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17.1 Enthalpy Changes, Calorimetry, Hess's Law & Born-Haber Cycle

Key Takeaways

  • Thermodynamic systems interact with surroundings across boundaries, where state functions like enthalpy (HH), internal energy (UU), and entropy (SS) depend solely on the initial and final states, yielding qp=ΔH=ΔU+PΔVq_p = \Delta H = \Delta U + P\Delta V at constant pressure.

  • Thermochemical reactions classify into exothermic (ΔH<0\Delta H < 0, heat released, products thermodynamically more stable) and endothermic (ΔH>0\Delta H > 0, heat absorbed), with standard enthalpy of formation ΔHf∘\Delta H_f^\circ defined as zero for pure elements in their standard reference states.

  • Neutralization of any strong monoprotic acid with a strong base in dilute aqueous solution releases an essentially constant enthalpy of ΔHn∘≈−57.4 kJ/mol\Delta H_n^\circ \approx -57.4\text{ kJ/mol} (−13.7 kcal/mol-13.7\text{ kcal/mol}), whereas weak acids or bases yield less negative values due to endothermic partial ionization.

  • Hess's Law of Constant Heat Summation states that the total enthalpy change is independent of the reaction pathway (ΔHrxn∘=∑nΔHf∘(products)−∑mΔHf∘(reactants)\Delta H^\circ_{\text{rxn}} = \sum n\Delta H_f^\circ(\text{products}) - \sum m\Delta H_f^\circ(\text{reactants})), enabling indirect calculation of ionic lattice energies via the Born-Haber cycle.

  • Spontaneous physical and chemical transformations require a net decrease in Gibbs free energy (ΔG=ΔH−TΔS<0\Delta G = \Delta H - T\Delta S < 0), where the balance between enthalpy and entropy dictates temperature-dependent spontaneity thresholds.

Last updated: October 2026

Enthalpy Changes, Calorimetry, Hess's Law & Born-Haber Cycle

In the Intermediate Chemistry Part I syllabus and the University of Engineering and Technology (UET) ECAT entry test, thermochemistry and chemical thermodynamics are foundational subjects. The ECAT Chemistry section places heavy emphasis on calculating standard enthalpy changes, interpreting potential energy profiles, utilizing calorimetry data to compute specific heat and heat capacities, and applying Hess's Law to thermochemical cycles such as the Born-Haber cycle for ionic crystals. Mastering the sign conventions, algebraic formulations, and spontaneity conditions under Gibbs free energy allows candidates to solve multi-step energetic problems rapidly without a calculator.


1. Thermodynamic Systems, Surroundings & State Functions

Thermodynamics examines the energy transformations accompanying physical changes and chemical reactions. To quantify these processes, classical thermodynamics divides the universe into two distinct parts:

  1. System: The specific portion of the universe under physical or chemical investigation (e.g., reactants and products in a reaction vessel, a gas cylinder with a movable piston, or an aqueous electrolyte solution).
  2. Surroundings: Everything in the universe outside the designated system that can interact with it by exchanging matter or energy (e.g., the solvent water, the container walls, the surrounding air, and measuring instruments).
  3. Boundary: The real or imaginary physical envelope separating the system from its surroundings. Boundaries may be rigid or movable, conducting (diathermic) or insulating (adiabatic).

Systems are classified into three thermodynamic categories based on permeability:

  • Open System: Can exchange both matter and energy with its surroundings (e.g., boiling water in an open beaker where water vapor escapes and heat enters).
  • Closed System: Can exchange energy (heat and work) but cannot exchange matter with its surroundings (e.g., a sealed flask containing reacting gases immersed in a water bath).
  • Isolated System: Can exchange neither matter nor energy with its surroundings (e.g., hot liquid contained inside a perfectly insulated, sealed thermos flask; the physical universe as a whole).

State Functions vs. Path Functions

A state function (or state variable) is a macroscopic thermodynamic property whose numerical value depends exclusively on the present physical state of the system (determined by temperature TT, pressure PP, volume VV, and composition) and is completely independent of the past history or the specific pathway taken to reach that state. If a system changes from State 1 to State 2, the net change in a state function XX is simply:

ΔX=Xfinal−Xinitial=X2−X1\Delta X = X_{\text{final}} - X_{\text{initial}} = X_2 - X_1

Crucial state functions in chemical thermodynamics include:

  • Internal Energy (UU or EE): The total sum of all microscopic kinetic energies (translational, rotational, vibrational) and potential energies (intermolecular forces, electronic bonds, nuclear binding forces) of the constituent particles.
  • Enthalpy (HH): Defined mathematically as the sum of internal energy and pressure-volume work potential: H=U+PVH = U + PV.
  • Entropy (SS): The quantitative measure of microscopic molecular disorder or energy dispersion.
  • Gibbs Free Energy (GG): The maximum thermodynamic energy available to do useful non-expansion work: G=H−TSG = H - TS.

In stark contrast, heat (qq) and work (ww) are path functions. Their values depend directly on the specific operational path or mechanism followed during the transformation. Heat and work represent energy in transit across a system boundary; neither heat nor work is a property possessed by a substance at rest.

