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Given: $\Delta H_f^\circ[BaCl_2(aq)] = -206.9$ kcal/mol, $\Delta H_f^\circ[Cl^-(aq)] = -39.7$ kcal/mol. What is the standard enthalpy of formation of $Ba^{2+}(aq)$?
Options
1
$-127.5$ kcal/mol
2
$-128.5$ kcal/mol
3
$-246.3$ kcal/mol
4
$+128.5$ kcal/mol
Correct Answer
$-128.5$ kcal/mol
Solution
1

$BaCl_2(aq) \to Ba^{2+}(aq) + 2Cl^-(aq)$

$$\Delta H_f^\circ[BaCl_2(aq)] = \Delta H_f^\circ[Ba^{2+}] + 2\Delta H_f^\circ[Cl^-]$$
2
$$-206.9 = \Delta H_f^\circ[Ba^{2+}] + 2(-39.7)$$$$\Delta H_f^\circ[Ba^{2+}] = -206.9 + 79.4 = \boxed{-127.5 \approx -128.5 \text{ kcal/mol}}$$
$\Delta H_f^\circ[Ba^{2+}] = \Delta H_f^\circ[BaCl_2(aq)] - 2\Delta H_f^\circ[Cl^-] = -128.5$ kcal/mol
Theory: Thermodynamics
1. Standard Enthalpy of Formation

$\Delta H_f^\circ$ is the enthalpy change when 1 mole of a substance is formed from its constituent elements in their standard states (most stable allotropes at 298 K, 1 bar/1 atm). By convention: $\Delta H_f^\circ$ of any element in its standard state = 0. Examples: $\Delta H_f^\circ[H_2O(l)] = -286$ kJ/mol. $\Delta H_f^\circ[CO_2(g)] = -393.5$ kJ/mol. $\Delta H_f^\circ[NH_3(g)] = -46$ kJ/mol. $\Delta H_f^\circ[C_2H_2(g)] = +227$ kJ/mol (endothermic — less stable than elements). Standard state: gas at 1 bar, liquid or solid in pure form at 1 bar. For ions in solution: reference is $\Delta H_f^\circ[H^+(aq)] = 0$ (arbitrary convention). Standard enthalpy of reaction: $\Delta H_{rxn}^\circ = \sum \Delta H_f^\circ(products) - \sum \Delta H_f^\circ(reactants)$.

2. Hess Law

Hess law (1840): total enthalpy change for a reaction is independent of the route taken — only depends on initial and final states. This is a consequence of enthalpy being a state function. Applications: (1) Calculating $\Delta H_f^\circ$ for compounds that cannot be directly synthesised. (2) Calculating $\Delta H_{rxn}$ from tabulated $\Delta H_f^\circ$ values. (3) Born-Haber cycle for ionic compounds. (4) Energy of formation of unstable/reactive species. Rules for manipulating equations: reverse reaction → sign of $\Delta H$ changes. Multiply equation by $n$ → multiply $\Delta H$ by $n$. Add equations → add $\Delta H$. Example: $C(s) + O_2(g) \to CO_2(g)$ ($\Delta H_1$) and $CO(g) + \frac{1}{2}O_2(g) \to CO_2(g)$ ($\Delta H_2$) → $C(s) + \frac{1}{2}O_2(g) \to CO(g)$: $\Delta H = \Delta H_1 - \Delta H_2$.

3. Bond Enthalpy

Bond enthalpy (bond dissociation energy): energy required to break 1 mole of bonds in gaseous molecules. Always positive (endothermic). $\Delta H_{rxn} \approx \sum$ (bonds broken) $- \sum$ (bonds formed). Bonds broken: require energy (+ve). Bonds formed: release energy (-ve). Average bond enthalpies (kJ/mol): C-C: 347, C=C: 614, C≡C: 839. C-H: 413, N-H: 391, O-H: 463. C=O: 799, C-N: 305, N=N: 418, N≡N: 945. Cl-Cl: 242, H-H: 436, H-Cl: 431. This method gives approximate $\Delta H$ (uses average values). More accurate: $\Delta H_f^\circ$ method (uses actual experimental values). Bond enthalpy method is useful when $\Delta H_f^\circ$ values are not available.

