Chemical Thermodynamics Formula Sheet
All key Chemical Thermodynamics formulas: first law, enthalpy, Hess's law, entropy, Gibbs energy, equilibrium, bond energy & Born-Haber. JEE Main & Advanced revision.
Last-minute revision sheet for the entire Chemical Thermodynamics chapter: systems and state functions, the First Law, work in different processes, enthalpy and Hess’s law, entropy and the Second Law, Gibbs energy and spontaneity, equilibrium relations, and thermochemistry (bond energy, Born-Haber, lattice energy). Every formula, relation, and value below is pulled straight from the chapter pages.
Chapter Map
graph TD
A[Chemical Thermodynamics] --> B[First Law: ΔU = q + w]
A --> C[Enthalpy: H = U + PV]
A --> D[Entropy: Second Law]
A --> E[Gibbs Energy: ΔG]
C --> C1[Hess's Law]
C --> C2[Bond / Born-Haber]
D --> D1[Spontaneity]
E --> E1[ΔG° = -RT ln K]
E --> E2[Van't Hoff]Systems, State & Path Functions
| Type of system | Matter exchange | Energy exchange |
|---|---|---|
| Open | Yes | Yes |
| Closed | No | Yes |
| Isolated | No | No |
| Function class | Members | Property |
|---|---|---|
| State functions | $U,\ H,\ S,\ G$ | Depend only on initial and final state |
| Path functions | $q,\ w$ | Depend on the route taken |
For a cyclic process: $\oint dU = 0$ but $\oint dq \neq 0,\ \oint dw \neq 0$.
“Heat IN, Work ON are positive.”
- Heat absorbed by system: $q > 0$ (endothermic); released: $q < 0$ (exothermic).
- Work done ON system (compression): $w > 0$; work done BY system (expansion): $w < 0$.
Physics often takes work done BY the system as positive — always check the convention in the question.
First Law of Thermodynamics
$$\boxed{\Delta U = q + w}$$$$\Delta U = U_{final} - U_{initial}$$For an ideal gas, internal energy depends only on temperature (no intermolecular forces):
$$U = f\cdot\frac{nRT}{2}\quad\left(\text{monatomic } U=\tfrac{3}{2}nRT,\ \text{diatomic } U=\tfrac{5}{2}nRT\right)$$where $f$ = degrees of freedom.
Work in Thermodynamic Processes
$$\boxed{w = -P_{ext}\,\Delta V}$$| Process | Work $w$ | Key results |
|---|---|---|
| Free expansion (into vacuum) | $0$ | $P_{ext}=0$; for ideal gas $q=0,\ \Delta U=0$ |
| Against constant $P_{ext}$ | $-P_{ext}\Delta V = -\Delta n_g RT$ | Irreversible |
| Reversible isothermal (ideal gas) | $-nRT\ln\dfrac{V_2}{V_1}$ | Maximum work |
| Adiabatic (ideal gas) | $\Delta U = nC_V\Delta T$ | $q=0$ |
Reversible isothermal work (ideal gas), all equivalent forms:
$$\boxed{w = -nRT\ln\frac{V_2}{V_1} = -2.303\,nRT\log\frac{V_2}{V_1} = -nRT\ln\frac{P_1}{P_2} = -2.303\,nRT\log\frac{P_1}{P_2}}$$Adiabatic work for ideal gas:
$$w = \Delta U = nC_V(T_2 - T_1) = \frac{nR(T_2 - T_1)}{\gamma - 1},\qquad \gamma = \frac{C_P}{C_V}$$For isothermal expansion of an ideal gas: $\Delta U = 0 \Rightarrow q = -w$, i.e. heat absorbed = work done by the gas.
Reversible expansion always extracts more work than irreversible between the same two states.
