4.4 Chemical Thermodynamics, Thermochemistry & Hess's Law

4.4 Chemical Thermodynamics, Thermochemistry & Hess's Law

Thermodynamics governs energy transformations and spontaneity in chemical systems. This section details energy conservation, enthalpy changes, thermochemical laws, entropy, and Gibbs Free Energy.

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Fundamental Concepts & Definitions

1. System: The specific portion of the universe under thermodynamic study.
   • Open System: Exchanges both energy and matter with surroundings.
   • Closed System: Exchanges energy but NOT matter.
   • Isolated System: Exchanges neither energy nor matter (e.g., liquid in an ideal thermos flask).
2. State Functions: Properties whose values depend only on the current state of the system, independent of the pathway taken to reach it. Examples: Pressure ($P$), Volume ($V$), Temperature ($T$), Internal Energy ($U$), Enthalpy ($H$), Entropy ($S$), Gibbs Free Energy ($G$).
   • Path Functions: Depend on the specific path taken (e.g., Heat $q$ and Work $w$).
3. Extensive vs Intensive Properties:
   • Extensive: Depend on the quantity of matter present (Mass, Volume, Enthalpy $H$, Internal Energy $U$).
   • Intensive: Independent of matter quantity (Temperature, Density, Pressure, Molar Volume, Refractive Index).

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First Law of Thermodynamics & Enthalpy

The First Law states that energy cannot be created or destroyed, only transformed: $\Delta U = q + w$

• Sign Conventions:
  • $q > 0$: Heat absorbed by system (endothermic); $q < 0$: Heat released by system (exothermic).
  • $w > 0$: Work done ON the system; $w < 0$: Work done BY the system ($w = -P \Delta V$).

Constant Volume vs Constant Pressure Processes

• At Constant Volume ($\Delta V = 0$): $w = 0 \implies q_v = \Delta U$.
• At Constant Pressure ($P = \text{const}$): $q_p = \Delta H = \Delta U + P \Delta V$.

Relationship Between $\Delta H$ and $\Delta U$

For reactions involving gases: $\Delta H = \Delta U + \Delta n_g R T$ where $\Delta n_g = (\text{moles of gaseous products}) - (\text{moles of gaseous reactants})$.

• If $\Delta n_g = 0 \implies \Delta H = \Delta U$.
• If $\Delta n_g > 0 \implies \Delta H > \Delta U$.
• If $\Delta n_g < 0 \implies \Delta H < \Delta U$.

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Thermochemistry & Standard Enthalpy Changes

• Exothermic Reaction: $\Delta H < 0$ (Heat released to surroundings; product enthalpy < reactant enthalpy).
• Endothermic Reaction: $\Delta H > 0$ (Heat absorbed from surroundings).

Standard Enthalpy Definitions (Standard State: $298.15 \text{ K}, 1 \text{ atm}$)

1. Standard Enthalpy of Formation ($\Delta H_f^\circ$): Enthalpy change when 1 mole of a compound is formed from its elements in their standard reference states.
   • By definition, $\Delta H_f^\circ$ of pure elements in their standard states is zero (e.g., $\text{O}{2(g)}, \text{C}{(graphite)}, \text{N}{2(g)}, \text{H}{2(g)} = 0$).
2. Standard Enthalpy of Combustion ($\Delta H_c^\circ$): Heat evolved when 1 mole of a substance is completely burned in excess oxygen ($\Delta H_c^\circ$ is always negative).
3. Standard Enthalpy of Neutralization ($\Delta H_{neut}^\circ$): Enthalpy change when 1 equivalent of an acid reacts with 1 equivalent of a base.
   • For ANY strong acid and strong base, $\Delta H_{neut}^\circ \approx -57.3 \text{ kJ/mol}$ (or $-13.7 \text{ kcal/mol}$) because the essential net reaction is simply: $\text{H}^+_{(aq)} + \text{OH}^-_{(aq)} \rightarrow \text{H}_2\text{O}_{(l)} \quad \Delta H^\circ = -57.3 \text{ kJ/mol}$

