Predicting Entropy Changes and Standard Molar Entropies

Intellectual Property and Course Context

• Course: Chem 1155 Class Activity 4

• Instructor: Brian Gute

• Intellectual Property Notice: This work is the intellectual property of the instructor, Brian Gute, and may not be altered, shared for commercial purposes, or distributed in any modified or unmodified form, either during or subsequent to enrollment in the course.

Fundamental Factors Governing Entropy Changes ($\Delta S$)

• Primary factors evaluated when predicting whether the entropy ($S$) of a physical or chemical process increases ($\Delta S > 0$) or decreases ($\Delta S < 0$):

  • Physical State / Phase Transitions: Gases possess significantly higher entropy than liquids, and liquids possess higher entropy than solids ($S_{\text{gas}} \gg S_{\text{liquid}} > S_{\text{solid}}$). Transitions toward more fluid or dispersed phases increase entropy.

  • Number of Moles of Gas ($\Delta n_{\text{gas}}$): A reaction that yields a net increase in the total moles of gas ($\Delta n_{\text{gas}} > 0$) results in an increase in entropy. A reaction that yields a net decrease in moles of gas ($\Delta n_{\text{gas}} < 0$) results in a decrease in entropy.

  • Dissolution and Solution Formation: The dissolution of a solid or liquid solute into a solvent generally increases entropy due to the increased spatial mobility of solute particles.

  • Temperature Changes: Increasing temperature expands the thermal energy distribution across available microstates, causing an increase in entropy.

• Process Sign and Rationale Analysis (Practice 1):

  • Process: Water freezes

  • Direction: Decreasing ($\Delta S < 0$)

  • Rationale: Liquid water converts into a highly ordered, rigid crystalline solid, restricting vibrational and rotational freedom.

  • Process: Water vaporizes

  • Direction: Increasing ($\Delta S > 0$)

  • Rationale: Liquid water transitions into the gaseous state, vastly increasing particle spatial distribution and microstates.

  • Process: $H_2O(s) \rightarrow H_2O(g)$

  • Direction: Increasing ($\Delta S > 0$)

  • Rationale: Solid ice undergoes sublimation directly into gaseous water, moving from a rigid crystalline structure to a highly dispersed phase with maximum microstates.

  • Process: $NaCl(s) \rightarrow Na^+(aq) + Cl^-(aq)$

  • Direction: Increasing ($\Delta S > 0$)

  • Rationale: Crystalline solid sodium chloride dissolves into mobile aqueous sodium and chloride ions, increasing solute structural disorder.

  • Process: $N_2(g) + 3 H_2(g) \rightarrow 2 NH_3(g)$

  • Direction: Decreasing ($\Delta S < 0$)

  • Rationale: The reaction combines $4\,mol$ of reactant gas into $2\,mol$ of product gas, reducing overall translational degrees of freedom.

Standard Molar Entropy ($S^\circ$) and Molecular Structure

• Standard State Conditions Definition:

  • Defined as $1\,mol$ of pure substance (or a $1.0\,M$ concentration for solutions) at a pressure of $1.0\,atm$ and at a specified temperature (typically $25\,^\circ\text{C}$ or $298.15\,K$).

• Thermodynamic Differences Between Enthalpy ($\Delta H^\circ$) and Entropy ($S^\circ$):

  • Standard Enthalpy of Formation ($\Delta H_f^\circ$): Defined relative to a reference state where any pure element in its most stable standard state has $\Delta H_f^\circ = 0\,kJ\,mol^{-1}$.

  • Standard Molar Entropy ($S^\circ$): Governed by the Third Law of Thermodynamics. The standard molar entropy for any pure crystalline substance at absolute zero ($0\,K$) is defined precisely as $0\,J\,mol^{-1}\,K^{-1}$. At all temperatures above $0\,K$, standard molar entropy is strictly greater than zero ($S^\circ > 0\,J\,mol^{-1}\,K^{-1}$).

• Primary Factors Influencing Standard Molar Entropy ($S^\circ$) Values (Practice 2):

  • Physical State (Phase): Gas molecules have far greater molar entropy than liquid or solid molecules of similar composition.

  • Molar Mass / Atomic Complexity: Heavier atoms and molecules possess closer energy level spacings, yielding a higher density of quantum microstates and higher $S^\circ$.

