JEE Main & Advanced

Chemical Thermodynamics

Chemical Thermodynamics for JEE Main & Advanced

1
Module 1

First Law and Enthalpy

System, Surroundings, Internal Energy, First LawTopic 1

System: Part of universe under study. Surroundings: Everything else. Boundary separates them.

Types of System:

TypeMass exchangeEnergy exchangeExample
OpenYesYesOpen beaker
ClosedNoYesSealed flask
IsolatedNoNoThermos flask (ideal)

State Function: Property depending only on state, not on path (e.g., $P, V, T, U, H, S, G$). Path Function: Depends on path taken ($q, W$).

Extensive Properties: Depend on amount ($V, n, U, H, S$). Intensive Properties: Independent of amount ($T, P, d, \chi$).

Internal Energy ($U$): Total energy (KE + PE + bond + nuclear) of system. Absolute $U$ not measurable; only $\Delta U$ measurable.

Work (W): $W = -P_{\text{ext}}\Delta V$ for $P$-$V$ work.

Sign Convention (IUPAC):

  • $q > 0$: heat absorbed by system; $q < 0$: heat released
  • $W > 0$: work done on system; $W < 0$: work done by system

First Law of Thermodynamics: Energy is conserved. $$\Delta U = q + W$$

For different processes:

ProcessCondition$\Delta U$$q$$W$
Isothermal$T$ const$0$$-W$$-q$
Adiabatic$q = 0$$W$$0$$\Delta U$
Isochoric$V$ const$q_V$$\Delta U$$0$
Isobaric$P$ const$q_P + W$$\Delta H$$-P\Delta V$
CyclicReturns to initial$0$$-W$$-q$

Reversible vs Irreversible:

  • Reversible: infinitely slow, equilibrium maintained
  • Irreversible: real, finite time, equilibrium not maintained

Isothermal reversible work (ideal gas): $W = -nRT\ln(V_2/V_1) = -nRT\ln(P_1/P_2)$.

Adiabatic reversible work: $W = (P_2V_2 - P_1V_1)/(\gamma - 1) = nC_V\Delta T$.

Worked Examples
1

$2$ moles of ideal gas expand isothermally and reversibly from $5$ L to $20$ L at $300$ K. Calculate $W$ and $q$.

Show solution

$$W = -nRT\ln(V_2/V_1) = -2 \times 8.314 \times 300 \times \ln(20/5)$$ $$= -2 \times 8.314 \times 300 \times \ln 4 = -2 \times 8.314 \times 300 \times 1.386$$ $$= -6914 \text{ J} \approx -6.91 \text{ kJ}$$ For isothermal: $\Delta U = 0$, so $q = -W = +6.91$ kJ.

Final Answer: $W = -6.91$ kJ; $q = +6.91$ kJ.

2

$1$ mole of gas is compressed adiabatically from $10$ L to $5$ L. Temperature increases from $300$ K to $400$ K. Find $\Delta U$. ($C_V = 20.5$ J/mol·K)

Show solution

$\Delta U = nC_V\Delta T = 1 \times 20.5 \times (400-300) = 2050$ J.

Final Answer: $\Delta U = +2050$ J.

✎ Self-Check — 5 questions0 / 5
Q1.

State function:

Q2.

First law of thermodynamics is:

Q3.

For isothermal process of ideal gas:

Q4.

Work done in cyclic process:

Q5.

Closed system can exchange:

Enthalpy, Hess's Law, Bond EnthalpyTopic 2

Enthalpy ($H$): State function defined as $H = U + PV$.

For change at constant $P$: $\Delta H = \Delta U + P\Delta V = q_P$.

Relation: $\Delta H = \Delta U + \Delta n_g RT$, where $\Delta n_g$ = (moles of gaseous products) - (moles of gaseous reactants).

Exothermic: $\Delta H < 0$ (heat released). Endothermic: $\Delta H > 0$ (heat absorbed).

Standard Enthalpy Changes:

TypeDefinitionSymbol
Formation$\Delta H$ for forming 1 mole from elements in standard states$\Delta_f H^\ominus$
Combustion$\Delta H$ for complete combustion of 1 mole$\Delta_c H^\ominus$
Neutralization$\Delta H$ when 1 equivalent of acid neutralizes 1 equivalent of base in dilute solution$\Delta_{\text{neut}} H^\ominus = -57.1$ kJ/mol (strong acid + strong base)
Solution$\Delta H$ when 1 mole solute is dissolved in large amount of solvent$\Delta_{\text{sol}} H$
Hydration$\Delta H$ when ions are hydrated by water$\Delta_{\text{hyd}} H$
Atomization$\Delta H$ to convert 1 mole substance into gaseous atoms$\Delta_a H$
Sublimation$\Delta H$ for solid → gas$\Delta_{\text{sub}} H$
Fusion$\Delta H$ for solid → liquid (at MP)$\Delta_{\text{fus}} H$
Vaporisation$\Delta H$ for liquid → gas (at BP)$\Delta_{\text{vap}} H$

Standard States are pure substance at $1$ bar pressure (specific $T$, usually $298$ K). $\Delta_f H^\ominus$ of element in standard state = $0$.

