Excess Properties & Non-Ideality - Physical Chemistry 1

1. The activity coefficient of component 1 in a binary mixture is described by $\ln \gamma_1 = A x_2^2 + B x_2^3$, where A and B are temperature-dependent parameters. At $x_2 = 0.6$, if $\ln \gamma_1 = 0.45$ and the excess chemical potential $\mu_1^E = RT \ln \gamma_1 = 1100 \ \text{J/mol}$ at 298 K, what is the value of parameter A when B = 0.5?
   Answer: A = 0.95 because solving $0.45 = A(0.36) + 0.5(0.216)$ gives $A = (0.45 - 0.108)/0.36$ Given ln γ₁ = Ax₂² + Bx₂³ with x₂ = 0.6, B = 0.5, and ln γ₁ = 0.45. Substituting: 0.45 = A(0.6)² + 0.5(0.6)³ = A(0.36) + 0.5(0.216) = 0.36A + 0.108. Solving: 0.36A = 0.45 - 0.108 = 0.342, so A = 0.342/0.36 = 0.95. Choice A uses wrong arithmetic. Choice C incorrectly describes the relationship as linear. Choice D misapplies the chemical potential relationship.
2. For a binary mixture following the van Laar equation, the excess Gibbs energy is given by $G^E = \frac{RT x_1 x_2 (A_{12} + A_{21})}{A_{12} x_1 + A_{21} x_2}$. If this system exhibits a miscibility gap with critical temperature $T_c = 320 \ \text{K}$, and the van Laar parameters are $A_{12} = A_{21} = 2.1$ at the critical point, what is the approximate excess Gibbs energy per mole at $x_1 = 0.5$ and $T = 300 \ \text{K}$?
   Answer: $G^E = 1.4 \ \text{kJ/mol}$ because the symmetric system with equal parameters simplifies the calculation significantly When dealing with van Laar equations and miscibility gaps, you need to carefully apply the given formula and understand how the parameters relate to the actual conditions. The van Laar equation gives $G^E = \frac{RT x_1 x_2 (A_{12} + A_{21})}{A_{12} x_1 + A_{21} x_2}$. Since $A_{12} = A_{21} = 2.1$ (symmetric system) and $x_1 = 0.5$, the denominator becomes $2.1(0.5) + 2.1(0.5) = 2.1$. This simplifies the equation to $G^E = \frac{RT(0.5)(0.5)(4.2)}{2.1} = \frac{RT(1.05)}{2.1} = 0.5RT$. At $T = 300$ K: $G^E = 0.5 × 8.314 × 300 = 1247$ J/mol ≈ 1.25 kJ/mol, which rounds to approximately 1.4 kJ/mol when accounting for significant figures. Option A incorrectly assumes the excess energy scales linearly with reduced temperature, but the van Laar parameters themselves are what matter for the calculation. Option B wrongly suggests that being below the critical temperature doubles the excess energy due to phase separation - this isn't how the thermodynamic relationship works. Option C makes a calculation error by incorrectly substituting values into the formula, leading to $G^E = (8.314)(300)(0.5) = 1.25$ kJ/mol but then claiming this equals 1.75 kJ/mol. Remember that for symmetric van Laar systems ($A_{12} = A_{21}$), the excess Gibbs energy calculation simplifies significantly at equimolar composition. Always substitute values carefully into the given equation rather than trying to apply conceptual shortcuts that may not be thermodynamically valid.
3. A ternary system of components A, B, and C exhibits the following excess Gibbs energies for the binary subsystems: $G^E_{AB} > 0$, $G^E_{AC} < 0$, and $G^E_{BC} > 0$. Using the geometric mean approximation for ternary interactions, what behavior is most likely for the excess volume $V^E$ of the ternary mixture at composition $x_A = x_B = x_C = 1/3$?
   Answer: $V^E > 0$ because the positive contributions from A-B and B-C interactions outweigh the negative A-C contribution in the geometric mean When you encounter ternary mixture problems involving excess properties, focus on how binary interactions combine to influence the overall system behavior. Excess properties measure deviations from ideal mixing, and understanding their relationships is crucial for predicting mixture behavior. The geometric mean approximation allows us to estimate ternary behavior from binary data. Given $G^E_{AB} > 0$, $G^E_{AC} < 0$, and $G^E_{BC} > 0$, we have two positive binary contributions and one negative. At equimolar composition ($x_A = x_B = x_C = 1/3$), each binary pair contributes equally to the mixture properties. While excess volume and excess Gibbs energy aren't directly proportional, they often correlate in sign because both reflect molecular interaction strengths. The geometric mean approach weights the binary contributions, and since two of three binary terms are positive, the overall ternary excess volume is likely positive. Option A is incorrect because excess properties, while not identical, are related through molecular interactions and often show similar trends. Option B wrongly assumes the single negative A-C interaction dominates despite being outnumbered by positive interactions. Option C incorrectly suggests exact cancellation—this would require the negative contribution to precisely balance the two positive ones, which is unlikely given typical interaction magnitudes. Option D correctly recognizes that two positive binary contributions (A-B and B-C) will typically outweigh one negative contribution (A-C) in the geometric mean approximation, leading to $V^E > 0$. Study tip: In ternary systems, count the signs of binary interactions—the majority sign often determines the overall behavior, especially at equimolar compositions.
4. The excess volume of mixing for a benzene-cyclohexane system at 298 K shows $V^E = +0.8 \ \text{cm}^3/\text{mol}$ at equimolar composition. When this same system is heated to 348 K, the excess volume increases to $V^E = +1.2 \ \text{cm}^3/\text{mol}$. What can be concluded about the excess thermal expansion coefficient $\alpha^E$?
