Class 11Physical ChemistryChapter 5: Chemical Thermodynamics

Chemical Thermodynamics

NCERT Syllabus Key Concepts:

Thermodynamic Systems (Open, Closed, Isolated) & Real vs Imaginary Boundaries

State Functions & Variables (p, V, T, n) vs Path Functions (q, w)

Extensive vs Intensive Properties & Container Partition Principle

Internal Energy (U), Adiabatic Work (Joule Equivalent) & First Law (ΔU = q + w)

IUPAC Sign Conventions for Heat (q) and Work (w)

Pressure-Volume Work: Irreversible (w = -Pex ΔV) vs Reversible Isothermal (w = -2.303 nRT log(Vf/Vi))

Enthalpy (H = U + pV), Heat at Constant Pressure (qp = ΔH), and Gas Expansion (ΔH = ΔU + Δng RT)

Heat Capacity (C), Specific Heat (c), Molar Heat Capacity (Cm), and Meyer Relation (Cp - Cv = R)

Calorimetry: Constant-Volume Bomb Calorimeter (ΔU = qV) vs Constant-Pressure Coffee Cup (ΔH = qp)

Standard States (1 bar, 298 K) & Standard Enthalpy of Formation (ΔfH° = 0 for reference elements)

Hess’s Law of Constant Heat Summation & Multi-Step Reaction Cycles

Enthalpies of Phase Changes (Fusion, Vaporization, Sublimation) & Combustion (ΔcH°)

Bond Dissociation Enthalpies & Mean Bond Enthalpy in Polyatomics: ΔrH° ≈ Σ BE(broken) - Σ BE(formed)

Lattice Enthalpy (ΔlatticeH°) & Born-Haber Cycle for Ionic Crystals (NaCl)

Enthalpy of Solution (ΔsolH° = ΔlatticeH° + ΔhydH°) & Enthalpy of Dilution

Spontaneity Criteria: Failure of Enthalpy Decrease Alone & Spontaneous Endothermic Reactions

Entropy (S) as Randomness/Disorder & Thermodynamic Definition (ΔS = qrev / T)

Second Law of Thermodynamics: ΔStotal = ΔSsys + ΔSsurr > 0 for Spontaneous Processes

Gibbs Free Energy (G = H - TS), Spontaneity Criterion (ΔG < 0), and Four Quadrants (Table 5.4)

Third Law of Thermodynamics: Absolute Zero Crystal Entropy (lim T->0 K S = 0)

Gibbs Energy and Chemical Equilibrium Constant (ΔrG° = -RT ln K = -2.303 RT log K)

Unit 5: Chemical Thermodynamics

NCERT Chemistry Class 11: Chapter 5
"It is the only physical theory of universal content concerning which I am convinced that, within the framework of the applicability of its basic concepts, it will never be overthrown."
Albert Einstein on Thermodynamics

The Big Picture: Why Study Chemical Thermodynamics?#

Chemical energy stored within molecular bonds can be transformed into thermal energy, mechanical work, or electrical work. When natural gas (methane) burns in air, chemical potential energy is liberated as heat. In an automobile engine or rocket thruster, that chemical energy is converted into directed mechanical work through gas expansion. In a galvanic cell, it produces electrical work to power external circuits.

The branch of physical science concerned with these quantitative energy transformations and the equilibrium states of matter is chemical thermodynamics.

Thermodynamics is governed by foundational characteristics that distinguish it from chemical kinetics and mechanics:

Macroscopic Scope: Thermodynamics deals exclusively with bulk systems comprising vast ensembles of particles (on the order of Avogadro's constant, $N_A \approx 6.022 \times 10^{23}$ ). It does not require knowledge of individual atomic trajectories, molecular velocities, or quantum eigenstate transitions.
Path-Independent State Functions: Thermodynamic analysis is concerned with the initial and final equilibrium states of a system. It does not depend on the specific pathway or mechanism connecting those states.
No Kinetic Information: Thermodynamics answers whether a process has the thermodynamic potential to occur spontaneously, how much heat or work will be exchanged, and where the chemical equilibrium position lies. It provides zero information regarding the reaction rate or time required to reach that equilibrium. A mixture of hydrogen and oxygen gas at $298\text{ K}$ is thermodynamically unstable with respect to water formation, yet the reaction proceeds at an imperceptible rate in the absence of a catalyst.

Thermodynamics addresses three primary questions:

How do we calculate the energy changes accompanying physical transformations and chemical reactions?
What fundamental physical principle dictates whether a reaction will proceed spontaneously without continuous external assistance?
To what extent will reactants convert into products before reaching dynamic equilibrium?

How to Use the Interactive Lab Modules:
Alongside this chapter, you have access to seven interactive simulation tools in the right-hand workspace:

System Boundary Simulator: Manipulate Open, Closed, and Isolated boundaries; observe real-time exchange of matter and energy.
Energy Transfer & Thermodynamic Sign Lab: Conduct Joule paddle-wheel churning and electrical coil heating; track $q$ , $w$ , and $\Delta U$ with IUPAC sign conventions.
Piston Gas & Indicator Lab: Compress and expand gases irreversibly or reversibly; visualize indicator $PV$ diagrams and evaluate $w = -\int p_{\text{ex}}\,dV$ .
Reaction Calorimeter & Temperature Lab: Measure heat evolved at constant volume in a bomb calorimeter ( $\Delta U = q_V$ ) versus constant pressure ( $\Delta H = q_p$ ).
Hess's Law Puzzle & Reaction Builder: Combine and scale intermediate reactions to construct target enthalpy cycles path-independently.
Bond Breaking & Making Lab: Sever reactant covalent bonds and forge product bonds to calculate approximate gas-phase enthalpies.
Molecular Disorder & Gibbs Spontaneity Simulator: Track particle microstates across solid, liquid, and gas phases; manipulate $\Delta H$ , $\Delta S$ , and temperature to map the four quadrants of spontaneity and evaluate equilibrium constants $K$ .

Follow the Interactive Simulation Experiment callouts integrated into each section.

5.1 Thermodynamic Terms#

To formulate thermodynamic laws quantitatively, precise terminology must be established for defining the portion of the universe under study and the nature of its boundaries.

5.1.1 The System, Surroundings, and Boundary#

Thermodynamic Entity	Formal Definition	Laboratory Example
System	The specific part of the universe chosen for thermodynamic investigation, containing reactants, products, and solvent.	The aqueous solution of reactants inside a conical flask.
Surroundings	The entire remainder of the universe outside the system boundaries that can exchange energy or matter with the system.	The water bath, laboratory bench, ambient air, and measuring instruments in thermal contact with the flask.
Boundary	The real physical or imaginary geometrical surface separating the system from its surroundings across which matter or energy can pass.	The borosilicate glass walls of the flask and the open liquid-air interface at the neck.

Mathematically, the relationship is expressed as:

$\text{Universe} = \text{System} + \text{Surroundings}$

In practical laboratory measurements, changes occurring within a chemical system affect only the immediate neighborhood of the system. Therefore, the effective surroundings comprise only that finite region capable of exchanging energy or matter with the system.

Boundaries are classified based on physical properties:

Diathermic (Conducting) Boundary: Permits heat exchange between system and surroundings driven by temperature gradients (for example, a copper vessel).
Adiabatic (Insulated) Boundary: Completely prevents heat transfer between system and surroundings ( $q = 0$ , such as the silvered vacuum walls of a Dewar flask).
Rigid Boundary: Possesses fixed volume and cannot be displaced, preventing pressure-volume work ( $dV = 0$ ).
Flexible Boundary: Allows volumetric expansion or contraction against external pressure (such as a gas cylinder fitted with a mobile piston).

5.1.2 Classification of Thermodynamic Systems#

Systems are classified according to whether their boundaries permit the exchange of matter, energy, or neither:

System Type	Exchange of Matter	Exchange of Energy	Laboratory Demonstration
Open System	Allowed	Allowed	A beaker of boiling water without a lid. Steam escapes to surroundings, and heat conducts through the base.
Closed System	Prohibited	Allowed	A sealed copper flask containing hot liquid. Mass cannot escape, but thermal energy dissipates into the surroundings.
Isolated System	Prohibited	Prohibited	Hot tea stored in a perfectly insulated thermos flask. Neither matter nor heat crosses the boundary.

Key Principle: The Energy of an Isolated System is Conserved
For any completely isolated system, because no mass, heat, or work can cross the boundary, the total energy remains strictly invariant: $\Delta U_{\text{isolated}} = 0$ .

Interactive Simulation Experiment: System Boundary Simulator
Open the System Boundary tab in the right-hand simulation suite:

Select the Open System preset. Introduce heat via the burner control. Observe water molecules evaporating across the boundary into the surroundings while energy transfers.
Switch to Closed System. Note that a sealed barrier prevents mass transfer, yet heat continues to equilibrate through diathermic walls.
Switch to Isolated System. Verify that both the mass counter and the internal energy monitor remain locked at fixed values regardless of external thermal fluctuations.

5.1.3 The State of the System and State Functions#

In classical mechanics, the state of a particle is fully specified by identifying its instantaneous position coordinates $(x, y, z)$ and momentum components $(p_x, p_y, p_z)$ . In thermodynamics, a vastly simpler macroscopic approach is utilized.

The state of a thermodynamic system is completely described by quoting a small set of measurable macroscopic bulk properties:

Pressure ( $p$ )
Temperature ( $T$ )
Volume ( $V$ )
Amount of substance ( $n$ )
Chemical composition

For a single-phase homogeneous pure substance, fixing any two independent macroscopic variables (such as $p$ and $T$ ) uniquely fixes all other variables according to an equation of state (for an ideal gas, $pV = nRT$ ).

State Functions versus Path Functions

Characteristic	State Function (State Variable)	Path Function
Dependence	Depends exclusively on the instantaneous thermodynamic state of the system, independent of the past history or pathway taken to reach that state.	Depends directly on the specific path, mechanism, or steps taken to transition between initial and final states.
Mathematical Property	Exact differential ( $df$ ); cyclic integral is zero: $\oint df = 0$ .	Inexact differential ( $\delta f$ ); cyclic integral is generally non-zero: $\oint \delta f \neq 0$ .
Finite Change	$\Delta f = f_{\text{final}} - f_{\text{initial}}$	Cannot be expressed as a difference of two state values: $q$ and $w$ .
Examples	Internal Energy ( $U$ ), Enthalpy ( $H$ ), Entropy ( $S$ ), Gibbs Free Energy ( $G$ ), Pressure ( $p$ ), Volume ( $V$ ), Temperature ( $T$ ).	Heat ( $q$ ), Mechanical Work ( $w$ ), Electrical Work ( $w_{\text{elec}}$ ).

Conceptual Analogy:
Elevation above sea level is a state function: whether you climb a mountain via a vertical cliff face or take a winding, gentle trail, the net change in altitude $\Delta h = h_{\text{summit}} - h_{\text{base}}$ is identical. The total distance walked and muscular energy expended depend directly on the route chosen and are path functions.

5.1.4 Extensive and Intensive Properties#

Macroscopic properties of matter are categorized based on their dependence on system size or amount of material:

Classification	Definition	Classic Properties	Partition Experiment
Extensive Property	A property whose magnitude depends directly on the total quantity, mass, or size of matter present in the system.	Mass ( $m$ ), Volume ( $V$ ), Internal Energy ( $U$ ), Enthalpy ( $H$ ), Heat Capacity ( $C$ ), Entropy ( $S$ ), Gibbs Energy ( $G$ ).	If a container of gas of volume $V$ is bisected by an impermeable partition, the volume in each half becomes $V/2$ .
Intensive Property	A property whose magnitude is completely independent of the total quantity or size of matter in the system.	Temperature ( $T$ ), Pressure ( $p$ ), Density ( $\rho$ ), Refractive Index, Molar Heat Capacity ( $C_m$ ), Specific Heat ( $c$ ), Surface Tension, Viscosity.	Upon bisecting the container of gas, the temperature $T$ and pressure $p$ in each half remain identical to the original system.

Critical Mathematical Property:
The ratio of any two extensive properties is an intensive property:

Mass (extensive) divided by Volume (extensive) yields Density (intensive): $\rho = \frac{m}{V}$ .
Heat Capacity (extensive) divided by Moles (extensive) yields Molar Heat Capacity (intensive): $C_m = \frac{C}{n}$ .
Enthalpy (extensive) divided by Moles (extensive) yields Molar Enthalpy (intensive): $H_m = \frac{H}{n}$ .

5.2 Internal Energy, Work, Heat, and the First Law#

5.2.1 Internal Energy ( $U$ ) as a State Function#

When a chemical or physical system exchanges energy with its surroundings, we introduce a quantity representing the total intrinsic energy residing within the system: the internal energy, denoted by $U$ .

At the microscopic atomic level, internal energy is the sum of:

Kinetic Energy Contributions: Translational, rotational, and vibrational motions of molecules, along with the orbital movements and spins of electrons.
Potential Energy Contributions: Electrostatic attractions and repulsions between atomic nuclei and electrons, chemical bonding energies, and intermolecular attractive forces (dispersion, dipole-dipole, hydrogen bonding).

$U = E_{\text{translational}} + E_{\text{rotational}} + E_{\text{vibrational}} + E_{\text{electronic}} + E_{\text{bonding}} + E_{\text{intermolecular}}$

Because of the extreme complexity of subatomic interactions and relativistic mass-energy equivalents, it is impossible to evaluate the absolute numerical value of $U$ for any macroscopic system. In chemical thermodynamics, an absolute value is never required. We measure only changes in internal energy:

$\Delta U = U_{\text{final}} - U_{\text{initial}}$

The internal energy of a system can change through three distinct interactions:

Heat passes into or out of the system.
Work is done on or by the system.
Matter enters or leaves the system.

5.2.2 Changing Internal Energy by Work: Adiabatic Processes#

Consider a system containing a quantity of liquid water enclosed in a rigid, thermally insulated container equipped with adiabatic walls ( $q = 0$ ).

Between 1840 and 1850, James Prescott Joule performed classic experiments to investigate mechanical and electrical equivalents of heat:

Pathway 1 (Mechanical Paddle Work): Mechanical work ( $w = 1.0\text{ kJ}$ ) was performed on the water by rotating a set of paddles, churning the viscous liquid. The temperature rose from $T_A$ to $T_B$ , causing a change in internal energy: $\Delta U = U_B - U_A$ .
Pathway 2 (Electrical Work): An identical quantity of electrical work ( $w = 1.0\text{ kJ}$ ) was supplied to the water by passing current through an immersion resistor. The recorded temperature increase was identical: $\Delta T = T_B - T_A$ .

Joule discovered that regardless of how adiabatic work is performed, a given quantity of work done on the system produces an identical change in state as measured by the temperature increment:

$\Delta U = U_2 - U_1 = w_{\text{ad}}$

This experimental invariance proves that internal energy $U$ is a characteristic state function: the adiabatic work depends solely on the initial and final states, independent of the path taken.

5.2.3 Changing Internal Energy by Heat#

Internal energy can also be altered without doing mechanical or electrical work, by putting the system in thermal contact with surroundings at a different temperature through thermally conducting (diathermic) walls.

Energy transferred across a boundary solely as a result of a temperature difference between the system and surroundings is defined as heat, denoted by $q$ .

If water at temperature $T_A$ is enclosed in a rigid, fixed-volume copper container and placed in a large water bath maintained at temperature $T_B$ ( $T_B > T_A$ ), heat flows into the system until thermal equilibrium is established. Because the vessel is rigid, no mechanical expansion work can occur ( $w = 0$ ). Under these conditions:

$\Delta U = q_V$

Where the subscript $V$ denotes heat transferred at constant volume.

5.2.4 The First Law of Thermodynamics#

When a closed system undergoes a change of state involving both heat exchange ( $q$ ) and mechanical/electrical work ( $w$ ), the net change in internal energy is given by the algebraic sum:

$\Delta U = q + w$

This formulation constitutes the mathematical statement of the First Law of Thermodynamics.

Statements of the First Law of Thermodynamics:

The total energy of an isolated system remains constant over time.
Energy can neither be created nor destroyed; it can only be transformed from one physical or chemical form into another.
A perpetual motion machine of the first kind (a machine that produces continuous mechanical work without consuming an equivalent quantity of energy) is physically impossible.

For an isolated system, no heat or work can be exchanged with the surroundings ( $q = 0, w = 0$ ). Therefore:

$\Delta U_{\text{isolated}} = 0$

Although $q$ and $w$ individually depend on the path chosen, their algebraic sum $\Delta U = q + w$ is strictly path-independent.

