas confined in a cylinder with a piston).
  • Surroundings: The rest of the universe outside the system that can interact with it by exchanging energy (heat/work) or matter. In practice, only the immediate environment affected by the system constitutes the operational surroundings.
  • Boundary: The real or imaginary surface separating the system from its surroundings. Boundaries may be:
    • Diathermic: Conducting boundary permitting heat transfer between system and surroundings.
    • Adiabatic: Thermally insulating boundary preventing all heat transfer ($q = 0$).
    • Rigid vs. Movable: Rigid boundaries prevent volume changes ($w = 0$), whereas movable boundaries allow mechanical expansion or compression work.

Classification of Thermodynamic Systems:

System Type Matter Exchange Energy Exchange Real-World Example
Open System Yes (Permitted) Yes (Permitted) Hot water boiling in an open beaker; living biological cells; human body.
Closed System No (Prevented) Yes (Permitted) Water enclosed in a sealed metallic flask or glass bulb; gas in a cylinder with sealed piston.
Isolated System No (Prevented) No (Prevented) Hot liquid contained in an ideally insulated Dewar (thermos) flask; the entire universe itself.
1.2 State Variables, State Functions & Path Functions

The thermodynamic state of a macroscopic system is uniquely defined by specifying its macroscopic state variables ($P, V, T, n$):

  • State Functions (State Variables): Properties whose numerical values depend solely on the current thermodynamic state of the system, completely independent of the path or method used to reach that state. Examples: Pressure ($P$), Volume ($V$), Temperature ($T$), Internal Energy ($U$), Enthalpy ($H$), Entropy ($S$), Gibbs Free Energy ($G$). $$\oint dX = 0 \quad (\text{Cyclic integral of any state function is identically zero})$$ The change in a state function depends only on initial and final states: $\Delta X = X_{\text{final}} - X_{\text{initial}}$.
  • Path Functions: Quantities whose values depend explicitly on the thermodynamic path or mechanism traversed during the transformation between states. Examples: Work ($w$) and Heat ($q$). They are inexact differentials ($\delta w, \delta q$), and their cyclic integrals are generally non-zero ($\oint \delta w \neq 0$).
1.3 Extensive vs. Intensive Properties
  • Extensive Properties: Properties that depend directly on the mass, volume, or quantity of matter present in the system. Examples: Mass ($m$), Volume ($V$), Total Internal Energy ($U$), Enthalpy ($H$), Entropy ($S$), Gibbs Free Energy ($G$), Heat Capacity ($C$).
    Additive Rule: If a system is divided into two halves, each half possesses half the original value of an extensive property.
  • Intensive Properties: Properties that are entirely independent of the amount or size of matter present in the system. Examples: Temperature ($T$), Pressure ($P$), Density ($d$), Molar Volume ($V_m$), Specific Heat Capacity ($c$), Molar Enthalpy, Refractive Index, Surface Tension, Viscosity, Electromotive Force ($E^\circ$).
    Ratio Rule: The ratio of two extensive properties is always an intensive property (e.g., $\text{Density} = \frac{\text{Mass}}{\text{Volume}}$, $\text{Molar Volume} = \frac{V}{n}$).
1.4 Thermodynamic Processes & Reversibility

A thermodynamic process occurs when a system transitions from one equilibrium state to another:

  • Isothermal Process: Temperature remains strictly constant throughout ($\Delta T = 0$, $dT = 0$). Diathermic walls permit heat flow. For an ideal gas, $\Delta U = 0$ and $\Delta H = 0$.
  • Adiabatic Process: System is completely insulated from surroundings so no heat is exchanged ($q = 0$, $\delta q = 0$). Expansion causes cooling; compression causes heating.
  • Isobaric Process: Pressure remains constant throughout ($\Delta P = 0$, $dP = 0$). Most open bench chemistry reactions are isobaric ($P = 1 \; \text{atm}$).
  • Isochoric Process: Volume remains constant throughout ($\Delta V = 0$, $dV = 0$). No expansion work is possible ($w = 0$). Reactions in rigid bomb calorimeters are isochoric.
  • Cyclic Process: A sequence of transformations that ultimately returns the system to its precise initial state. For any cyclic process: $\Delta U_{\text{cycle}} = 0, \; \Delta H_{\text{cycle}} = 0, \; \Delta S_{\text{cycle}} = 0$.
Reversible vs. Irreversible Processes:
A Reversible Process is an idealized quasistatic process carried out infinitely slowly through an unbroken succession of thermodynamic equilibrium states, where the driving force exceeds the opposing force by only an infinitesimal amount ($dP \to 0$). It produces the maximum possible work during expansion. An Irreversible (Spontaneous) Process occurs rapidly under finite driving force differences, accompanied by dissipative friction; it cannot be reversed without leaving permanent changes in the surroundings. All natural spontaneous processes are irreversible.

Module 2: First Law of Thermodynamics, Work, Heat & Internal Energy

2.1 Internal Energy ($U$) and Heat ($q$)

Internal Energy ($U$): The total microscopic energy possessed by a thermodynamic system, encompassing molecular translational, rotational, and vibrational kinetic energies, alongside electronic, nuclear, and intermolecular potential energies. $$U = E_{\text{trans}} + E_{\text{rot}} + E_{\text{vib}} + E_{\text{electronic}} + E_{\text{bonding}} + E_{\text{intermolecular}}$$ The absolute value of $U$ cannot be measured directly, but the change $\Delta U = U_2 - U_1$ is an experimentally measurable state function.

Heat ($q$): Energy transferred across a boundary solely as a result of a temperature difference between the system and surroundings. Heat flows spontaneously from higher to lower temperature until thermal equilibrium is attained.

2.2 Derivation of Pressure-Volume ($P-V$) Work

Consider a frictionless, weightless piston of cross-sectional area $A$ confining a gas inside a cylinder. The gas expands against an external pressure $P_{\text{ext}}$ through an infinitesimal displacement $dl$:

$$\text{Opposing Force: } F_{\text{ext}} = P_{\text{ext}} \times A$$ $$\text{Infinitesimal Work: } \delta w = -F_{\text{ext}} \times dl = -P_{\text{ext}} \times (A \times dl) = -P_{\text{ext}} dV$$

Integrating from initial volume $V_1$ to final volume $V_2$ gives the general expression for mechanical expansion work:

$$w = -\int_{V_1}^{V_2} P_{\text{ext}} dV$$

Special Cases of Work:

  1. Irreversible Expansion against Constant External Pressure: $$w = -P_{\text{ext}} \int_{V_1}^{V_2} dV = -P_{\text{ext}} (V_2 - V_1) = -P_{\text{ext}} \Delta V$$
  2. Free Expansion into Vacuum: $$P_{\text{ext}} = 0 \implies w = 0 \quad (\text{No work is done during free expansion, whether reversible or irreversible})$$
  3. Isothermal Reversible Expansion of an Ideal Gas:

    In a reversible expansion, the internal gas pressure balances external pressure at every infinitesimal step ($P_{\text{ext}} = P_{\text{gas}} - dP \approx P_{\text{gas}}$). From the ideal gas law, $P = \frac{nRT}{V}$:

    $$w_{\text{rev}} = -\int_{V_1}^{V_2} \frac{nRT}{V} dV = -nRT \ln\left(\frac{V_2}{V_1}\right) = -2.303 nRT \log_{10}\left(\frac{V_2}{V_1}\right)$$

    Since temperature is constant, Boyle's Law dictates $P_1 V_1 = P_2 V_2 \implies \frac{V_2}{V_1} = \frac{P_1}{P_2}$:

    $$w_{\text{rev}} = -2.303 nRT \log_{10}\left(\frac{P_1}{P_2}\right)$$

    Note: For expansion ($V_2 > V_1$), $\log_{10}(V_2/V_1) > 0$, so $w < 0$ (work done BY system). For compression ($V_2 < V_1$), $w > 0$ (work done ON system). Furthermore, $|w_{\text{rev}}| > |w_{\text{irrev}}|$ for expansion.

