PCA216X: Physical Chemistry – Enthalpy, Entropy, and Gibbs Free Energy Study Guide

Course and Document Information

• Unit: Unit 5 Enthalpies, Entropies and Gibbs Free energy.

• Department: Faculty of Science, Department of Chemistry.

• Subject: Physical Chemistry IIIA (PCA216X).

• Lecturer: Dr Maphoru MV.

The Variations and Types of Enthalpy

• Enthalpy varies with temperature and is categorized throughout various physical and chemical processes:   - Vaporization   - Fusion   - Sublimation   - Atomization   - Dissociation   - Ionization (comparable to Ionization energy)   - Formation   - Reaction   - Combustion   - Hydrogenation   - Bond Strength/bond enthalpy   - Bond dissociation enthalpy

Standard Enthalpy of Formation ($\Delta H_f^\circ$)

• Definition: This is the enthalpy change when a compound in its standard state is formed from its elements in their reference states.

• Reference States: The standard enthalpy of formation of any element in its reference state is defined as zero.

• Reference States of Specific Molecules/Metals:   - Monoatomic and diatomic molecules such as $O_2(g)$, $N_2(g)$, $I_2(s)$, $H_2(g)$, $Br_2(l)$, $F_2(g)$, and $Cl_2(g)$ have a standard enthalpy of formation of zero.   - Metals in their reference states also have a value of zero.   - Exceptions: For $Br_2(g)$ and $I_2(g)$, the standard enthalpy of formation is not zero because these are not their reference states at standard conditions.

• Organic Compounds: Enthalpy changes of formation for organic compounds are commonly obtained via calculation from enthalpies of combustion and the application of Hess's law.

Calculation of $\Delta H^\circ$ Using Standard Enthalpy Tables

• Hess's Law Formula:   - $\Delta H^\circ = [\sum n\Delta H^\circ(\text{products})] - [\sum m\Delta H^\circ(\text{reactants})]$   - Where $n$ and $m$ are the stoichiometric coefficients of the products and reactants respectively.

• Example: Calculating $\Delta H^\circ$ of Combustion for Benzene ($C_6H_6$)   - Balanced Reaction: $C_6H_6(l) + \frac{15}{2}O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l)$   - Given Standard Enthalpies of Formation ($\Delta H_f^\circ$):     - $CO_2(g) = -393.51\,kJ/mol$     - $C_6H_6(l) = +48.7\,kJ/mol$     - $H_2O(l) = -285.53\,kJ/mol$   - Calculation:     - $\Delta H_{\text{comb}} = [6\,mol\,CO_2 \times (-393.51\,kJ/mol) + 3\,mol\,H_2O \times (-285.53\,kJ/mol)] - [1\,mol\,C_6H_6 \times (+48.7\,kJ/mol) + \frac{15}{2}\,mol\,O_2 \times (0)]$     - $\Delta H_{\text{comb}} = -3267\,kJ/mol$

Calculation of $\Delta H^\circ$ Using Multi-Step Reactions (Hess's Law)

• Example: Combustion of Propane ($C_3H_8$)   - Balanced Reaction: $C_3H_8(g) + 5O_2(g) \rightarrow 3CO_2(g) + 4H_2O(l)$   - Conceptual Steps:     - 1. Decomposing Propane: $C_3H_8(g) \rightarrow 3C(\text{graphite}) + 4H_2(g)$     - 2. Forming Dioxide: $3C(\text{graphite}) + 3O_2(g) \rightarrow 3CO_2(g)$     - 3. Forming Water: $4H_2(g) + 2O_2(g) \rightarrow 4H_2O(l)$   - Enthalpy Calculation:     - $\Delta H = [3(-393.5\,kJ) + 4(-285.8\,kJ)] - [1(-103.85\,kJ) + 5(0\,kJ)]$     - $\Delta H = [(-1180.5\,kJ) + (-1143.2\,kJ)] - [-103.85\,kJ]$     - $\Delta H = (-2323.7\,kJ) - (-103.85\,kJ)$     - $\Delta H = -2219.9\,kJ$

Determination of Enthalpy of Formation via Multiple Reactions

• Target Reaction: Forming Diborane ($B_2H_6$) at $298\,K$   - $2B(s) + 3H_2(g) \rightarrow B_2H_6(g)$

• Methodology for Manipulating Reactions:   - If you reverse a reaction, multiply its enthalpy change by $-1$.   - If you multiply the stoichiometric ratio of reactants and products by a number, multiply the enthalpy by that same number.