The First Law of Thermodynamics in Thermochemistry

The First Law of Thermodynamics is the law of conservation of energy: energy can neither be created nor destroyed, although it may be converted from one form to another. Mathematically:

ΔU=q+w\Delta U = q + w

where ΔU\Delta U is the change in internal energy, qq is heat absorbed by the system from the surroundings (q>0q > 0 for endothermic, q<0q < 0 for exothermic), and ww is work done on the system (w=−PΔVw = -P\Delta V when expansion work is done by the system against an opposing external pressure PP). Thus:

ΔU=q−PΔV\Delta U = q - P\Delta V

Two critical operational constraints govern chemical reactions:

  1. Constant Volume Processes (V=constant  ⟹  ΔV=0V = \text{constant} \implies \Delta V = 0): Because no mechanical pressure-volume work can occur (w=0w = 0), the heat absorbed or evolved at constant volume (qvq_v) is exactly equal to the change in internal energy: qv=ΔUq_v = \Delta U
  2. Constant Pressure Processes (P=constantP = \text{constant}): Most bench chemistry and biological reactions occur in open vessels under constant atmospheric pressure (P=1 atmP = 1\text{ atm}). Rearranging the First Law yields: qp=ΔU+PΔVq_p = \Delta U + P\Delta V Because H=U+PVH = U + PV, taking the differential at constant pressure gives ΔH=ΔU+PΔV\Delta H = \Delta U + P\Delta V. Therefore: qp=ΔHq_p = \Delta H The heat evolved or absorbed by a chemical system at constant atmospheric pressure is identically equal to its change in enthalpy (ΔH\Delta H).

For reactions involving ideal gases, PΔV=ΔngRTP\Delta V = \Delta n_g R T, where Δng=∑ngas(products)−∑ngas(reactants)\Delta n_g = \sum n_{\text{gas}}(\text{products}) - \sum n_{\text{gas}}(\text{reactants}). Hence:

ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g R T

  • If Δng=0\Delta n_g = 0 (e.g., H2(g)+Cl2(g)→2HCl(g)\text{H}_2(g) + \text{Cl}_2(g) \rightarrow 2\text{HCl}(g)), then PΔV=0P\Delta V = 0 and ΔH=ΔU\Delta H = \Delta U.
  • If Δng>0\Delta n_g > 0 (gas-producing reactions), work is done by the system against the atmosphere, so ΔH>ΔU\Delta H > \Delta U.
  • If Δng<0\Delta n_g < 0 (gas-consuming reactions), work is done on the system by the atmosphere, so ΔH<ΔU\Delta H < \Delta U.

2. Exothermic vs. Endothermic Reactions & Potential Energy Profiles

Every chemical reaction involves the simultaneous breaking of chemical bonds in reactant molecules (which always requires energy input, an endothermic process) and the formation of new chemical bonds in product molecules (which always releases energy, an exothermic process). The net enthalpy change ΔH=Hproducts−Hreactants\Delta H = H_{\text{products}} - H_{\text{reactants}} reflects the energy difference between these complementary events:

  1. Exothermic Reactions (ΔH<0\Delta H < 0):
    • Energy released during product bond formation exceeds energy consumed in reactant bond cleavage.
    • Heat is evolved to the surroundings, causing the temperature of the reaction mixture and container to rise (ΔT>0\Delta T > 0).
    • The total enthalpy of the products is strictly less than the total enthalpy of the reactants (Hproducts<HreactantsH_{\text{products}} < H_{\text{reactants}}).
    • Products reside at a lower potential energy minimum, making them thermodynamically more stable than the reactants.
    • Examples: Combustion of hydrocarbons (CH4+2O2→CO2+2H2O\text{CH}_4 + 2\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O}, ΔH=−890.4 kJ/mol\Delta H = -890.4\text{ kJ/mol}), acid-base neutralizations, respiration, synthesis of ammonia via the Haber process.
  2. Endothermic Reactions (ΔH>0\Delta H > 0):
    • Energy required to break reactant bonds exceeds energy liberated when product bonds form.
    • Heat is absorbed from the surroundings, resulting in a temperature decrease of the surrounding medium unless external heating is applied (ΔT<0\Delta T < 0).
    • The total enthalpy of the products is strictly greater than the total enthalpy of the reactants (Hproducts>HreactantsH_{\text{products}} > H_{\text{reactants}}).
    • Products reside at a higher potential energy level and are thermodynamically less stable than reactants.
    • Examples: Thermal decomposition of limestone (CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g), ΔH=+178 kJ/mol\Delta H = +178\text{ kJ/mol}), photosynthesis, thermal cracking of alkanes, formation of nitric oxide from air (N2+O2→2NO\text{N}_2 + \text{O}_2 \rightarrow 2\text{NO}, ΔH=+180 kJ/mol\Delta H = +180\text{ kJ/mol}).