2.3 The First Law of Thermodynamics

The First Law is the universal thermodynamic statement of the Law of Conservation of Energy: Energy can neither be created nor destroyed, although it can be transformed from one form into another. The total energy of an isolated system remains constant.

Mathematical Formulation of First Law: $$\Delta U = q + w$$ $$dU = \delta q + \delta w = \delta q - P_{\text{ext}} dV$$

IUPAC Sign Conventions (Mandatory for Chemistry):

  • Heat absorbed by the system from surroundings: $q > 0$ ($+q$, endothermic).
  • Heat released by the system to surroundings: $q < 0$ ($-q$, exothermic).
  • Work done ON the system by surroundings (compression): $w > 0$ ($+w$).
  • Work done BY the system on surroundings (expansion): $w < 0$ ($-w$).
2.4 Heat at Constant Volume ($q_v$)

For a closed system undergoing a process at constant volume ($\Delta V = 0$):

$$w = -P_{\text{ext}} \Delta V = 0 \implies \Delta U = q_v$$

This demonstrates that the heat exchanged by a system at constant volume ($q_v$) is exactly equal to the change in its internal energy $\Delta U$. Because $U$ is a state function, $q_v$ becomes independent of path for an isochoric transformation.

Module 3: Enthalpy ($H$), Heat Capacities ($C_p, C_v$) & Mayer's Relation

3.1 Definition and Physical Meaning of Enthalpy ($H$)

Most chemical laboratory and industrial processes occur under constant atmospheric pressure ($P_{\text{ext}} = P = \text{constant}$) rather than constant volume. From the First Law:

$$\Delta U = q_p + w = q_p - P \Delta V \implies q_p = \Delta U + P \Delta V = (U_2 - U_1) + P(V_2 - V_1)$$ $$q_p = (U_2 + P V_2) - (U_1 + P V_1)$$

To quantify this constant-pressure heat exchange, the thermodynamic state function Enthalpy ($H$), also called heat content, is defined:

$$H = U + PV$$ $$\Delta H = H_2 - H_1 = q_p$$

The change in enthalpy $\Delta H$ is precisely equal to the heat absorbed or evolved by the system at constant pressure ($q_p$). Because $U, P,$ and $V$ are all state functions, Enthalpy $H$ is strictly a state function and an extensive property.

3.2 Relationship Between $\Delta H$ and $\Delta U$ in Gaseous Reactions

For any process involving ideal gases, $PV = nRT$. Substituting into $\Delta H = \Delta U + \Delta(PV)$ at constant temperature $T$:

$$\Delta(PV) = \Delta(n_g RT) = (\Delta n_g) RT$$
$$\Delta H = \Delta U + \Delta n_g RT$$

where $\Delta n_g$ is the stoichiometric difference between moles of gaseous products and gaseous reactants:

$$\Delta n_g = \sum n_g(\text{products}) - \sum n_g(\text{reactants})$$

Physical Implications:

  • If $\Delta n_g = 0$ (e.g., $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$): $\Delta H = \Delta U$.
  • If $\Delta n_g > 0$ (e.g., $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$): $\Delta H > \Delta U$ (expansion work is performed against the atmosphere).
  • If $\Delta n_g < 0$ (e.g., $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$): $\Delta H < \Delta U$ (compression work is done by the atmosphere on the system).
  • For reactions involving only liquids and solids: $\Delta V \approx 0 \implies \Delta H \approx \Delta U$.
3.3 Heat Capacities: $C_p$, $C_v$ and Mayer's Relation

Heat Capacity ($C$): The quantity of heat required to raise the temperature of a given mass of substance by one degree ($1 \; \text{K}$ or $1^\circ\text{C}$): $C = \frac{q}{\Delta T} = \frac{\delta q}{dT}$.

  • Molar Heat Capacity at Constant Volume ($C_v$): $$C_v = \left(\frac{\partial U}{\partial T}\right)_v \implies dU = n C_v dT \implies \Delta U = n C_v \Delta T$$
  • Molar Heat Capacity at Constant Pressure ($C_p$): $$C_p = \left(\frac{\partial H}{\partial T}\right)_p \implies dH = n C_p dT \implies \Delta H = n C_p \Delta T$$

Derivation of Mayer's Relation ($C_p - C_v = R$):

For 1 mole of an ideal gas: $H = U + PV = U + RT$. Differentiating with respect to temperature $T$:

$$\frac{dH}{dT} = \frac{dU}{dT} + R \implies C_p = C_v + R \implies C_p - C_v = R$$

Physical Rationale for $C_p > C_v$: At constant volume, all supplied heat is utilized exclusively to raise molecular kinetic energy (increase temperature). At constant pressure, the gas expands against the atmosphere; hence, supplied heat must not only raise internal kinetic energy but also supply external expansion work ($P \Delta V = R \Delta T$).

Heat Capacity Ratio ($\gamma = C_p / C_v$):

  • Monoatomic gas ($\text{He, Ne, Ar}$): $C_v = \frac{3}{2}R, \; C_p = \frac{5}{2}R \implies \gamma = \frac{5}{3} \approx 1.67$.
  • Diatomic gas ($\text{N}_2, \text{O}_2, \text{HCl}$ at room temp): $C_v = \frac{5}{2}R, \; C_p = \frac{7}{2}R \implies \gamma = \frac{7}{5} = 1.40$.
  • Triatomic / Polyatomic non-linear gas ($\text{H}_2\text{O, SO}_2$): $C_v = 3R, \; C_p = 4R \implies \gamma = \frac{4}{3} \approx 1.33$.
3.4 Adiabatic Reversible Expansion of an Ideal Gas

In an adiabatic process, $q = 0$. From the First Law, $\Delta U = w_{\text{ad}}$. Therefore:

$$n C_v dT = -P dV = -\frac{nRT}{V} dV \implies \frac{dT}{T} = -\frac{R}{C_v} \frac{dV}{V} = -(\gamma - 1) \frac{dV}{V}$$

Integrating yields the Poisson relations for reversible adiabatic processes:

$$P V^\gamma = \text{constant}, \quad T V^{\gamma - 1} = \text{constant}, \quad T^\gamma P^{1 - \gamma} = \text{constant}$$ $$w_{\text{adiabatic}} = \Delta U = n C_v (T_2 - T_1) = \frac{n R (T_2 - T_1)}{\gamma - 1} = \frac{P_2 V_2 - P_1 V_1}{\gamma - 1}$$

Module 4: Thermochemistry, Standard Enthalpies, Hess's Law & Bond Enthalpies

4.1 Standard States and Types of Enthalpy Changes

Thermochemistry deals with the heat energy changes accompanying chemical reactions and phase transformations. The Standard State of a substance at a specified temperature (usually $298.15 \; \text{K} = 25^\circ\text{C}$) is its purest and most stable physical form at a pressure of exactly $1 \; \text{bar}$ ($10^5 \; \text{Pa}$).

