Chemical Thermodynamics revision notes

• First Law of Thermodynamics (FLOT) / Law of Conservation of Energy: The energy of an isolated system is constant. Energy can neither be created nor destroyed. Mathematical form: $\Delta U = q + w$.
• Hess's Law of Constant Heat Summation: If a reaction takes place in several steps, its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions at the same temperature. Enthalpy change is independent of the path.
• Second Law of Thermodynamics: In an isolated system, there is always a tendency for the system's energy to become more disordered. For a spontaneous process, the entropy of the universe always increases ($\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr} > 0$).
• Third Law of Thermodynamics: The entropy of any perfectly pure crystalline substance approaches zero as the temperature approaches absolute zero ($0 \text{ K}$).

Thermochemical Equation Conventions

• Coefficients strictly represent moles, not molecules.
• The physical state must be specified because enthalpy changes with phase.
• If you multiply/divide the coefficients of an equation, you must multiply/divide the $\Delta_r H^\ominus$ by the same factor.
• Reversing the reaction reverses the sign of $\Delta_r H^\ominus$.

• Internal Energy ($U$): Sum of all types of energy (chemical, electrical, mechanical) associated with a system. Absolute $U$ cannot be calculated; only $\Delta U$ is measured.
• Heat ($q$): Energy transfer due to a temperature difference.
• Work ($w$): Mechanical transfer of energy. Pressure-volume work is given by $w = -p_{ex} \Delta V$.

IUPAC Sign Conventions

JEE Tip Always use these strictly in Chemistry to avoid sign errors in numericals!

• $q > 0$ (+): Heat absorbed by the system (Endothermic).
• $q < 0$ (-): Heat released from the system (Exothermic).
• $w > 0$ (+): Work done on the system (Compression).
• $w < 0$ (-): Work done by the system (Expansion).

Reversible vs. Irreversible Processes

• Reversible: Process proceeds infinitely slowly by a series of equilibrium states such that system and surroundings are always in near equilibrium. $p_{ex} \approx p_{in} \pm dp$.
• Irreversible: Process carried out rapidly or against a constant external pressure.
• Free Expansion: Expansion of a gas into a vacuum ($p_{ex} = 0$). JEE Tip Work done in free expansion is ALWAYS zero ($w = 0$), regardless of whether the process is reversible or irreversible. For isothermal free expansion of an ideal gas: $\Delta U = 0, q = 0, w = 0$.

• Enthalpy ($H$): Defined as $H = U + pV$. It represents the total heat content of a system.

Heat Transfer Conditions

• Heat transferred at constant volume ($q_v$) = $\Delta U$.
• Heat transferred at constant pressure ($q_p$) = $\Delta H$.

Relation between $\Delta H$ and $\Delta U$

• General: $\Delta H = \Delta U + p\Delta V$ (at constant $p$).
• For Ideal Gases: $\Delta H = \Delta U + \Delta n_g RT$.
• JEE Tip $\Delta n_g$ ONLY includes gaseous species! $\Delta n_g = (\sum \text{moles of gaseous products}) - (\sum \text{moles of gaseous reactants})$. Treat solids and liquids as having negligible volume changes ($\Delta V \approx 0$).

Heat Capacity ($C$)

The heat required to raise the temperature of a system by $1^\circ C$ or $1 \text{ K}$.

• Specific Heat Capacity ($c$): per unit mass ($q = mc\Delta T$).
• Molar Heat Capacity ($C_m$): per mole ($q = nC_m\Delta T$).

$C_p$ and $C_v$

Molar heat capacity at constant pressure ($C_p$) and constant volume ($C_v$).

• For an ideal gas: $C_p - C_v = R$.

• Bomb Calorimeter: Used to measure heat of combustion. The steel vessel is sealed (constant volume, $\Delta V = 0$). JEE Tip Bomb calorimetry measures $\Delta U$ ($q_v$), NOT $\Delta H$. You must use $\Delta H = \Delta U + \Delta n_g RT$ to find the enthalpy of reaction afterward.
• Constant Pressure Calorimeter: A simple calorimeter open to atmospheric pressure. Measures $\Delta H$ ($q_p$).

• Standard Enthalpy of Reaction ($\Delta_r H^\ominus$): Enthalpy change when all participating substances are in standard states (pure form at 1 bar pressure).

Standard Enthalpy of Formation ($\Delta_f H^\ominus$)

Enthalpy change when one mole of a compound is formed from its constituent elements in their most stable reference states.

• JEE Tip The $\Delta_f H^\ominus$ of elements in their standard reference states (e.g., $O_2(g)$, $C(graphite)$, $S(rhombic)$, $H_2(g)$) is exactly ZERO.