Heat, Heat Capacity & Process Map
| Process | $\Delta T$ | $\Delta U$ | $w$ | $q$ |
|---|---|---|---|---|
| Isothermal (ideal gas) | $0$ | $0$ | $-nRT\ln\frac{V_2}{V_1}$ | $-w$ |
| Adiabatic | changes | $nC_V\Delta T$ | $\Delta U$ | $0$ |
| Isochoric (const $V$) | changes | $nC_V\Delta T$ | $0$ | $\Delta U$ |
| Isobaric (const $P$) | changes | $nC_V\Delta T$ | $-P\Delta V$ | $nC_P\Delta T$ |
| Free expansion (ideal) | $0$ | $0$ | $0$ | $0$ |
| Cyclic | $0$ | $0$ | $-q$ | $q$ |
Heat capacities and the key relation:
$$C_V = \left(\frac{dU}{dT}\right)_V,\qquad C_P = \left(\frac{dH}{dT}\right)_P,\qquad q_V = nC_V\Delta T,\qquad q_P = nC_P\Delta T$$$$\boxed{C_P - C_V = R\quad(\text{ideal gas})}$$At constant volume $q_V = \Delta U$; at constant pressure $q_P = \Delta H$.
Enthalpy
$$\boxed{H = U + PV},\qquad \Delta H = q_P$$At constant pressure, and for gas-phase reactions (ideal gas $PV = nRT$):
$$\boxed{\Delta H = \Delta U + P\Delta V = \Delta U + \Delta n_g RT}$$where $\Delta n_g$ = (moles of gaseous products) − (moles of gaseous reactants). Count gaseous moles only.
| $\Delta n_g$ | Relation |
|---|---|
| $>0$ | $\Delta H > \Delta U$ |
| $<0$ | $\Delta H < \Delta U$ |
| $0$ | $\Delta H = \Delta U$ |
| Reaction type | $\Delta H$ sign |
|---|---|
| Exothermic (e.g. combustion, freezing) | $\Delta H < 0$ |
| Endothermic (e.g. melting, photosynthesis) | $\Delta H > 0$ |
Types of Enthalpy Change
| Type | Symbol | Definition |
|---|---|---|
| Formation | $\Delta_f H°$ | 1 mol compound from elements in standard states |
| Combustion | $\Delta_c H°$ | 1 mol burned completely in O₂ (always $<0$) |
| Neutralization | $\Delta_{neut} H°$ | 1 mol H⁺ + 1 mol OH⁻ → H₂O |
| Atomization | $\Delta_a H°$ | 1 mol converted to gaseous atoms |
| Bond dissociation | $\Delta_{bond} H°$ | Break 1 mol of a specific bond |
| Sublimation | $\Delta_{sub} H°$ | Solid → gas |
| Fusion | $\Delta_{fus} H°$ | Solid → liquid |
| Vaporization | $\Delta_{vap} H°$ | Liquid → gas |
| Ionization | $\Delta_{ion} H$ | Remove electron from gaseous atom |
| Electron gain | $\Delta_{eg} H$ | Add electron to gaseous atom |
| Hydration | $\Delta_{hyd} H$ | Gaseous ion → aqueous ion |
| Solution | $\Delta_{sol} H$ | Dissolve substance in solvent |
Standard state: 298 K (unless stated), 1 bar pressure, 1 M for solutions, most stable form of the substance.
$\Delta_f H°[\text{element in standard state}] = 0$ — e.g. O₂(g), H₂(g), C(graphite), Br₂(l) are zero; O₃(g) and C(diamond) are not. Strong acid + strong base neutralization is a near-constant $\Delta_{neut} H° = -57.1$ kJ/mol (the real reaction is just H⁺ + OH⁻ → H₂O).