Potential Energy Reaction Profiles

In both exothermic and endothermic reactions, reactants must acquire a minimum kinetic energy barrier, the activation energy (EaE_a), to reach the high-energy, unstable transitional arrangement known as the activated complex or transition state:

  • In an exothermic profile, the reactants begin at an elevated baseline HRH_R, ascend an energy hill of height Ea,forwardE_{a,\text{forward}}, and descend past the activated complex to a lower baseline HPH_P. The net enthalpy release is ΔH=HP−HR=Ea,forward−Ea,reverse<0\Delta H = H_P - H_R = E_{a,\text{forward}} - E_{a,\text{reverse}} < 0.
  • In an endothermic profile, the reactants begin at a lower baseline HRH_R, ascend an energy hill Ea,forwardE_{a,\text{forward}}, and settle at an elevated product plateau HPH_P. Here, Ea,forward>Ea,reverseE_{a,\text{forward}} > E_{a,\text{reverse}}, yielding ΔH=HP−HR>0\Delta H = H_P - H_R > 0.

3. Standard Enthalpy Changes Defined & Quantified

Thermochemical measurements depend on temperature, pressure, and the physical states of reactants and products. To establish universal tables of thermodynamic constants, IUPAC specifies standard thermodynamic states:

  • Standard Temperature: Exactly 298.15 K298.15\text{ K} (25∘C25^\circ\text{C}).
  • Standard Pressure: 1 atm1\text{ atm} (101.325 kPa101.325\text{ kPa}) in most Intermediate textbooks; IUPAC now recommends 1 bar1\text{ bar} (100 kPa100\text{ kPa}), a difference too small to change typical answers.
  • Standard State of a Substance: The most stable, pure physical state (solid, liquid, or gas) of the pure substance under standard conditions (298 K,1 atm298\text{ K}, 1\text{ atm}).
  • Standard Concentration: Exactly 1.0 mol⋅dm−31.0\text{ mol}\cdot\text{dm}^{-3} (1.0 M1.0\text{ M}) for dissolved aqueous solutes.

Standard enthalpy changes carry the degree superscript symbol ΔH∘\Delta H^\circ.

Standard Enthalpy of Formation (ΔHf∘\Delta H_f^\circ)

The standard enthalpy of formation (ΔHf∘\Delta H_f^\circ) is the enthalpy change that occurs when exactly one mole of a pure compound is synthesized directly from its constituent chemical elements in their standard reference physical states at 298 K298\text{ K} and 1 atm1\text{ atm}:

C(graphite)+O2(g)→CO2(g)ΔHf∘=−393.5 kJ/mol\text{C}(\text{graphite}) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H_f^\circ = -393.5\text{ kJ/mol}

H2(g)+12O2(g)→H2O(l)ΔHf∘=−285.8 kJ/mol\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l) \quad \Delta H_f^\circ = -285.8\text{ kJ/mol}

ECAT Axiom: By universal thermodynamic convention, the standard enthalpy of formation of every pure chemical element in its standard reference state is assigned a value of identically zero: ΔHf∘[O2(g)]=ΔHf∘[N2(g)]=ΔHf∘[H2(g)]=ΔHf∘[Cgraphite]=ΔHf∘[Na(s)]=ΔHf∘[Br2(l)]=0\Delta H_f^\circ[\text{O}_2(g)] = \Delta H_f^\circ[\text{N}_2(g)] = \Delta H_f^\circ[\text{H}_2(g)] = \Delta H_f^\circ[\text{C}_{\text{graphite}}] = \Delta H_f^\circ[\text{Na}(s)] = \Delta H_f^\circ[\text{Br}_2(l)] = 0 However, allotropes or states other than the reference state have non-zero enthalpies of formation. For example, ΔHf∘[Cdiamond]=+1.9 kJ/mol\Delta H_f^\circ[\text{C}_{\text{diamond}}] = +1.9\text{ kJ/mol}, ΔHf∘[O3(g)]=+142.7 kJ/mol\Delta H_f^\circ[\text{O}_3(g)] = +142.7\text{ kJ/mol}, and ΔHf∘[Br2(g)]=+30.9 kJ/mol\Delta H_f^\circ[\text{Br}_2(g)] = +30.9\text{ kJ/mol}.

Standard Enthalpy of Combustion (ΔHc∘\Delta H_c^\circ)

The standard enthalpy of combustion (ΔHc∘\Delta H_c^\circ) is the enthalpy change accompanying the complete combustion of exactly one mole of a substance in an excess of pure oxygen gas under standard conditions (298 K,1 atm298\text{ K}, 1\text{ atm}):

CH4(g)+2O2(g)→CO2(g)+2H2O(l)ΔHc∘=−890.4 kJ/mol\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l) \quad \Delta H_c^\circ = -890.4\text{ kJ/mol}

C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l)ΔHc∘=−1367 kJ/mol\text{C}_2\text{H}_5\text{OH}(l) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l) \quad \Delta H_c^\circ = -1367\text{ kJ/mol}

Because combustion reactions invariably release heat, ΔHc∘\Delta H_c^\circ is always negative (exothermic).

