14.3 Hess's Law & Standard Enthalpies of Formation

14.3 Hess's Law & Standard Enthalpies of Formation

Quick Summary: Enthalpy ($H$) is a state function; hence, the net enthalpy change of a chemical transformation is independent of pathway. Hess's Law of Heat Summation permits determination of reaction enthalpies by algebraically summing elementary thermochemical steps. Standard states are defined at $1\text{ bar}$ (or $1\text{ atm}$) and $298.15\text{ K}$ ($25\ ^\circ\text{C}$). The standard enthalpy of formation ($\Delta H^\circ_f$) measures enthalpy when $1\text{ mole}$ of compound forms from pure elements in standard states (with $\Delta H^\circ_f = 0\text{ kJ/mol}$ for stable reference elements). Reaction enthalpies are calculated via $\Delta H^\circ_{\text{rxn}} = \sum n \Delta H^\circ_f(\text{products}) - \sum m \Delta H^\circ_f(\text{reactants})$. For gas-phase reactions, average bond dissociation enthalpies estimate reaction energetics via $\Delta H^\circ_{\text{rxn}} \approx \sum D(\text{bonds broken}) - \sum D(\text{bonds formed})$.

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1. Enthalpy Path Independence & Hess's Law of Heat Summation

Hess's Law of Heat Summation states that if a chemical reaction can be expressed as the algebraic sum of two or more steps, the standard enthalpy change for the overall process equals the sum of the enthalpy changes of the individual steps: $\Delta H_{\text{overall}} = \sum_{i} \Delta H_i = \Delta H_1 + \Delta H_2 + \Delta H_3 + \dots$

Algebraic Rules for Thermochemical Equations

When manipulating chemical equations to construct an overall reaction:

1. Reversing an equation changes the sign of $\Delta H$: $\text{A} \longrightarrow \text{B} \quad \Delta H = +X \implies \text{B} \longrightarrow \text{A} \quad \Delta H = -X$
2. Multiplying stoichiometric coefficients by factor $n$ multiplies $\Delta H$ by $n$: $n\text{A} \longrightarrow n\text{B} \quad \Delta H = n \cdot (\Delta H_1)$
3. Dividing stoichiometric coefficients divides $\Delta H$ accordingly: Halving coefficients halves $\Delta H$.
4. Summing equations adds their enthalpies: Intermediates appearing identically on reactant and product sides cancel.

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2. Standard Thermodynamic States & Formation Enthalpy ($\Delta H^\circ_f$)

Standard thermodynamic states (denoted by $^\circ$) establish reproducible conditions:

• Gases: Pure gas at a partial pressure of $1\text{ bar}$ ($1\text{ atm}$).
• Solutions: Dissolved solute at a concentration of exactly $1.00\text{ M}$.
• Pure Substances: The most stable physical form at $1\text{ atm}$ and specified temperature (usually $298.15\text{ K} = 25\ ^\circ\text{C}$).

The Formation Reaction & Zero Reference Baseline

The standard enthalpy of formation ($\Delta H^\circ_f$) is the enthalpy change accompanying the synthesis of one mole of a substance from its constituent elements in their most stable standard states. By international convention, the standard enthalpy of formation of any pure chemical element in its most stable standard allotrope at $298.15\text{ K}$ and $1\text{ atm}$ is defined as zero: $\Delta H^\circ_f [\text{element in reference state}] = 0\text{ kJ/mol}$

Standard Reference Elements versus Metastable Allotropes

• Carbon: $\Delta H^\circ_f [\text{C(graphite)}] = 0\text{ kJ/mol}$, whereas $\Delta H^\circ_f [\text{C(diamond)}] = +1.9\text{ kJ/mol}$.
• Oxygen: $\Delta H^\circ_f [\text{O}_2(g)] = 0\text{ kJ/mol}$, whereas $\Delta H^\circ_f [\text{O}_3(g)] = +142.7\text{ kJ/mol}$.
• Bromine: $\Delta H^\circ_f [\text{Br}_2(l)] = 0\text{ kJ/mol}$, whereas $\Delta H^\circ_f [\text{Br}_2(g)] = +30.9\text{ kJ/mol}$.