5.2.5 IUPAC Sign Conventions in Chemical Thermodynamics#

To maintain mathematical consistency across all equations, chemical thermodynamics adopts the standard IUPAC Sign Convention:

Quantity	Algebraic Sign	Physical Interpretation
Heat absorbed by system ( $q > 0$ )	Positive ( $+$ )	Thermal energy enters the system from surroundings; internal energy increases.
Heat released by system ( $q < 0$ )	Negative ( $-$ )	Thermal energy exits the system to surroundings; internal energy decreases.
Work done ON the system ( $w > 0$ )	Positive ( $+$ )	Surroundings compress the system or perform work; internal energy increases.
Work done BY the system ( $w < 0$ )	Negative ( $-$ )	System expands against surroundings or does work; internal energy decreases.

Important Caution: Physics vs. Chemistry Sign Convention for Work:
In many classical physics textbooks, work done by the system is defined with a positive sign, yielding $\Delta U = q - w_{\text{physics}}$ . In modern chemical thermodynamics and IUPAC standards, all energy inputs into the system carry a positive sign, so $\Delta U = q + w$ , where $w = -p_{\text{ex}}\,\Delta V$ .

NCERT Problem 5.1 (Step-by-Step Solution):
Express the change in internal energy of a system when:
(i) No heat is absorbed by the system from the surroundings, but work ( $w$ ) is done on the system. What type of wall does the system have?
(ii) No work is done on the system, but $q$ amount of heat is taken out from the system and given to the surroundings. What type of wall does the system have?
(iii) $w$ amount of work is done by the system and $q$ amount of heat is supplied to the system. What type of system would it be?

Step 1: Given Quantities & Unit Harmonization

Case (i): $q = 0$ , work done on system $= +w$ .
Case (ii): $w = 0$ , heat released to surroundings $= -q$ .
Case (iii): Work done by system $= -w$ , heat absorbed $= +q$ .

Step 2: Fundamental Governing Formula
$\Delta U = q + w$

Step 3: Direct Substitution & Arithmetic Computation

(i) $\Delta U = 0 + w = w_{\text{ad}}$ . Because $q = 0$ , the container possesses an adiabatic wall.
(ii) $\Delta U = -q + 0 = -q$ . Because heat transfers across the boundary in response to temperature differences, the container possesses a thermally conducting (diathermic) wall.
(iii) $\Delta U = q - w$ . Because energy transfers across the boundary as both heat and work, but matter is contained, this represents a closed system.

Step 4: Physical Significance & Laboratory Insight
The physical nature of the boundary governs which energy transfer modes are accessible to a system during a state transition.

NCERT Problem 5.7 (Step-by-Step Solution):
In a process, $701\text{ J}$ of heat is absorbed by a system and $394\text{ J}$ of work is done by the system. What is the change in internal energy for the process?

Step 1: Given Quantities & Unit Harmonization

Heat absorbed by system: $q = +701\text{ J}$
Work done by system: $w = -394\text{ J}$

Step 2: Fundamental Governing Formula
$\Delta U = q + w$

Step 3: Direct Substitution & Arithmetic Computation
$\Delta U = (+701\text{ J}) + (-394\text{ J}) = +307\text{ J}$

Step 4: Physical Significance & Laboratory Insight
Although the system expanded against its surroundings and expended $394\text{ J}$ of mechanical work, the thermal energy influx of $701\text{ J}$ exceeded the work done, producing a net increase of $307\text{ J}$ in the internal energy of the system.

5.3 Pressure-Volume Work and Reversible Processes#

In chemical processes, the most common form of mechanical work is pressure-volume ( $pV$ ) work, occurring when gases expand or contract against an external resisting pressure.

5.3.1 Derivation of Expansion Work#

Consider a cylinder fitted with a frictionless, massless mobile piston of cross-sectional area $A$ , containing an ideal gas at pressure $p$ .

Suppose the external opposing pressure exerted on the outer face of the piston is $p_{\text{ex}}$ .

The opposing force acting on the piston is: $F_{\text{opp}} = p_{\text{ex}} \cdot A$
If the gas expands and displaces the piston outward by a distance $l$ , the work done by the system against the resisting force is: $w = -\text{Force} \times \text{Distance} = -(p_{\text{ex}} \cdot A) \cdot l$
The change in volume is the area multiplied by displacement: $\Delta V = A \cdot l = V_f - V_i$ .
Substituting gives the fundamental expression for mechanical work against a constant external pressure: $w = -p_{\text{ex}}\,\Delta V = -p_{\text{ex}}\,(V_f - V_i)$

Verification of Sign Conventions:

During Gas Compression: $V_f < V_i \implies \Delta V < 0$ .
$w = -p_{\text{ex}}(-\Delta V) = +p_{\text{ex}}\,|\Delta V| > 0$ .
Work is done on the system, and $w$ is mathematically positive.
During Gas Expansion: $V_f > V_i \implies \Delta V > 0$ .
$w = -p_{\text{ex}}(+\Delta V) < 0$ .
Work is done by the system, and $w$ is mathematically negative.

5.3.2 Multi-Step and Reversible Processes#

If compression or expansion occurs in a single sudden step against a constant external pressure, the magnitude of work equals the area of the rectangle $p_{\text{ex}} \cdot (V_f - V_i)$ on an indicator $PV$ diagram.

If the external pressure is adjusted in a series of finite intermediate stages:

$w = -\sum p_{\text{ex}}\,\Delta V$

When the pressure changes in infinite infinitesimal increments, the process becomes thermodynamically reversible:

Reversible Process Definition:
A process is said to be reversible if it is conducted infinitely slowly through a continuous succession of equilibrium states, such that the driving force exceeds the opposing force by only an infinitesimal amount ( $dp$ ), and the process can be reversed in direction at any instant by an infinitesimal modification of external parameters.

For a reversible expansion:

$p_{\text{ex}} = p_{\text{in}} - dp$

For a reversible compression:

$p_{\text{ex}} = p_{\text{in}} + dp$

Substituting into the work integral:

$w_{\text{rev}} = -\int_{V_i}^{V_f} (p_{\text{in}} \pm dp)\,dV$

Because the product of two infinitesimal differentials $dp \cdot dV$ is negligibly small, this reduces to:

$w_{\text{rev}} = -\int_{V_i}^{V_f} p_{\text{in}}\,dV$

5.3.3 Isothermal Reversible Expansion of an Ideal Gas#

For $n$ moles of an ideal gas undergoing an isothermal process ( $T = \text{constant}$ ), the ideal gas law gives:

$p = \frac{nRT}{V}$

Substituting into the work integral:

$w_{\text{rev}} = -\int_{V_i}^{V_f} \frac{nRT}{V}\,dV = -nRT \ln\left(\frac{V_f}{V_i}\right)$

Converting from natural logarithm to common base-10 logarithm:

$w_{\text{rev}} = -2.303\,nRT \log_{10}\left(\frac{V_f}{V_i}\right)$

According to Boyle's Law, at constant temperature $p_i V_i = p_f V_f \implies \frac{V_f}{V_i} = \frac{p_i}{p_f}$ . Thus:

$w_{\text{rev}} = -2.303\,nRT \log_{10}\left(\frac{p_i}{p_f}\right)$

Maximum Work Theorem:
For any isothermal expansion between specified initial and final volumes, a reversible pathway extracts the maximum possible work from the gas: $|w_{\text{rev}}| > |w_{\text{irrev}}|$ . Conversely, compressing a gas reversibly requires the minimum possible work.

5.3.4 Free Expansion of an Ideal Gas#

Expansion of a gas into an evacuated vessel (vacuum) is termed free expansion. Because the opposing external pressure is zero ( $p_{\text{ex}} = 0$ ):

$w = -p_{\text{ex}}\,\Delta V = -0 \times \Delta V = 0$

In 1843, Joule verified experimentally that when an ideal gas undergoes free expansion inside an isolated water bath, no temperature change occurs. Therefore:

$\Delta T = 0 \implies \Delta U = 0$

Applying the First Law of Thermodynamics:

$\Delta U = q + w \implies 0 = q + 0 \implies q = 0$

Thus, during isothermal free expansion of an ideal gas into vacuum, no work is performed and no heat is absorbed or evolved.

Thermodynamic Process	Conditions Imposed	Heat ( $q$ )	Work ( $w$ )	Internal Energy Change ( $\Delta U$ )
Isothermal Reversible Expansion	$T = \text{const}, p_{\text{ex}} = p \pm dp$	$q = +2.303 nRT \log(V_f/V_i)$	$w = -2.303 nRT \log(V_f/V_i)$	$\Delta U = 0$
Isothermal Irreversible Expansion	$T = \text{const}, p_{\text{ex}} = \text{const}$	$q = +p_{\text{ex}}(V_f - V_i)$	$w = -p_{\text{ex}}(V_f - V_i)$	$\Delta U = 0$
Isothermal Free Expansion	$p_{\text{ex}} = 0$ (vacuum)	$q = 0$	$w = 0$	$\Delta U = 0$
Adiabatic Reversible Process	$q = 0$	$q = 0$	$w = \Delta U$	$\Delta U = n C_V (T_f - T_i)$
Isochoric Process	$V = \text{const}, \Delta V = 0$	$q_V = \Delta U$	$w = 0$	$\Delta U = q_V$

NCERT Problem 5.2 (Step-by-Step Solution):
Two litres of an ideal gas at a pressure of $10\text{ atm}$ expands isothermally at $25^{\circ}\text{C}$ into a vacuum until its total volume is $10\text{ litres}$ . How much heat is absorbed and how much work is done in the expansion?

Step 1: Given Quantities & Unit Harmonization

Initial volume: $V_i = 2\text{ L}$
Final volume: $V_f = 10\text{ L}$
External resisting pressure: $p_{\text{ex}} = 0\text{ atm}$ (vacuum)
Temperature: $T = 25^{\circ}\text{C} = 298.15\text{ K} = \text{constant}$

Step 2: Fundamental Governing Formula
$w = -p_{\text{ex}}\,(V_f - V_i), \quad \Delta U = q + w$

Step 3: Direct Substitution & Arithmetic Computation
$w = -0 \times (10 - 2)\text{ L} = 0\text{ J}$
For an isothermal expansion of an ideal gas, $\Delta U = 0$ :
$q = -w = 0\text{ J}$

Step 4: Physical Significance & Laboratory Insight
Because the opposing pressure is zero, the expanding gas meets no resistance. Consequently, no work is performed, and no thermal energy is drawn from the surroundings.

NCERT Problem 5.3 (Step-by-Step Solution):
Consider the same expansion as Problem 5.2, but this time conducted against a constant external pressure of $1\text{ atm}$ .

Step 1: Given Quantities & Unit Harmonization

$V_i = 2\text{ L}, \quad V_f = 10\text{ L}, \quad \Delta V = 8\text{ L}$
External pressure: $p_{\text{ex}} = 1\text{ atm}$
Unit conversion: $1\text{ L}\cdot\text{atm} = 101.325\text{ J}$

Step 2: Fundamental Governing Formula
$w = -p_{\text{ex}}\,\Delta V, \quad q = -w\text{ (for isothermal ideal gas)}$

Step 3: Direct Substitution & Arithmetic Computation
$w = -1\text{ atm} \times (10 - 2)\text{ L} = -8\text{ L}\cdot\text{atm}$
Converting to joules:
$w = -8 \times 101.325\text{ J} = -810.6\text{ J}$
Heat absorbed:
$q = -w = +8\text{ L}\cdot\text{atm} = +810.6\text{ J}$

Step 4: Physical Significance & Laboratory Insight
Expansion against a finite resisting pressure forces the gas to do work on the surroundings. To maintain constant temperature, the gas absorbs an equivalent quantity of heat ( $810.6\text{ J}$ ) from the thermal reservoir.

NCERT Problem 5.4 (Step-by-Step Solution):
Consider the expansion given in Problem 5.2, for $1\text{ mol}$ of an ideal gas conducted reversibly.

Step 1: Given Quantities & Unit Harmonization

$n = 1.0\text{ mol}$
$T = 298.15\text{ K}$
$V_i = 2.0\text{ L}, \quad V_f = 10.0\text{ L} \implies \frac{V_f}{V_i} = 5.0$
$R = 8.314\text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$ (or $0.08206\text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$ )
$\log_{10}(5) = 0.6990$

Step 2: Fundamental Governing Formula
$w_{\text{rev}} = -2.303\,nRT \log_{10}\left(\frac{V_f}{V_i}\right)$

Step 3: Direct Substitution & Arithmetic Computation
In $\text{L}\cdot\text{atm}$ :
$w_{\text{rev}} = -2.303 \times (1.0) \times (0.08206) \times (298.15) \times 0.6990 = -39.37\text{ L}\cdot\text{atm}$
In Joules:
$w_{\text{rev}} = -2.303 \times (1.0\text{ mol}) \times (8.314\text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}) \times (298.15\text{ K}) \times 0.6990$
$w_{\text{rev}} = -3988.4\text{ J} = -3.988\text{ kJ}$
Because the process is isothermal ( $\Delta U = 0$ ):
$q = -w_{\text{rev}} = +39.37\text{ L}\cdot\text{atm} = +3.988\text{ kJ}$

Step 4: Physical Significance & Laboratory Insight
Notice that reversible work ( $-39.37\text{ L}\cdot\text{atm}$ ) is nearly five times greater in magnitude than single-step irreversible work ( $-8.0\text{ L}\cdot\text{atm}$ ). An infinite number of infinitesimal steps extracts maximum mechanical work from the gas.

Interactive Simulation Experiment: Piston Gas & Indicator Lab
Open the Piston & Gas tab in the simulation suite:

Set initial volume to $2.0\text{ L}$ and pressure to $10.0\text{ atm}$ .
Choose Single-Step Irreversible Expansion against $1.0\text{ atm}$ . Note the shaded rectangular work area ( $w = -8.0\text{ L}\cdot\text{atm}$ ).
Reset and select Reversible Isothermal Expansion. Watch the piston move smoothly through multiple equilibrium points. Observe that the work area under the hyperbolic $PV$ curve expands to $-39.37\text{ L}\cdot\text{atm}$ .

5.4 Enthalpy ( $H$ ) and Heat Capacity#

5.4.1 Definition of Enthalpy as a State Function#

In a chemistry laboratory, the majority of chemical reactions are carried out in open vessels (beakers, flasks, test tubes) under constant atmospheric pressure ( $p = \text{constant}$ ), rather than in sealed constant-volume bombs.

Applying the First Law under constant pressure conditions:

$\Delta U = q_p + w = q_p - p\,\Delta V$

Rearranging for heat absorbed at constant pressure:

$q_p = \Delta U + p\,\Delta V$

Writing $\Delta U = U_2 - U_1$ and $\Delta V = V_2 - V_1$ :

$q_p = (U_2 - U_1) + p\,(V_2 - V_1) = (U_2 + pV_2) - (U_1 + pV_1)$

This prompts the introduction of a new thermodynamic state function, the Enthalpy ( $H$ ), derived from the Greek word enthalpien (meaning to heat or warm):

$H = U + pV$

Because $U$ , $p$ , and $V$ are all state functions, $H$ is strictly a state function. Therefore, the enthalpy change for a finite process is:

$\Delta H = H_2 - H_1 = q_p$

Key Principle:
The change in enthalpy of a system during a chemical or physical transformation is quantitatively equal to the heat absorbed or released at constant pressure.

Exothermic Reaction: System releases heat to surroundings $\implies q_p < 0 \implies \Delta H < 0$ .
Endothermic Reaction: System absorbs heat from surroundings $\implies q_p > 0 \implies \Delta H > 0$ .