  • Standard Enthalpy of Formation ($\Delta_f H^\circ$): The enthalpy change accompanying the formation of 1 mole of a compound directly from its constituent elements in their standard reference states.
    By IUPAC Convention: The standard enthalpy of formation of every pure element in its most stable reference physical state is arbitrarily assigned as zero (e.g., $\Delta_f H^\circ[\text{C}_{\text{graphite}}] = 0$, $\Delta_f H^\circ[\text{O}_2(g)] = 0$, $\Delta_f H^\circ[\text{S}_{\text{rhombic}}] = 0$, $\Delta_f H^\circ[\text{P}_{\text{white}}] = 0$, $\Delta_f H^\circ[\text{Br}_2(l)] = 0$). $$\Delta_r H^\circ = \sum \nu_p \Delta_f H^\circ(\text{products}) - \sum \nu_r \Delta_f H^\circ(\text{reactants})$$
  • Standard Enthalpy of Combustion ($\Delta_c H^\circ$): The enthalpy change occurring when 1 mole of a substance undergoes complete oxidation/combustion in excess oxygen under standard conditions (always exothermic, $\Delta_c H^\circ < 0$). $$\Delta_r H^\circ = \sum \nu_r \Delta_c H^\circ(\text{reactants}) - \sum \nu_p \Delta_c H^\circ(\text{products})$$
  • Enthalpy of Neutralization: The heat liberated when 1 gram-equivalent of an acid is completely neutralized by 1 gram-equivalent of a base in dilute aqueous solution. For any strong acid and strong base: $$\text{H}^+(aq) + \text{OH}^-(aq) \longrightarrow \text{H}_2\text{O}(l) \quad \Delta_{\text{neut}} H^\circ = -57.1 \; \text{kJ/mol} \; (-13.7 \; \text{kcal/mol})$$ For weak acids or weak bases (e.g., $\text{CH}_3\text{COOH}$ or $\text{NH}_4\text{OH}$), $|\Delta_{\text{neut}} H^\circ| < 57.1 \; \text{kJ}$ because part of the evolved heat is consumed in the endothermic ionization of the weak electrolyte.
4.2 Hess's Law of Constant Heat Summation

Formulated empirically by Germain Hess in 1840, this law is a direct manifestation of the First Law of Thermodynamics and the state function nature of enthalpy:

Hess's Law Statement: The total enthalpy change for a chemical reaction is identical whether the reaction takes place in a single step or in a series of intermediate steps. $$\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 + \dots = \sum_{i=1}^m \Delta H_i$$

Applications: Determination of standard enthalpies of formation for compounds that cannot be synthesized directly from elements (e.g., $\text{CO}, \text{CH}_4, \text{C}_2\text{H}_2$), calculation of lattice enthalpies via the Born-Haber cycle, and determination of resonance energies.

4.3 Bond Dissociation Enthalpy and Reaction Enthalpies
  • Bond Dissociation Enthalpy: The enthalpy required to break 1 mole of a specific covalent bond in a gaseous molecule into isolated gaseous atoms or radicals (e.g., $\text{H}_2(g) \to 2\text{H}(g), \; \Delta_{\text{bond}} H = +435.8 \; \text{kJ/mol}$).
  • Mean (Average) Bond Enthalpy: In polyatomic molecules like methane ($\text{CH}_4$), breaking successive $\text{C-H}$ bonds requires different energies due to changing electronic environments. The mean bond enthalpy is the average: $$\epsilon(\text{C-H}) = \frac{1}{4} \Delta_a H^\circ[\text{CH}_4(g)] = \frac{1665}{4} = 416.25 \; \text{kJ/mol}$$
  • Calculating Reaction Enthalpy from Bond Enthalpies:
    $$\Delta_r H^\circ = \sum \text{B.E.}(\text{reactants broken}) - \sum \text{B.E.}(\text{products formed})$$
    Critical Examination Warning: Notice that for bond energies, reactants come FIRST (energy absorbed to break bonds minus energy released when new bonds form). Do not confuse this with enthalpies of formation where products come first!
4.4 Temperature Dependence of Reaction Enthalpy: Kirchhoff's Equations

The variation of reaction enthalpy with temperature at constant pressure is governed by Kirchhoff's Law:

$$\left(\frac{\partial \Delta_r H}{\partial T}\right)_p = \Delta C_p \implies \Delta_r H_{T_2} - \Delta_r H_{T_1} = \int_{T_1}^{T_2} \Delta C_p dT = \Delta C_p (T_2 - T_1)$$

where $\Delta C_p = \sum \nu_p C_p(\text{products}) - \sum \nu_r C_p(\text{reactants})$. Similarly, at constant volume: $\Delta_r U_{T_2} - \Delta_r U_{T_1} = \Delta C_v (T_2 - T_1)$.

Module 5: Second Law of Thermodynamics, Spontaneity & Entropy ($S$)

5.1 Limitations of the First Law & Need for the Second Law

The First Law of Thermodynamics affirms the conservation of energy during any process, but it fails completely to answer two fundamental questions:

  1. Why do natural processes proceed spontaneously in one direction only (e.g., heat flows from hot to cold, gas expands into vacuum, water flows downhill)?
  2. Is a decrease in enthalpy (exothermicity, $\Delta H < 0$) a sufficient criterion for spontaneity? No! Many endothermic processes proceed spontaneously at room temperature, such as the dissolution of ammonium nitrate in water ($\Delta H > 0$), evaporation of water, and the melting of ice above $0^\circ\text{C}$. Conversely, many exothermic reactions do not proceed without external initiation.
5.2 Concept of Entropy ($S$) and Molecular Disorder

Rudolf Clausius (1865) introduced the thermodynamic state function Entropy ($S$) as a quantitative measure of the degree of randomness, molecular disorder, or spatial dispersal of matter and thermal energy in a system.

Mathematical Definition of Entropy: $$dS = \frac{\delta q_{\text{rev}}}{T}$$ $$\Delta S = S_2 - S_1 = \int_1^2 \frac{\delta q_{\text{rev}}}{T}$$

where $\delta q_{\text{rev}}$ is the heat exchanged reversibly at absolute thermodynamic temperature $T$. The SI unit of entropy is $\text{J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}$.