• Standard Enthalpy of Combustion ($\Delta_c H^\ominus$): Enthalpy change when one mole of a substance undergoes complete combustion in excess oxygen. Always highly exothermic ($\Delta_c H^\ominus < 0$).

Enthalpies of Phase Transformations

• Standard Enthalpy of Fusion ($\Delta_{fus}H^\ominus$): Enthalpy change when 1 mole of solid melts in standard state. Always positive (endothermic).

• Standard Enthalpy of Vaporization ($\Delta_{vap}H^\ominus$): Heat required to vaporize 1 mole of liquid at constant T and 1 bar. Magnitude depends on intermolecular forces (e.g., higher for water due to H-bonding compared to acetone).

• Standard Enthalpy of Sublimation ($\Delta_{sub}H^\ominus$): Direct conversion of 1 mole of solid to vapor at constant T and 1 bar (e.g., solid $CO_2$ or naphthalene).

• Enthalpy of Atomization ($\Delta_a H^\ominus$): Enthalpy change on breaking one mole of bonds completely to obtain individual atoms in the gas phase.

Bond Enthalpy ($\Delta_{bond} H^\ominus$)

• Diatomic molecules: Bond dissociation enthalpy equals enthalpy of atomization.
• Polyatomic molecules: Successive bond breaking requires different energies (e.g., the 4 C-H bonds in $CH_4$). Hence, we use Mean Bond Enthalpy.
• Lattice Enthalpy ($\Delta_{lattice} H^\ominus$): Enthalpy change when one mole of an ionic compound dissociates into its gaseous ions (Endothermic: $\Delta H > 0$). Calculated via the Born-Haber Cycle.
• Enthalpy of Solution ($\Delta_{sol} H^\ominus$): $\Delta_{sol} H^\ominus = \Delta_{lattice} H^\ominus + \Delta_{hyd} H^\ominus$. If lattice enthalpy is very high compared to hydration enthalpy, the salt is insoluble.
• Enthalpy of Dilution: The heat withdrawn from the surroundings when additional solvent is added to a solution. It depends on the original concentration of the solution and the amount of solvent added. Approaches a limiting value at infinite dilution.
• Temperature Dependence of Ionization Energy and Electron Affinity (Kirchhoff's equation application): Ionization energy and electron affinity are technically defined at absolute zero (0 K). At any other temperature $T$, heat capacities ($C_p$) must be accounted for. For ionization ($M(g) \rightarrow M^+(g) + e^-$), $\Delta_r C_p^\ominus = +5/2 R$. Thus, $\Delta_{ion} H^\ominus = E_0 (\text{ionization energy}) + 5/2 RT$.

• Entropy ($S$): State function measuring the degree of randomness or disorder. ($S_{gas} > S_{liquid} > S_{solid}$).
• Entropy of Mixing (Diffusion): When a partition between two gases is removed, they spontaneously diffuse. The driving force is not an energy decrease ($\Delta H = 0$), but rather the system becoming less predictable and more chaotic (increase in entropy).
• Gibbs Free Energy ($G$): $G = H - TS$. An extensive state function. $\Delta G$ gives the net energy available to do useful work.

Criteria for Spontaneity (at constant $T$ and $P$)

• $\Delta G < 0$: Spontaneous.
• $\Delta G > 0$: Non-spontaneous.
• $\Delta G = 0$: Equilibrium.

Effect of Temperature on Spontaneity ($\Delta G = \Delta H - T\Delta S$)

• $\Delta H < 0, \Delta S > 0$: Spontaneous at ALL temperatures ($\Delta G < 0$).
• $\Delta H > 0, \Delta S < 0$: Non-spontaneous at ALL temperatures ($\Delta G > 0$).
• $\Delta H > 0, \Delta S > 0$: Spontaneous at HIGH temperatures (entropy driven).
• $\Delta H < 0, \Delta S < 0$: Spontaneous at LOW temperatures (enthalpy driven).
• Gibbs Energy and Equilibrium: JEE Tip At equilibrium, $\Delta G = 0$, but standard Gibbs energy $\Delta G^\ominus$ is NOT zero unless $K = 1$. The relation is $\Delta G^\ominus = -RT \ln K$.