5.4.2 Relationship Between $Delta H$ and $Delta U$ #

Differentiating the enthalpy equation for a finite transformation:

$\Delta H = \Delta U + \Delta(pV)$

If pressure is maintained constant:

$\Delta H = \Delta U + p\,\Delta V$

For chemical reactions involving only condensed phases (solids and liquids), volume changes upon heating or reacting are minuscule ( $\Delta V \approx 0$ ). Consequently, for solid-liquid reactions:

$\Delta H \approx \Delta U$

However, when chemical reactions involve gases, the volume occupied by gaseous products can differ substantially from gaseous reactants. Assuming ideal gas behavior for all gaseous components ( $pV = nRT$ ):

$p V_A = n_A R T \quad \text{and} \quad p V_B = n_B R T$

Subtracting the reactant state from the product state:

$p(V_B - V_A) = (n_B - n_A)RT \implies p\,\Delta V = \Delta n_g\,RT$

Where $\Delta n_g$ is defined as:

$\Delta n_g = \sum n_{\text{gaseous products}} - \sum n_{\text{gaseous reactants}}$

Substituting into the enthalpy equation yields the core relationship:

$\Delta H = \Delta U + \Delta n_g\,RT$

Reaction Type	Stoichiometric Change	Enthalpy vs. Internal Energy	Classic Chemical Example
$\Delta n_g = 0$	Gaseous moles conserved	$\Delta H = \Delta U$	$\mathrm{H}_2\text{(g)} + \mathrm{Cl}_2\text{(g)} \rightarrow 2\mathrm{HCl}\text{(g)}$
$\Delta n_g > 0$	Gaseous moles expand	$\Delta H > \Delta U$	$\mathrm{PCl}_5\text{(g)} \rightarrow \mathrm{PCl}_3\text{(g)} + \mathrm{Cl}_2\text{(g)}$
$\Delta n_g < 0$	Gaseous moles contract	$\Delta H < \Delta U$	$\mathrm{N}_2\text{(g)} + 3\mathrm{H}_2\text{(g)} \rightarrow 2\mathrm{NH}_3\text{(g)}$

NCERT Problem 5.5 (Step-by-Step Solution):
If water vapour is assumed to be a perfect gas, molar enthalpy change for vaporisation of $1\text{ mol}$ of water at $1\text{ bar}$ and $100^{\circ}\text{C}$ is $41.00\text{ kJ}\cdot\text{mol}^{-1}$ . Calculate the internal energy change when $1\text{ mol}$ of water is vaporised at $1\text{ bar}$ pressure and $100^{\circ}\text{C}$ .

Step 1: Given Quantities & Unit Harmonization

Chemical transformation: $\mathrm{H}_2\mathrm{O}\text{(l)} \rightarrow \mathrm{H}_2\mathrm{O}\text{(g)}$
Enthalpy of vaporization: $\Delta_{\text{vap}}H = +41.00\text{ kJ}\cdot\text{mol}^{-1}$
Temperature: $T = 100^{\circ}\text{C} = 373.15\text{ K}$
Gas constant: $R = 8.314\text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1} = 8.314 \times 10^{-3}\text{ kJ}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$
Change in gaseous moles: $\Delta n_g = 1\text{ mol (gas)} - 0\text{ mol (liquid)} = +1.0\text{ mol}$

Step 2: Fundamental Governing Formula
$\Delta H = \Delta U + \Delta n_g\,RT \implies \Delta U = \Delta H - \Delta n_g\,RT$

Step 3: Direct Substitution & Arithmetic Computation
Compute the work of expansion:
$\Delta n_g\,RT = (1.0\text{ mol}) \times (8.314 \times 10^{-3}\text{ kJ}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}) \times (373.15\text{ K}) = 3.102\text{ kJ}\cdot\text{mol}^{-1}$
Compute internal energy change:
$\Delta U = 41.00\text{ kJ}\cdot\text{mol}^{-1} - 3.102\text{ kJ}\cdot\text{mol}^{-1} = 37.898\text{ kJ}\cdot\text{mol}^{-1} \approx 37.90\text{ kJ}\cdot\text{mol}^{-1}$

Step 4: Physical Significance & Laboratory Insight
Of the $41.00\text{ kJ}$ of thermal energy supplied to vaporize one mole of liquid water at $100^{\circ}\text{C}$ , $37.90\text{ kJ}$ is utilized internally to overcome intermolecular hydrogen bonding networks, while $3.10\text{ kJ}$ is expended pushing back the ambient atmosphere ( $p\Delta V$ work).

5.4.3 Heat Capacity and Specific Heat#

When heat is absorbed by a substance, its temperature increases. The temperature increment $\Delta T$ is directly proportional to the quantity of heat transferred:

$q = C\,\Delta T$

The proportionality factor $C$ is termed the heat capacity of the system:

It is an extensive property with units of $\text{J}\cdot\text{K}^{-1}$ .
When $C$ is large, absorbing a large quantity of thermal energy produces only a small temperature change (for instance, liquid water).

To establish an intensive property independent of sample size:

Specific Heat Capacity ( $c$ ): The quantity of heat required to raise the temperature of one unit mass ( $1\text{ g}$ or $1\text{ kg}$ ) of a substance by one Kelvin: $q = m \cdot c \cdot \Delta T$ Units: $\text{J}\cdot\text{g}^{-1}\cdot\text{K}^{-1}$ or $\text{J}\cdot\text{kg}^{-1}\cdot\text{K}^{-1}$ .
Molar Heat Capacity ( $C_m$ ): The quantity of heat required to raise the temperature of one mole of a substance by one Kelvin: $C_m = \frac{C}{n}, \quad q = n \cdot C_m \cdot \Delta T$ Units: $\text{J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$ .

5.4.4 Derivation of the Meyer Relation ( $C_p - C_V = R$ ) for an Ideal Gas#

Because the state of a gas can be altered under conditions of constant volume or constant pressure, two distinct heat capacities are defined:

Heat Capacity at Constant Volume ( $C_V$ ):
$q_V = \Delta U = C_V\,\Delta T \implies C_V = \left(\frac{\partial U}{\partial T}\right)_V$
Heat Capacity at Constant Pressure ( $C_p$ ):
$q_p = \Delta H = C_p\,\Delta T \implies C_p = \left(\frac{\partial H}{\partial T}\right)_p$

To derive their relationship for $1\text{ mole}$ of an ideal gas:

From the definition of enthalpy: $H = U + pV$
For one mole of an ideal gas, $pV = RT$ . Substituting gives: $H = U + RT$
Differentiating with respect to temperature: $\frac{dH}{dT} = \frac{dU}{dT} + R$
Substituting $C_p = \frac{dH}{dT}$ and $C_V = \frac{dU}{dT}$ : $C_p = C_V + R \implies C_p - C_V = R$

Physical Reason Why $C_p > C_V$ :
When an ideal gas is heated at constant volume, all supplied heat is converted into molecular kinetic energy, raising the temperature directly (no work is performed). When heated at constant pressure, the gas expands against the atmosphere, so extra energy must be supplied to perform mechanical expansion work ( $p\Delta V = R\Delta T$ ) while achieving the same temperature rise.

NCERT Problem 5.9 (Step-by-Step Solution):
Calculate the number of $\text{kJ}$ of heat necessary to raise the temperature of $60.0\text{ g}$ of aluminium from $35^{\circ}\text{C}$ to $55^{\circ}\text{C}$ . Molar heat capacity of $\mathrm{Al}$ is $24\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$ . Molar mass of $\mathrm{Al} = 27.0\text{ g/mol}$ .

Step 1: Given Quantities & Unit Harmonization

Mass of aluminium: $m = 60.0\text{ g}$
Molar mass: $M = 27.0\text{ g/mol}$
Initial temperature: $T_1 = 35^{\circ}\text{C}$
Final temperature: $T_2 = 55^{\circ}\text{C} \implies \Delta T = (55 - 35) = 20\text{ K}$
Molar heat capacity: $C_m = 24.0\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$

Step 2: Fundamental Governing Formula
$n = \frac{m}{M}, \quad q = n \cdot C_m \cdot \Delta T$

Step 3: Direct Substitution & Arithmetic Computation
$n = \frac{60.0\text{ g}}{27.0\text{ g/mol}} = 2.222\text{ mol}$
$q = (2.222\text{ mol}) \times (24.0\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}) \times (20\text{ K}) = 1066.67\text{ J} = 1.067\text{ kJ}$

Step 4: Physical Significance & Laboratory Insight
Supplying $1.067\text{ kJ}$ of energy elevates the lattice vibrational kinetic energies of $2.222\text{ moles}$ of aluminium atoms across the $20\text{ K}$ temperature interval.

5.5 Measurement of $Delta U$ and $Delta H$ : Calorimetry#

Energy changes accompanying physical transformations or chemical reactions are determined experimentally using calorimetry.

5.5.1 Constant-Volume Bomb Calorimetry ( $Delta U$ Measurement)#

To measure the heat of combustion at constant volume, a bomb calorimeter is employed:

A known mass of combustible sample is placed in a platinum cup inside a heavy-walled, sealed steel pressure vessel (the bomb).
The bomb is charged with pure dioxygen under high pressure ( $\approx 25\text{ to }30\text{ atm}$ ) and submerged in a thermally insulated water bath equipped with a precision thermometer and mechanical stirrer.
The sample is ignited electrically. The heat liberated passes into the surrounding water and calorimeter components.
Because the steel bomb is rigid and completely sealed: $\Delta V = 0 \implies w = -p_{\text{ex}}\,\Delta V = 0$
From the First Law: $\Delta U = q_V$
The total heat absorbed by the calorimeter is: $q_{\text{calorimeter}} = C_V \cdot \Delta T$
By conservation of energy, the heat liberated by the combustion reaction is: $q_{\text{reaction}} = -q_{\text{calorimeter}} = -C_V \cdot \Delta T$

5.5.2 Constant-Pressure Calorimetry ( $Delta H$ Measurement)#

For reactions conducted in aqueous solutions (such as acid-base neutralization or precipitation), heat is measured in a coffee cup calorimeter operating under ambient atmospheric pressure:

The polystyrene cup provides thermal insulation while remaining open to the atmosphere.
Pressure remains constant ( $p = p_{\text{atm}}$ ).
The heat measured directly reflects the enthalpy of reaction: $q_p = \Delta_r H = -m_{\text{solution}} \cdot c_s \cdot \Delta T$

NCERT Problem 5.6 (Step-by-Step Solution):
$1.0\text{ g}$ of graphite is burnt in a bomb calorimeter in excess of oxygen at $298\text{ K}$ and $1\text{ atm}$ pressure according to the equation:
$\mathrm{C}\text{(graphite)} + \mathrm{O}_2\text{(g)} \rightarrow \mathrm{CO}_2\text{(g)}$
During the reaction, temperature rises from $298.0\text{ K}$ to $299.0\text{ K}$ . If the heat capacity of the bomb calorimeter is $20.7\text{ kJ/K}$ , what is the enthalpy change for the above reaction at $298\text{ K}$ and $1\text{ atm}$ ?

Step 1: Given Quantities & Unit Harmonization

Mass of graphite: $m = 1.0\text{ g}$
Molar mass of carbon: $M = 12.01\text{ g/mol}$
Temperature rise: $\Delta T = (299.0 - 298.0) = 1.0\text{ K}$
Calorimeter heat capacity: $C_V = 20.7\text{ kJ/K}$

Step 2: Fundamental Governing Formula
$q_{\text{reaction}} = -C_V \cdot \Delta T, \quad \Delta U = \frac{q_{\text{reaction}}}{n}, \quad \Delta H = \Delta U + \Delta n_g\,RT$

Step 3: Direct Substitution & Arithmetic Computation
Heat absorbed by calorimeter:
$q = C_V \times \Delta T = 20.7\text{ kJ/K} \times 1.0\text{ K} = 20.7\text{ kJ}$
Heat of reaction for $1.0\text{ g}$ :
$q_{\text{reaction}} = -20.7\text{ kJ}$
Moles of carbon burned:
$n = \frac{1.0\text{ g}}{12.01\text{ g/mol}} = 0.08326\text{ mol}$
Molar internal energy change:
$\Delta U = \frac{-20.7\text{ kJ}}{1.0\text{ g}} \times 12.01\text{ g/mol} = -2.486 \times 10^2\text{ kJ/mol} = -248.6\text{ kJ/mol}$
Evaluate $\Delta n_g$ for $\mathrm{C}\text{(s)} + \mathrm{O}_2\text{(g)} \rightarrow \mathrm{CO}_2\text{(g)}$ :
$\Delta n_g = 1\text{ mol (CO}_2\text{)} - 1\text{ mol (O}_2\text{)} = 0$
Therefore:
$\Delta H = \Delta U + (0)RT = \Delta U = -2.486 \times 10^2\text{ kJ/mol} = -248.6\text{ kJ/mol}$

Step 4: Physical Significance & Laboratory Insight
Because the number of moles of gas is identically conserved during complete combustion of graphite to carbon dioxide, no pressure-volume expansion work is performed. Hence, the constant-volume energy change $\Delta U$ and constant-pressure enthalpy change $\Delta H$ are numerically identical.

NCERT Problem 5.8 (Step-by-Step Solution):
The reaction of cyanamide, $\mathrm{NH}_2\mathrm{CN}\text{(s)}$ , with dioxygen was carried out in a bomb calorimeter, and $\Delta U$ was found to be $-742.7\text{ kJ}\cdot\text{mol}^{-1}$ at $298\text{ K}$ . Calculate enthalpy change for the reaction at $298\text{ K}$ :
$\mathrm{NH}_2\mathrm{CN}\text{(s)} + \frac{3}{2}\mathrm{O}_2\text{(g)} \rightarrow \mathrm{N}_2\text{(g)} + \mathrm{CO}_2\text{(g)} + \mathrm{H}_2\mathrm{O}\text{(l)}$

Step 1: Given Quantities & Unit Harmonization

$\Delta U = -742.7\text{ kJ}\cdot\text{mol}^{-1}$
$T = 298.15\text{ K}$
$R = 8.314 \times 10^{-3}\text{ kJ}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$
Gaseous products: $1\text{ mol }\mathrm{N}_2 + 1\text{ mol }\mathrm{CO}_2 = 2\text{ mol}$
Gaseous reactants: $\frac{3}{2}\text{ mol }\mathrm{O}_2 = 1.5\text{ mol}$
Gaseous change: $\Delta n_g = 2 - 1.5 = +0.5\text{ mol}$

Step 2: Fundamental Governing Formula
$\Delta H = \Delta U + \Delta n_g\,RT$

Step 3: Direct Substitution & Arithmetic Computation
$\Delta n_g\,RT = (0.5\text{ mol}) \times (8.314 \times 10^{-3}\text{ kJ}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}) \times (298.15\text{ K}) = +1.240\text{ kJ}\cdot\text{mol}^{-1}$
$\Delta H = -742.7\text{ kJ}\cdot\text{mol}^{-1} + 1.240\text{ kJ}\cdot\text{mol}^{-1} = -741.46\text{ kJ}\cdot\text{mol}^{-1} \approx -741.5\text{ kJ}\cdot\text{mol}^{-1}$

Step 4: Physical Significance & Laboratory Insight
Because half a mole of gas is generated during combustion, the system performs expansion work against the surroundings at constant pressure, making the enthalpy change slightly less negative (less exothermic) than the constant-volume internal energy change.

Interactive Simulation Experiment: Reaction Calorimeter & Temperature Lab
Open the Calorimeter tab in the simulation suite:

Select the Combustion of Graphite preset in the Bomb Calorimeter mode.
Observe the virtual temperature sensor rise from $298.0\text{ K}$ to $299.0\text{ K}$ .
Verify that the computer-logged heat output matches $q_V = -20.7\text{ kJ}$ .
Switch to Reaction Temperature Lab and compare an exothermic profile (neutralization of $\mathrm{HCl} + \mathrm{NaOH}$ , temperature rises, $\Delta H < 0$ ) against an endothermic profile (dissolution of $\mathrm{NH}_4\mathrm{NO}_3$ , temperature drops, $\Delta H > 0$ ).

5.6 Standard Enthalpy of Reactions and Thermochemical Laws#

5.6.1 Standard State Convention#

Because the enthalpy change of a reaction varies with temperature and pressure, standard thermodynamic reference conditions are established:

Standard State Definition:
The standard state of a chemical substance at a specified temperature (customarily $298.15\text{ K} = 25^{\circ}\text{C}$ ) is its pure, stable form of aggregation at a pressure of exactly $1\text{ bar}$ ( $10^5\text{ Pa}$ ).

For a gas: Pure gas behaving ideally at $p = 1\text{ bar}$ .
For a liquid or solid: Pure substance at $p = 1\text{ bar}$ .
For a solute in solution: Standard concentration of $1.0\text{ mol}\cdot\text{L}^{-1}$ ( $1\text{ M}$ ) at $p = 1\text{ bar}$ .

Standard state quantities are designated by the superscript symbol $^{\circ}$ (or $^{\ominus}$ ), for example: $\Delta_r H^{\circ}$ .