Physical Characteristics of Entropy:

  • Entropy is a state function and an extensive property.
  • Entropy increases upon phase transitions from ordered to disordered states: $$S_{\text{solid}} < S_{\text{liquid}} \ll S_{\text{gas}}$$
  • Entropy increases when the number of gaseous molecules increases during a chemical reaction ($\Delta n_g > 0$).
  • Entropy increases with increasing temperature, expansion of volume, and dissolution of crystalline solutes in solvents.
5.3 The Second Law of Thermodynamics & Universal Spontaneity Criterion

The Second Law can be stated in several equivalent formulations:

  • Clausius Statement: It is impossible to construct a cyclic machine whose sole effect is the transfer of heat from a cooler body to a hotter body without external work.
  • Kelvin-Planck Statement: It is impossible to construct a heat engine operating in a cycle that absorbs heat from a reservoir and converts it completely into equivalent mechanical work without discharging heat to a cold reservoir.
  • Entropy Formulation (Universal Criterion for Spontaneity): In any natural, spontaneous (irreversible) process, the total entropy of the universe (system plus surroundings) always increases:
    $$\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0 \quad (\text{Spontaneous / Irreversible})$$ $$\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} = 0 \quad (\text{Reversible / Equilibrium})$$ $$\Delta S_{\text{universe}} < 0 \quad (\text{Non-spontaneous / Impossible})$$
5.4 Entropy Calculations in Phase Transitions and Gas Expansions
  1. Phase Transitions at Constant Temperature:
    • Entropy of Fusion: $\Delta S_{\text{fus}} = \frac{\Delta H_{\text{fus}}}{T_f}$ (where $T_f$ is melting point in Kelvin).
    • Entropy of Vaporization: $\Delta S_{\text{vap}} = \frac{\Delta H_{\text{vap}}}{T_b}$ (where $T_b$ is boiling point in Kelvin).
    • Trouton's Rule: For most non-polar, non-associated liquids (like benzene, $\text{CCl}_4$, hexane), the entropy of vaporization at normal boiling point is nearly constant: $\Delta S_{\text{vap}} \approx 88 \; \text{J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}$ (or $10.5 \; R$). Water and ethanol deviate due to hydrogen bonding.
  2. Entropy Change of an Ideal Gas:

    Combining the First Law ($dq = dU + P dV$) and ideal gas relations ($dU = n C_v dT$, $P = nRT/V$):

    $$dS = \frac{dU + P dV}{T} = n C_v \frac{dT}{T} + n R \frac{dV}{V}$$ $$\Delta S = n C_v \ln\left(\frac{T_2}{T_1}\right) + n R \ln\left(\frac{V_2}{V_1}\right) = n C_p \ln\left(\frac{T_2}{T_1}\right) - n R \ln\left(\frac{P_2}{P_1}\right)$$
    • For an Isothermal Process ($T_1 = T_2$): $\Delta S = n R \ln\left(\frac{V_2}{V_1}\right) = 2.303 n R \log_{10}\left(\frac{V_2}{V_1}\right) = 2.303 n R \log_{10}\left(\frac{P_1}{P_2}\right)$.
    • For an Isochoric Process ($V_1 = V_2$): $\Delta S = n C_v \ln\left(\frac{T_2}{T_1}\right) = 2.303 n C_v \log_{10}\left(\frac{T_2}{T_1}\right)$.
    • For an Isobaric Process ($P_1 = P_2$): $\Delta S = n C_p \ln\left(\frac{T_2}{T_1}\right) = 2.303 n C_p \log_{10}\left(\frac{T_2}{T_1}\right)$.

Module 6: Gibbs Free Energy ($G$), Spontaneity Criteria, Equilibrium & Third Law

6.1 Gibbs Free Energy ($G$) and the Gibbs-Helmholtz Equation

While the Second Law establishes $\Delta S_{\text{universe}} > 0$ as the ultimate test of spontaneity, evaluating $\Delta S_{\text{surroundings}}$ is practically inconvenient. Josiah Willard Gibbs (1873) resolved this by defining a thermodynamic potential that depends entirely on system variables at constant temperature and pressure:

$$G = H - TS$$

Because $H, T,$ and $S$ are state functions, Gibbs Free Energy ($G$) is a state function and extensive property. For an isothermal change:

$$\Delta G_{\text{system}} = \Delta H_{\text{system}} - T \Delta S_{\text{system}}$$

At constant temperature and pressure, the heat transferred to surroundings is $q_{\text{surr}} = -q_p = -\Delta H_{\text{sys}}$. Thus:

$$\Delta S_{\text{surr}} = \frac{q_{\text{surr}}}{T} = -\frac{\Delta H_{\text{sys}}}{T}$$ $$\Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} = \Delta S_{\text{sys}} - \frac{\Delta H_{\text{sys}}}{T}$$

Multiplying through by $-T$ yields:

$$-T \Delta S_{\text{univ}} = \Delta H_{\text{sys}} - T \Delta S_{\text{sys}} = \Delta G_{\text{sys}}$$
Spontaneity Criterion at Constant $T$ and $P$: $$\Delta G < 0 \iff \Delta S_{\text{univ}} > 0 \quad (\text{Spontaneous / Exergonic process})$$ $$\Delta G = 0 \iff \Delta S_{\text{univ}} = 0 \quad (\text{Thermodynamic Equilibrium})$$ $$\Delta G > 0 \iff \Delta S_{\text{univ}} < 0 \quad (\text{Non-spontaneous / Endergonic process; reverse is spontaneous})$$

Physical Significance: $-\Delta G$ represents the maximum useful (non-expansion) work obtainable from a system at constant temperature and pressure: $-\Delta G = w_{\text{useful}} = w_{\text{non-}PV}$ (e.g., electrical work generated by a galvanic cell: $\Delta G = -n F E_{\text{cell}}$).

6.2 Spontaneity Matrix: Interplay of $\Delta H$ and $\Delta S$ Across Temperatures

The sign of $\Delta G = \Delta H - T \Delta S$ depends on the algebraic signs of $\Delta H$ and $\Delta S$ and the magnitude of $T$:

$\Delta H$ $\Delta S$ Sign of $\Delta G = \Delta H - T\Delta S$ Spontaneity Behavior Example
Negative ($-$) Positive ($+$) Always Negative ($-$) Spontaneous at all temperatures. Enthalpy and entropy both favor spontaneity. $2\text{O}_3(g) \to 3\text{O}_2(g)$
Positive ($+$) Negative ($-$) Always Positive ($+$) Non-spontaneous at all temperatures. Reverse reaction is spontaneous at all $T$. $3\text{O}_2(g) \to 2\text{O}_3(g)$
Negative ($-$) Negative ($-$) Negative at Low $T$; Positive at High $T$ Spontaneous at low temperatures (Enthalpy driven). Non-spontaneous at high temperatures ($T > \Delta H / \Delta S$). $\text{N}_2(g) + 3\text{H}_2(g) \to 2\text{NH}_3(g)$
Positive ($+$) Positive ($+$) Positive at Low $T$; Negative at High $T$ Spontaneous at high temperatures (Entropy driven, $T > \Delta H / \Delta S$). Non-spontaneous at low temperatures. $\text{CaCO}_3(s) \to \text{CaO}(s) + \text{CO}_2(g)$
6.3 Standard Gibbs Free Energy Change and Equilibrium Constant ($K$)

For any reversible reaction $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$, the actual free energy change $\Delta_r G$ is related to the standard free energy change $\Delta_r G^\circ$ and the reaction quotient $Q$ by:

$$\Delta_r G = \Delta_r G^\circ + RT \ln Q$$

At dynamic chemical equilibrium, $\Delta_r G = 0$ and $Q = K$ (the equilibrium constant). Substituting these conditions:

$$\Delta_r G^\circ = -RT \ln K = -2.303 RT \log_{10} K$$ $$K = e^{-\frac{\Delta_r G^\circ}{RT}} = 10^{-\frac{\Delta_r G^\circ}{2.303 RT}}$$
  • If $\Delta_r G^\circ < 0$: $K > 1$, equilibrium heavily favors products (reaction proceeds to near completion).
  • If $\Delta_r G^\circ = 0$: $K = 1$, reactants and products are present in equal effective thermodynamic concentrations.
  • If $\Delta_r G^\circ > 0$: $K < 1$, equilibrium heavily favors reactants (reaction barely proceeds).
6.4 Third Law of Thermodynamics & Absolute Entropies