Hess's Law of Constant Heat Summation: Enthalpy change for a reaction depends only on initial and final states, not on the pathway. Allows calculation of $\Delta H$ for reactions hard to measure directly.

$$\Delta H_{\text{rxn}} = \sum\nu \Delta_f H^\ominus_{\text{products}} - \sum\nu \Delta_f H^\ominus_{\text{reactants}}$$

Bond Enthalpy (Bond Dissociation Energy): Energy needed to break $1$ mole of a particular bond in gaseous state.

$$\Delta H_{\text{rxn}} = \sum(\text{Bond enthalpies of reactants}) - \sum(\text{Bond enthalpies of products})$$

(Note: works only for gaseous species; mean bond enthalpies are used for poly-atomic molecules.)

Lattice Enthalpy: $\Delta H$ when 1 mole of an ionic solid dissociates into gaseous ions: $$\text{NaCl}(s) \to \text{Na}^+(g) + \text{Cl}^-(g); \quad \Delta_L H = +788 \text{ kJ/mol}$$

Kirchhoff's Equation: $\Delta H_2 - \Delta H_1 = \Delta C_P(T_2 - T_1)$.

Worked Examples
1

Given: $\Delta_f H$ (CO₂) = $-393.5$, $\Delta_f H$ (H₂O) = $-285.8$, $\Delta_f H$ (CH₄) = $-74.8$ kJ/mol. Find $\Delta_c H$ of methane.

Show solution

CH₄ + 2O₂ → CO₂ + 2H₂O. $\Delta H = [\Delta_f H(\text{CO}_2) + 2\Delta_f H(\text{H}_2O)] - [\Delta_f H(\text{CH}_4) + 2\Delta_f H(\text{O}_2)]$ $= [-393.5 + 2(-285.8)] - [-74.8 + 0]$ $= -393.5 - 571.6 + 74.8 = -890.3$ kJ/mol.

Final Answer: $\Delta_c H = -890.3$ kJ/mol.

2

Calculate $\Delta H$ for: $H_2(g) + Cl_2(g) \to 2HCl(g)$. Given bond enthalpies: H-H = $436$, Cl-Cl = $242$, H-Cl = $431$ kJ/mol.

Show solution

$$\Delta H = (E_{H-H} + E_{Cl-Cl}) - 2E_{H-Cl} = (436 + 242) - 2(431)$$ $$= 678 - 862 = -184 \text{ kJ/mol}$$

Final Answer: $\Delta H = -184$ kJ/mol (exothermic).

✎ Self-Check — 5 questions0 / 5
Q1.

Enthalpy of formation of element in standard state:

Q2.

Exothermic reaction:

Q3.

Hess's law is based on:

Q4.

$\Delta H$ for neutralization of strong acid by strong base in dilute solution is approximately:

Q5.

For a reaction where $\Delta n_g = -1$, $T = 300$ K:

2
Module 2

Entropy, Gibbs Free Energy and Spontaneity

Entropy and Second LawTopic 1

Entropy ($S$): Measure of disorder/randomness. State function. SI unit: J/(mol·K).

Definition: For reversible process at temperature $T$: $dS = dq_{\text{rev}}/T$.

Second Law of Thermodynamics: For an isolated system, entropy never decreases. $$\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surr}} \geq 0$$

  • Spontaneous (irreversible): $\Delta S_{\text{universe}} > 0$
  • Reversible (equilibrium): $\Delta S_{\text{universe}} = 0$

Entropy Changes:

Phase change$\Delta S$
Fusion (s→l)$+$
Vaporisation (l→g)$+$ (large)
Sublimation (s→g)$+$ (largest)
Solidification (l→s)$-$

Trends:

  • $S(\text{gas}) > S(\text{liquid}) > S(\text{solid})$
  • $\Delta S$ positive for: number of gaseous moles increases, mixing, dissolving solids
  • $\Delta S$ negative for: gas → liquid → solid, gas moles decrease

Standard Entropy of Reaction: $$\Delta S^\ominus_{\text{rxn}} = \sum\nu S^\ominus_{\text{products}} - \sum\nu S^\ominus_{\text{reactants}}$$

Surroundings: Heat lost by system is gained by surroundings. $$\Delta S_{\text{surr}} = -\Delta H_{\text{sys}}/T$$ (for constant $T, P$).

Third Law of Thermodynamics: Entropy of a perfectly crystalline solid at absolute zero ($0$ K) is zero. $$S(0 \text{ K}) = 0 \text{ (perfect crystal)}$$ Allows absolute entropy values to be defined for substances.

Entropy for ideal gas (isothermal): $\Delta S = nR\ln(V_2/V_1) = -nR\ln(P_2/P_1)$.

For temperature change at const $V$: $\Delta S = nC_V \ln(T_2/T_1)$. At const $P$: $\Delta S = nC_P \ln(T_2/T_1)$.