Standard Enthalpy of Atomization (ΔHat∘\Delta H_{\text{at}}^\circ)

The standard enthalpy of atomization (ΔHat∘\Delta H_{\text{at}}^\circ) is the enthalpy change required to dissociate an element or compound into exactly one mole of separated, gaseous atoms from its standard state:

12Cl2(g)→Cl(g)ΔHat∘=+121 kJ/mol\frac{1}{2}\text{Cl}_2(g) \rightarrow \text{Cl}(g) \quad \Delta H_{\text{at}}^\circ = +121\text{ kJ/mol}

Na(s)→Na(g)ΔHat∘=+108 kJ/mol\text{Na}(s) \rightarrow \text{Na}(g) \quad \Delta H_{\text{at}}^\circ = +108\text{ kJ/mol}

Atomization invariably requires breaking metallic, covalent, or van der Waals interactions; thus, ΔHat∘\Delta H_{\text{at}}^\circ is always positive (endothermic).

Standard Enthalpy of Neutralization (ΔHn∘\Delta H_n^\circ)

The standard enthalpy of neutralization (ΔHn∘\Delta H_n^\circ) is the enthalpy change when one mole of water (H2O\text{H}_2\text{O}) is produced by the neutralization reaction between an acid and a base under standard aqueous conditions:

H+(aq)+OH−(aq)→H2O(l)ΔHn∘≈−57.4 kJ/mol(−13.7 kcal/mol)\text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l) \quad \Delta H_n^\circ \approx -57.4\text{ kJ/mol} \quad (-13.7\text{ kcal/mol})

  • Strong Acid + Strong Base: Strong mineral acids (HCl,HNO3,HClO4\text{HCl}, \text{HNO}_3, \text{HClO}_4) and strong alkali hydroxides (NaOH,KOH\text{NaOH}, \text{KOH}) are completely (100%) ionized in dilute aqueous solutions. Consequently, the spectator ions (Na+,K+,Cl−,NO3−\text{Na}^+, \text{K}^+, \text{Cl}^-, \text{NO}_3^-) play no role, and the net ionic reaction is universally H+(aq)+OH−(aq)→H2O(l)\text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l). For all such pairs, ΔHn∘\Delta H_n^\circ remains constant at approximately −57.4 kJ/mol-57.4\text{ kJ/mol}.
  • Weak Acid or Weak Base Neutralization: When a weak acid (e.g., acetic acid, CH3COOH\text{CH}_3\text{COOH}; hydrocyanic acid, HCN\text{HCN}) or a weak base (e.g., ammonium hydroxide, NH4OH\text{NH}_4\text{OH}) is neutralized, the measured enthalpy release is noticeably less negative (e.g., −55.2 kJ/mol-55.2\text{ kJ/mol} for CH3COOH+NaOH\text{CH}_3\text{COOH} + \text{NaOH} and only −12.1 kJ/mol-12.1\text{ kJ/mol} for HCN+NaOH\text{HCN} + \text{NaOH}). Weak electrolytes exist largely as undissociated molecules in water; before neutralization can occur, a fraction of the liberated heat must be consumed endothermically to ionize the weak bonds (known as the enthalpy of ionization, ΔHion\Delta H_{\text{ion}}): ΔHobserved=ΔHn∘(strong)+ΔHion=−57.4 kJ/mol+ΔHion\Delta H_{\text{observed}} = \Delta H_n^\circ(\text{strong}) + \Delta H_{\text{ion}} = -57.4\text{ kJ/mol} + \Delta H_{\text{ion}}

4. Experimental Calorimetry: Glass Calorimeter vs. Bomb Calorimeter

Calorimetry is the experimental laboratory technique employed to measure heat absorbed or released during physical or chemical processes.

1. Glass Calorimeter (Coffee-Cup Calorimeter)

The glass calorimeter consists of an insulated glass beaker or nested expanded polystyrene (styrofoam) cups fitted with a cork stopper, a precision sensitive thermometer (e.g., Beckman thermometer reading to 0.01∘C0.01^\circ\text{C}), and a wire stirrer. Because the cup is open to atmospheric pressure through a loose vent hole:

  • It operates at constant pressure (P=constantP = \text{constant}). Hence, the measured heat flow directly equals the enthalpy change: qp=ΔHq_p = \Delta H
  • The heat gained or lost by the aqueous solution is determined from the mass of the solution mm, its specific heat capacity cc (taken as 4.184 J⋅g−1⋅K−14.184\text{ J}\cdot\text{g}^{-1}\cdot\text{K}^{-1} for dilute aqueous solutions, equal to water), and the temperature change ΔT=Tfinal−Tinitial\Delta T = T_{\text{final}} - T_{\text{initial}}: q=m⋅c⋅ΔTq = m \cdot c \cdot \Delta T
  • The molar enthalpy change is computed by dividing qq by the moles of limiting reactant nn: ΔH=−qn=−m⋅c⋅ΔTn\Delta H = -\frac{q}{n} = -\frac{m \cdot c \cdot \Delta T}{n}
  • The glass calorimeter is ideally suited for measuring heats of neutralization, heats of solution (dissolution of salts), and precipitation reactions in aqueous media.