1. First Law: $\Delta U = q + w$,
2. Irreversible Work: $w = -p_{ex} \Delta V = -p_{ex}(V_f - V_i)$
3. Reversible Isothermal Work: $w_{rev} = -2.303 nRT \log \left(\frac{V_f}{V_i}\right) = -2.303 nRT \log \left(\frac{p_i}{p_f}\right)$,
4. Enthalpy Definition: $H = U + pV$
5. Enthalpy-Internal Energy Relation: $\Delta H = \Delta U + \Delta n_g RT$
6. Heat Capacity: $q = C\Delta T = nC_m\Delta T = mc\Delta T$,,
7. Relation between $C_p$ and $C_v$: $C_p - C_v = R$ (for ideal gas)
8. Enthalpy of Reaction (from formation enthalpies): $\Delta_r H^\ominus = \sum a_i \Delta_f H^\ominus(\text{products}) - \sum b_i \Delta_f H^\ominus(\text{reactants})$,
9. Enthalpy of Reaction (from bond enthalpies): $\Delta_r H^\ominus = \sum \text{Bond Enthalpies(Reactants)} - \sum \text{Bond Enthalpies(Products)}$,
10. Entropy Change: $\Delta S = \frac{q_{rev}}{T}$
11. Total Entropy Change: $\Delta S_{total} = \Delta S_{sys} + \Delta S_{surr}$
12. Gibbs Energy Change: $\Delta G = \Delta H - T\Delta S$,
13. Standard Gibbs Energy & Equilibrium: $\Delta_r G^\ominus = -RT \ln K = -2.303 RT \log K$

• Exception in the Third Law of Thermodynamics: The entropy of a perfectly pure crystalline solid approaches zero at absolute zero (0 K). Why is this an exception space? Theoretical arguments and practical evidence show that the entropy of solutions and supercooled liquids is NOT zero at 0 K. They are frozen in a disordered state.
• Anomaly in Polyatomic Bond Enthalpies: In a polyatomic molecule like $CH_4$, breaking the four C-H bonds successively requires different amounts of energy (e.g., 427, 439, 452, and 347 kJ/mol). Why? Each time a bond is broken, the remaining fragment changes its electronic and structural environment. Therefore, we use a Mean Bond Enthalpy for polyatomic molecules.
• Exception to $\Delta H \neq \Delta U$: The difference between $\Delta H$ and $\Delta U$ is usually not significant for systems consisting solely of solids and liquids. Why? Because solids and liquids do not suffer any significant volume changes upon heating or reacting ($\Delta V \approx 0$), making $p\Delta V \approx 0$.
• Exception to Enthalpy Driving Spontaneity: While a decrease in enthalpy (exothermic processes) is a major contributory factor for spontaneity, endothermic reactions can also be strictly spontaneous. Why? Processes like the dissolution of salts, melting of ice, or mixing of two gases occur spontaneously despite absorbing heat because the driving force is a massive increase in entropy (disorder) that overrides the enthalpy cost.
• Anomaly in Fluoride Solubility: Fluorides tend to be significantly less soluble in water than their corresponding chlorides. Why? Solubility depends on the balance between lattice enthalpy and hydration enthalpy. Fluorides have exceptionally high lattice enthalpies that often cannot be overcome by their hydration enthalpies.
• Anomaly in defining Ionization Energy and Electron Affinity terms: Earlier definitions used "Ionization energy" and "Electron affinity" purely as energy terms. Thermodynamic principles strictly differentiate them now as Enthalpy of Ionization and Electron Gain Enthalpy. Why? Because true Ionization Energy is measured at absolute zero (0 K). At room temperature (298 K), the thermal heat capacities of the gases ($5/2 RT$) alter the total enthalpy.

• Isothermal Reversible vs Irreversible Work Calculation: Extremely frequent. Direct application of formula vs equating area under the curve.
• Calculation of $\Delta H$ from $\Delta U$ ($\Delta n_g RT$ relation): Usually embedded in bomb calorimeter word problems.
• Hess's Law and Enthalpy of Formation/Combustion: Combining 3-4 thermochemical equations to find the $\Delta H$ of a target equation.
• Born-Haber Cycle: Calculating lattice enthalpy from sublimation, ionization, dissociation, and electron gain enthalpies. Watch out for stoichiometry (e.g., taking half of the $Cl_2$ dissociation energy).
• Spontaneity and Gibbs Free Energy: Determining the temperature range at which a reaction becomes spontaneous by setting $\Delta G = 0$ and solving for $T = \Delta H / \Delta S$.
• Equilibrium Constant and $\Delta G^\ominus$: Utilizing $\Delta_r G^\ominus = -2.303 RT \log K$ and applying antilogarithms to find $K$. Ensure units match (R = 8.314 J/K mol, and $\Delta G^\ominus$ usually in kJ/mol must be converted to J/mol).