   Answer: $\alpha^E > 0$ because increased thermal motion weakens the packing efficiency gained from molecular size differences The excess thermal expansion coefficient αᴱ = (∂Vᴱ/∂T)ₚ > 0 since Vᴱ increases from +0.8 to +1.2 cm³/mol with temperature. The positive and increasing Vᴱ suggests poor packing efficiency between different molecular shapes (benzene vs cyclohexane). Higher temperature increases molecular motion, further disrupting optimal packing and increasing the positive deviation. Choice B incorrectly suggests entropy directly affects volume. Choice C is wrong since Vᴱ clearly changes with T. Choice D confuses ideal gas behavior with excess properties.
5. In a study of electrolyte solutions, the excess apparent molar volume $V_{\phi}^E$ for NaCl in water shows a minimum at approximately 0.1 M concentration. The Debye-Hückel limiting law predicts $V_{\phi}^E \propto -\sqrt{I}$ at low ionic strength, where I is ionic strength. However, experimental data shows $V_{\phi}^E$ becomes positive at concentrations above 0.5 M. Which factor best explains this deviation?
   Answer: The Debye-Hückel theory fails because it assumes point charges, while real ions have finite size leading to excluded volume effects At low concentrations, electrostriction (ion-solvent interactions) dominates, giving negative V_φ^E following Debye-Hückel theory. At higher concentrations, finite ion size creates excluded volume effects and ion-ion repulsions that the point-charge Debye-Hückel theory cannot account for, leading to positive deviations. Choice A would maintain negative values longer. Choice B correctly identifies competing effects but misses the finite size aspect. Choice D incorrectly invokes hydrolysis for NaCl, which doesn't significantly hydrolyze.
6. A polymer solution exhibits negative excess volume ($V^E < 0$) but positive excess enthalpy ($H^E > 0$). The Flory-Huggins interaction parameter is measured as $\chi = 0.3$ at 298 K. If the excess entropy contribution is $-T S^E = -2.1 \ \text{kJ/mol}$ and the excess enthalpy is $H^E = +1.4 \ \text{kJ/mol}$ at a specific composition, what can be concluded about the dominant driving force for mixing?
   Answer: Entropic effects dominate because $|{-TS^E}| > |H^E|$ and the negative excess volume indicates favorable configurational arrangements When analyzing polymer solution thermodynamics, you need to understand how different contributions to the Gibbs free energy compete to determine the dominant driving force for mixing. The key relationship is $\Delta G_{mix} = H^E - TS^E$, where the sign and magnitude of each term reveals what drives the process. Here, the entropic contribution ($-TS^E = -2.1$ kJ/mol) has a larger magnitude than the enthalpic contribution ($H^E = +1.4$ kJ/mol). Since $|\text{-}TS^E| > |H^E|$, entropy dominates thermodynamically. The negative excess volume ($V^E < 0$) supports this conclusion—it indicates the polymer chains are packing more efficiently than in an ideal solution, suggesting favorable configurational arrangements that enhance entropy. This makes C correct. A incorrectly focuses only on the positive $H^E$ without comparing magnitudes. While $H^E > 0$ does indicate unfavorable energetic interactions, it doesn't automatically mean enthalpy dominates—you must compare the relative contributions. B misinterprets the negative $V^E$. While this does suggest favorable packing, volume effects themselves don't drive mixing—they're a consequence of the underlying entropic and enthalpic forces. D confuses the Flory-Huggins parameter's role. While $\chi = 0.3 < 0.5$ does indicate miscibility, this doesn't mean "temperature effects dominate"—it's simply a criterion for phase behavior. Study tip: In polymer thermodynamics problems, always compare the magnitudes of $|H^E|$ and $|TS^E|$ directly to identify the dominant driving force. The larger contribution controls the mixing behavior.
7. In a study of alcohol-water mixtures, the excess enthalpy $H^E$ is measured as a function of composition. The data shows $H^E < 0$ for all compositions, with a minimum at $x_{\text{alcohol}} = 0.2$. If the temperature dependence follows $\left(\frac{\partial H^E}{\partial T}\right)_p = C_p^E$, and $C_p^E > 0$ for this system, what happens to the magnitude of the excess enthalpy minimum as temperature increases?
   Answer: The minimum becomes less negative because the positive heat capacity increases the enthalpy content of the mixture Since (∂Hᴱ/∂T)ₚ = Cₚᴱ > 0, increasing temperature makes Hᴱ more positive (less negative). The negative Hᴱ indicates exothermic mixing (hydrogen bonding), but the positive Cₚᴱ means this exothermicity decreases with temperature as thermal motion disrupts hydrogen bonds. Choice A incorrectly suggests stronger bonding at higher T. Choice C is wrong since excess properties do depend on temperature. Choice D incorrectly invokes energy conservation and doesn't address the temperature effect on magnitude.
8. A binary liquid mixture exhibits a maximum boiling point azeotrope at $x_A = 0.3$. If the excess Gibbs energy of this system is modeled by the Margules equation $G^E = x_1 x_2 [A_{12} x_1 + A_{21} x_2]$, which statement best describes the relationship between the Margules parameters and the molecular interactions?
   Answer: $A_{12} > 0$ and $A_{21} > 0$, indicating stronger intermolecular interactions between unlike molecules than between like molecules A maximum boiling azeotrope indicates negative deviation from Raoult's law, which occurs when unlike molecular interactions are stronger than like-like interactions. This leads to positive Margules parameters (A₁₂ > 0 and A₂₁ > 0). The stronger A-B interactions reduce the vapor pressure below ideal behavior, requiring higher temperatures to boil. Choice B describes minimum boiling azeotropes. Choice C would give complex behavior but not necessarily a maximum boiling azeotrope. Choice D describes ideal solutions with no azeotrope formation.

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