  • Molecular Complexity: Molecules containing more atoms have a larger number of vibrational and rotational degrees of freedom, increasing $S^\circ$.

  • Allotropic Structure / Rigid Lattice: Less constrained allotropes have higher entropy than rigid covalent network structures.

Comparative Analysis of Standard Molar Entropy ($S^\circ$)

• Practice 3: Ranking Physical States of Water

  • Ranking (Highest $S^\circ$ to Lowest $S^\circ$): $H_2O(g) > H_2O(l) > H_2O(s)$

  • Most Influential Factor: Physical state (phase) of matter.

  • Explanation: Gaseous water ($H_2O(g)$) has completely unrestricted spatial and translational motion. Liquid water ($H_2O(l)$) has hindered fluidity due to dynamic hydrogen bonding. Solid ice ($H_2O(s)$) is immobilized within a hexagonal crystal lattice, yielding the lowest entropy.

• Practice 4: Ranking Nitrogen Oxide Gas Compounds

  • Ranking (Highest $S^\circ$ to Lowest $S^\circ$): $N_2O_4(g) > NO_2(g) > NO(g)$

  • Most Influential Factor: Molecular complexity and molar mass (number of atoms per molecule).

  • Explanation: All three substances exist in the gas phase. Dinitrogen tetroxide ($N_2O_4(g)$) contains $6$ atoms per molecule and the largest mass, providing maximum vibrational modes. Nitrogen dioxide ($NO_2(g)$) contains $3$ atoms per molecule. Nitric oxide ($NO(g)$) is a simple diatomic gas with $2$ atoms per molecule, giving it the fewest available microstates.

• Practice 5: Ranking Distinct Gaseous Species

  • Ranking (Highest $S^\circ$ to Lowest $S^\circ$): $C_2H_4(g) > NO(g) > CO(g)$

  • Most Influential Factor: Molecular complexity (number of atoms) combined with molar mass.

  • Explanation: Ethylene ($C_2H_4(g)$) contains $6$ atoms per molecule, giving it substantially greater structural complexity and rotational/vibrational modes compared to the diatomic gases. When comparing the two diatomic gases, $NO(g)$ has a molar mass of approximately $30.01\,g\,mol^{-1}$, whereas $CO(g)$ has a molar mass of approximately $28.01\,g\,mol^{-1}$. The greater mass of $NO(g)$ results in a slightly higher standard molar entropy than $CO(g)$.

Predicting Reaction Standard Entropy Changes ($\Delta S^\circ$)

• Practice 6: Analysis of $5 H_2S(g) + 3 SO_2(g) \rightarrow S_8(s) + 5 H_2O(g)$

  • Entropy Direction: Decreasing ($\Delta S^\circ < 0$)

  • Chemical Equation Analysis: Reactants consist of $5\,mol$ of $H_2S(g)$ and $3\,mol$ of $SO_2(g)$, yielding a total of $8\,mol$ of gaseous reactants. Products consist of $1\,mol$ of solid octasulfur ($S_8(s)$) and $5\,mol$ of gaseous water ($H_2O(g)$), yielding $5\,mol$ of gaseous products and $1\,mol$ of solid.

  • Explanation: The total amount of gas decreases from $8\,mol$ to $5\,mol$ ($\Delta n_{\text{gas}} = -3\,mol$), and a solid precipitate is produced. A net loss of gaseous species significantly reduces system spatial disorder.

• Practice 7: Analysis of $2 NaCl(s) + H_2SO_4(l) \rightarrow 2 HCl(g) + Na_2SO_4(s)$

  • Entropy Direction: Increasing ($\Delta S^\circ > 0$)

  • Chemical Equation Analysis: Reactants consist of $2\,mol$ of solid $NaCl(s)$ and $1\,mol$ of liquid $H_2SO_4(l)$ ($0\,mol$ of gas). Products consist of $2\,mol$ of gaseous $HCl(g)$ and $1\,mol$ of solid $Na_2SO_4(s)$ ($2\,mol$ of gas).

  • Explanation: The reaction generates $2\,mol$ of gaseous product ($\Delta n_{\text{gas}} = +2\,mol$) from solid and liquid reactants, leading to a substantial gain in translational microstates and system entropy.

