Chapter 17 Equilibrium and Entropy Math Practice Guide

Conceptual Overview of Entropy (\Delta S): * Entropy is a thermodynamic quantity representing the degree of disorder or randomness in a chemical system. * A positive entropy change ( $+\Delta S$ ) indicates an increase in disorder (system becomes more chaotic). * A negative entropy change ( $-\Delta S$ ) indicates a decrease in disorder (system becomes more ordered).
Criteria for Determining the Direction of Entropy Change (Positive vs. Negative): * Phase Changes: Transitioning from solid to liquid, or liquid to gas, results in a positive change ( $+\Delta S$ ) because particles move more freely. Conversely, gas to liquid or liquid to solid is negative ( $-\Delta S$ ). * Dissolution: Dissolving a solid into a solvent typically increases entropy ( $+\Delta S$ ). * Number of Particles: Reactions that produce a greater number of product molecules than reactant molecules (especially in the gas phase) generally have a positive entropy change ( $+\Delta S$ ). * Temperature Changes: An increase in temperature results in increased kinetic energy and higher entropy ( $+\Delta S$ ).

Calculation Methodology for Enthalpy (\Delta H), Entropy (\Delta S), and Gibbs Free Energy (\Delta G): * To calculate standard changes for a reaction, use the summation formula: $\Delta X^\circ_{\text{rxn}} = \sum n \Delta X^\circ_{f, \text{products}} - \sum m \Delta X^\circ_{f, \text{reactants}}$ * Where $X$ represents $H$ , $S$ , or $G$ , and $n$ and $m$ represent the stoichiometric coefficients from the balanced chemical equation.
Problem 2/3 Analysis (Standard Table Practice): * Reaction A: Requires identifying the standard values from an appendix for the provided reaction components. * Reaction B: Follows the same summation protocol. * Reaction C (White Phosphorus): Specifically notes the inclusion of white phosphorus ( $P_4$ ). For white phosphorus in its standard state, $\Delta H^\circ_f = 0\,kJ\,mol^{-1}$ , but its standard absolute entropy ( $S^\circ$ ) must be retrieved from the table for calculations.

Definition of the Equilibrium Expression: * For a reversible reaction $aA + bB \rightleftharpoons cC + dD$ , the equilibrium constant expression is written as: $K_{eq} = \frac{[C]^c [D]^d}{[A]^a [B]^b}$ * Pure Solids and Liquids Exception: Pure solids ( $s$ ) and pure liquids ( $l$ ) are omitted from the equilibrium expression because their concentrations remain essentially constant during the reaction. Their activity is defined as $1$ .
Practice Problem 4: Writing Keq Expressions: * a. $2H_2S(g) \rightleftharpoons 2H_2(g) + S_2(g)$ * Expression: $K_{eq} = \frac{[H_2]^2 [S_2]}{[H_2S]^2}$ * b. $4NO(g) + 2O2(g) \rightleftharpoons 2N_2O_4(g)$ * Expression: $K_{eq} = \frac{[N_2O_4]^2}{[NO]^4 [O_2]^2}$ * c. $2NO(g) + Br_2(g) \rightleftharpoons 2NOBr(g)$ * Expression: $K_{eq} = \frac{[NOBr]^2}{[NO]^2 [Br_2]}$ * d. $CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(l)$ * Note: Methanol is a liquid and is excluded from the numerator. * Expression: $K_{eq} = \frac{1}{[CO] [H_2]^2}$ * e. $2HI(g) \rightleftharpoons H_2(g) + I_2(g)$ * Expression: $K_{eq} = \frac{[H_2] [I_2]}{[HI]^2}$

Mechanical Stress via Pressure: * According to Le Chatelier's Principle, if the total pressure of a system at equilibrium is increased (often by decreasing the volume), the system will shift its equilibrium position to favor the side with the fewer total moles of gas to relieve the pressure. * If the moles of gas are equal on both sides, pressure changes have no effect on the equilibrium position.
Practice Problem 5 Analysis (Applied to Problem 4 Reactions): * a. $2H_2S(g) \rightleftharpoons 2H_2(g) + S_2(g)$ : Reactants = $2\,moles$ , Products = $3\,moles$ . Increase in pressure shifts equilibrium to the Left (toward reactants). * b. $4NO(g) + 2O_2(g) \rightleftharpoons 2N_2O_4(g)$ : Reactants = $6\,moles$ , Products = $2\,moles$ . Increase in pressure shifts equilibrium to the Right (toward products). * c. $2NO(g) + Br_2(g) \rightleftharpoons 2NOBr(g)$ : Reactants = $3\,moles$ , Products = $2\,moles$ . Increase in pressure shifts equilibrium to the Right. * d. $CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(l)$ : Reactants = $3\,moles\,gas$ , Products = $0\,moles\,gas$ (liquid does not contribute to gas pressure). Increase in pressure shifts equilibrium to the Right. * e. $2HI(g) \rightleftharpoons H_2(g) + I_2(g)$ : Reactants = $2\,moles$ , Products = $2\,moles$ . Increase in pressure has No Effect on the equilibrium position.

Problem 6: Calculation at 1400°C: * Reaction: $2H_2S(g) \rightleftharpoons 2H_2(g) + S_2(g)$ * System Volume: $1.0\,liter$ * Equilibrium Concentrations: * $[H_2S] = \frac{0.18\,mol}{1.0\,L} = 0.18\,M$ * $[H_2] = \frac{0.014\,mol}{1.0\,L} = 0.014\,M$ * $[S_2] = \frac{0.035\,mol}{1.0\,L} = 0.035\,M$ * Execution: $K_{eq} = \frac{(0.014)^2 \times (0.035)}{(0.18)^2}$ $K_{eq} = \frac{1.96 \times 10^{-4} \times 0.035}{0.0324}$ $K_{eq} = \frac{6.86 \times 10^{-6}}{0.0324} \approx 2.12 \times 10^{-4}$
Problem 7: Comparative Calculation: * New Conditions: $[H_2S] = 0.25\,M$ , $[H_2] = 0.88\,M$ , and $[S_2] = 0.44\,M$ * Execution: $K_{eq} = \frac{[H_2]^2 [S_2]}{[H_2S]^2} = \frac{(0.88)^2 \times (0.44)}{(0.25)^2}$ $K_{eq} = \frac{0.7744 \times 0.44}{0.0625}$ $K_{eq} = \frac{0.340736}{0.0625} = 5.451776$

Problem 8: Solving for [NO2]: * Reaction: $2NO_2(g) \rightleftharpoons N_2O_4(g)$ * Given Constants: $K_{eq} = 5.6$ * Given Concentration: $[N_2O_4] = 0.66\,M$ * Equilibrium Expression Setup: $K_{eq} = \frac{[N_2O_4]}{[NO_2]^2}$ $5.6 = \frac{0.66}{[NO_2]^2}$ * Algebraic Solution: $[NO_2]^2 = \frac{0.66}{5.6}$ $[NO_2]^2 = 0.117857…$ $[NO_2] = \sqrt{0.117857…} \approx 0.343\,M$