Worked Examples
1

$2$ moles of ideal gas expand isothermally from $1$ L to $10$ L at $300$ K. Find $\Delta S$ of system.

Show solution

$$\Delta S = nR\ln(V_2/V_1) = 2 \times 8.314 \times \ln 10$$ $$= 2 \times 8.314 \times 2.303 = 38.29 \text{ J/K}$$

Final Answer: $\Delta S = +38.29$ J/K.

2

$1$ mole of water at $373$ K vaporizes. $\Delta H_{\text{vap}} = 40.65$ kJ/mol. Find $\Delta S$.

Show solution

At BP, process is reversible: $$\Delta S = \Delta H/T = 40650/373 \approx 109 \text{ J/K mol}$$

Final Answer: $\Delta S \approx 109$ J/(K·mol).

✎ Self-Check — 5 questions0 / 5
Q1.

Entropy of a system at $0$ K (perfect crystal):

Q2.

Among solids, liquids, gases — entropy is highest in:

Q3.

In an irreversible process:

Q4.

Spontaneous endothermic reactions are favored by:

Q5.

Surroundings entropy change at constant $T, P$:

Gibbs Free Energy and SpontaneityTopic 2

Gibbs Free Energy ($G$): Combined criterion of spontaneity at constant $T, P$: $$G = H - TS$$ $$\Delta G = \Delta H - T\Delta S$$

Spontaneity Criteria at constant $T, P$:

$\Delta G$ProcessReversibility
$< 0$Spontaneous (favorable)Irreversible
$> 0$Non-spontaneousReverse is spontaneous
$= 0$At equilibriumReversible

$\Delta G$ Analysis Based on Signs:

$\Delta H$$\Delta S$$\Delta G$Spontaneity
$-$ (exothermic)$+$ (entropy increases)$-$ alwaysSpontaneous at all $T$
$-$$-$$-$ at low $T$, $+$ at high $T$Spontaneous at low $T$
$+$$+$$+$ at low $T$, $-$ at high $T$Spontaneous at high $T$
$+$$-$$+$ alwaysNon-spontaneous at all $T$

Temperature of equilibrium (for case where $\Delta G$ changes sign): $$T = \Delta H/\Delta S$$

Relation with Equilibrium Constant: $$\Delta G^\ominus = -RT\ln K = -2.303 RT\log K$$

If $K > 1$: $\Delta G^\ominus < 0$ (favors products); $K < 1$: $\Delta G^\ominus > 0$ (favors reactants); $K = 1$: $\Delta G^\ominus = 0$.

Free Energy Change Under Non-standard Conditions: $$\Delta G = \Delta G^\ominus + RT\ln Q$$ where $Q$ = reaction quotient.

Relation with Electrochemistry: $\Delta G^\ominus = -nFE^\ominus_{\text{cell}}$, where $n$ = electrons, $F$ = Faraday constant ($96500$ C/mol).

Maximum useful work (other than P-V): $W_{\text{useful}} = -\Delta G$ (at constant $T, P$, reversible).

Standard $\Delta G^\ominus$ of formation: Like $\Delta_f H^\ominus$, $\Delta_f G^\ominus$ of element in standard state = $0$. For a reaction: $$\Delta G^\ominus_{\text{rxn}} = \sum\nu\Delta_f G^\ominus_{\text{products}} - \sum\nu\Delta_f G^\ominus_{\text{reactants}}$$

Worked Examples
1

For a reaction at $300$ K, $\Delta H = -10$ kJ/mol, $\Delta S = -20$ J/(K·mol). Predict spontaneity.

Show solution

$\Delta G = \Delta H - T\Delta S = -10000 - 300(-20) = -10000 + 6000 = -4000$ J/mol. Since $\Delta G < 0$, reaction is spontaneous at $300$ K.

Temperature at equilibrium: $T = \Delta H/\Delta S = -10000/(-20) = 500$ K. Above $500$ K, $\Delta G > 0$ → non-spontaneous.

Final Answer: Spontaneous; equilibrium at $T = 500$ K.

2

$K = 1.0 \times 10^4$ for a reaction at $298$ K. Find $\Delta G^\ominus$.

Show solution

$$\Delta G^\ominus = -2.303 RT\log K = -2.303 \times 8.314 \times 298 \times \log(10^4)$$ $$= -2.303 \times 8.314 \times 298 \times 4 = -22,826 \text{ J/mol} \approx -22.83 \text{ kJ/mol}$$

Final Answer: $\Delta G^\ominus = -22.83$ kJ/mol.

✎ Self-Check — 5 questions0 / 5
Q1.

For spontaneous process at const $T, P$:

Q2.

At equilibrium:

Q3.

The relation $\Delta G^\ominus = -RT\ln K$ holds for:

Q4.

$\Delta G = \Delta H - T\Delta S$ implies spontaneity at high $T$ if:

Q5.

If $\Delta H = +50$ kJ, $\Delta S = +100$ J/K. Temperature above which reaction is spontaneous:

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