5.6.2 Standard Enthalpy of Formation ( $\Delta_f H^{\circ}$ )#

Standard Molar Enthalpy of Formation Definition:
The standard enthalpy change accompanying the formation of exactly one mole of a chemical compound in its standard state from its constituent elements in their most stable reference states of aggregation at $298.15\text{ K}$ and $1\text{ bar}$ pressure.

By international thermodynamic convention:

$\Delta_f H^{\circ}[\text{element in its reference state}] = 0.00\text{ kJ}\cdot\text{mol}^{-1}$

Element	Most Stable Reference State at $298\text{ K}$ , $1\text{ bar}$	Standard Enthalpy of Formation ( $\Delta_f H^{\circ}$ )
Carbon	Graphite, $\mathrm{C}\text{(graphite, s)}$	$0.00\text{ kJ/mol}$
Carbon	Diamond, $\mathrm{C}\text{(diamond, s)}$	$+1.89\text{ kJ/mol}$ (Not reference state)
Oxygen	Dioxygen gas, $\mathrm{O}_2\text{(g)}$	$0.00\text{ kJ/mol}$
Oxygen	Ozone gas, $\mathrm{O}_3\text{(g)}$	$+142.7\text{ kJ/mol}$ (Not reference state)
Hydrogen	Dihydrogen gas, $\mathrm{H}_2\text{(g)}$	$0.00\text{ kJ/mol}$
Bromine	Liquid dibromine, $\mathrm{Br}_2\text{(l)}$	$0.00\text{ kJ/mol}$
Chlorine	Dichlorine gas, $\mathrm{Cl}_2\text{(g)}$	$0.00\text{ kJ/mol}$
Sulfur	Rhombic sulfur, $\mathrm{S}_{\alpha}\text{(rhombic, s)}$	$0.00\text{ kJ/mol}$

Enthalpy of Reaction from Standard Enthalpies of Formation

For any generalized chemical reaction:

$\sum b_i\,\text{Reactants}_i \rightarrow \sum a_i\,\text{Products}_i$

The standard reaction enthalpy is computed via:

$\Delta_r H^{\circ} = \sum a_i\,\Delta_f H^{\circ}(\text{products}) - \sum b_i\,\Delta_f H^{\circ}(\text{reactants})$

Where $a_i$ and $b_i$ represent stoichiometric coefficients from the balanced chemical equation.

5.6.3 Hess's Law of Constant Heat Summation#

Because enthalpy is a state function, the net enthalpy change between an initial state (reactants) and a final state (products) is completely path-independent.

Hess's Law of Constant Heat Summation:
If a chemical reaction can be expressed as the algebraic sum of a sequence of intermediate chemical reactions, the standard reaction enthalpy for the overall process is equal to the algebraic sum of the standard reaction enthalpies of each individual intermediate step at the same temperature:
$\Delta_r H^{\circ} = \Delta_r H_1^{\circ} + \Delta_r H_2^{\circ} + \Delta_r H_3^{\circ} + \cdots$

Rules for Manipulating Thermochemical Equations:

Reversing an Equation: When a thermochemical reaction is written in reverse, the algebraic sign of $\Delta_r H^{\circ}$ must be inverted: $\mathrm{N}_2\text{(g)} + 3\mathrm{H}_2\text{(g)} \rightarrow 2\mathrm{NH}_3\text{(g)}, \quad \Delta_r H^{\circ} = -92.4\text{ kJ/mol}$ $2\mathrm{NH}_3\text{(g)} \rightarrow \mathrm{N}_2\text{(g)} + 3\mathrm{H}_2\text{(g)}, \quad \Delta_r H^{\circ} = +92.4\text{ kJ/mol}$
Scaling Coefficients: Multiplying all stoichiometric coefficients in a balanced equation by a factor $c$ scales the magnitude of $\Delta_r H^{\circ}$ by the same factor $c$ .
Algebraic Addition: When adding equations, chemical species appearing identically on both the reactant and product sides cancel out as spectator species.

NCERT Problem 5.9 (Step-by-Step Solution):
The combustion of one mole of benzene takes place at $298\text{ K}$ and $1\text{ atm}$ . After combustion, $\mathrm{CO}_2\text{(g)}$ and $\mathrm{H}_2\mathrm{O}\text{(l)}$ are produced and $3267.0\text{ kJ}$ of heat is liberated. Calculate the standard enthalpy of formation, $\Delta_f H^{\circ}$ of benzene. Standard enthalpies of formation of $\mathrm{CO}_2\text{(g)}$ and $\mathrm{H}_2\mathrm{O}\text{(l)}$ are $-393.5\text{ kJ}\cdot\text{mol}^{-1}$ and $-285.83\text{ kJ}\cdot\text{mol}^{-1}$ respectively.

Step 1: Given Quantities & Unit Harmonization
Target reaction (formation of benzene):
$6\mathrm{C}\text{(graphite)} + 3\mathrm{H}_2\text{(g)} \rightarrow \mathrm{C}_6\mathrm{H}_6\text{(l)}, \quad \Delta_f H^{\circ} = ?$
Given thermochemical data:
(1) $\mathrm{C}_6\mathrm{H}_6\text{(l)} + \frac{15}{2}\mathrm{O}_2\text{(g)} \rightarrow 6\mathrm{CO}_2\text{(g)} + 3\mathrm{H}_2\mathrm{O}\text{(l)}, \quad \Delta_c H^{\circ} = -3267.0\text{ kJ/mol}$
(2) $\mathrm{C}\text{(graphite)} + \mathrm{O}_2\text{(g)} \rightarrow \mathrm{CO}_2\text{(g)}, \quad \Delta_f H^{\circ} = -393.5\text{ kJ/mol}$
(3) $\mathrm{H}_2\text{(g)} + \frac{1}{2}\mathrm{O}_2\text{(g)} \rightarrow \mathrm{H}_2\mathrm{O}\text{(l)}, \quad \Delta_f H^{\circ} = -285.83\text{ kJ/mol}$

Step 2: Fundamental Governing Formula
Method A (Hess's Law Summation):
Target $= 6 \times (2) + 3 \times (3) - (1)$
Method B (Formation Formula):
$\Delta_r H^{\circ} = \sum \Delta_f H^{\circ}(\text{products}) - \sum \Delta_f H^{\circ}(\text{reactants})$

Step 3: Direct Substitution & Arithmetic Computation
Using Method B on the combustion reaction (1):
$\Delta_c H^{\circ} = [6\,\Delta_f H^{\circ}(\mathrm{CO}_2) + 3\,\Delta_f H^{\circ}(\mathrm{H}_2\mathrm{O})] - [\Delta_f H^{\circ}(\mathrm{C}_6\mathrm{H}_6) + \frac{15}{2}\,\Delta_f H^{\circ}(\mathrm{O}_2)]$
Substituting known values ( $\Delta_f H^{\circ}(\mathrm{O}_2) = 0$ ):
$-3267.0 = [6(-393.5) + 3(-285.83)] - [\Delta_f H^{\circ}(\mathrm{C}_6\mathrm{H}_6) + 0]$
$-3267.0 = [-2361.0 - 857.49] - \Delta_f H^{\circ}(\mathrm{C}_6\mathrm{H}_6)$
$-3267.0 = -3218.49 - \Delta_f H^{\circ}(\mathrm{C}_6\mathrm{H}_6)$
$\Delta_f H^{\circ}(\mathrm{C}_6\mathrm{H}_6) = -3218.49 + 3267.0 = +48.51\text{ kJ}\cdot\text{mol}^{-1}$

Step 4: Physical Significance & Laboratory Insight
The standard enthalpy of formation of liquid benzene is positive ( $+48.51\text{ kJ/mol}$ ), establishing that benzene is an endothermic compound with higher potential energy than its constituent elements in their reference states.

Interactive Simulation Experiment: Hess's Law Puzzle & Reaction Builder
Open the Hess's Law tab in the simulation suite:

Select the Benzene Formation puzzle.
Use the multiplier slider to scale the carbon combustion equation by $6$ . Observe the enthalpy card dynamically scale to $-2361.0\text{ kJ}$ .
Scale the water formation equation by $3$ (enthalpy becomes $-857.49\text{ kJ}$ ).
Click Invert Direction on the benzene combustion equation; observe the sign invert from $-3267.0\text{ kJ}$ to $+3267.0\text{ kJ}$ .
Click Sum Reactions. Verify spectator $\mathrm{O}_2, \mathrm{CO}_2, \mathrm{H}_2\mathrm{O}$ cancel out, leaving the net target reaction with $\Delta_f H^{\circ} = +48.51\text{ kJ/mol}$ .

5.7 Enthalpies of Different Types of Reactions#

5.7.1 Enthalpy of Phase Transformations#

Transformation	Process Description	NCERT Example	Standard Enthalpy Value
Enthalpy of Fusion ( $\Delta_{\text{fus}}H^{\circ}$ )	Heat required to melt one mole of a solid substance at its melting point under standard pressure ( $1\text{ bar}$ ).	$\mathrm{H}_2\mathrm{O}\text{(s)} \rightarrow \mathrm{H}_2\mathrm{O}\text{(l)}$ at $273.15\text{ K}$	$+6.01\text{ kJ}\cdot\text{mol}^{-1}$
Enthalpy of Vaporization ( $\Delta_{\text{vap}}H^{\circ}$ )	Heat required to vaporize one mole of a liquid at its boiling point under standard pressure ( $1\text{ bar}$ ).	$\mathrm{H}_2\mathrm{O}\text{(l)} \rightarrow \mathrm{H}_2\mathrm{O}\text{(g)}$ at $373.15\text{ K}$	$+40.79\text{ kJ}\cdot\text{mol}^{-1}$
Enthalpy of Sublimation ( $\Delta_{\text{sub}}H^{\circ}$ )	Enthalpy change when one mole of a solid transforms directly into vapor under standard pressure ( $1\text{ bar}$ ).	$\mathrm{CO}_2\text{(s, dry ice)} \rightarrow \mathrm{CO}_2\text{(g)}$ at $195\text{ K}$	$+25.2\text{ kJ}\cdot\text{mol}^{-1}$

Phase transformations obey Hess's Law:

$\Delta_{\text{sub}}H^{\circ} = \Delta_{\text{fus}}H^{\circ} + \Delta_{\text{vap}}H^{\circ}$

The magnitude of $\Delta_{\text{vap}}H^{\circ}$ directly reflects the strength of intermolecular forces in the condensed phase. Water exhibits a high vaporization enthalpy ( $40.79\text{ kJ/mol}$ ) due to extensive three-dimensional hydrogen bonding networks, whereas acetone requires only $29.1\text{ kJ/mol}$ due to weaker dipole-dipole attractions.

5.7.2 Enthalpy of Combustion ( $\Delta_c H^{\circ}$ )#

Standard Enthalpy of Combustion Definition:
The enthalpy change accompanying the complete combustion of one mole of a substance in its standard state with excess dioxygen, with all reactants and products measured in their standard states at the specified temperature.

Combustion reactions are invariably exothermic ( $\Delta_c H^{\circ} < 0$ ). They represent the energetic foundation of industrial manufacturing, heating, and biological metabolism:

Complete combustion of butane in LPG cylinders: $\mathrm{C}_4\mathrm{H}_{10}\text{(g)} + \frac{13}{2}\mathrm{O}_2\text{(g)} \rightarrow 4\mathrm{CO}_2\text{(g)} + 5\mathrm{H}_2\mathrm{O}\text{(l)}, \quad \Delta_c H^{\circ} = -2658.0\text{ kJ}\cdot\text{mol}^{-1}$
Cellular respiration of D-glucose in living organisms: $\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6\text{(s)} + 6\mathrm{O}_2\text{(g)} \rightarrow 6\mathrm{CO}_2\text{(g)} + 6\mathrm{H}_2\mathrm{O}\text{(l)}, \quad \Delta_c H^{\circ} = -2802.0\text{ kJ}\cdot\text{mol}^{-1}$

5.7.3 Enthalpy of Atomization ( $\Delta_a H^{\circ}$ )#

The enthalpy change accompanying the complete cleavage of all chemical bonds in one mole of a substance to obtain isolated atoms in the gas phase:

For diatomic molecules, it is equal to the bond dissociation enthalpy: $\mathrm{H}_2\text{(g)} \rightarrow 2\mathrm{H}\text{(g)}, \quad \Delta_a H^{\circ} = +435.0\text{ kJ}\cdot\text{mol}^{-1}$
For metals, it equals the enthalpy of sublimation: $\mathrm{Na}\text{(s)} \rightarrow \mathrm{Na}\text{(g)}, \quad \Delta_a H^{\circ} = +108.4\text{ kJ}\cdot\text{mol}^{-1}$
For polyatomic molecules, it equals the sum of all bond dissociation energies: $\mathrm{CH}_4\text{(g)} \rightarrow \mathrm{C}\text{(g)} + 4\mathrm{H}\text{(g)}, \quad \Delta_a H^{\circ} = +1665.0\text{ kJ}\cdot\text{mol}^{-1}$

5.7.4 Bond Enthalpy ( $\Delta_{\text{bond}}H^{\circ}$ )#

In chemical reactions, old bonds in reactant molecules are severed (requiring energy input, an endothermic process) and new bonds in product molecules are forged (liberating energy, an exothermic process).

Two distinct terms are utilized depending on molecular complexity:

Bond Dissociation Enthalpy: The enthalpy required to break one mole of a specific covalent bond in a gaseous diatomic molecule. $\mathrm{Cl}_2\text{(g)} \rightarrow 2\mathrm{Cl}\text{(g)}, \quad \Delta_{\mathrm{Cl}-\mathrm{Cl}}H^{\circ} = +242.0\text{ kJ}\cdot\text{mol}^{-1}$
Mean (Average) Bond Enthalpy: In polyatomic molecules, stepwise bond cleavage reveals that each successive bond requires a different quantity of energy due to electronic redistribution in the remaining radical fragments:
- Step 1: $\mathrm{CH}_4\text{(g)} \rightarrow \mathrm{CH}_3\text{(g)} + \mathrm{H}\text{(g)}, \quad \Delta H_1 = +427\text{ kJ/mol}$
- Step 2: $\mathrm{CH}_3\text{(g)} \rightarrow \mathrm{CH}_2\text{(g)} + \mathrm{H}\text{(g)}, \quad \Delta H_2 = +439\text{ kJ/mol}$
- Step 3: $\mathrm{CH}_2\text{(g)} \rightarrow \mathrm{CH}\text{(g)} + \mathrm{H}\text{(g)}, \quad \Delta H_3 = +452\text{ kJ/mol}$
- Step 4: $\mathrm{CH}\text{(g)} \rightarrow \mathrm{C}\text{(g)} + \mathrm{H}\text{(g)}, \quad \Delta H_4 = +347\text{ kJ/mol}$

Total atomization enthalpy: $\Delta_a H^{\circ} = 427 + 439 + 452 + 347 = 1665\text{ kJ/mol}$ .
The mean bond enthalpy for the $\mathrm{C}-\mathrm{H}$ bond is the arithmetic average:

$\Delta_{\mathrm{C}-\mathrm{H}}H^{\circ} = \frac{1}{4}\,\Delta_a H^{\circ} = \frac{1665}{4} = 416.25\text{ kJ}\cdot\text{mol}^{-1}$

Estimating Enthalpy of Reaction from Bond Enthalpies:

For all-gas-phase reactions, the standard enthalpy of reaction can be approximated:

$\Delta_r H^{\circ} \approx \sum \text{Bond Enthalpies of Reactants (bonds broken)} - \sum \text{Bond Enthalpies of Products (bonds formed)}$

Interactive Simulation Experiment: Bond Breaking & Making Lab
Open the Bond Enthalpy tab in the simulation suite:

Select the reaction $\mathrm{CH}_4\text{(g)} + 2\mathrm{O}_2\text{(g)} \rightarrow \mathrm{CO}_2\text{(g)} + 2\mathrm{H}_2\mathrm{O}\text{(g)}$ .
Click Break Reactant Bonds. The tool severs $4 \times \mathrm{C}-\mathrm{H}$ ( $4 \times 414 = 1656\text{ kJ}$ ) and $2 \times \mathrm{O}=\mathrm{O}$ ( $2 \times 498 = 996\text{ kJ}$ ). Energy absorbed: $+2652\text{ kJ}$ .
Click Form Product Bonds. The tool creates $2 \times \mathrm{C}=\mathrm{O}$ ( $2 \times 741 = 1482\text{ kJ}$ ) and $4 \times \mathrm{O}-\mathrm{H}$ ( $4 \times 464 = 1856\text{ kJ}$ ). Energy liberated: $-3338\text{ kJ}$ .
Observe the net energy balance: $\Delta_r H^{\circ} \approx 2652 - 3338 = -686\text{ kJ/mol}$ .