2. Bomb Calorimeter

The bomb calorimeter is a specialized, rigid, high-pressure reaction vessel constructed of heavy, corrosion-resistant stainless steel. It is specifically designed to measure heats of combustion for fuels, explosive compounds, food samples, and hydrocarbons that require pure oxygen under pressure:

  • The sample of known mass is placed in a platinum or silica crucible inside the bomb.
  • The bomb is pressurized with pure oxygen to 20 to 30 atmospheres20\text{ to }30\text{ atmospheres} to ensure instantaneous and complete combustion.
  • An electrical ignition wire (iron or platinum) is submerged into the sample and connected to an external battery circuit.
  • The sealed steel bomb is immersed in an insulated outer water bath containing a known mass of water, equipped with an electric motorized stirrer and a precision Beckman thermometer.
  • An electric pulse fires the ignition wire, combusting the sample. Heat released transfers through the steel walls into the surrounding water jacket.

Because the heavy steel walls prevent any volume change (ΔV=0\Delta V = 0):

  • No pressure-volume work is done (w=−PΔV=0w = -P\Delta V = 0).
  • The heat released at constant volume (qvq_v) represents the internal energy change: qv=ΔUq_v = \Delta U
  • Heat released is determined using the calibrated heat capacity of the calorimeter system (CC): q=C⋅ΔTq = C \cdot \Delta T where C=W⋅cC = W \cdot c (WW is the water equivalent of the calorimeter vessel plus contents, and cc is the specific heat capacity).
  • The molar internal energy of combustion is: ΔUc=−C⋅ΔTnsample\Delta U_c = -\frac{C \cdot \Delta T}{n_{\text{sample}}}
  • To obtain the standard enthalpy of combustion ΔHc∘\Delta H_c^\circ, apply the thermodynamic conversion: ΔHc∘=ΔUc+ΔngRT\Delta H_c^\circ = \Delta U_c + \Delta n_g R T where Δng\Delta n_g is the net change in stoichiometric moles of gaseous reactants and products.

5. Hess's Law of Constant Heat Summation

Formulated empirically by Germain Henri Hess in 1840, Hess's Law of Constant Heat Summation is a direct corollary of the First Law of Thermodynamics and the state function nature of enthalpy:

Statement: If a chemical reaction can be carried out in a single step or in a sequence of multiple independent steps, the overall enthalpy change (ΔH\Delta H) is identical, regardless of the pathway or the intermediate stages traversed, provided initial reactants and final products are in the identical thermodynamic states.

If a direct reaction A→BA \rightarrow B has enthalpy ΔH\Delta H, and the alternative multi-step route proceeds through intermediates CC and DD with enthalpies ΔH1\Delta H_1, ΔH2\Delta H_2, and ΔH3\Delta H_3 (A→C→D→BA \rightarrow C \rightarrow D \rightarrow B):

ΔH=ΔH1+ΔH2+ΔH3\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3

          ΔH (direct path)
    A ───────────────────────► B
    │                         ▲
ΔH₁ │                         │ ΔH₃
    ▼                         │
    C ───────────────────────► D
              ΔH₂

Crucial Applications in ECAT Calculations

Hess's Law is indispensable because many chemical reactions cannot be measured directly in a calorimeter. For instance, the complete oxidation of graphite cannot be halted cleanly at carbon monoxide (CO\text{CO}); it inevitably produces a mixture of CO\text{CO} and CO2\text{CO}_2. By combining the easily measured combustions of graphite and carbon monoxide:

  1. C(graphite)+O2(g)→CO2(g)ΔH1∘=−393.5 kJ/mol\text{C}(\text{graphite}) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H_1^\circ = -393.5\text{ kJ/mol}
  2. CO(g)+12O2(g)→CO2(g)ΔH2∘=−283.0 kJ/mol\text{CO}(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H_2^\circ = -283.0\text{ kJ/mol}

Reversing equation (2) and adding it to equation (1):

C(graphite)+O2(g)+CO2(g)→CO2(g)+CO(g)+12O2(g)\text{C}(\text{graphite}) + \text{O}_2(g) + \text{CO}_2(g) \rightarrow \text{CO}_2(g) + \text{CO}(g) + \frac{1}{2}\text{O}_2(g)

C(graphite)+12O2(g)→CO(g)\text{C}(\text{graphite}) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g)

ΔHf∘[CO]=ΔH1∘−ΔH2∘=−393.5−(−283.0)=−110.5 kJ/mol\Delta H_f^\circ[\text{CO}] = \Delta H_1^\circ - \Delta H_2^\circ = -393.5 - (-283.0) = -110.5\text{ kJ/mol}

General Formula from Standard Enthalpies of Formation

For any generalized reaction aA+bB→cC+dDaA + bB \rightarrow cC + dD, Hess's Law yields:

ΔHreaction∘=∑nΔHf∘(products)−∑mΔHf∘(reactants)\Delta H^\circ_{\text{reaction}} = \sum n\Delta H_f^\circ(\text{products}) - \sum m\Delta H_f^\circ(\text{reactants})

ΔHreaction∘=[cΔHf∘(C)+dΔHf∘(D)]−[aΔHf∘(A)+bΔHf∘(B)]\Delta H^\circ_{\text{reaction}} = \left[c\Delta H_f^\circ(C) + d\Delta H_f^\circ(D)\right] - \left[a\Delta H_f^\circ(A) + b\Delta H_f^\circ(B)\right]


6. The Born-Haber Cycle & Lattice Energy of Ionic Solids

Ionic solids such as sodium chloride (NaCl\text{NaCl}) and magnesium oxide (MgO\text{MgO}) consist of infinite three-dimensional crystal lattices stabilized by strong electrostatic attractions between oppositely charged ions.