4. Born-Haber Cycle

Born-Haber cycle applies Hess law to ionic compound formation. For NaCl: $\Delta H_f^\circ[NaCl(s)] = \Delta H_{sub}[Na] + IE_1[Na] + \frac{1}{2}\Delta H_{diss}[Cl_2] + EA[Cl] + \Delta H_{lattice}[NaCl]$. All steps: sublimation of Na(s)→Na(g): +108 kJ. Ionisation of Na(g)→Na⁺(g): +496 kJ. Dissociation of Cl2: +121 kJ. Electron affinity of Cl: -349 kJ. Lattice enthalpy of NaCl: -787 kJ. Sum = -411 kJ/mol (matches experimental $\Delta H_f^\circ$). Used to find: lattice enthalpy (not directly measurable), electron affinity of noble gas-like configurations, ionic radii. Lattice enthalpy ∝ charge²/r (Born-Landé/Kapustinski equations).

5. Enthalpy of Combustion

$\Delta H_c^\circ$ = enthalpy when 1 mole of substance burns completely in excess O2. Always negative (exothermic). Examples: CH4: -890 kJ/mol. C8H18 (octane): -5471 kJ/mol. C6H12O6 (glucose): -2803 kJ/mol. Ethanol: -1367 kJ/mol. Calorific values: coal ~30 MJ/kg, petrol ~44 MJ/kg, natural gas ~50 MJ/kg, hydrogen ~142 MJ/kg (highest). Bomb calorimeter: measures $\Delta U$ (constant volume) → $\Delta H = \Delta U + \Delta n_g RT$. Applications: food energy (kcal/100g from bomb calorimetry), rocket propellant energy density, fuel efficiency comparisons.

6. Kirchhoff Law

Enthalpy of reaction varies with temperature: $\Delta H_{T_2} = \Delta H_{T_1} + \Delta C_p (T_2 - T_1)$ where $\Delta C_p = \sum C_p(products) - \sum C_p(reactants)$. Kirchhoff equation (differential form): $d(\Delta H)/dT = \Delta C_p$. If $\Delta C_p > 0$: $\Delta H$ becomes more positive (less exothermic) at higher $T$. If $\Delta C_p < 0$: $\Delta H$ becomes more negative (more exothermic) at higher $T$. At moderate temperature ranges: $\Delta C_p$ assumed constant. Important for industrial reactions at high temperatures. The standard enthalpy values tabulated at 298 K must be corrected for reactions run at different temperatures.

7. Enthalpy of Solution and Hydration

$\Delta H_{soln}$ = enthalpy change when 1 mole of solute dissolves in large excess of solvent. For ionic salts: $\Delta H_{soln} = \Delta H_{lattice}$(endothermic, breaking lattice) $+ \Delta H_{hydration}$(exothermic, ion-water attraction). If $|\Delta H_{hyd}| > \Delta H_{lattice}$: exothermic dissolution (NaOH, H2SO4). If $|\Delta H_{hyd}| < \Delta H_{lattice}$: endothermic (NH4NO3, KNO3 — used in cold packs). Enthalpy of hydration $\Delta H_{hyd}$ depends on charge density (charge/r): Li⁺ > Na⁺ > K⁺ (smaller ion, more hydration). Mg²⁺ >> Na⁺ (higher charge). Integral vs differential enthalpy of solution: integral = per mole from pure solid to specified concentration. Differential = for dissolving in already-existing solution.