Calculating Reaction Enthalpy
Using standard enthalpies of formation (multiply each by its stoichiometric coefficient):
$$\boxed{\Delta_r H° = \sum \Delta_f H°(\text{products}) - \sum \Delta_f H°(\text{reactants})}$$Using combustion data (note the reversed sign relative to formation):
$$\Delta_r H° = \sum \Delta_c H°(\text{reactants}) - \sum \Delta_c H°(\text{products})$$Formation from combustion data:
$$\Delta_f H°(\text{compound}) = \sum \Delta_c H°(\text{elements}) - \Delta_c H°(\text{compound})$$Hess’s Law
$$\boxed{\Delta H_{overall} = \sum \Delta H_{steps}}$$Because $H$ is a state function. Rules when manipulating equations:
- Reverse a reaction → change sign of $\Delta H$.
- Multiply a reaction by a factor → multiply $\Delta H$ by the same factor.
- Add reactions → add their $\Delta H$ values.
Bond Enthalpy & Thermochemistry
$$\boxed{\Delta H_{rxn} = \sum(\text{bonds broken}) - \sum(\text{bonds formed})}$$equivalently $\sum(\text{BE of reactants}) - \sum(\text{BE of products})$.
- Breaking bonds: endothermic, positive (“Break Bad”).
- Forming bonds: exothermic, negative.
- Bond dissociation energy (specific bond, specific molecule) vs average bond energy (mean across many molecules); average O–H from H₂O $= \tfrac{499+428}{2} = 464$ kJ/mol.
Common Average Bond Energies (kJ/mol)
| Single | BE | Double | BE | Triple | BE |
|---|---|---|---|---|---|
| H–H | 436 | C=C | 611 | N≡N | 946 |
| C–C | 347 | C=O | 741 | C≡C | 837 |
| C–H | 414 | O=O | 498 | C≡O | 1072 |
| O–H | 464 | C=N | 615 | C≡N | 891 |
Trends: triple > double > single; smaller atoms form stronger bonds (H–F 568 > H–Cl 431 > H–Br 366 > H–I 298).
Born-Haber Cycle (lattice energy)
$$\boxed{\Delta_f H° = \Delta_{sub} H + \Delta_{diss} H + \Delta_{ion} H + \Delta_{eg} H + U}$$(Match stoichiometry: e.g. $\tfrac{1}{2}\text{Cl}_2 \to \text{Cl}$ uses half the Cl–Cl bond energy.) Mnemonic “Some Dumb Idiots Eat Lettuce” = Sublimation, Dissociation, Ionization, Electron gain, Lattice.
Lattice energy ($\text{M}^{n+}(g) + \text{X}^{n-}(g) \to \text{MX}(s)$, exothermic so $U<0$):
$$U \propto \frac{|z_+|\,|z_-|}{r_+ + r_-}$$Higher ionic charge and smaller ions → more negative $U$ (e.g. MgO ≈ −3791 vs NaCl ≈ −788 kJ/mol).
Resonance Energy
$$E_{res} = \Delta H_{\text{calculated (Kekulé/bonds)}} - \Delta H_{\text{experimental}}$$A more stable (resonance-stabilized) molecule releases less heat on combustion. Benzene resonance energy ≈ −150 kJ/mol.