Definition of Lattice Energy (ΔHL\Delta H_L)

The lattice energy (ΔHL\Delta H_L or U0U_0) of an ionic crystalline compound is defined as the enthalpy change that occurs when one mole of the solid crystalline ionic compound is formed from its constituent gaseous ions separated to infinite distance under standard conditions:

Mz+(g)+Xz−(g)→MX(s)ΔHL<0(exothermic convention)\text{M}^{z+}(g) + \text{X}^{z-}(g) \rightarrow \text{MX}(s) \quad \Delta H_L < 0 \quad (\text{exothermic convention})

(Alternatively, in some physical contexts, lattice dissociation energy is defined as the energy required to completely separate one mole of the solid crystal into isolated gaseous ions, which is equal in magnitude but positive: ΔHL,diss>0\Delta H_{L,\text{diss}} > 0. On the ECAT, lattice formation is treated as exothermic, yielding large negative values.)

Because gaseous ions cannot be combined directly in a calorimeter without solvent or competing phases, lattice energy cannot be measured directly. Max Born and Fritz Haber devised a closed thermochemical cyclic application of Hess's Law known as the Born-Haber Cycle.

Stepwise Dissection of the Born-Haber Cycle for NaCl(s)\text{NaCl}(s)

The synthesis of solid sodium chloride from its standard elemental states represents the standard enthalpy of formation:

Na(s)+12Cl2(g)→NaCl(s)ΔHf∘=−411 kJ/mol\text{Na}(s) + \frac{1}{2}\text{Cl}_2(g) \rightarrow \text{NaCl}(s) \quad \Delta H_f^\circ = -411\text{ kJ/mol}

The alternative indirect route consists of five discrete thermodynamic steps:

  1. Sublimation of Solid Sodium (ΔHs∘\Delta H_s^\circ or ΔHat∘[Na]\Delta H_{\text{at}}^\circ[\text{Na}]): Solid sodium metal is vaporized into isolated gaseous sodium atoms: Na(s)→Na(g)ΔHs∘=+108 kJ/mol(endothermic)\text{Na}(s) \rightarrow \text{Na}(g) \quad \Delta H_s^\circ = +108\text{ kJ/mol} \quad (\text{endothermic})
  2. First Ionization Energy of Sodium (IE1IE_1): One mole of electrons is removed from gaseous sodium atoms to generate gaseous sodium cations: Na(g)→Na+(g)+e−IE1=+496 kJ/mol(endothermic)\text{Na}(g) \rightarrow \text{Na}^+(g) + e^- \quad IE_1 = +496\text{ kJ/mol} \quad (\text{endothermic})
  3. Dissociation / Atomization of Chlorine Gas (12ΔHdiss∘\frac{1}{2}\Delta H_{\text{diss}}^\circ or ΔHat∘[Cl]\Delta H_{\text{at}}^\circ[\text{Cl}]): Half a mole of diatomic chlorine gas molecules is cleaved into one mole of gaseous chlorine free radicals: 12Cl2(g)→Cl(g)12ΔHdiss∘=+121 kJ/mol(endothermic, as bond energy DCl-Cl=+242 kJ/mol)\frac{1}{2}\text{Cl}_2(g) \rightarrow \text{Cl}(g) \quad \frac{1}{2}\Delta H_{\text{diss}}^\circ = +121\text{ kJ/mol} \quad (\text{endothermic, as bond energy } D_{\text{Cl-Cl}} = +242\text{ kJ/mol})
  4. First Electron Affinity of Chlorine (EA1EA_1): One mole of gaseous chlorine atoms captures one mole of electrons to produce gaseous chloride anions: Cl(g)+e−→Cl−(g)EA1=−349 kJ/mol(exothermic)\text{Cl}(g) + e^- \rightarrow \text{Cl}^-(g) \quad EA_1 = -349\text{ kJ/mol} \quad (\text{exothermic})
  5. Lattice Formation Enthalpy (ΔHL\Delta H_L): Gaseous sodium cations and gaseous chloride anions condense to form one mole of rigid crystalline sodium chloride: Na+(g)+Cl−(g)→NaCl(s)ΔHL\text{Na}^+(g) + \text{Cl}^-(g) \rightarrow \text{NaCl}(s) \quad \Delta H_L

Algebraic Formulation of the Cycle

By Hess's Law, the direct formation enthalpy equals the algebraic sum of all steps in the indirect route:

ΔHf∘=ΔHs∘+IE1+12ΔHdiss∘+EA1+ΔHL\Delta H_f^\circ = \Delta H_s^\circ + IE_1 + \frac{1}{2}\Delta H_{\text{diss}}^\circ + EA_1 + \Delta H_L