5.7.5 Lattice Enthalpy and the Born-Haber Cycle#

Lattice Enthalpy ( $\Delta_{\text{lattice}}H^{\circ}$ ) Definition:
The enthalpy change accompanying the complete dissociation of one mole of a solid ionic crystalline compound into its constituent isolated gaseous ions:
$\mathrm{MX}\text{(s)} \rightarrow \mathrm{M}^+\text{(g)} + \mathrm{X}^-\text{(g)}, \quad \Delta_{\text{lattice}}H^{\circ} > 0$

Because gaseous ions cannot be pulled apart directly in a calorimeter without solvent or surface interactions, lattice enthalpies are evaluated indirectly using a thermodynamic closed cycle devised by Max Born and Fritz Haber: the Born-Haber Cycle.

Deconstruction of the Born-Haber Cycle for Sodium Chloride ( $mathrm{NaCl}$ ):

Sublimation of Metallic Sodium:
$\mathrm{Na}\text{(s)} \rightarrow \mathrm{Na}\text{(g)}, \quad \Delta_{\text{sub}}H^{\circ} = +108.4\text{ kJ/mol}$
Ionization of Gaseous Sodium Atoms:
$\mathrm{Na}\text{(g)} \rightarrow \mathrm{Na}^+\text{(g)} + e^-, \quad \Delta_i H^{\circ} = +496.0\text{ kJ/mol}$
Dissociation of Dichlorine Molecules:
$\frac{1}{2}\mathrm{Cl}_2\text{(g)} \rightarrow \mathrm{Cl}\text{(g)}, \quad \frac{1}{2}\,\Delta_{\text{bond}}H^{\circ} = +121.0\text{ kJ/mol}$
Electron Gain by Gaseous Chlorine Atoms:
$\mathrm{Cl}\text{(g)} + e^- \rightarrow \mathrm{Cl}^-\text{(g)}, \quad \Delta_{\text{eg}}H^{\circ} = -348.6\text{ kJ/mol}$
Condensation of Gaseous Ions into Crystal Lattice:
$\mathrm{Na}^+\text{(g)} + \mathrm{Cl}^-\text{(g)} \rightarrow \mathrm{NaCl}\text{(s)}, \quad -\Delta_{\text{lattice}}H^{\circ}$
Direct Standard Formation Reaction:
$\mathrm{Na}\text{(s)} + \frac{1}{2}\mathrm{Cl}_2\text{(g)} \rightarrow \mathrm{NaCl}\text{(s)}, \quad \Delta_f H^{\circ} = -411.2\text{ kJ/mol}$

Applying Hess's Law around the closed cycle:

$\Delta_f H^{\circ} = \Delta_{\text{sub}}H^{\circ} + \Delta_i H^{\circ} + \frac{1}{2}\,\Delta_{\text{bond}}H^{\circ} + \Delta_{\text{eg}}H^{\circ} - \Delta_{\text{lattice}}H^{\circ}$

Rearranging to solve for the lattice dissociation enthalpy of $\mathrm{NaCl}$ :

$\Delta_{\text{lattice}}H^{\circ} = \Delta_{\text{sub}}H^{\circ} + \Delta_i H^{\circ} + \frac{1}{2}\,\Delta_{\text{bond}}H^{\circ} + \Delta_{\text{eg}}H^{\circ} - \Delta_f H^{\circ}$

$\Delta_{\text{lattice}}H^{\circ} = 108.4 + 496.0 + 121.0 - 348.6 - (-411.2) = +788.0\text{ kJ}\cdot\text{mol}^{-1}$

Enthalpy of Solution and Hydration:

When an ionic solid dissolves in water, two competing energetic processes take place:

The ionic lattice must be torn apart into gaseous ions: requires input of $\Delta_{\text{lattice}}H^{\circ}$ .
The separated gaseous ions are solvated by polar water dipoles: releases $\Delta_{\text{hyd}}H^{\circ}$ .

$\Delta_{\text{sol}}H^{\circ} = \Delta_{\text{lattice}}H^{\circ} + \Delta_{\text{hyd}}H^{\circ}$

For $\mathrm{NaCl}$ :
$\Delta_{\text{sol}}H^{\circ} = (+788.0\text{ kJ/mol}) + (-784.0\text{ kJ/mol}) = +4.0\text{ kJ}\cdot\text{mol}^{-1}$
The net dissolution process is slightly endothermic ( $+4.0\text{ kJ/mol}$ ), causing a slight drop in temperature when table salt dissolves in water.

5.8 Spontaneity and the Second Law of Thermodynamics#

5.8.1 The Physical Limitation of the First Law#

The First Law of Thermodynamics establishes the strict conservation of energy ( $\Delta U = q + w$ ), but it places zero restriction on the direction in which natural processes unfold:

When a hot copper block is brought into contact with a cold copper block, heat flows spontaneously from the hot body to the cold body until thermal equilibrium is attained. The reverse process (heat spontaneously flowing from the cold block to make the hot block hotter) would conserve total energy perfectly, yet it never occurs in nature.
When the valve between an evacuated flask and a gas-filled flask is opened, gas expands spontaneously to occupy the total available volume. The gas never spontaneously compresses itself back into a single flask.
A drop of ink placed in water diffuses until uniformly distributed; it never un-diffuses back into an isolated droplet.

A process that has the natural potential to occur without the continuous intervention of an external driving agency is termed a spontaneous process. Spontaneous processes are inherently irreversible.

5.8.2 Is Decrease in Enthalpy the Criterion for Spontaneity?#

In classical mechanics, physical systems spontaneously minimize their potential energy: water flows downhill, a stretched spring contracts, and a falling stone strikes the ground.

By analogy, early thermochemists postulated that chemical reactions might proceed spontaneously solely in the direction of lower enthalpy (exothermic reactions, $\Delta H < 0$ ): $\frac{1}{2}\mathrm{N}_2\text{(g)} + \frac{3}{2}\mathrm{H}_2\text{(g)} \rightarrow \mathrm{NH}_3\text{(g)}, \quad \Delta_r H^{\circ} = -46.1\text{ kJ/mol} \quad \text{(Spontaneous)}$ $\mathrm{H}_2\text{(g)} + \frac{1}{2}\mathrm{O}_2\text{(g)} \rightarrow \mathrm{H}_2\mathrm{O}\text{(l)}, \quad \Delta_r H^{\circ} = -285.8\text{ kJ/mol} \quad \text{(Spontaneous)}$

However, numerous endothermic processes ( $\Delta H > 0$ ) occur spontaneously in nature:

Evaporation of liquid water at room temperature: $\mathrm{H}_2\mathrm{O}\text{(l)} \rightarrow \mathrm{H}_2\mathrm{O}\text{(g)}, \quad \Delta_{\text{vap}}H^{\circ} = +44.01\text{ kJ/mol} \quad \text{(Spontaneous)}$
Dissolution of ammonium nitrate in water: $\mathrm{NH}_4\mathrm{NO}_3\text{(s)} + \mathrm{aq} \rightarrow \mathrm{NH}_4^+\text{(aq)} + \mathrm{NO}_3^-\text{(aq)}, \quad \Delta H > 0 \quad \text{(Spontaneous)}$
Thermal decomposition of limestone at high temperatures: $\mathrm{CaCO}_3\text{(s)} \rightarrow \mathrm{CaO}\text{(s)} + \mathrm{CO}_2\text{(g)}, \quad \Delta_r H^{\circ} = +178.3\text{ kJ/mol} \quad \text{(Spontaneous above 1100 K)}$

Furthermore, when two inert ideal gases diffuse into each other inside an isolated chamber, the enthalpy change is exactly zero ( $\Delta H = 0$ ), yet the mixing process is completely spontaneous and irreversible.

Conclusion:
A decrease in enthalpy is a contributing factor, but not the sole criterion for spontaneity. A second thermodynamic driving force is required.

5.8.3 The Concept of Entropy ( $S$ )#

Consider the spontaneous interdiffusion of two gases ( $A$ and $B$ ) initially separated by a partition. Before removal of the partition, gas $A$ is confined to the left half and gas $B$ to the right half. The state is highly ordered and predictable: any molecule sampled from the left is guaranteed to be $A$ .

When the partition is withdrawn, molecules diffuse randomly until uniformly dispersed. Sampling a molecule now carries statistical uncertainty. The system has transitioned from an ordered state to a disordered, chaotic state.

To quantify the degree of disorder, randomness, and energetic dispersal in a system, we introduce a thermodynamic state function: Entropy, denoted by $S$ .

Thermodynamic Definition of Entropy Change

When thermal energy is added to a system, it increases random molecular motions (translation, rotation, vibration), increasing disorder. The randomizing effect of heat depends directly on the temperature at which it is introduced:

Adding $100\text{ J}$ of heat to a cold system (low temperature, where particles are quiet) introduces a dramatic increase in disorder.
Adding the same $100\text{ J}$ of heat to a hot system (high temperature, where particles are already violently chaotic) produces only a negligible percentage increase in disorder.

Therefore, the change in entropy is inversely proportional to temperature. For a reversible transformation:

$\Delta S = \frac{q_{\text{rev}}}{T}$

Units of entropy: $\text{J}\cdot\text{K}^{-1}$ (or $\text{J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$ for molar entropy).

5.8.4 The Second Law of Thermodynamics#

The Second Law of Thermodynamics:
In any spontaneous (naturally occurring) process, the total entropy of the universe (system plus surroundings) must strictly increase:
$\Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0$

At thermodynamic equilibrium:

$\Delta S_{\text{total}} = 0$

For a non-spontaneous process:

$\Delta S_{\text{total}} < 0$

Entropy of Physical States:

For any substance, the degree of spatial and energetic disorder follows the sequence:

$S_{\text{solid}} \ll S_{\text{liquid}} \ll S_{\text{gas}}$

In a crystalline solid, atoms are locked into rigid lattice sites with minimal entropy.
In a liquid, molecules translate and tumble while remaining in close contact.
In a gas, molecules fly freely across vast intermolecular volumes, maximizing spatial disorder and entropy.

NCERT Problem 5.10 (Step-by-Step Solution):
Predict in which of the following processes entropy increases or decreases:
(i) A liquid crystallizes into a solid.
(ii) Temperature of a crystalline solid is raised from $0\text{ K}$ to $115\text{ K}$ .
(iii) $2\mathrm{NaHCO}_3\text{(s)} \rightarrow \mathrm{Na}_2\mathrm{CO}_3\text{(s)} + \mathrm{CO}_2\text{(g)} + \mathrm{H}_2\mathrm{O}\text{(g)}$
(iv) $\mathrm{H}_2\text{(g)} \rightarrow 2\mathrm{H}\text{(g)}$

Step 1: Given Quantities & Qualitative Criteria
State change criteria: Formation of gases or increased temperature increases entropy; crystallization or decreasing temperature decreases entropy.

Step 2: Fundamental Governing Formula
$\Delta S = S_{\text{final}} - S_{\text{initial}}$

Step 3: Direct Analysis & Reasoning

(i) Entropy Decreases ( $\Delta S < 0$ ): Upon freezing, mobile liquid molecules become locked into a highly ordered crystalline lattice, reducing randomness.
(ii) Entropy Increases ( $\Delta S > 0$ ): At $0\text{ K}$ , constituent particles are static. As temperature rises to $115\text{ K}$ , thermal energy excites lattice vibrations, increasing disorder.
(iii) Entropy Increases ( $\Delta S > 0$ ): Two moles of solid reactant decompose to yield one mole of solid plus two moles of gas ( $\Delta n_g = +2$ ). Gaseous products possess vastly higher entropy than solids.
(iv) Entropy Increases ( $\Delta S > 0$ ): One mole of diatomic molecules dissociates into two moles of independent gaseous atoms. The doubling of particles increases configurational microstates.

Step 4: Physical Significance & Laboratory Insight
Tracking changes in the number of gaseous moles ( $\Delta n_g$ ) provides a reliable qualitative indicator of the sign of $\Delta_r S^{\circ}$ .

NCERT Problem 5.11 (Step-by-Step Solution):
For the oxidation of iron:
$4\mathrm{Fe}\text{(s)} + 3\mathrm{O}_2\text{(g)} \rightarrow 2\mathrm{Fe}_2\mathrm{O}_3\text{(s)}$
The entropy change of the system is $-549.4\text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$ at $298\text{ K}$ . In spite of the negative entropy change of the system, why is this reaction spontaneous? (Given $\Delta_r H^{\circ} = -1648.0 \times 10^3\text{ J}\cdot\text{mol}^{-1}$ ).

Step 1: Given Quantities & Unit Harmonization

$\Delta S_{\text{system}} = -549.4\text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$
$\Delta_r H^{\circ} = -1648.0 \times 10^3\text{ J}\cdot\text{mol}^{-1}$
$T = 298.15\text{ K}$

Step 2: Fundamental Governing Formula
The heat liberated by the exothermic reaction is absorbed by the surroundings at constant pressure:
$q_{\text{surroundings}} = -\Delta_r H^{\circ}$
$\Delta S_{\text{surroundings}} = \frac{q_{\text{surroundings}}}{T} = \frac{-\Delta_r H^{\circ}}{T}$
$\Delta S_{\text{total}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}}$

Step 3: Direct Substitution & Arithmetic Computation
$\Delta S_{\text{surroundings}} = \frac{-(-1648.0 \times 10^3\text{ J}\cdot\text{mol}^{-1})}{298.15\text{ K}} = +5527.4\text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$
$\Delta S_{\text{total}} = -549.4 + 5527.4 = +4978.0\text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$

Step 4: Physical Significance & Laboratory Insight
Although iron and oxygen gas become locked into an ordered crystalline oxide lattice (decreasing the system entropy), the massive quantity of heat released into the surroundings ( $1648\text{ kJ}$ ) creates enormous thermal agitation in the ambient environment. Because $\Delta S_{\text{surroundings}} \gg |\Delta S_{\text{system}}|$ , the net entropy change of the universe is overwhelmingly positive ( $+4978\text{ J/K}$ ), driving the spontaneous rusting of iron.

5.9 Gibbs Free Energy ( $G$ ) and Reaction Spontaneity#

Evaluating the spontaneity of a chemical reaction using $\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}}$ requires calculating entropy changes for both the reaction vessel and the entire surrounding universe.

To evaluate spontaneity using properties pertaining exclusively to the system, the American mathematical physicist Josiah Willard Gibbs defined a new thermodynamic function: the Gibbs Energy (or Gibbs Free Energy), denoted by $G$ :

$G = H - TS$

Because $H$ , $T$ , and $S$ are state functions, $G$ is an extensive property and a state function.

5.9.1 Derivation of the Gibbs Equation#

For an isothermal transformation ( $T = \text{constant}$ ):

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

We know from Section 5.8 that the entropy change of the surroundings is:

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

Substituting into the total entropy expression:

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

Multiplying through by absolute temperature $T$ :

$T\,\Delta S_{\text{total}} = T\,\Delta S_{\text{sys}} - \Delta H_{\text{sys}} = -(\Delta H_{\text{sys}} - T\,\Delta S_{\text{sys}})$

Recognizing the right-hand term as $-\Delta G_{\text{sys}}$ :

$T\,\Delta S_{\text{total}} = -\Delta G_{\text{sys}} \implies \Delta G_{\text{sys}} = -T\,\Delta S_{\text{total}}$

5.9.2 Criterion for Spontaneity at Constant $T$ and $p$ #

Because absolute temperature $T$ is always positive ( $T > 0\text{ K}$ ), the algebraic sign of $\Delta G_{\text{sys}}$ is strictly opposite to that of $\Delta S_{\text{total}}$ :

Value of $\Delta G$	Value of $\Delta S_{\text{total}}$	Spontaneity Status	Physical Interpretation
$\mathbf{\Delta G < 0}$	$\Delta S_{\text{total}} > 0$	Spontaneous (Exergonic)	Reaction proceeds spontaneously in the forward direction as written.
$\mathbf{\Delta G > 0}$	$\Delta S_{\text{total}} < 0$	Non-Spontaneous (Endergonic)	Forward reaction cannot occur; reverse reaction is spontaneous.
$\mathbf{\Delta G = 0}$	$\Delta S_{\text{total}} = 0$	Dynamic Equilibrium	System is at chemical equilibrium; no net reaction occurs.