Formulated by Walther Nernst (Nernst Heat Theorem, 1906) and generalized by Max Planck:

Third Law Statement: The entropy of a perfectly crystalline pure chemical substance approaches zero as the absolute temperature approaches absolute zero ($0 \; \text{K}$). $$\lim_{T \to 0} S = 0$$

Significance of the Third Law:

  • Unlike enthalpy or internal energy (where only changes $\Delta H$ and $\Delta U$ are measurable), the Third Law enables the determination of Absolute Standard Molar Entropies ($S^\circ$) for any substance at temperature $T$: $$S_T^\circ = \int_0^T \frac{C_p}{T} dT$$
  • Standard entropy of reaction is then directly calculated: $\Delta_r S^\circ = \sum \nu_p S^\circ(\text{products}) - \sum \nu_r S^\circ(\text{reactants})$.
  • Residual Entropy: Certain substances do not possess zero entropy at $0 \; \text{K}$ because their molecules freeze into disordered orientations in the crystal lattice. Examples include $\text{CO}$ (dipole orientation disorder: $\text{C-O}$ vs $\text{O-C}$), $\text{N}_2\text{O}$, $\text{NO}$, and ice (hydrogen bond orientation disorder). The residual entropy of $\text{CO}$ is $S_{\text{res}} = R \ln 2 \approx 5.76 \; \text{J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}$.

Key Formulas, Reactions & Definitions

First Law of Thermodynamics & Pressure-Volume Work
$$\Delta U = q + w, \quad w_{\text{rev}} = -2.303 nRT \log_{10}\left(\frac{V_2}{V_1}\right), \quad w_{\text{irrev}} = -P_{\text{ext}}\Delta V$$
Enthalpy & Gaseous Reaction Relations
$$H = U + PV, \quad \Delta H = \Delta U + \Delta n_g RT, \quad q_p = \Delta H, \quad q_v = \Delta U$$
Heat Capacities, Mayer's Relation & Adiabatic Process
$$C_p - C_v = R, \quad \gamma = \frac{C_p}{C_v}, \quad P V^\gamma = \text{const}, \quad w_{\text{ad}} = \frac{nR(T_2 - T_1)}{\gamma - 1}$$
Hess's Law & Bond Enthalpy Formulations
$$\Delta_r H^\circ = \sum \nu_p \Delta_f H^\circ(\text{prod}) - \sum \nu_r \Delta_f H^\circ(\text{react}) = \sum \text{B.E.}(\text{react}) - \sum \text{B.E.}(\text{prod})$$
Entropy Change Calculations & Second Law
$$\Delta S = \frac{q_{\text{rev}}}{T}, \quad \Delta S_{\text{vap}} = \frac{\Delta H_{\text{vap}}}{T_b}, \quad \Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} > 0$$
Gibbs-Helmholtz Equation & Equilibrium Constant
$$\Delta G = \Delta H - T \Delta S, \quad \Delta_r G^\circ = -2.303 RT \log_{10} K, \quad -\Delta G = w_{\text{useful}}$$

Conceptual Solved Examples & Case Studies

Example 1
Two moles of an ideal gas expand isothermally and reversibly from an initial volume of 10.0 L to a final volume of 100.0 L at 27°C. (a) Calculate the work done by the gas in joules. (b) What would be the work done if the gas expanded freely into vacuum? [R = 8.314 J·K⁻¹·mol⁻¹]
Step-by-Step Solution:
Part (a): Reversible Isothermal Expansion Work:
State variables: $$n = 2.0 \; \text{mol}, \quad T = 27 + 273.15 = 300.15 \; \text{K} \approx 300 \; \text{K}$$ $$V_1 = 10.0 \; \text{L}, \quad V_2 = 100.0 \; \text{L}, \quad R = 8.314 \; \text{J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}$$
Apply the formula for reversible isothermal expansion of an ideal gas: $$w_{\text{rev}} = -2.303 nRT \log_{10}\left(\frac{V_2}{V_1}\right)$$ $$w_{\text{rev}} = -2.303 \times 2.0 \times 8.314 \times 300 \times \log_{10}\left(\frac{100.0}{10.0}\right)$$ Since $\log_{10}(10) = 1.0$: $$w_{\text{rev}} = -2.303 \times 2.0 \times 8.314 \times 300 \times 1 = -11488.29 \; \text{J} \approx -11.49 \; \text{kJ}$$
The negative sign signifies that work is done by the system on the surroundings.

Part (b): Free Expansion into Vacuum:
During free expansion, the opposing external pressure is zero ($P_{\text{ext}} = 0$): $$w_{\text{free}} = -\int P_{\text{ext}} dV = -0 \times \Delta V = 0 \; \text{J}$$
Final Answer: (a) $w_{\text{rev}} = -11.49 \; \text{kJ}$; (b) $w_{\text{free}} = 0 \; \text{J}$.
Example 2
The standard enthalpy of combustion of liquid benzene (C₆H₆) at 298 K is -3267.0 kJ/mol. Calculate the standard internal energy change (ΔU°) for the combustion reaction at 298 K. [Given: R = 8.314 J·K⁻¹·mol⁻¹]
Step-by-Step Solution:
Step 1: Write the balanced thermochemical equation for the combustion of benzene:
$$\text{C}_6\text{H}_6(l) + \frac{15}{2}\text{O}_2(g) \longrightarrow 6\text{CO}_2(g) + 3\text{H}_2\text{O}(l) \quad \Delta_c H^\circ = -3267.0 \; \text{kJ/mol}$$
Step 2: Calculate $\Delta n_g$ (moles of gaseous products minus gaseous reactants):
Important: Liquid benzene and liquid water are omitted from $\Delta n_g$! $$\Delta n_g = n_g(\text{products}) - n_g(\text{reactants}) = 6 - \frac{15}{2} = 6 - 7.5 = -1.5 \; \text{mol}$$
Step 3: Apply the relationship $\Delta H^\circ = \Delta U^\circ + \Delta n_g RT$:
$$\Delta U^\circ = \Delta H^\circ - \Delta n_g RT$$ Convert $R$ to $\text{kJ} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}$: $$R = 8.314 \times 10^{-3} \; \text{kJ} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}$$ $$T = 298 \; \text{K}$$ $$\Delta n_g RT = (-1.5 \; \text{mol}) \times (8.314 \times 10^{-3} \; \text{kJ/mol} \cdot \text{K}) \times 298 \; \text{K} = -3.716 \; \text{kJ}$$
Step 4: Compute $\Delta U^\circ$:
$$\Delta U^\circ = -3267.0 - (-3.716) = -3267.0 + 3.716 = -3263.284 \; \text{kJ/mol} \approx -3263.3 \; \text{kJ/mol}$$
Final Answer: $\Delta U^\circ = -3263.3 \; \text{kJ/mol}$.
Example 3
Using Hess's Law of Constant Heat Summation, calculate the standard enthalpy of formation of acetylene gas [C₂H₂(g)] from the following thermochemical data: (i) C(graphite) + O₂(g) → CO₂(g); ΔH₁ = -393.5 kJ/mol (ii) H₂(g) + ½O₂(g) → H₂O(l); ΔH₂ = -285.8 kJ/mol (iii) C₂H₂(g) + ⁵/₂O₂(g) → 2CO₂(g) + H₂O(l); ΔH₃ = -1300.0 kJ/mol
Step-by-Step Solution:

Step 1: Write the target thermochemical equation:
The formation of 1 mole of acetylene from its standard elements:

$$2\text{C}(\text{graphite}) + \text{H}_2(g) \longrightarrow \text{C}_2\text{H}_2(g) \quad \Delta_f H^\circ = ?$$


Step 2: Manipulate the given thermochemical equations:

  1. We need $2\text{C}(\text{graphite})$ on the left, so multiply equation (i) by 2:

$$2\text{C}(\text{graphite}) + 2\text{O}_2(g) \longrightarrow 2\text{CO}_2(g) \quad \Delta H_A = 2 \times (-393.5) = -787.0 \; \text{kJ}$$


2. We need $1\text{H}_2(g)$ on the left, so keep equation (ii) as is:

$$\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \longrightarrow \text{H}_2\text{O}(l) \quad \Delta H_B = -285.8 \; \text{kJ}$$


3. We need $1\text{C}_2\text{H}_2(g)$ on the right, so reverse equation (iii):

$$2\text{CO}_2(g) + \text{H}_2\text{O}(l) \longrightarrow \text{C}_2\text{H}_2(g) + \frac{5}{2}\text{O}_2(g) \quad \Delta H_C = -(-1300.0) = +1300.0 \; \text{kJ}$$


Step 3: Sum the manipulated equations:

$$[2\text{C} + 2\text{O}_2] + [\text{H}_2 + \frac{1}{2}\text{O}_2] + [2\text{CO}_2 + \text{H}_2\text{O}] \longrightarrow 2\text{CO}_2 + \text{H}_2\text{O} + \text{C}_2\text{H}_2 + \frac{5}{2}\text{O}_2$$

Cancelling common terms ($2\text{CO}_2, \text{H}_2\text{O}$, and $\frac{5}{2}\text{O}_2$) yields the target equation:

$$2\text{C}(\text{graphite}) + \text{H}_2(g) \longrightarrow \text{C}_2\text{H}_2(g)$$


Step 4: Compute $\Delta_f H^\circ$:

$$\Delta_f H^\circ = \Delta H_A + \Delta H_B + \Delta H_C = -787.0 + (-285.8) + 1300.0$$

$$\Delta_f H^\circ = -1072.8 + 1300.0 = +227.2 \; \text{kJ/mol}$$


Final Answer: Standard enthalpy of formation $\Delta_f H^\circ[\text{C}_2\text{H}_2(g)] = +227.2 \; \text{kJ/mol}$ (endothermic compound).
Example 4
Calculate the standard enthalpy change (Δ_r H°) for the hydrogenation of ethene to ethane: C₂H₄(g) + H₂(g) → C₂H₆(g) Given the following average bond enthalpies: B.E.(C=C) = 606 kJ/mol; B.E.(C-C) = 347 kJ/mol; B.E.(C-H) = 414 kJ/mol; B.E.(H-H) = 436 kJ/mol.
Step-by-Step Solution:
Step 1: Write structural formulas identifying all chemical bonds:
Reactants:
  • $\text{C}_2\text{H}_4$: One $\text{C=C}$ double bond and four $\text{C-H}$ single bonds.
  • $\text{H}_2$: One $\text{H-H}$ single bond.
Products:
  • $\text{C}_2\text{H}_6$: One $\text{C-C}$ single bond and six $\text{C-H}$ single bonds.

Step 2: Tabulate total bond energies for reactants (bonds broken):
$$\sum \text{B.E.}(\text{reactants}) = 1 \times \text{B.E.}(\text{C=C}) + 4 \times \text{B.E.}(\text{C-H}) + 1 \times \text{B.E.}(\text{H-H})$$ $$\sum \text{B.E.}(\text{reactants}) = 606 + 4(414) + 436 = 606 + 1656 + 436 = 2698 \; \text{kJ}$$
Step 3: Tabulate total bond energies for products (bonds formed):
$$\sum \text{B.E.}(\text{products}) = 1 \times \text{B.E.}(\text{C-C}) + 6 \times \text{B.E.}(\text{C-H})$$ $$\sum \text{B.E.}(\text{products}) = 347 + 6(414) = 347 + 2484 = 2831 \; \text{kJ}$$
Step 4: Apply the bond enthalpy formula:
$$\Delta_r H^\circ = \sum \text{B.E.}(\text{bonds broken}) - \sum \text{B.E.}(\text{bonds formed})$$ $$\Delta_r H^\circ = 2698 - 2831 = -133.0 \; \text{kJ/mol}$$
Alternative shortcut: Net bonds broken = $1 \; (\text{C=C}) + 1 \; (\text{H-H}) = 606 + 436 = 1042 \; \text{kJ}$.
Net bonds formed = $1 \; (\text{C-C}) + 2 \; (\text{C-H}) = 347 + 2(414) = 347 + 828 = 1175 \; \text{kJ}$.
$$\Delta_r H^\circ = 1042 - 1175 = -133.0 \; \text{kJ/mol}$$
Final Answer: $\Delta_r H^\circ = -133.0 \; \text{kJ/mol}$ (exothermic addition reaction).
Example 5
For a chemical reaction: 2A(g) + B(g) → 2D(g), ΔH° = -40.0 kJ/mol and ΔS° = -100.0 J·K⁻¹·mol⁻¹ at 298 K. (a) Calculate ΔG° at 298 K and predict whether the reaction is spontaneous at this temperature. (b) Above what temperature will the reaction become non-spontaneous?
Step-by-Step Solution:
Part (a): Calculation of $\Delta G^\circ$ at 298 K:
Given: $$\Delta H^\circ = -40.0 \; \text{kJ/mol} = -40000 \; \text{J/mol}$$ $$\Delta S^\circ = -100.0 \; \text{J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}$$ $$T = 298 \; \text{K}$$
Apply the Gibbs-Helmholtz equation: $$\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$$ $$\Delta G^\circ = -40000 - [298 \times (-100.0)] = -40000 - (-29800)$$ $$\Delta G^\circ = -40000 + 29800 = -10200 \; \text{J/mol} = -10.2 \; \text{kJ/mol}$$
Since $\Delta G^\circ < 0$ (negative), the reaction is spontaneous at $298 \; \text{K}$.