Rearranging to isolate lattice energy ΔHL\Delta H_L:

ΔHL=ΔHf∘−[ΔHs∘+IE1+12ΔHdiss∘+EA1]\Delta H_L = \Delta H_f^\circ - \left[\Delta H_s^\circ + IE_1 + \frac{1}{2}\Delta H_{\text{diss}}^\circ + EA_1\right]

Substituting standard experimental values:

ΔHL=−411−[108+496+121+(−349)]=−411−[376]=−787 kJ/mol\Delta H_L = -411 - [108 + 496 + 121 + (-349)] = -411 - [376] = -787\text{ kJ/mol}

Factors Governing the Magnitude of Lattice Energy

Electrostatic lattice energy is approximated by the Born-Landé or Kapustinskii equation derived from Coulomb's Law:

ΔHL∝−z+⋅z−r++r−\Delta H_L \propto -\frac{z^+ \cdot z^-}{r_+ + r_-}

  1. Ionic Charges (z+,z−z^+, z^-): Lattice energy increases dramatically as ionic charges increase. For example, in MgO\text{MgO}, z+=+2z^+ = +2 and z−=−2z^- = -2 (z+z−=4z^+ z^- = 4), giving a massive lattice energy of ΔHL=−3791 kJ/mol\Delta H_L = -3791\text{ kJ/mol}, nearly five times greater than that of NaCl\text{NaCl} (z+z−=1,ΔHL=−787 kJ/molz^+ z^- = 1, \Delta H_L = -787\text{ kJ/mol}).
  2. Internuclear Separation / Ionic Radii (r++r−r_+ + r_-): Smaller ionic radii allow ions to pack closer together, intensifying electrostatic attractions and increasing lattice energy: LiF(−1030 kJ/mol)>NaCl(−787 kJ/mol)>KBr(−671 kJ/mol)\text{LiF} (-1030\text{ kJ/mol}) > \text{NaCl} (-787\text{ kJ/mol}) > \text{KBr} (-671\text{ kJ/mol}).

7. Reaction Spontaneity & Gibbs Free Energy (ΔG\Delta G)

A spontaneous process is a natural process that proceeds on its own without requiring persistent external driving forces (e.g., water flowing downhill, heat diffusing from hot to cold bodies, gas expanding into a vacuum). While many spontaneous processes are exothermic (ΔH<0\Delta H < 0), some endothermic processes (e.g., dissolution of NH4NO3\text{NH}_4\text{NO}_3 in water, melting of ice above 0∘C0^\circ\text{C}) occur spontaneously because of an increase in molecular disorder (entropy, SS).

The Second Law of Thermodynamics establishes that for any spontaneous process, the total entropy of the universe must strictly increase:

ΔSuniverse=ΔSsystem+ΔSsurroundings>0\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0

Josiah Willard Gibbs combined enthalpy and entropy into a single thermodynamic state function called Gibbs Free Energy (GG):

G=H−TSG = H - TS

For an isothermal, isobaric process, the change in Gibbs free energy is:

ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S

Spontaneity Criteria

  • ΔG<0\Delta G < 0 (Negative): The process is spontaneous (exergonic). Useful work can be extracted.
  • ΔG=0\Delta G = 0 (Zero): The system is in a state of dynamic equilibrium (Teq=ΔHΔST_{\text{eq}} = \frac{\Delta H}{\Delta S}).
  • ΔG>0\Delta G > 0 (Positive): The process is non-spontaneous (endergonic). Energy must be continuously supplied; however, the reverse process is spontaneous.

Temperature-Dependent Spontaneity Matrix

ΔH\Delta HΔS\Delta S−TΔS-T\Delta SSign of ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta SSpontaneity ConditionPhysical Example
Negative (−-)Positive (++)Negative (−-)Always Negative (−-)Spontaneous at all temperatures2O3(g)→3O2(g)2\text{O}_3(g) \rightarrow 3\text{O}_2(g)
Positive (++)Negative (−-)Positive (++)Always Positive (++)Non-spontaneous at all temperatures3O2(g)→2O3(g)3\text{O}_2(g) \rightarrow 2\text{O}_3(g)
Negative (−-)Negative (−-)Positive (++)Negative at low TT; Positive at high TTSpontaneous only at low temperatures (T<ΔHΔST < \frac{\Delta H}{\Delta S})Condensation of steam, freezing of water
Positive (++)Positive (++)Negative (−-)Positive at low TT; Negative at high TTSpontaneous only at high temperatures (T>ΔHΔST > \frac{\Delta H}{\Delta S})Thermal decomposition of CaCO3(s)\text{CaCO}_3(s), evaporation