Physical Meaning of "Free" Energy:
In any process, the total enthalpy change $\Delta H$ represents total heat exchanged. The quantity $T\Delta S$ represents bound energy that must be dissipated into molecular disorder and cannot be harnessed. Therefore, $\Delta G = \Delta H - T\Delta S$ represents the net usable energy available to perform non-expansion (useful) work:
$-\Delta G = w_{\text{useful, max}}$

5.9.3 Effect of Temperature on Reaction Spontaneity#

The Gibbs-Helmholtz equation $\Delta_r G^{\circ} = \Delta_r H^{\circ} - T\Delta_r S^{\circ}$ involves a competition between the enthalpy factor (favoring $\Delta H < 0$ ) and the entropy factor (favoring $\Delta S > 0$ ):

Case	$\Delta_r H^{\circ}$	$\Delta_r S^{\circ}$	Sign of $\Delta_r G^{\circ}$	Temperature Dependence & Spontaneity (NCERT Table 5.4)
Case 1	Negative ( $-$ )	Positive ( $+$ )	Negative ( $-$ )	Spontaneous at all temperatures. Enthalpy and entropy factors both favor spontaneity. ( $2\mathrm{O}_3\text{(g)} \rightarrow 3\mathrm{O}_2\text{(g)}$ )
Case 2	Negative ( $-$ )	Negative ( $-$ )	Negative ( $-$ ) at low $T$ <br>Positive ( $+$ ) at high $T$	Spontaneous at low temperatures; non-spontaneous at high temperatures. ( $T\Delta S$ term dominates at elevated $T$ ). (Exothermic syntheses: $\mathrm{N}_2 + 3\mathrm{H}_2 \rightarrow 2\mathrm{NH}_3$ )
Case 3	Positive ( $+$ )	Positive ( $+$ )	Positive ( $+$ ) at low $T$ <br>Negative ( $-$ ) at high $T$	Non-spontaneous at low temperatures; becomes spontaneous at high temperatures. ( $T\Delta S$ outweighs $\Delta H$ ). (Melting of ice, decomposition of $\mathrm{CaCO}_3$ )
Case 4	Positive ( $+$ )	Negative ( $-$ )	Positive ( $+$ )	Non-spontaneous at all temperatures. Both enthalpy and entropy oppose reaction. Forward reaction impossible. ( $3\mathrm{O}_2\text{(g)} \rightarrow 2\mathrm{O}_3\text{(g)}$ )

Equilibrium Crossover Temperature ( $T_{\text{eq}}$ ):

For Cases 2 and 3, setting $\Delta_r G^{\circ} = 0$ identifies the transition temperature where the reaction switches between spontaneity and non-spontaneity:

$0 = \Delta_r H^{\circ} - T_{\text{eq}}\,\Delta_r S^{\circ} \implies T_{\text{eq}} = \frac{\Delta_r H^{\circ}}{\Delta_r S^{\circ}}$

5.9.4 The Third Law of Thermodynamics#

As the temperature of a substance decreases, molecular thermal agitation (translation, rotation, vibration) slows down.

In 1906, Walther Nernst formulated the principle governing the low-temperature limit of entropy:

The Third Law of Thermodynamics:
The entropy of any pure, perfectly crystalline substance approaches zero as the absolute temperature approaches zero Kelvin:
$\lim_{T \rightarrow 0\text{ K}} S = 0$

Physical Foundation:

At absolute zero ( $0\text{ K}$ ), all thermal motion ceases. For a perfect crystal, every constituent atom is located at its unique, lowest-energy lattice coordinate. In statistical mechanics, this corresponds to a single thermodynamic microstate ( $W = 1$ ). Applying Boltzmann's entropy formula:

$S = k_B \ln W = k_B \ln(1) = 0$

The Third Law enables evaluation of the absolute molar entropy ( $S^{\circ}$ ) of pure substances by integrating heat capacity data from $0\text{ K}$ to temperature $T$ :

$S^{\circ}(T) = \int_0^T \frac{C_p}{T}\,dT + \sum \frac{\Delta H_{\text{phase}}}{T_{\text{phase}}}$

5.10 Gibbs Energy Change and Chemical Equilibrium#

For a reversible chemical reaction proceeding toward equilibrium:

$\mathrm{A} + \mathrm{B} \rightleftharpoons \mathrm{C} + \mathrm{D}$

As reactants convert to products, the Gibbs energy of the reaction mixture decreases continuously until it reaches a minimum value. At this minimum, the slope of Gibbs energy with respect to reaction extent is zero:

$\Delta_r G = 0$

5.10.1 Relationship Between $\Delta_r G^{\circ}$ and Equilibrium Constant ( $K$ )#

The actual Gibbs energy change $\Delta_r G$ at any arbitrary non-equilibrium composition is related to the standard Gibbs energy change $\Delta_r G^{\circ}$ by the relation:

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

Where $Q$ is the reaction quotient. At dynamic chemical equilibrium, $\Delta_r G = 0$ and $Q = K$ :

$0 = \Delta_r G^{\circ} + RT \ln K$

Rearranging yields the fundamental thermodynamic equation connecting thermodynamics to chemical equilibrium:

$\Delta_r G^{\circ} = -RT \ln K$

Converting to common base-10 logarithms:

$\Delta_r G^{\circ} = -2.303\,RT \log_{10} K$

Solving for the equilibrium constant $K$ :

$K = e^{-\frac{\Delta_r G^{\circ}}{RT}} = 10^{-\frac{\Delta_r G^{\circ}}{2.303 RT}}$

Standard Gibbs Energy	Equilibrium Constant	Reaction Extent at Equilibrium
$\mathbf{\Delta_r G^{\circ} < 0}$ (strongly negative)	$K \gg 1$	Products predominate heavily; reaction proceeds to near completion.
$\mathbf{\Delta_r G^{\circ} = 0}$	$K = 1$	Reactants and products are present in comparable concentrations.
$\mathbf{\Delta_r G^{\circ} > 0}$ (strongly positive)	$K \ll 1$	Reactants predominate heavily; negligible product formation.

NCERT Problem 5.12 (Step-by-Step Solution):
Calculate $\Delta_r G^{\circ}$ for conversion of oxygen to ozone, $\frac{3}{2}\mathrm{O}_2\text{(g)} \rightarrow \mathrm{O}_3\text{(g)}$ at $298\text{ K}$ , if $K_p$ for this conversion is $2.47 \times 10^{-29}$ .

Step 1: Given Quantities & Unit Harmonization

Reaction: $\frac{3}{2}\mathrm{O}_2\text{(g)} \rightarrow \mathrm{O}_3\text{(g)}$
Temperature: $T = 298.15\text{ K}$
Equilibrium constant: $K_p = 2.47 \times 10^{-29}$
Gas constant: $R = 8.314\text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$

Step 2: Fundamental Governing Formula
$\Delta_r G^{\circ} = -2.303\,RT \log_{10}(K_p)$

Step 3: Direct Substitution & Arithmetic Computation
$\log_{10}(2.47 \times 10^{-29}) = \log_{10}(2.47) - 29 = 0.3927 - 29 = -28.6073$
$\Delta_r G^{\circ} = -2.303 \times (8.314\text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}) \times (298.15\text{ K}) \times (-28.6073)$
$\Delta_r G^{\circ} = -5708.8 \times (-28.6073) = +163313\text{ J}\cdot\text{mol}^{-1} = +163.31\text{ kJ}\cdot\text{mol}^{-1} \approx +163\text{ kJ}\cdot\text{mol}^{-1}$

Step 4: Physical Significance & Laboratory Insight
The enormous positive value of $\Delta_r G^{\circ}$ ( $+163.3\text{ kJ/mol}$ ) aligns with the minuscule equilibrium constant ( $10^{-29}$ ). Ozone will not form spontaneously from atmospheric dioxygen at room temperature without electrical discharge or ultraviolet photolysis.

NCERT Problem 5.13 (Step-by-Step Solution):
Find out the value of the equilibrium constant for the following reaction at $298\text{ K}$ :
$2\mathrm{NH}_3\text{(g)} + \mathrm{CO}_2\text{(g)} \rightleftharpoons \mathrm{NH}_2\mathrm{CONH}_2\text{(aq)} + \mathrm{H}_2\mathrm{O}\text{(l)}$
Standard Gibbs energy change, $\Delta_r G^{\circ}$ at the given temperature is $-13.6\text{ kJ}\cdot\text{mol}^{-1}$ .

Step 1: Given Quantities & Unit Harmonization

$\Delta_r G^{\circ} = -13.6\text{ kJ}\cdot\text{mol}^{-1} = -13600\text{ J}\cdot\text{mol}^{-1}$
$T = 298.15\text{ K}$
$R = 8.314\text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$

Step 2: Fundamental Governing Formula
$\log_{10} K = \frac{-\Delta_r G^{\circ}}{2.303\,RT}$

Step 3: Direct Substitution & Arithmetic Computation
$\log_{10} K = \frac{-(-13600\text{ J}\cdot\text{mol}^{-1})}{2.303 \times (8.314\text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}) \times (298.15\text{ K})} = \frac{13600}{5708.8} = 2.3823$
Taking the antilogarithm:
$K = 10^{2.3823} = 10^{0.3823} \times 10^2 \approx 2.41 \times 10^2$

Step 4: Physical Significance & Laboratory Insight
Because $\Delta_r G^{\circ}$ is negative, $K > 1$ ( $2.4 \times 10^2$ ), confirming that the industrial synthesis of urea from ammonia and carbon dioxide is thermodynamically favorable at $298\text{ K}$ .

NCERT Problem 5.14 (Step-by-Step Solution):
At $60^{\circ}\text{C}$ , dinitrogen tetroxide is $50\%$ dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.
$\mathrm{N}_2\mathrm{O}_4\text{(g)} \rightleftharpoons 2\mathrm{NO}_2\text{(g)}$

Step 1: Given Quantities & Unit Harmonization

Total pressure: $p_{\text{total}} = 1.0\text{ atm}$
Temperature: $T = 60^{\circ}\text{C} = 333.15\text{ K}$
Degree of dissociation: $\alpha = 0.50$

Step 2: Fundamental Governing Formula
For the dissociation $\mathrm{N}_2\mathrm{O}_4 \rightleftharpoons 2\mathrm{NO}_2$ :

Initial moles: $1$ mole of $\mathrm{N}_2\mathrm{O}_4$ , $0$ mole of $\mathrm{NO}_2$ .
Equilibrium moles: $(1 - \alpha)$ moles of $\mathrm{N}_2\mathrm{O}_4$ , $2\alpha$ moles of $\mathrm{NO}_2$ .
Total equilibrium moles: $n_{\text{total}} = (1 - \alpha) + 2\alpha = 1 + \alpha$ .
Mole fractions:
$x_{\mathrm{N}_2\mathrm{O}_4} = \frac{1 - \alpha}{1 + \alpha}, \quad x_{\mathrm{NO}_2} = \frac{2\alpha}{1 + \alpha}$
Partial pressures ( $p_i = x_i \cdot p_{\text{total}}$ ):
$K_p = \frac{(p_{\mathrm{NO}_2})^2}{p_{\mathrm{N}_2\mathrm{O}_4}} = \frac{4\alpha^2}{(1 - \alpha^2)}\,p_{\text{total}}$
$\Delta_r G^{\circ} = -RT \ln K_p = -2.303\,RT \log_{10} K_p$

Step 3: Direct Substitution & Arithmetic Computation
For $\alpha = 0.50$ :
$x_{\mathrm{N}_2\mathrm{O}_4} = \frac{1 - 0.5}{1 + 0.5} = \frac{0.5}{1.5} = \frac{1}{3}$
$x_{\mathrm{NO}_2} = \frac{2(0.5)}{1 + 0.5} = \frac{1.0}{1.5} = \frac{2}{3}$
Partial pressures: $p_{\mathrm{N}_2\mathrm{O}_4} = \frac{1}{3}\text{ atm}$ , $p_{\mathrm{NO}_2} = \frac{2}{3}\text{ atm}$ .
$K_p = \frac{(2/3)^2}{1/3} = \frac{4/9}{1/3} = \frac{4}{3} = 1.333\text{ atm}$
Compute $\Delta_r G^{\circ}$ :
$\Delta_r G^{\circ} = -2.303 \times (8.314\text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}) \times (333.15\text{ K}) \times \log_{10}(1.333)$
$\log_{10}(1.333) = 0.1249$
$\Delta_r G^{\circ} = -2.303 \times 8.314 \times 333.15 \times 0.1249 = -796.8\text{ J}\cdot\text{mol}^{-1} \approx -0.797\text{ kJ}\cdot\text{mol}^{-1}$

Step 4: Physical Significance & Laboratory Insight
Because $\Delta_r G^{\circ}$ is slightly negative at $60^{\circ}\text{C}$ ( $-0.80\text{ kJ/mol}$ ), the equilibrium constant $K_p > 1$ , reflecting a moderate tendency for $\mathrm{N}_2\mathrm{O}_4$ to dissociate into brown nitrogen dioxide vapor at elevated temperatures.

Interactive Simulation Experiment: Molecular Disorder & Gibbs Spontaneity Simulator
Open the Entropy & Spontaneity tab in the simulation suite:

In the Molecular Disorder mode, drag the temperature slider from $0\text{ K}$ $0 K$ to $500\text{ K}$ $500 K$ .
- At $0\text{ K}$ , observe complete crystal lattice immobility ( $S = 0$ , verifying Third Law).
- Increase temperature to view vibrational motion, melting to liquid, and gas vaporization with dispersed microstates.
Switch to Spontaneity Simulator. Set $\Delta H = +40\text{ kJ/mol}$ and $\Delta S = +120\text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$ .
Observe the real-time $\Delta G$ indicator gauge. At $T = 250\text{ K}$ , $\Delta G = +10\text{ kJ/mol}$ (non-spontaneous, red gauge).
Slide temperature above the crossover $T_{\text{eq}} = \frac{40000}{120} = 333.3\text{ K}$ . Observe $\Delta G$ turn negative (spontaneous, green gauge), illustrating Case 3 from Table 5.4.

Solutions to End-of-Chapter NCERT Exercises (5.1 to 5.22)#

Exercise 5.1#

Question: Choose the correct answer. A thermodynamic state function is a quantity:
(i) used to determine heat changes
(ii) whose value is independent of path
(iii) used to determine pressure-volume work
(iv) whose value depends on temperature only
Solution:
(ii) whose value is independent of path.
Rationale: By definition, a state function depends exclusively on the current equilibrium state of the system and not on the pathway or sequence of steps taken to reach that state.

Exercise 5.2#

Question: For the process to occur under adiabatic conditions, the correct condition is:
(i) $\Delta T = 0$
(ii) $\Delta p = 0$
(iii) $q = 0$
(iv) $w = 0$
Solution:
(iii) $q = 0$ .
Rationale: An adiabatic process is strictly defined as one in which no heat transfers across the system boundary ( $q = 0$ ). The temperature can and usually does change due to internal work.

Exercise 5.3#

Question: The enthalpies of all elements in their standard states are:
(i) unity
(ii) zero
(iii) $< 0$
(iv) different for each element
Solution:
(ii) zero.
Rationale: By international thermodynamic convention, the standard molar enthalpy of formation of an element in its most stable reference form of aggregation at $298.15\text{ K}$ and $1\text{ bar}$ is assigned a value of zero.

Exercise 5.4#

Question: $\Delta U^{\circ}$ of combustion of methane is $-X\text{ kJ}\cdot\text{mol}^{-1}$ . The value of $\Delta H^{\circ}$ is:
(i) $= \Delta U^{\circ}$
(ii) $> \Delta U^{\circ}$
(iii) $< \Delta U^{\circ}$
(iv) $= 0$
Solution:
(iii) $< \Delta U^{\circ}$ .
Rationale: The thermochemical combustion reaction is:
$\mathrm{CH}_4\text{(g)} + 2\mathrm{O}_2\text{(g)} \rightarrow \mathrm{CO}_2\text{(g)} + 2\mathrm{H}_2\mathrm{O}\text{(l)}$
$\Delta n_g = 1\text{ mol (CO}_2\text{)} - (1 + 2)\text{ mol (reactants)} = 1 - 3 = -2$
$\Delta H^{\circ} = \Delta U^{\circ} + \Delta n_g\,RT = -X + (-2RT) = -X - 2RT$
Because $2RT > 0$ , $\Delta H^{\circ}$ is more negative than $\Delta U^{\circ}$ , so $\Delta H^{\circ} < \Delta U^{\circ}$ .