Part (b): Temperature Threshold for Spontaneity:
At thermodynamic equilibrium, $\Delta G^\circ = 0$: $$\Delta H^\circ - T_{\text{eq}} \Delta S^\circ = 0 \implies T_{\text{eq}} = \frac{\Delta H^\circ}{\Delta S^\circ}$$ $$T_{\text{eq}} = \frac{-40000 \; \text{J/mol}}{-100.0 \; \text{J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}} = 400.0 \; \text{K} \; (126.85^\circ\text{C})$$
Because both $\Delta H^\circ$ and $\Delta S^\circ$ are negative:
  • For $T < 400 \; \text{K}$: $|\Delta H^\circ| > |T\Delta S^\circ| \implies \Delta G^\circ < 0$ (spontaneous).
  • For $T > 400 \; \text{K}$: $|T\Delta S^\circ| > |\Delta H^\circ| \implies \Delta G^\circ > 0$ (non-spontaneous).
Therefore, the reaction becomes non-spontaneous above $400.0 \; \text{K}$.
Final Answer: (a) $\Delta G^\circ = -10.2 \; \text{kJ/mol}$ (spontaneous); (b) Non-spontaneous above $400.0 \; \text{K}$.
Example 6
(a) For the equilibrium: N₂O₄(g) ⇌ 2NO₂(g) at 298 K, standard free energies of formation are Δ_f G°(N₂O₄) = 97.8 kJ/mol and Δ_f G°(NO₂) = 51.3 kJ/mol. Calculate: (i) the standard Gibbs free energy change (Δ_r G°), and (ii) the equilibrium constant (K_p) at 298 K. [R = 8.314 J·K⁻¹·mol⁻¹]. (b) State the Third Law of Thermodynamics and explain why carbon monoxide (CO) exhibits residual entropy at absolute zero.
Step-by-Step Solution:
Part (a): $\Delta_r G^\circ$ and Equilibrium Constant ($K_p$):
Reaction: $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$
1. Standard Free Energy Change ($\Delta_r G^\circ$):
$$\Delta_r G^\circ = \sum \nu_p \Delta_f G^\circ(\text{products}) - \sum \nu_r \Delta_f G^\circ(\text{reactants})$$ $$\Delta_r G^\circ = 2 \times \Delta_f G^\circ[\text{NO}_2] - 1 \times \Delta_f G^\circ[\text{N}_2\text{O}_4]$$ $$\Delta_r G^\circ = 2(51.3) - 97.8 = 102.6 - 97.8 = +4.80 \; \text{kJ/mol} = +4800 \; \text{J/mol}$$
2. Equilibrium Constant ($K_p$):
$$\Delta_r G^\circ = -2.303 RT \log_{10} K_p \implies \log_{10} K_p = -\frac{\Delta_r G^\circ}{2.303 RT}$$ $$\log_{10} K_p = -\frac{4800}{2.303 \times 8.314 \times 298} = -\frac{4800}{5705.85} = -0.8412$$ Taking antilogarithms: $$K_p = 10^{-0.8412} = 0.144$$
Since $K_p < 1$, the equilibrium lies predominantly to the left (favoring $\text{N}_2\text{O}_4$) at $298 \; \text{K}$.

Part (b): Third Law of Thermodynamics and Residual Entropy:
  • Third Law Statement: The entropy of a perfectly crystalline, pure homogeneous chemical substance is zero at the absolute zero of temperature ($0 \; \text{K}$): $$\lim_{T \to 0} S = 0$$
  • Residual Entropy of Carbon Monoxide: In a perfectly crystalline substance, all molecules must occupy completely ordered lattice sites with identical orientations. However, the carbon monoxide molecule ($\text{C}\equiv\text{O}$) has an extremely small dipole moment ($\mu \approx 0.11 \; \text{D}$). As liquid $\text{CO}$ is cooled and solidifies near $0 \; \text{K}$, the energy difference between parallel ($\dots \text{C-O} \; \text{C-O} \dots$) and antiparallel ($\dots \text{C-O} \; \text{O-C} \dots$) orientations is negligible. The molecules freeze into random orientations with two equally probable arrangements per molecule ($W = 2^{N_A}$). According to Boltzmann entropy formula: $$S_{\text{residual}} = k_B \ln W = k_B \ln(2^{N_A}) = N_A k_B \ln 2 = R \ln 2 = 8.314 \times 0.69315 \approx 5.76 \; \text{J} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}$$ Because of this frozen structural disorder, crystalline carbon monoxide does not achieve zero entropy at $0 \; \text{K}$.
Final Answer: (a)(i) $\Delta_r G^\circ = +4.80 \; \text{kJ/mol}$; (a)(ii) $K_p = 0.144$; (b) Third Law defined; residual entropy explained via Boltzmann formula $S = R \ln 2 \approx 5.76 \; \text{J}/(\text{mol}\cdot\text{K})$.

Common Misconceptions & Examiner Traps

Common Misconception

Confusing IUPAC chemical sign conventions with classical physics definitions for work and heat.

Scientific Reality & Correction

Under IUPAC chemical thermodynamics, work is defined from the perspective of the system: $w = -P_{\text{ext}} \Delta V$ and $\Delta U = q + w$. When the system expands, work is done BY the system on surroundings, so $w < 0$.

Common Misconception

Including solids and liquids when calculating delta n_g in the delta H vs delta U equation.

Scientific Reality & Correction

The term $\Delta n_g$ in $\Delta H = \Delta U + \Delta n_g RT$ stands strictly for the change in moles of GASEOUS species only: $\Delta n_g = \sum n_g(\text{gaseous products}) - \sum n_g(\text{gaseous reactants})$. Condensed phases occupy negligible volume.

Common Misconception

Reversing the order of reactants and products when calculating reaction enthalpies from bond energies.

Scientific Reality & Correction

Bond breaking absorbs energy ($+\text{B.E.}$, reactants) while bond formation releases energy ($-\text{B.E.}$, products). Therefore: $\Delta_r H = \sum \text{B.E.}(\text{reactants broken}) - \sum \text{B.E.}(\text{products formed})$.

Common Misconception

Mixing mismatched energy units (kJ vs J) in the Gibbs-Helmholtz equation.

Scientific Reality & Correction

Always convert $\Delta S$ to $\text{kJ} \cdot \text{K}^{-1}$ by dividing by 1000, or convert $\Delta H$ to Joules by multiplying by 1000 before evaluating $\Delta G$.

Common Misconception

Assuming that a process with negative system entropy change cannot be spontaneous.

Scientific Reality & Correction

The Second Law mandates that the entropy of the UNIVERSE must increase ($\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0$). A process with $\Delta S_{\text{system}} < 0$ is spontaneous if it is sufficiently exothermic that $\Delta S_{\text{surroundings}} = -\Delta H / T > |\Delta S_{\text{system}}|$ (e.g., freezing of water below $0^\circ\text{C}$).

Chemical Thermodynamics: First Law, Hess Law, Entropy & Gibbs Spontaneity Matrix

Thermodynamics — Master Concept Architecture First Law & P-V Work • Hess's Law & Enthalpy • Entropy (S) • Gibbs Free Energy (ΔG) WBCHSE CLASS 11 1. First Law of Thermodynamics & P-V Work P V Reversible Work |w_rev| is Maximum Irrev: -P_ext·ΔV ΔU = q + w (IUPAC) First Law: ΔU = q + w [Conservat ion of Energy] Reversible Work: w_rev = -2.303 nRT log(V₂/V₁) P-V Expansion Work: w = -∫ P_ext d V Free Expansion (Vacuum): P_ext = 0 → w = 0 2. Thermochemistry & Hess's Law of Heat Summation React Prod ΔH Intermed ΔH₁ ΔH₂ ΔH = ΔH₁ + ΔH₂ Hess's Law: ΔH = ΔH₁ + ΔH₂ (State function path independence) Reaction Enthalpy: Δ_r H° = ∑Δ_f H °(products) - ∑Δ_f H°(reactants) Bond Enthalpy: Δ_r H = ∑B.E.(bro ken) - ∑B.E.(formed) qp = ΔH, qv = ΔU | Relation: ΔH = ΔU + Δn_g RT 3. Second Law & Entropy (Molecular Disorder) Solid (S_low) Liquid Gas (S_high) Entropy ΔS > 0 (Spontaneous) ΔS_univ = ΔS_sys + ΔS_surr Spontaneous: ΔS_univ > 0 Entropy Definition: dS = dq_rev / T (J·K⁻¹·mol⁻¹) Second Law: ΔS_universe = ΔS_syste m + ΔS_surr > 0 Phase Change: ΔS_vap = ΔH_vap / T_ b | Trouton's Rule Solid → Liquid → Gas: Molecular di sorder & entropy increase 4. Gibbs Free Energy (ΔG) & Spontaneity Matrix ΔG = ΔH - TΔS Spontaneity Matrix ΔH<0, ΔS>0 Always Spont (All Temp) ΔH<0, ΔS<0 Low T Spont (Enthalpy driven) ΔH>0, ΔS>0 High T Spont (T > ΔH/ΔS) ΔH>0, ΔS<0 Never Spont (Reverse Spont) Equilibrium: ΔG = 0, T = ΔH/ΔS Gibbs Equation: ΔG = ΔH - TΔS [Use ful work = -ΔG] Spontaneous: ΔG < 0 | Equilibrium: ΔG = 0 | Non-spontaneous: ΔG > 0 Temperature Effect: ΔH<0, ΔS>0 → S pontaneous at all T Equilibrium Constant: Δ_r G° = -2. K > 1 → ΔG° < 0 (Favored)