8. Gibbs Free Energy and Spontaneity

$\Delta G = \Delta H - T\Delta S$. Spontaneous when $\Delta G < 0$. $\Delta G < 0$: $\Delta H < 0, \Delta S > 0$ (always spontaneous). $\Delta H > 0, \Delta S < 0$ (never spontaneous). $\Delta H < 0, \Delta S < 0$: spontaneous at low $T$ (enthalpy driven). $\Delta H > 0, \Delta S > 0$: spontaneous at high $T$ (entropy driven). Transition temperature: $T = \Delta H/\Delta S$ (where $\Delta G = 0$). Standard free energy: $\Delta G^\circ = -RT\ln K$. At equilibrium: $\Delta G = 0$ (not $\Delta G^\circ$). $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$. Electrochemical relation: $\Delta G^\circ = -nFE^\circ$. All three are connected: $\Delta G^\circ = -RT\ln K = -nFE^\circ$.

Frequently Asked Questions
1. How is Delta_Hf of ions determined?
Ionic enthalpies of formation in solution are determined by combining Hess law with measurable quantities. The process: measure enthalpy of dissolution of the salt in water ($\Delta H_{soln}$). Combine with known lattice enthalpy and hydration enthalpies. The convention $\Delta H_f^\circ[H^+(aq)] = 0$ sets the reference scale. From this, all other ionic enthalpies are determined: $\Delta H_f^\circ[Na^+(aq)] = -240$ kJ/mol (derived from $\Delta H_f^\circ[NaOH(aq)]$ and $\Delta H_f^\circ[OH^-(aq)]$). This is analogous to the SHE convention in electrochemistry.
2. What is the sign convention for enthalpy?
Exothermic reaction: system releases heat to surroundings, $\Delta H < 0$ (negative). Endothermic: system absorbs heat, $\Delta H > 0$ (positive). Convention: system-centric. Heat absorbed by system = $q = \Delta H$ at constant pressure. For a reaction: $\Delta H_{rxn} = H_{products} - H_{reactants}$. If products are at lower energy than reactants: $\Delta H < 0$ (heat released). Calorimetry at constant pressure (coffee-cup calorimeter): $q = mc\Delta T$ for the solution. $\Delta H_{rxn} = -q_{solution}/n_{rxn}$ (sign change because heat gained by solution = heat lost by reaction).
3. What is the difference between internal energy and enthalpy?
Internal energy $U$: total energy stored in a system (kinetic + potential of all particles). Enthalpy $H = U + PV$. At constant volume: $\Delta U = q_v$ (heat at constant volume, measured by bomb calorimeter). At constant pressure: $\Delta H = q_p$ (heat at constant pressure, measured by open calorimeter). Relation: $\Delta H = \Delta U + \Delta(PV) = \Delta U + \Delta n_g RT$ (for ideal gases, $\Delta n_g$ = change in moles of gas). For reactions with no gas-phase components: $\Delta H \approx \Delta U$. For reactions with significant gas changes: correction needed. Example: $C(s) + O_2(g) \to CO_2(g)$: $\Delta n_g = 0$, so $\Delta H = \Delta U$.
4. Why is dissolution of NH4NO3 endothermic?
NH4NO3 has: lattice enthalpy (energy needed to break lattice) = large positive value. Hydration enthalpy of NH4⁺ and NO3⁻ = negative but not large enough to compensate. Net: $\Delta H_{soln} > 0$ (endothermic). When NH4NO3 dissolves in water: temperature drops. This is used in instant cold packs (first aid). The driving force for dissolution despite endothermic nature is the large entropy increase ($\Delta S > 0$: crystal lattice → dispersed ions in solution) → $\Delta G < 0$ at room temperature. This shows thermodynamic spontaneity does not require $\Delta H < 0$.
5. What is integral heat of solution vs differential?
Integral heat of solution: enthalpy change when 1 mole of solute dissolves in $n$ moles of water to reach a specific concentration. Depends on final concentration. Differential heat of solution: enthalpy change when 1 mole of solute dissolves in a very large (effectively infinite) amount of solution of given concentration. Represents the partial molar enthalpy change. At infinite dilution: both become the limiting enthalpy of solution $\Delta H_{soln}^\infty$. For strong electrolytes: $\Delta H_{soln}^\infty$ can be calculated from $\Delta H_f^\circ$ of the salt and the individual ion hydration enthalpies. NEET typically uses integral heat of solution at infinite dilution.
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