Exercise 5.5#

Question: The enthalpy of combustion of methane, graphite, and dihydrogen at $298\text{ K}$ are $-890.3\text{ kJ}\cdot\text{mol}^{-1}$ , $-393.5\text{ kJ}\cdot\text{mol}^{-1}$ , and $-285.8\text{ kJ}\cdot\text{mol}^{-1}$ respectively. Enthalpy of formation of $\mathrm{CH}_4\text{(g)}$ will be:
(i) $-74.8\text{ kJ}\cdot\text{mol}^{-1}$
(ii) $-52.27\text{ kJ}\cdot\text{mol}^{-1}$
(iii) $+74.8\text{ kJ}\cdot\text{mol}^{-1}$
(iv) $+52.26\text{ kJ}\cdot\text{mol}^{-1}$
Solution:
(i) $-74.8\text{ kJ}\cdot\text{mol}^{-1}$ .
Step-by-Step Breakdown:
Target formation equation:
$\mathrm{C}\text{(graphite)} + 2\mathrm{H}_2\text{(g)} \rightarrow \mathrm{CH}_4\text{(g)}, \quad \Delta_f H^{\circ} = ?$
Given combustion reactions:
(1) $\mathrm{CH}_4\text{(g)} + 2\mathrm{O}_2\text{(g)} \rightarrow \mathrm{CO}_2\text{(g)} + 2\mathrm{H}_2\mathrm{O}\text{(l)}, \quad \Delta H_1 = -890.3\text{ kJ/mol}$
(2) $\mathrm{C}\text{(graphite)} + \mathrm{O}_2\text{(g)} \rightarrow \mathrm{CO}_2\text{(g)}, \quad \Delta H_2 = -393.5\text{ kJ/mol}$
(3) $\mathrm{H}_2\text{(g)} + \frac{1}{2}\mathrm{O}_2\text{(g)} \rightarrow \mathrm{H}_2\mathrm{O}\text{(l)}, \quad \Delta H_3 = -285.8\text{ kJ/mol}$
Target equation $= (2) + 2 \times (3) - (1)$ :
$\Delta_f H^{\circ} = \Delta H_2 + 2\,\Delta H_3 - \Delta H_1$
$\Delta_f H^{\circ} = -393.5 + 2(-285.8) - (-890.3) = -393.5 - 571.6 + 890.3 = -965.1 + 890.3 = -74.8\text{ kJ}\cdot\text{mol}^{-1}$

Exercise 5.6#

Question: A reaction, $\mathrm{A} + \mathrm{B} \rightarrow \mathrm{C} + \mathrm{D} + q$ is found to have a positive entropy change. The reaction will be:
(i) possible at high temperature
(ii) possible only at low temperature
(iii) not possible at any temperature
(iv) possible at any temperature
Solution:
(iv) possible at any temperature.
Rationale: The liberation of heat ( $+q$ ) indicates an exothermic reaction: $\Delta H < 0$ . The entropy change is positive: $\Delta S > 0$ . In the Gibbs equation:
$\Delta G = \Delta H - T\Delta S = (\text{negative}) - T(\text{positive}) < 0$
Because both terms are negative, $\Delta G$ is negative at all values of absolute temperature $T$ . The reaction is spontaneous at any temperature.

Exercise 5.7#

Question: In a process, $701\text{ J}$ of heat is absorbed by a system and $394\text{ J}$ of work is done by the system. What is the change in internal energy for the process?
Solution:

$q = +701\text{ J}$ (absorbed)
$w = -394\text{ J}$ (done by system)
$\Delta U = q + w = (+701\text{ J}) + (-394\text{ J}) = +307\text{ J}$

Exercise 5.8#

Question: The reaction of cyanamide, $\mathrm{NH}_2\mathrm{CN}\text{(s)}$ , with dioxygen was carried out in a bomb calorimeter, and $\Delta U$ was found to be $-742.7\text{ kJ}\cdot\text{mol}^{-1}$ at $298\text{ K}$ . Calculate enthalpy change for the reaction at $298\text{ K}$ :
$\mathrm{NH}_2\mathrm{CN}\text{(s)} + \frac{3}{2}\mathrm{O}_2\text{(g)} \rightarrow \mathrm{N}_2\text{(g)} + \mathrm{CO}_2\text{(g)} + \mathrm{H}_2\mathrm{O}\text{(l)}$
Solution:
$\Delta n_g = (1 + 1) - 1.5 = +0.5\text{ mol}$
$\Delta H = \Delta U + \Delta n_g\,RT = -742.7\text{ kJ/mol} + (0.5 \times 8.314 \times 10^{-3} \times 298.15)\text{ kJ/mol}$
$\Delta H = -742.7 + 1.24 = -741.46\text{ kJ}\cdot\text{mol}^{-1} \approx -741.5\text{ kJ}\cdot\text{mol}^{-1}$

Exercise 5.9#

Question: Calculate the number of $\text{kJ}$ of heat necessary to raise the temperature of $60.0\text{ g}$ of aluminium from $35^{\circ}\text{C}$ to $55^{\circ}\text{C}$ . Molar heat capacity of $\mathrm{Al}$ is $24\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$ .
Solution:
$n = \frac{60.0\text{ g}}{27.0\text{ g/mol}} = 2.222\text{ mol}$
$\Delta T = 55 - 35 = 20\text{ K}$
$q = n \cdot C_m \cdot \Delta T = (2.222\text{ mol}) \times (24\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}) \times (20\text{ K}) = 1066.67\text{ J} = 1.067\text{ kJ}$

Exercise 5.10#

Question: Calculate the enthalpy change on freezing of $1.0\text{ mol}$ of water at $10.0^{\circ}\text{C}$ to ice at $-10.0^{\circ}\text{C}$ . Given $\Delta_{\text{fus}}H = 6.03\text{ kJ}\cdot\text{mol}^{-1}$ at $0^{\circ}\text{C}$ , $C_p[\mathrm{H}_2\mathrm{O}\text{(l)}] = 75.3\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$ , and $C_p[\mathrm{H}_2\mathrm{O}\text{(s)}] = 36.8\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$ .
Solution:
The transformation is divided into three thermodynamic steps:

Step 1: Cooling liquid water from $10.0^{\circ}\text{C}$ ( $283.15\text{ K}$ ) to $0^{\circ}\text{C}$ ( $273.15\text{ K}$ ):
$\Delta H_1 = n \cdot C_p[\mathrm{H}_2\mathrm{O}\text{(l)}] \cdot (0 - 10) = (1.0\text{ mol}) \times (75.3\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}) \times (-10\text{ K}) = -753\text{ J} = -0.753\text{ kJ}$
Step 2: Freezing water at $0^{\circ}\text{C}$ into ice (reverse of fusion):
$\Delta H_2 = -\Delta_{\text{fus}}H = -6.03\text{ kJ}\cdot\text{mol}^{-1}$
Step 3: Cooling ice from $0^{\circ}\text{C}$ ( $273.15\text{ K}$ ) to $-10.0^{\circ}\text{C}$ ( $263.15\text{ K}$ ):
$\Delta H_3 = n \cdot C_p[\mathrm{H}_2\mathrm{O}\text{(s)}] \cdot (-10 - 0) = (1.0\text{ mol}) \times (36.8\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}) \times (-10\text{ K}) = -368\text{ J} = -0.368\text{ kJ}$
Total enthalpy change:
$\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 = -0.753 - 6.03 - 0.368 = -7.151\text{ kJ}\cdot\text{mol}^{-1}$

Exercise 5.11#

Question: Enthalpy of combustion of carbon to $\mathrm{CO}_2$ is $-393.5\text{ kJ}\cdot\text{mol}^{-1}$ . Calculate the heat released upon formation of $35.2\text{ g}$ of $\mathrm{CO}_2$ from carbon and dioxygen gas.
Solution:
Reaction: $\mathrm{C}\text{(s)} + \mathrm{O}_2\text{(g)} \rightarrow \mathrm{CO}_2\text{(g)}, \quad \Delta H = -393.5\text{ kJ/mol}$
Molar mass of $\mathrm{CO}_2 = 12.01 + 2(16.00) = 44.01\text{ g/mol}$ .
Moles of $\mathrm{CO}_2$ formed:
$n = \frac{35.2\text{ g}}{44.01\text{ g/mol}} = 0.7998\text{ mol} \approx 0.80\text{ mol}$
Heat released:
$q = n \times |\Delta H| = 0.7998\text{ mol} \times 393.5\text{ kJ/mol} = 314.73\text{ kJ} \approx 315\text{ kJ}$

Exercise 5.12#

Question: Enthalpies of formation of $\mathrm{CO}\text{(g)}$ , $\mathrm{CO}_2\text{(g)}$ , $\mathrm{N}_2\mathrm{O}\text{(g)}$ , and $\mathrm{N}_2\mathrm{O}_4\text{(g)}$ are $-110$ , $-393$ , $+81$ , and $+9.7\text{ kJ}\cdot\text{mol}^{-1}$ respectively. Find the value of $\Delta_r H$ for the reaction:
$\mathrm{N}_2\mathrm{O}_4\text{(g)} + 3\mathrm{CO}\text{(g)} \rightarrow \mathrm{N}_2\mathrm{O}\text{(g)} + 3\mathrm{CO}_2\text{(g)}$
Solution:
$\Delta_r H = [\Delta_f H(\mathrm{N}_2\mathrm{O}) + 3\,\Delta_f H(\mathrm{CO}_2)] - [\Delta_f H(\mathrm{N}_2\mathrm{O}_4) + 3\,\Delta_f H(\mathrm{CO})]$
$\Delta_r H = [81 + 3(-393)] - [9.7 + 3(-110)]$
$\Delta_r H = [81 - 1179] - [9.7 - 330] = -1098 - (-320.3) = -1098 + 320.3 = -777.7\text{ kJ}\cdot\text{mol}^{-1}$

Exercise 5.13#

Question: Given:
$\mathrm{N}_2\text{(g)} + 3\mathrm{H}_2\text{(g)} \rightarrow 2\mathrm{NH}_3\text{(g)}, \quad \Delta_r H^{\circ} = -92.4\text{ kJ}\cdot\text{mol}^{-1}$
What is the standard enthalpy of formation of $\mathrm{NH}_3\text{ gas}$ ?
Solution:
Standard enthalpy of formation is defined per one mole of substance formed from reference elements:
$\Delta_f H^{\circ}(\mathrm{NH}_3) = \frac{\Delta_r H^{\circ}}{2} = \frac{-92.4\text{ kJ/mol}}{2} = -46.2\text{ kJ}\cdot\text{mol}^{-1}$

Exercise 5.14#

Question: Calculate the standard enthalpy of formation of $\mathrm{CH}_3\mathrm{OH}\text{(l)}$ from the following data:
(1) $\mathrm{CH}_3\mathrm{OH}\text{(l)} + \frac{3}{2}\mathrm{O}_2\text{(g)} \rightarrow \mathrm{CO}_2\text{(g)} + 2\mathrm{H}_2\mathrm{O}\text{(l)}, \quad \Delta_r H^{\circ} = -726\text{ kJ}\cdot\text{mol}^{-1}$
(2) $\mathrm{C}\text{(graphite)} + \mathrm{O}_2\text{(g)} \rightarrow \mathrm{CO}_2\text{(g)}, \quad \Delta_c H^{\circ} = -393\text{ kJ}\cdot\text{mol}^{-1}$
(3) $\mathrm{H}_2\text{(g)} + \frac{1}{2}\mathrm{O}_2\text{(g)} \rightarrow \mathrm{H}_2\mathrm{O}\text{(l)}, \quad \Delta_f H^{\circ} = -286\text{ kJ}\cdot\text{mol}^{-1}$
Solution:
Target equation:
$\mathrm{C}\text{(graphite)} + 2\mathrm{H}_2\text{(g)} + \frac{1}{2}\mathrm{O}_2\text{(g)} \rightarrow \mathrm{CH}_3\mathrm{OH}\text{(l)}, \quad \Delta_f H^{\circ} = ?$
Target $= (2) + 2 \times (3) - (1)$ :
$\Delta_f H^{\circ} = (-393) + 2(-286) - (-726)$
$\Delta_f H^{\circ} = -393 - 572 + 726 = -965 + 726 = -239\text{ kJ}\cdot\text{mol}^{-1}$

Exercise 5.15#

Question: Calculate the enthalpy change for the process:
$\mathrm{CCl}_4\text{(g)} \rightarrow \mathrm{C}\text{(g)} + 4\mathrm{Cl}\text{(g)}$
and calculate bond enthalpy of $\mathrm{C}-\mathrm{Cl}$ in $\mathrm{CCl}_4\text{(g)}$ .
Given:

$\Delta_{\text{vap}}H^{\circ}(\mathrm{CCl}_4) = 30.5\text{ kJ}\cdot\text{mol}^{-1}$
$\Delta_f H^{\circ}(\mathrm{CCl}_4) = -135.5\text{ kJ}\cdot\text{mol}^{-1}$
$\Delta_a H^{\circ}(\mathrm{C}) = 715.0\text{ kJ}\cdot\text{mol}^{-1}$
$\Delta_a H^{\circ}(\mathrm{Cl}_2) = 242.0\text{ kJ}\cdot\text{mol}^{-1}$
Solution:
Standard enthalpy of formation reaction for liquid carbon tetrachloride:
$\mathrm{C}\text{(graphite)} + 2\mathrm{Cl}_2\text{(g)} \rightarrow \mathrm{CCl}_4\text{(l)}, \quad \Delta_f H^{\circ} = -135.5\text{ kJ/mol}$
Vaporizing liquid $\mathrm{CCl}_4$ :
$\mathrm{CCl}_4\text{(l)} \rightarrow \mathrm{CCl}_4\text{(g)}, \quad \Delta_{\text{vap}}H^{\circ} = +30.5\text{ kJ/mol}$
Adding gives formation of gaseous $\mathrm{CCl}_4$ :
$\mathrm{C}\text{(graphite)} + 2\mathrm{Cl}_2\text{(g)} \rightarrow \mathrm{CCl}_4\text{(g)}, \quad \Delta_f H^{\circ}[\mathrm{CCl}_4\text{(g)}] = -135.5 + 30.5 = -105.0\text{ kJ/mol}$
Now consider atomizing the constituent elements into isolated gaseous atoms:
Atomization of carbon: $\mathrm{C}\text{(graphite)} \rightarrow \mathrm{C}\text{(g)}, \quad \Delta H = +715.0\text{ kJ/mol}$
Atomization of chlorine: $2\mathrm{Cl}_2\text{(g)} \rightarrow 4\mathrm{Cl}\text{(g)}, \quad \Delta H = 2 \times 242.0 = +484.0\text{ kJ/mol}$
Total enthalpy of gaseous atoms from elements:
$\Delta H_{\text{atoms}} = 715.0 + 484.0 = +1199.0\text{ kJ/mol}$
For the atomization reaction $\mathrm{CCl}_4\text{(g)} \rightarrow \mathrm{C}\text{(g)} + 4\mathrm{Cl}\text{(g)}$ :
$\Delta_a H^{\circ} = \Delta H_{\text{atoms}} - \Delta_f H^{\circ}[\mathrm{CCl}_4\text{(g)}] = 1199.0 - (-105.0) = 1199.0 + 105.0 = 1304.0\text{ kJ}\cdot\text{mol}^{-1}$
Mean bond enthalpy of $\mathrm{C}-\mathrm{Cl}$ :
$\Delta_{\mathrm{C}-\mathrm{Cl}}H^{\circ} = \frac{\Delta_a H^{\circ}}{4} = \frac{1304.0}{4} = 326.0\text{ kJ}\cdot\text{mol}^{-1}$

Exercise 5.16#

Question: For an isolated system, $\Delta U = 0$ , what will be $\Delta S$ ?
Solution:
For an isolated system, no energy or matter can be exchanged with the surroundings. For any spontaneous or natural irreversible process occurring within an isolated system, the entropy must increase according to the Second Law:
$\Delta S > 0$
At thermodynamic equilibrium, $\Delta S = 0$ . Hence, for any spontaneous change in an isolated system, $\Delta S$ is strictly positive.