• Data Provided:   - 1. $H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(g)$, $\Delta H_1 = -241.8\,kJ\,mol^{-1}$   - 2. $B_2O_3(s) + 3H_2O(g) \rightarrow B_2H_6(g) + 3O_2(g)$, $\Delta H_2 = +1941\,kJ\,mol^{-1}$   - 3. $2B(s) + \frac{3}{2}O_2(g) \rightarrow B_2O_3(s)$, $\Delta H_3 = -2368\,kJ\,mol^{-1}$

• Calculated Steps:   - Multiply reaction (1) by 3: $3H_2(g) + \frac{3}{2}O_2(g) \rightarrow 3H_2O(g)$, $\Delta H = 3(-241.8)\,kJ\,mol^{-1}$   - Add reaction (2) and reaction (3).

• Final Result:   - $\Delta_f H^\circ = -1152.4\,kJ/mol$

Enthalpy Calculations Using Bond Energy Methods

• General Formula:   - $\Delta H = \sum(BE)<em>{\text{reactants}} - \sum(BE)</em>{\text{products}}$   - Note: Stoichiometric coefficients must be considered.

• Example 1: Chlorination of Methane   - Reaction: $CH_4(g) + Cl_2(g) \rightarrow CH_3Cl(g) + HCl(g)$   - Bond Energies ($kJ/mol$): $C-H = 414$, $Cl-Cl = 239$, $C-Cl = 339$, $Cl-H = 431$.   - Calculation:     - $\Delta H = [4(C-H) + 1(Cl-Cl)] - [3(C-H) + 1(C-Cl) + 1(Cl-H)]$     - $\Delta H = [4(414\,kJ/mol) + 1(239\,kJ/mol)] - [3(414\,kJ/mol) + 1(339\,kJ/mol) + 1(431\,kJ/mol)]$

• Example 2: Combustion of an Unsaturated Hydrocarbon   - Reaction: $C_4H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(g)$   - Bond Energies ($kJ/mol$): $C=C = 614$, $C-H = 413$, $O=O = 493$, $C=O = 799$, $O-H = 466$.   - Calculation:     - $\Delta H = [1(C=C) + 4(C-H) + 3(O=O)] - [4(C=O) + 4(O-H)]$     - $\Delta H = [1(614\,kJ/mol) + 4(413\,kJ/mol) + 3(493\,kJ/mol)] - [4(799\,kJ/mol) + 4(466\,kJ/mol)]$     - $\Delta H = -1309\,kJ/mol$   - Conclusion: The reaction is exothermic; energy is given off.

Entropy ($S$) and Standard Entropy Calculations

• Hess's Law for Entropy:   - $\Delta S^\circ = [\sum n S^\circ(\text{products})] - [\sum m S^\circ(\text{reactants})]$

• Example: Formation of Sulfuric Acid   - Balanced Reaction: $SO_3(g) + H_2O(l) \rightarrow H_2SO_4(aq)$   - Standard Entropies ($J\,mol^{-1}\,K^{-1}$):     - $H_2SO_4(aq) = 157$     - $H_2O(l) = 69.96$     - $SO_3(g) = 256.2$   - Calculation:     - $\Delta S_{\text{rxn}} = [1 \times S(H_2SO_4)] - [1 \times S(SO_3) + 1 \times S(H_2O)]$     - $\Delta S_{\text{rxn}} = (1 \times 157\,J\,mol^{-1}\,K^{-1}) - [(1 \times 256.2\,J\,mol^{-1}\,K^{-1}) + (1 \times 69.96\,J\,mol^{-1}\,K^{-1})]$     - $\Delta S_{\text{rxn}} = -169\,J\,mol^{-1}\,K^{-1}$   - Observations: $\Delta S^\circ$ is negative, indicating a decrease in entropy. This occurs because two molecules react to produce one molecule.