Entropy & the Second Law
Statistical (Boltzmann) and thermodynamic definitions:
$$S = k_B \ln W,\qquad k_B = 1.38\times10^{-23}\ \text{J/K}$$$$\boxed{\Delta S = \frac{q_{rev}}{T}}\quad(\text{units J/K or J·mol}^{-1}\text{K}^{-1})$$$$\boxed{\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings} \geq 0}$$$$\Delta S_{surroundings} = -\frac{\Delta H_{system}}{T}\quad(\text{const } T,P)$$| $\Delta S_{universe}$ | Process |
|---|---|
| $>0$ | Spontaneous |
| $=0$ | Equilibrium / reversible |
| $<0$ | Non-spontaneous |
Entropy in Specific Processes
| Process | Formula | Condition |
|---|---|---|
| Phase transition | $\Delta S = \dfrac{\Delta H_{trans}}{T_{trans}}$ | Use $T_m$ or $T_b$ |
| Heating (const $P$) | $\Delta S = nC_P\ln\dfrac{T_2}{T_1}$ | — |
| Heating (const $V$) | $\Delta S = nC_V\ln\dfrac{T_2}{T_1}$ | — |
| Isothermal expansion (ideal gas) | $\Delta S = nR\ln\dfrac{V_2}{V_1} = -nR\ln\dfrac{P_2}{P_1}$ | $T$ constant |
| Mixing of ideal gases | $\Delta S_{mix} = -nR(x_1\ln x_1 + x_2\ln x_2)$ | Always $>0$ |
| Reaction | $\Delta S°_{rxn} = \sum S°(\text{prod}) - \sum S°(\text{react})$ | Use absolute $S°$ |
Entropy Trends
| Factor | Effect on $S$ |
|---|---|
| Solid → liquid → gas | Increases ($S_g \gg S_l > S_s$) |
| Temperature ↑ | Increases |
| Volume ↑ | Increases |
| Number of particles/moles ↑ | Increases |
| Molecular complexity ↑ | Increases |
| Dissolution (usually) | Increases |
$\Delta S > 0$ when $n_{products} > n_{reactants}$ (especially when gas is produced).
Gibbs Free Energy & Spontaneity
$$\boxed{G = H - TS}\qquad\boxed{\Delta G = \Delta H - T\Delta S}$$$$\boxed{\Delta G = -T\,\Delta S_{universe}},\qquad w_{max} = -\Delta G$$| $\Delta G$ | Meaning |
|---|---|
| $<0$ | Spontaneous (forward) |
| $=0$ | Equilibrium |
| $>0$ | Non-spontaneous (reverse favored) |
The Four Cases ($\Delta H$ vs $\Delta S$)
| $\Delta H$ | $\Delta S$ | $\Delta G$ | Spontaneity | Example |
|---|---|---|---|---|
| − | + | always − | Always spontaneous | Combustion |
| + | − | always + | Never spontaneous | Reverse combustion |
| − | − | − at low $T$ | Spontaneous at low $T$ | Freezing water, Haber |
| + | + | − at high $T$ | Spontaneous at high $T$ | Melting ice, CaCO₃ decomposition |
Crossover temperature (where $\Delta G = 0$):
$$\boxed{T_{eq} = \frac{\Delta H}{\Delta S}}$$“Give Her The Stuff”: $\Delta G = \Delta H - T\Delta S$.
Keep units consistent — $\Delta H$ in kJ and $\Delta S$ in J/K is the classic blunder; convert one before subtracting.
Standard Gibbs Energy & Equilibrium
$$\Delta_r G° = \sum \Delta_f G°(\text{products}) - \sum \Delta_f G°(\text{reactants}),\qquad \Delta_f G°[\text{element}] = 0$$$$\Delta G° = \Delta H° - T\Delta S°$$$$\boxed{\Delta G° = -RT\ln K = -2.303\,RT\log K}$$$$\boxed{\Delta G = \Delta G° + RT\ln Q}$$At equilibrium $Q = K$ and $\Delta G = 0$. Rearrangements:
$$K = e^{-\Delta G°/RT},\qquad \log K = -\frac{\Delta G°}{2.303RT}$$At 298 K (with $\Delta G°$ in kJ/mol):
$$\boxed{\log K = -\frac{\Delta G°\ (\text{kJ/mol})}{5.7}}$$so every 5.7 kJ/mol change in $\Delta G°$ shifts $K$ by a factor of 10.
| $Q$ vs $K$ | $\Delta G$ | Direction |
|---|---|---|
| $Q < K$ | $<0$ | Forward (toward products) |
| $Q = K$ | $0$ | At equilibrium |
| $Q > K$ | $>0$ | Backward (toward reactants) |
Van’t Hoff Equation ($K$ vs $T$)
$$\ln K = -\frac{\Delta H°}{RT} + \frac{\Delta S°}{R},\qquad \frac{d\ln K}{dT} = \frac{\Delta H°}{RT^2}$$$$\boxed{\ln\frac{K_2}{K_1} = -\frac{\Delta H°}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)}$$- Exothermic ($\Delta H° < 0$): $K$ decreases as $T$ increases.