8. Thermochemical Parameters & Standard Definitions Reference Table

ParameterSymbolOperational DefinitionSign ConventionStandard Units
Internal Energy ChangeΔU\Delta UHeat exchanged at constant volume (ΔU=qv\Delta U = q_v)++ for endothermic, −- for exothermickJ⋅mol−1\text{kJ}\cdot\text{mol}^{-1}
Enthalpy ChangeΔH\Delta HHeat exchanged at constant pressure (qp=ΔU+PΔVq_p = \Delta U + P\Delta V)++ for endothermic, −- for exothermickJ⋅mol−1\text{kJ}\cdot\text{mol}^{-1}
Standard Enthalpy of FormationΔHf∘\Delta H_f^\circEnthalpy change to form 1 mol1\text{ mol} compound from standard reference elementsAssigned 00 for pure reference elementskJ⋅mol−1\text{kJ}\cdot\text{mol}^{-1}
Standard Enthalpy of CombustionΔHc∘\Delta H_c^\circEnthalpy change when 1 mol1\text{ mol} substance burns completely in excess O2\text{O}_2Invariably negative (−-)kJ⋅mol−1\text{kJ}\cdot\text{mol}^{-1}
Standard Enthalpy of AtomizationΔHat∘\Delta H_{\text{at}}^\circEnthalpy change to produce 1 mol1\text{ mol} gaseous atoms from standard stateInvariably positive (++)kJ⋅mol−1\text{kJ}\cdot\text{mol}^{-1}
Standard Enthalpy of NeutralizationΔHn∘\Delta H_n^\circEnthalpy change when 1 mol1\text{ mol} H2O\text{H}_2\text{O} forms from acid-base neutralizationInvariably negative (−57.4 kJ/mol-57.4\text{ kJ/mol} for strong pairs)kJ⋅mol−1\text{kJ}\cdot\text{mol}^{-1}
Lattice EnergyΔHL\Delta H_LEnthalpy change when 1 mol1\text{ mol} ionic crystal forms from gaseous ionsExothermic convention (large negative values)kJ⋅mol−1\text{kJ}\cdot\text{mol}^{-1}
Gibbs Free Energy ChangeΔG\Delta GNet change in usable driving energy at constant T,PT, P (ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S)Spontaneous if ΔG<0\Delta G < 0; Equilibrium if ΔG=0\Delta G = 0kJ⋅mol−1\text{kJ}\cdot\text{mol}^{-1}
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Born-Haber Thermochemical Cycle for Sodium Chloride (NaCl)
Test Your Knowledge

Using the Born-Haber cycle thermochemical data, calculate the lattice energy (ΔHL\Delta H_L) of magnesium fluoride (MgF2(s)\text{MgF}_2(s)):

  • Sublimation enthalpy of Mg(s)\text{Mg}(s): +148 kJ/mol+148\text{ kJ/mol}
  • First ionization energy of Mg(g)\text{Mg}(g): +738 kJ/mol+738\text{ kJ/mol}
  • Second ionization energy of Mg+(g)\text{Mg}^+(g): +1451 kJ/mol+1451\text{ kJ/mol}
  • Bond dissociation enthalpy of F2(g)\text{F}_2(g): +158 kJ/mol+158\text{ kJ/mol}
  • Electron affinity of F(g)\text{F}(g): −328 kJ/mol-328\text{ kJ/mol}
  • Standard enthalpy of formation ΔHf∘[MgF2(s)]\Delta H_f^\circ[\text{MgF}_2(s)]: −1124 kJ/mol-1124\text{ kJ/mol}
A

−2963 kJ/mol-2963\text{ kJ/mol}

B

−2635 kJ/mol-2635\text{ kJ/mol}

C

−3121 kJ/mol-3121\text{ kJ/mol}

D

−1839 kJ/mol-1839\text{ kJ/mol}

Test Your Knowledge

When 100 cm3100\text{ cm}^3 of 0.5 M HCl0.5\text{ M } \text{HCl} is mixed with 100 cm3100\text{ cm}^3 of 0.5 M NaOH0.5\text{ M } \text{NaOH} in a well-insulated glass calorimeter, the temperature of the solution rises by 3.43∘C3.43^\circ\text{C}. Assuming the resulting solution has a total volume of 200 cm3200\text{ cm}^3, density of 1.0 g/cm31.0\text{ g/cm}^3, and specific heat capacity of 4.184 J⋅g−1⋅K−14.184\text{ J}\cdot\text{g}^{-1}\cdot\text{K}^{-1}, what is the experimental standard enthalpy of neutralization per mole of water formed?

A

−28.7 kJ/mol-28.7\text{ kJ/mol}

B

−57.4 kJ/mol-57.4\text{ kJ/mol}

C

−114.8 kJ/mol-114.8\text{ kJ/mol}

D

−68.2 kJ/mol-68.2\text{ kJ/mol}

Test Your Knowledge

For the thermal decomposition reaction CaCO3(s)→CaO(s)+CO2(g)\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g), the standard enthalpy change is ΔH∘=+178.3 kJ/mol\Delta H^\circ = +178.3\text{ kJ/mol} and standard entropy change is ΔS∘=+160.5 J⋅mol−1⋅K−1\Delta S^\circ = +160.5\text{ J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}. At what minimum temperature will this endothermic reaction become spontaneous under standard atmospheric pressure?

A

834∘C834^\circ\text{C}

B

910∘C910^\circ\text{C}

C

838∘C838^\circ\text{C}

D

1111∘C1111^\circ\text{C}

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