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Hess's Law of Constant Heat Summation

Hess's Law states that if a chemical reaction takes place in one step or in several steps, the total enthalpy change is identical regardless of the route taken. $\Delta H_{\text{net}} = \Delta H_1 + \Delta H_2 + \Delta H_3 + \dots$

General Hess's Law Calculation Formula

$\Delta H^\circ_{\text{rxn}} = \sum n \Delta H_f^\circ(\text{products}) - \sum m \Delta H_f^\circ(\text{reactants})$

Born-Haber Cycle Application

Hess's Law is used to calculate ionic lattice energy ($U$) via the Born-Haber cycle for $\text{NaCl}_{(s)}$: $\Delta H_f^\circ(\text{NaCl}) = \Delta H_{at}^\circ(\text{Na}) + IE_1(\text{Na}) + \frac{1}{2} D(\text{Cl}_2) + EA_1(\text{Cl}) + U_{\text{lattice}}$

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Spontaneity, Entropy & Gibbs Free Energy

Second Law of Thermodynamics & Entropy ($S$)

For a spontaneous process, the total entropy of the universe increases: $\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0$ Entropy measures system disorder/randomness (units: $\text{J}\cdot\text{K}^{-1}\cdot\text{mol}^{-1}$). Entropy increases during phase transitions: $\text{Solid} \rightarrow \text{Liquid} \rightarrow \text{Gas}$.

Gibbs Free Energy Equation

$\Delta G = \Delta H - T \Delta S$

Spontaneity Criteria at Constant $T$ and $P$

• $\Delta G < 0$: Process is spontaneous in the forward direction.
• $\Delta G = 0$: System is at dynamic equilibrium.
• $\Delta G > 0$: Process is non-spontaneous in the forward direction.

Temperature Dependence of Spontaneity

Free Energy & Equilibrium Constant

$\Delta G^\circ = -R T \ln K_c = -2.303 R T \log_{10} K_c$

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Worked Numerical Examples

Example 1: Hess's Law Reaction Enthalpy Calculation

Problem: Calculate $\Delta H^\circ_{\text{rxn}}$ for the combustion of Methane: $\text{CH}_{4(g)} + 2\text{O}_{2(g)} \rightarrow \text{CO}_{2(g)} + 2\text{H}_2\text{O}_{(l)}$ Given standard formation enthalpies: $\Delta H_f^\circ(\text{CH}{4(g)}) = -74.8 \text{ kJ/mol}$, $\Delta H_f^\circ(\text{CO}{2(g)}) = -393.5 \text{ kJ/mol}$, and $\Delta H_f^\circ(\text{H}2\text{O}{(l)}) = -285.8 \text{ kJ/mol}$.

Solution: $\Delta H^\circ_{\text{rxn}} = [\Delta H_f^\circ(\text{CO}_2) + 2 \Delta H_f^\circ(\text{H}_2\text{O})] - [\Delta H_f^\circ(\text{CH}_4) + 2 \Delta H_f^\circ(\text{O}_2)]$ Note that $\Delta H_f^\circ(\text{O}_{2(g)}) = 0 \text{ kJ/mol}$. $\Delta H^\circ_{\text{rxn}} = [-393.5 + 2(-285.8)] - [-74.8 + 0]$ $\Delta H^\circ_{\text{rxn}} = [-393.5 - 571.6] + 74.8 = -965.1 + 74.8 = -890.3 \text{ kJ/mol}$

Example 2: Relationship Between $\Delta H$ and $\Delta U$

Problem: For the synthesis of Ammonia: $\text{N}{2(g)} + 3\text{H}{2(g)} \rightarrow 2\text{NH}_{3(g)}$, express $\Delta H$ in terms of $\Delta U$.

Solution:

1. Moles of gaseous products $= 2$.
2. Moles of gaseous reactants $= 1 + 3 = 4$.
3. $\Delta n_g = 2 - 4 = -2$.
4. Substituting into $\Delta H = \Delta U + \Delta n_g RT$ gives: $\Delta H = \Delta U - 2 R T$