- Endothermic ($\Delta H° > 0$): $K$ increases as $T$ increases.
Reference Data Tables
Values quoted in the chapter (298 K unless noted).
Standard Enthalpies of Formation $\Delta_f H°$ (kJ/mol)
| Compound | $\Delta_f H°$ | Compound | $\Delta_f H°$ |
|---|---|---|---|
| H₂O(l) | −285.8 | NH₃(g) | −46 |
| H₂O(g) | −241.8 | NO(g) | +90 |
| CO₂(g) | −393.5 | NO₂(g) | +33 |
| CO(g) | −110.5 | SO₂(g) | −297 |
| CH₄(g) | −74.8 | SO₃(g) | −396 |
| C₂H₆(g) | −85 | HCl(g) | −92 |
| C₂H₄(g) | +52 | NaCl(s) | −411 |
| C₂H₂(g) | +227 | CaO(s) | −635 |
Standard Entropies $S°$ (J·mol⁻¹·K⁻¹)
| Substance | $S°$ | Substance | $S°$ |
|---|---|---|---|
| H₂(g) | 131 | H₂O(l) | 70 |
| O₂(g) | 205 | H₂O(g) | 189 |
| N₂(g) | 192 | CO₂(g) | 214 |
| CH₄(g) | 186 |
Typical ranges: solids 50–100, liquids 100–200, gases 150–300 J·mol⁻¹·K⁻¹.
Heats of Combustion $\Delta_c H°$ (kJ/mol)
| Substance | $\Delta_c H°$ | Substance | $\Delta_c H°$ |
|---|---|---|---|
| H₂(g) | −286 | C₃H₈(g) | −2220 |
| C(graphite) | −394 | C₄H₁₀(g) | −2878 |
| CH₄(g) | −890 | C₂H₅OH(l) | −1367 |
| C₂H₆(g) | −1560 | C₆H₁₂O₆(s) | −2808 |
| C₈H₁₈(l) | −5471 |
Lattice Energies $U$ (kJ/mol)
| Compound | $U$ | Compound | $U$ |
|---|---|---|---|
| LiF | −1037 | MgO | −3791 |
| NaCl | −788 | CaO | −3520 |
| KBr | −689 | Al₂O₃ | −15916 |
| CsI | −604 |
Useful Constants
| Constant | Value |
|---|---|
| Gas constant $R$ | 8.314 J·mol⁻¹·K⁻¹ |
| Boltzmann constant $k_B$ | $1.38\times10^{-23}$ J/K |
| $2.303\,R$ | 19.15 J·mol⁻¹·K⁻¹ |
| $2.303\,RT$ at 298 K | 5.708 kJ/mol |
Quick-Fire Recall
- First law: $\Delta U = q + w$ — Heat IN, Work ON positive.
- Constant V → $q_V = \Delta U$; constant P → $q_P = \Delta H$.
- $\Delta H = \Delta U + \Delta n_g RT$ — gaseous moles only.
- $C_P - C_V = R$ (ideal gas).
- Hess: $\Delta H$ is path-independent (state function).
- Bond method: broken − formed; breaking is positive.
- $\Delta S_{universe} \geq 0$ decides spontaneity; $\Delta S_{surr} = -\Delta H/T$.
- $\Delta G = \Delta H - T\Delta S = -T\Delta S_{universe}$; $T_{eq} = \Delta H/\Delta S$.
- $\Delta G° = -RT\ln K$; at 298 K $\log K = -\Delta G°/5.7$.
- $\Delta G = \Delta G° + RT\ln Q$; Van’t Hoff for $K$ vs $T$.