Gibbs Free Energy ($G$)

• Standard Gibbs Free Energy Calculation:   - $\Delta G^\circ = [\sum n\Delta G^\circ(\text{products})] - [\sum m\Delta G^\circ(\text{reactants})]$

• Example: Combustion of Methane   - Reaction: $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$   - Free Energy Values ($kJ\,mol^{-1}$): $CH_4 = -50.8$, $O_2 = 0$, $CO_2 = -394.4$, $H_2O = -237.2$.   - Calculation:     - $\Delta G_{\text{comb}} = [(1 \times -394.4) + (2 \times -237.2)] - [(1 \times -50.8) + (2 \times 0)]$     - $\Delta G_{\text{comb}} = -818.8\,kJ\,mol^{-1}$   - Conclusion: As $\Delta G^\circ$ is negative, the reaction is spontaneous.

• Rules for $\Delta G^\circ$ Values:   - Large negative value: The reaction mixture contains large amounts of products at equilibrium.   - Large positive value: The reaction mixture contains insignificant amounts of products at equilibrium.

Fundamental Gibbs Relationships and Temperature

• The Gibbs-Helmholtz Type Equation:   - $\Delta G = \Delta H - T\Delta S$   - Interpretation Table for $\Delta G = \Delta H - T\Delta S$:     - If $\Delta H$ is negative and $\Delta S$ is positive: $\Delta G$ is negative at all temperatures.     - If $\Delta H$ is positive and $\Delta S$ is negative: $\Delta G$ is positive at all temperatures.     - If both are positive: High temperatures favor spontaneity.     - If both are negative: Low temperatures favor spontaneity.

• Non-Standard Conditions:   - $\Delta G = \Delta G^\circ + RT \ln(Q)$   - $R = 8.314\,J\,mol^{-1}\,K^{-1}$   - $Q$ is the reaction quotient (calculated using partial pressures in atm for gases or molarity for solutions).

• Relationship with Equilibrium Constant ($K$):   - At equilibrium, $\Delta G = 0$ and $Q = K$.   - $\Delta G^\circ = -RT \ln(K)$   - $K = e^{-\frac{\Delta G^\circ}{RT}}$

Equilibrium and Directionality Examples

• Direction of Reaction (Nitrogen Dioxide System):   - Reaction: $2NO_2(g) \rightleftharpoons N_2O_4(g)$   - Given: $\Delta G_{298}^\circ = -5.40\,kJ/mole$ of $N_2O_4$. $P(NO_2) = 0.25\,atm$, $P(N_2O_4) = 0.60\,atm$.   - Calculation:     - $\Delta G = \Delta G^\circ + RT \ln(\frac{P(N_2O_4)}{P(NO_2)^2})$     - $\Delta G = -5.40 \times 10^{3}\,J/mol + (8.314\,J/mol\,K \times 298\,K \times \ln(\frac{0.60}{0.25^2}))$     - $\Delta G = +2.0 \times 10^{2}\,J/mol$   - Result: $\Delta G$ is positive; therefore, the forward reaction is non-spontaneous. The reaction will proceed in the reverse direction to reach equilibrium.

• Calculating $\Delta G^\circ$ from $K_p$ (Air Pollution Study):   - Context: Brownish haze from $NO_2$. Reaction: $2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g)$.   - Data: $K_p = 1.7 \times 10^{12}$ at $25.00^\circ C$.   - Calculation:     - $\Delta G^\circ = -8.314\,J/mol\,K \times 298\,K \times \ln(1.7 \times 10^{12})$     - $\Delta G^\circ = -6.980 \times 10^{4}\,J$

Combined Spontaneity Problem

• Scenario: Calculate $\Delta G^\circ$ for a reaction at $298\,K$ with $\Delta H^\circ = -17.86\,kJ/mol$ and $\Delta S^\circ = 18.65\,J/mol\,K$.

• Step 1: Convert Entropy Units:   - $\Delta S^\circ = 0.01865\,kJ/mol\,K$

• Step 2: Apply Formula:   - $\Delta G = \Delta H - T\Delta S$   - $\Delta G = -17.86\,kJ/mol - (298\,K \times 0.01865\,kJ/mol\,K)$   - $\Delta G = -23.42\,kJ/mol$

• Conclusion: Because $\Delta G$ is negative (less than zero), the reaction is spontaneous at $298\,K$.