Exercise 5.17#

Question: For the reaction at $298\text{ K}$ ,
$2\mathrm{A} + \mathrm{B} \rightarrow \mathrm{C}$
$\Delta H = 400\text{ kJ}\cdot\text{mol}^{-1}$ and $\Delta S = 0.2\text{ kJ}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$ . At what temperature will the reaction become spontaneous considering $\Delta H$ and $\Delta S$ to be constant over the temperature range?
Solution:
For spontaneity, $\Delta G < 0$ :
$\Delta G = \Delta H - T\Delta S < 0 \implies \Delta H < T\Delta S \implies T > \frac{\Delta H}{\Delta S}$
Substituting values:
$T > \frac{400\text{ kJ}\cdot\text{mol}^{-1}}{0.2\text{ kJ}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}} = 2000\text{ K}$
The reaction will become spontaneous at temperatures strictly above $2000\text{ K}$ .

Exercise 5.18#

Question: For the reaction,
$2\mathrm{Cl}\text{(g)} \rightarrow \mathrm{Cl}_2\text{(g)}$
what are the signs of $\Delta H$ and $\Delta S$ ?
Solution:

Sign of $\Delta H$ is Negative ( $-$ ): Chemical bond formation releases energy (exothermic process).
Sign of $\Delta S$ is Negative ( $-$ ): Two moles of randomly moving isolated gaseous chlorine atoms combine to form one mole of structured diatomic chlorine molecules, reducing randomness and translational microstates.

Exercise 5.19#

Question: For the reaction:
$2\mathrm{A}\text{(g)} + \mathrm{B}\text{(g)} \rightarrow 2\mathrm{D}\text{(g)}$
$\Delta U^{\circ} = -10.5\text{ kJ}$ and $\Delta S^{\circ} = -44.1\text{ J}\cdot\text{K}^{-1}$ at $298\text{ K}$ . Calculate $\Delta G^{\circ}$ for the reaction, and predict whether the reaction may occur spontaneously.
Solution:
$\Delta n_g = 2 - (2 + 1) = -1.0\text{ mol}$
$\Delta H^{\circ} = \Delta U^{\circ} + \Delta n_g\,RT = -10.5\text{ kJ} + (-1.0\text{ mol}) \times (8.314 \times 10^{-3}\text{ kJ}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}) \times (298.15\text{ K})$
$\Delta H^{\circ} = -10.5\text{ kJ} - 2.479\text{ kJ} = -12.979\text{ kJ}$
Now evaluate $\Delta G^{\circ}$ :
$\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$
$T\Delta S^{\circ} = (298.15\text{ K}) \times (-44.1 \times 10^{-3}\text{ kJ}\cdot\text{K}^{-1}) = -13.148\text{ kJ}$
$\Delta G^{\circ} = -12.979\text{ kJ} - (-13.148\text{ kJ}) = -12.979 + 13.148 = +0.169\text{ kJ}$
Because $\Delta G^{\circ} > 0$ ( $+0.17\text{ kJ}$ ), the reaction is non-spontaneous at standard conditions and $298\text{ K}$ .

Exercise 5.20#

Question: The equilibrium constant for a reaction is $10$ . What will be the value of $\Delta G^{\circ}$ ? ( $R = 8.314\text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}, T = 300\text{ K}$ ).
Solution:
$\Delta_r G^{\circ} = -2.303\,RT \log_{10} K$
$\Delta_r G^{\circ} = -2.303 \times (8.314\text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}) \times (300\text{ K}) \times \log_{10}(10)$
Because $\log_{10}(10) = 1$ :
$\Delta_r G^{\circ} = -5744.14\text{ J}\cdot\text{mol}^{-1} = -5.744\text{ kJ}\cdot\text{mol}^{-1}$

Exercise 5.21#

Question: Comment on the thermodynamic stability of $\mathrm{NO}\text{(g)}$ , given:
$\frac{1}{2}\mathrm{N}_2\text{(g)} + \frac{1}{2}\mathrm{O}_2\text{(g)} \rightarrow \mathrm{NO}\text{(g)}, \quad \Delta_r H^{\circ} = +90\text{ kJ}\cdot\text{mol}^{-1}$
$\mathrm{NO}\text{(g)} + \frac{1}{2}\mathrm{O}_2\text{(g)} \rightarrow \mathrm{NO}_2\text{(g)}, \quad \Delta_r H^{\circ} = -74\text{ kJ}\cdot\text{mol}^{-1}$
Solution:

The standard enthalpy of formation of $\mathrm{NO}\text{(g)}$ is positive ( $+90\text{ kJ/mol}$ ), meaning it is an endothermic compound and thermodynamically unstable with respect to decomposition into elemental $\mathrm{N}_2$ and $\mathrm{O}_2$ .
Furthermore, the oxidation of $\mathrm{NO}\text{(g)}$ to $\mathrm{NO}_2\text{(g)}$ is exothermic ( $\Delta_r H^{\circ} = -74\text{ kJ/mol}$ ), meaning $\mathrm{NO}\text{(g)}$ is unstable in the presence of excess oxygen and readily oxidizes to reddish-brown $\mathrm{NO}_2$ gas.

Exercise 5.22#

Question: Calculate the entropy change in surroundings when $1.00\text{ mol}$ of $\mathrm{H}_2\mathrm{O}\text{(l)}$ is formed under standard conditions. Given $\Delta_f H^{\circ} = -286.0\text{ kJ}\cdot\text{mol}^{-1}$ .
Solution:
The standard formation of liquid water:
$\mathrm{H}_2\text{(g)} + \frac{1}{2}\mathrm{O}_2\text{(g)} \rightarrow \mathrm{H}_2\mathrm{O}\text{(l)}, \quad \Delta_r H^{\circ} = -286.0\text{ kJ}\cdot\text{mol}^{-1}$
Because the process is exothermic, heat absorbed by the surroundings at constant pressure and $298.15\text{ K}$ is:
$q_{\text{surr}} = -\Delta_r H^{\circ} = -(-286.0\text{ kJ}) = +286.0\text{ kJ} = +286000\text{ J}$
Entropy change in the surroundings:
$\Delta S_{\text{surr}} = \frac{q_{\text{surr}}}{T} = \frac{+286000\text{ J}}{298.15\text{ K}} = +959.25\text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1} \approx +959.3\text{ J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$

Complete NCERT Chapter Summary & Formula Master Sheet#

Concept / Quantity	Governing Mathematical Equation	Physical Meaning & IUPAC Sign Convention
First Law of Thermodynamics	$\Delta U = q + w$	Total energy conservation. $q > 0$ when absorbed; $w > 0$ when work is done on the system.
Irreversible Expansion Work	$w = -p_{\text{ex}}\,\Delta V = -p_{\text{ex}}(V_f - V_i)$	Work done against constant resisting external pressure $p_{\text{ex}}$ .
Reversible Isothermal Work	$w_{\text{rev}} = -2.303\,nRT \log_{10}\left(\frac{V_f}{V_i}\right)$	Maximum work extracted through infinite equilibrium expansion steps.
Free Expansion into Vacuum	$p_{\text{ex}} = 0 \implies w = 0, q = 0, \Delta U = 0$	Gas expands into vacuum without work or heat exchange.
Enthalpy State Function	$H = U + pV$	Total heat content. At constant pressure: $\Delta H = q_p$ .
Enthalpy and Internal Energy	$\Delta H = \Delta U + \Delta n_g\,RT$	Relationship for ideal gas reactions where $\Delta n_g = \sum n_{g,\text{prod}} - \sum n_{g,\text{react}}$ .
Molar Heat Capacity Relation	$C_p - C_V = R$	Difference between constant-pressure and constant-volume heat capacities for $1\text{ mol}$ of ideal gas.
Bomb Calorimeter (Constant $V$ )	$q_V = \Delta U = -C_V \cdot \Delta T$	Heat evolved at rigid constant volume measures change in internal energy directly.
Reaction Enthalpy from Formation	$\Delta_r H^{\circ} = \sum a_i\Delta_f H^{\circ}(\text{prod}) - \sum b_i\Delta_f H^{\circ}(\text{react})$	Standard reaction enthalpy from element formation enthalpies (reference elements $= 0$ ).
Hess's Law of Heat Summation	$\Delta_r H^{\circ} = \Delta_r H_1^{\circ} + \Delta_r H_2^{\circ} + \cdots$	Path-independence of enthalpy allows algebraic addition of intermediate thermochemical equations.
Reaction Enthalpy from Bond Enthalpies	$\Delta_r H^{\circ} \approx \sum \text{BE}(\text{reactants}) - \sum \text{BE}(\text{products})$	Approximate gas-phase reaction enthalpy from bond cleavage and formation energies.
Lattice Enthalpy (Born-Haber Cycle)	$\Delta_f H^{\circ} = \Delta_{\text{sub}}H^{\circ} + \Delta_i H^{\circ} + \frac{1}{2}\Delta_{\text{bond}}H^{\circ} + \Delta_{\text{eg}}H^{\circ} - \Delta_{\text{lattice}}H^{\circ}$	Thermodynamic cycle relating ionic solid lattice energy to elemental formation and gaseous ion properties.
Enthalpy of Solution	$\Delta_{\text{sol}}H^{\circ} = \Delta_{\text{lattice}}H^{\circ} + \Delta_{\text{hyd}}H^{\circ}$	Balance between lattice breakdown and ionic hydration energy.
Entropy of Reversible Process	$\Delta S = \frac{q_{\text{rev}}}{T}$	Quantitative measure of thermal energy dispersal and molecular randomness.
Second Law of Thermodynamics	$\Delta S_{\text{total}} = \Delta S_{\text{sys}} + \Delta S_{\text{surr}} > 0$	Spontaneous processes strictly increase the total entropy of the universe.
Gibbs Free Energy Equation	$\Delta G = \Delta H - T\Delta S$	Single-system criterion for spontaneity at constant temperature and pressure.
Spontaneity Criterion	$\Delta G < 0 \implies \text{Spontaneous}$ <br> $\Delta G = 0 \implies \text{Equilibrium}$ <br> $\Delta G > 0 \implies \text{Non-Spontaneous}$	Master condition for chemical and physical spontaneity at constant $T, p$ .
Third Law of Thermodynamics	$\lim_{T \rightarrow 0\text{ K}} S = 0$	Entropy of a pure, perfect crystalline solid approaches zero at absolute zero.
Gibbs Energy and Equilibrium Constant	$\Delta_r G^{\circ} = -RT \ln K = -2.303\,RT \log_{10} K$	Link between standard free energy and the chemical equilibrium position.

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Chemical Bonding and Molecular Structure Equilibrium

Chemical Thermodynamics

Unit 5: Chemical Thermodynamics

The Big Picture: Why Study Chemical Thermodynamics?#

5.1 Thermodynamic Terms#

5.1.1 The System, Surroundings, and Boundary#

5.1.2 Classification of Thermodynamic Systems#

5.1.3 The State of the System and State Functions#

State Functions versus Path Functions

5.1.4 Extensive and Intensive Properties#

5.2 Internal Energy, Work, Heat, and the First Law#

5.2.1 Internal Energy (UUU) as a State Function#

5.2.2 Changing Internal Energy by Work: Adiabatic Processes#

5.2.3 Changing Internal Energy by Heat#

5.2.4 The First Law of Thermodynamics#

5.2.5 IUPAC Sign Conventions in Chemical Thermodynamics#

5.3 Pressure-Volume Work and Reversible Processes#

5.3.1 Derivation of Expansion Work#

Verification of Sign Conventions:

5.3.2 Multi-Step and Reversible Processes#

5.3.3 Isothermal Reversible Expansion of an Ideal Gas#

5.3.4 Free Expansion of an Ideal Gas#

5.4 Enthalpy (HHH) and Heat Capacity#

5.4.1 Definition of Enthalpy as a State Function#

5.4.2 Relationship Between DeltaHDelta HDeltaH and DeltaUDelta UDeltaU#

5.4.3 Heat Capacity and Specific Heat#

5.4.4 Derivation of the Meyer Relation (Cp−CV=RC_p - C_V = RCp​−CV​=R) for an Ideal Gas#

5.5 Measurement of DeltaUDelta UDeltaU and DeltaHDelta HDeltaH: Calorimetry#

5.5.1 Constant-Volume Bomb Calorimetry (DeltaUDelta UDeltaU Measurement)#

5.5.2 Constant-Pressure Calorimetry (DeltaHDelta HDeltaH Measurement)#

5.6 Standard Enthalpy of Reactions and Thermochemical Laws#

5.6.1 Standard State Convention#

5.6.2 Standard Enthalpy of Formation (ΔfH∘\Delta_f H^{\circ}Δf​H∘)#

Enthalpy of Reaction from Standard Enthalpies of Formation

5.6.3 Hess's Law of Constant Heat Summation#

Rules for Manipulating Thermochemical Equations:

5.7 Enthalpies of Different Types of Reactions#

5.7.1 Enthalpy of Phase Transformations#

5.7.2 Enthalpy of Combustion (ΔcH∘\Delta_c H^{\circ}Δc​H∘)#

5.7.3 Enthalpy of Atomization (ΔaH∘\Delta_a H^{\circ}Δa​H∘)#

5.7.4 Bond Enthalpy (ΔbondH∘\Delta_{\text{bond}}H^{\circ}Δbond​H∘)#

Estimating Enthalpy of Reaction from Bond Enthalpies:

5.7.5 Lattice Enthalpy and the Born-Haber Cycle#

Deconstruction of the Born-Haber Cycle for Sodium Chloride (mathrmNaClmathrm{NaCl}mathrmNaCl):

Enthalpy of Solution and Hydration:

5.8 Spontaneity and the Second Law of Thermodynamics#

5.8.1 The Physical Limitation of the First Law#

5.8.2 Is Decrease in Enthalpy the Criterion for Spontaneity?#

5.8.3 The Concept of Entropy (SSS)#

Thermodynamic Definition of Entropy Change

5.8.4 The Second Law of Thermodynamics#

Entropy of Physical States:

5.9 Gibbs Free Energy (GGG) and Reaction Spontaneity#

5.9.1 Derivation of the Gibbs Equation#

5.9.2 Criterion for Spontaneity at Constant TTT and ppp#

5.9.3 Effect of Temperature on Reaction Spontaneity#

Equilibrium Crossover Temperature (TeqT_{\text{eq}}Teq​):

5.9.4 The Third Law of Thermodynamics#

Physical Foundation:

5.10 Gibbs Energy Change and Chemical Equilibrium#

5.10.1 Relationship Between ΔrG∘\Delta_r G^{\circ}Δr​G∘ and Equilibrium Constant (KKK)#

Solutions to End-of-Chapter NCERT Exercises (5.1 to 5.22)#

Exercise 5.1#

Exercise 5.2#

Exercise 5.3#

Exercise 5.4#

Exercise 5.5#

Exercise 5.6#

Exercise 5.7#

Exercise 5.8#

Exercise 5.9#

Exercise 5.10#

Exercise 5.11#

Exercise 5.12#

Exercise 5.13#

Exercise 5.14#

Exercise 5.15#

Exercise 5.16#

Exercise 5.17#

Exercise 5.18#

Exercise 5.19#

5.2.1 Internal Energy ( $U$ ) as a State Function#

5.4 Enthalpy ( $H$ ) and Heat Capacity#

5.4.2 Relationship Between $Delta H$ and $Delta U$ #

5.4.4 Derivation of the Meyer Relation ( $C_p - C_V = R$ ) for an Ideal Gas#

5.5 Measurement of $Delta U$ and $Delta H$ : Calorimetry#

5.5.1 Constant-Volume Bomb Calorimetry ( $Delta U$ Measurement)#

5.5.2 Constant-Pressure Calorimetry ( $Delta H$ Measurement)#

5.6.2 Standard Enthalpy of Formation ( $\Delta_f H^{\circ}$ )#

5.7.2 Enthalpy of Combustion ( $\Delta_c H^{\circ}$ )#

5.7.3 Enthalpy of Atomization ( $\Delta_a H^{\circ}$ )#

5.7.4 Bond Enthalpy ( $\Delta_{\text{bond}}H^{\circ}$ )#

Deconstruction of the Born-Haber Cycle for Sodium Chloride ( $mathrm{NaCl}$ ):

5.8.3 The Concept of Entropy ( $S$ )#

5.9 Gibbs Free Energy ( $G$ ) and Reaction Spontaneity#

5.9.2 Criterion for Spontaneity at Constant $T$ and $p$ #

Equilibrium Crossover Temperature ( $T_{\text{eq}}$ ):

5.10.1 Relationship Between $\Delta_r G^{\circ}$ and Equilibrium Constant ( $K$ )#