• Practice 8: Analysis of $N_2(g) + 3 H_2(g) \rightarrow 2 NH_3(g)$

  • Entropy Direction: Decreasing ($\Delta S^\circ < 0$)

  • Chemical Equation Analysis: Reactants consist of $1\,mol$ of $N_2(g)$ and $3\,mol$ of $H_2(g)$, totaling $4\,mol$ of gas. Products consist of $2\,mol$ of $NH_3(g)$, totaling $2\,mol$ of gas.

  • Explanation: The conversion of $4\,mol$ of gas into $2\,mol$ of gas ($\Delta n_{\text{gas}} = -2\,mol$) reduces the total number of independently moving gaseous particles, decreasing system disorder.

Quantitative Calculation of Reaction Standard Molar Entropy ($\Delta S^\circ_{\text{rxn}}$)

• Mathematical Governing Equation: $\Delta S^\circ_{\text{rxn}} = \sum n S^\circ(\text{products}) - \sum m S^\circ(\text{reactants})$ Where $n$ and $m$ represent the stoichiometric coefficients of products and reactants.

• Example 1: Synthesis of Liquid Water

  • Chemical Equation: $2 H_2(g) + O_2(g) \rightarrow 2 H_2O(l)$

  • Standard Molar Entropy Reference Table Data:

  • $H_2(g)$: $130.7\,J\,mol^{-1}\,K^{-1}$

  • $O_2(g)$: $205.2\,J\,mol^{-1}\,K^{-1}$

  • $H_2O(g)$: $188.8\,J\,mol^{-1}\,K^{-1}$

  • $H_2O(l)$: $70.0\,J\,mol^{-1}\,K^{-1}$

  • Calculation Steps: $\Delta S^\circ_{\text{rxn}} = [2 \times S^\circ(H_2O(l))] - [2 \times S^\circ(H_2(g)) + 1 \times S^\circ(O_2(g))]$ $\Delta S^\circ_{\text{rxn}} = [2 \times 70.0\,J\,mol^{-1}\,K^{-1}] - [2 \times 130.7\,J\,mol^{-1}\,K^{-1} + 205.2\,J\,mol^{-1}\,K^{-1}]$ $\Delta S^\circ_{\text{rxn}} = 140.0\,J\,K^{-1} - [261.4\,J\,K^{-1} + 205.2\,J\,K^{-1}]$ $\Delta S^\circ_{\text{rxn}} = 140.0\,J\,K^{-1} - 466.6\,J\,K^{-1}$ $\Delta S^\circ_{\text{rxn}} = -326.6\,J\,K^{-1}$

  • Calculated Standard Entropy Change: $-326.6\,J\,K^{-1}$

• Practice 9: Synthesis of Ammonia Gas

  • Chemical Equation: $N_2(g) + 3 H_2(g) \rightarrow 2 NH_3(g)$

  • Standard Molar Entropy Reference Table Data:

  • $H_2(g)$: $130.7\,J\,mol^{-1}\,K^{-1}$

  • $N_2(g)$: $191.6\,J\,mol^{-1}\,K^{-1}$

  • $NH_3(g)$: $192.8\,J\,mol^{-1}\,K^{-1}$

  • $NH_3(aq)$: $111.3\,J\,mol^{-1}\,K^{-1}$

  • Calculation Steps: $\Delta S^\circ_{\text{rxn}} = [2 \times S^\circ(NH_3(g))] - [1 \times S^\circ(N_2(g)) + 3 \times S^\circ(H_2(g))]$ $\Delta S^\circ_{\text{rxn}} = [2 \times 192.8\,J\,mol^{-1}\,K^{-1}] - [191.6\,J\,mol^{-1}\,K^{-1} + 3 \times 130.7\,J\,mol^{-1}\,K^{-1}]$ $\Delta S^\circ_{\text{rxn}} = 385.6\,J\,K^{-1} - [191.6\,J\,K^{-1} + 392.1\,J\,K^{-1}]$ $\Delta S^\circ_{\text{rxn}} = 385.6\,J\,K^{-1} - 583.7\,J\,K^{-1}$ $\Delta S^\circ_{\text{rxn}} = -198.1\,J\,K^{-1}$

  • Calculated Standard Entropy Change: $-198.1\,J\,K^{-1}$