Reference Table: Selected Standard Enthalpies of Formation

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3. Calculating Reaction Enthalpies from Formation Data

Any chemical transformation can be conceptualized as decomposing reactants into elemental reference constituents ($-\sum m \Delta H^\circ_f(\text{reactants})$) and recombining those elements into products ($+\sum n \Delta H^\circ_f(\text{products})$): $\Delta H^\circ_{\text{rxn}} = \sum n \Delta H^\circ_f(\text{products}) - \sum m \Delta H^\circ_f(\text{reactants})$ where $n$ and $m$ are stoichiometric coefficients from the balanced chemical equation.

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4. Average Bond Dissociation Enthalpies

For gas-phase covalent reactions, reaction enthalpies can be estimated from average bond dissociation enthalpies ($D$), defined as the energy required to break one mole of a specific bond in the gas phase:

• Breaking bonds is ALWAYS endothermic: Energy is required to overcome electrostatic attraction ($D > 0$).
• Forming bonds is ALWAYS exothermic: Energy is released as atoms bond in lower potential energy states ($\Delta H < 0$).

$\Delta H^\circ_{\text{rxn}} \approx \sum D(\text{bonds broken, reactants}) - \sum D(\text{bonds formed, products})$ Contrast: Bond calculations use Reactants minus Products (broken minus formed), opposite the order of formation enthalpy calculations.

Limitations of Bond Enthalpies

1. Phase Constraint: Restricted strictly to gas-phase species; non-gas species require separate phase change adjustments.
2. Averaged Quantities: Bond strengths vary depending on local chemical environment; tabulated values represent averages across diverse compounds.

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5. Worked Problems: Hess's Law & Bond Energies

Example 1: Multistep Hess's Law Problem

Problem: Determine $\Delta H^\circ$ for: $2\text{ B}(s) + 3\text{ H}_2(g) \longrightarrow \text{B}_2\text{H}_6(g)$ given:

1. $2\text{ B}(s) + \frac{3}{2}\text{ O}_2(g) \longrightarrow \text{B}_2\text{O}_3(s) \quad \Delta H_1 = -1273\text{ kJ}$
2. $\text{B}_2\text{H}_6(g) + 3\text{ O}_2(g) \longrightarrow \text{B}_2\text{O}_3(s) + 3\text{ H}_2\text{O}(g) \quad \Delta H_2 = -2035\text{ kJ}$
3. $\text{H}_2(g) + \frac{1}{2}\text{ O}_2(g) \longrightarrow \text{H}_2\text{O}(g) \quad \Delta H_3 = -242\text{ kJ}$

Solution Strategy:

• Keep equation (1) as written: $\Delta H = -1273\text{ kJ}$
• Reverse equation (2): $\Delta H = +2035\text{ kJ}$
• Multiply equation (3) by 3: $\Delta H = 3(-242\text{ kJ}) = -726\text{ kJ}$

Summing gives: $\Delta H^\circ = -1273\text{ kJ} + 2035\text{ kJ} - 726\text{ kJ} = +36\text{ kJ}$

Example 2: Combustion from Enthalpies of Formation

Problem: Calculate $\Delta H^\circ_{\text{rxn}}$ for: $\text{C}_3\text{H}_8(g) + 5\text{ O}_2(g) \longrightarrow 3\text{ CO}_2(g) + 4\text{ H}_2\text{O}(l)$ $\Delta H^\circ_{\text{rxn}} = [3(-393.5) + 4(-285.8)] - [1(-103.8) + 5(0)]$ $\Delta H^\circ_{\text{rxn}} = [-1180.5 - 1143.2] - [-103.8] = -2323.7 + 103.8 = -2219.9\text{ kJ/mol}$