Chapter Summary & 10 Key Takeaways

Takeaway 1
Thermodynamic state functions depend solely on the initial and final states of a system (U, H, S, G, P, V, T), whereas work and heat are path-dependent quantities.
Takeaway 2
The First Law of Thermodynamics establishes the conservation of energy: delta U = q + w, where IUPAC defines heat absorbed (q > 0) and work done on the system (w > 0) as positive.
Takeaway 3
Isothermal reversible expansion of an ideal gas produces maximum work: w_rev = -2.303 n R T log10(V2 / V1), whereas free expansion into vacuum yields zero work.
Takeaway 4
Enthalpy represents heat transferred at constant pressure (q_p = delta H), related to internal energy by delta H = delta U + delta n_g R T for gaseous reactions.
Takeaway 5
Molar heat capacity at constant pressure exceeds that at constant volume by the universal gas constant for ideal gases: Cp - Cv = R.
Takeaway 6
Hess law of constant heat summation states that total enthalpy change is identical whether a reaction occurs in a single step or multiple steps.
Takeaway 7
Reaction enthalpy can be computed from bond enthalpies via: delta_r H = sum of bond energies of reactants broken minus sum of bond energies of products formed.
Takeaway 8
The Second Law of Thermodynamics requires that the total entropy of the universe increases in every spontaneous process: delta S_universe = delta S_system + delta S_surroundings > 0.
Takeaway 9
Gibbs free energy change delta G = delta H - T delta S serves as the master criterion for spontaneity at constant temperature and pressure: delta G less than 0 is spontaneous, delta G equal to 0 is equilibrium, and delta G greater than 0 is non-spontaneous.
Takeaway 10
The Third Law of Thermodynamics states that the entropy of a perfectly crystalline pure substance approaches zero at absolute zero (0 K), enabling calculation of absolute molar entropies.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
Differentiate between an extensive property and an intensive property with two examples of each. Is molar heat capacity extensive or intensive?
Reveal Answer & Explanation
Answer: An extensive property depends on the mass or quantity of matter present in the system (e.g., total mass, volume, internal energy, enthalpy). If the system is divided into two halves, the value of an extensive property is halved. An intensive property is completely independent of the mass or size of the system (e.g., temperature, pressure, density, refractive index). While total heat capacity ($C$) is an extensive property, molar heat capacity ($C_m = C/n$) is the ratio of two extensive properties (heat capacity divided by moles), making it an intensive property.
2
State the First Law of Thermodynamics in mathematical form. Why is the work done in a reversible isothermal expansion of an ideal gas greater than that in an irreversible expansion between the same volume limits?
Reveal Answer & Explanation
Answer: The First Law of Thermodynamics states: $\Delta U = q + w$. In a reversible expansion, the opposing external pressure balances internal gas pressure at every infinitesimal step ($P_{\text{ext}} = P_{\text{int}} - dP$), maintaining the maximum possible opposing force throughout the expansion. In an irreversible expansion, the external pressure is dropped suddenly to the final pressure ($P_{\text{ext}} = P_2$), which is much lower than the gas pressure during expansion. Because work equals the area under the $P-V$ curve ($\int P_{\text{ext}} dV$), the reversible path covers a significantly larger area, producing maximum work: $|w_{\text{rev}}| > |w_{\text{irrev}}|$.
3
Explain why the enthalpy of neutralization of any strong acid by any strong base in dilute aqueous solution is constant (-57.1 kJ/equiv), whereas it is less for a weak acid like acetic acid.
Reveal Answer & Explanation
Answer: Strong acids and strong bases are completely ionized in dilute aqueous solution. The overall neutralization reaction simplifies identically to the combination of hydrogen ions and hydroxide ions to form liquid water: $\text{H}^+(aq) + \text{OH}^-(aq) \to \text{H}_2\text{O}(l)$, for which $\Delta H^\circ = -57.1 \; \text{kJ/mol}$. In contrast, a weak acid like acetic acid is only partially ionized. When neutralized, heat is first consumed endothermically to ionize un-dissociated acid molecules (enthalpy of ionization, $\Delta_{\text{ion}} H > 0$). Consequently, the net measured heat evolved is diminished: $\Delta_{\text{neut}} H = -57.1 + \Delta_{\text{ion}} H$ (for $\text{CH}_3\text{COOH}$ and $\text{NaOH}$, $\Delta_{\text{neut}} H \approx -55.2 \; \text{kJ/mol}$, with $\Delta_{\text{ion}} H \approx +1.9 \; \text{kJ/mol}$).
4
Define entropy. Why does the entropy of the universe increase during an irreversible process while remaining constant in a reversible process?
Reveal Answer & Explanation
Answer: Entropy ($S$) is a thermodynamic state function measuring the degree of randomness, molecular disorder, or spatial dispersal of matter and energy in a system ($dS = dq_{\text{rev}}/T$). In a reversible process, system and surroundings remain in virtual equilibrium at every instant; any entropy gained by the system is exactly equal to that lost by the surroundings: $\Delta S_{\text{sys}} = -\Delta S_{\text{surr}} \implies \Delta S_{\text{univ}} = 0$. In an irreversible process, finite gradients exist and energy is dissipated irreversibly into random thermal motion, generating net entropy: $\Delta S_{\text{univ}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} > 0$.
5
How does Gibbs free energy change (ΔG) serve as a criterion for spontaneity and equilibrium at constant temperature and pressure? State the conditions under which an endothermic reaction can be spontaneous.
Reveal Answer & Explanation
Answer: At constant temperature and pressure, $\Delta G = \Delta H - T\Delta S = -T\Delta S_{\text{univ}}$. If $\Delta G < 0$, $\Delta S_{\text{univ}} > 0$, making the process spontaneous. If $\Delta G = 0$, the system is in dynamic equilibrium. If $\Delta G > 0$, the forward process is non-spontaneous. An endothermic reaction ($\Delta H > 0$) can be spontaneous if it is accompanied by an increase in entropy ($\Delta S > 0$) and the temperature is sufficiently high such that the $T\Delta S$ term outweighs $\Delta H$, giving a net negative $\Delta G$ ($T > \Delta H / \Delta S$). Examples include evaporation of water and melting of ice above $0^\circ\text{C}$.
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