Chemical Potential - Physical Chemistry 1

For a binary solution at constant temperature and pressure, the chemical potential of component A is given by $\mu_A = \mu_A^* + RT \ln(x_A \gamma_A)$ , where $\gamma_A$ is the activity coefficient. If the partial molar enthalpy of A is $\bar{H}_A = 8500$ J/mol and the partial molar entropy is $\bar{S}_A = 25.0$ J/(mol·K) at 298 K, what is the relationship between the chemical potential and these partial molar quantities?
Answer: $\mu_A = \bar{H}_A - T\bar{S}_A$ is valid regardless of solution ideality since it represents the partial molar Gibbs energy definition The chemical potential is defined as the partial molar Gibbs energy: $\mu_A = \bar{G}_A = \bar{H}_A - T\bar{S}_A$ . This fundamental thermodynamic relationship holds regardless of whether the solution is ideal or non-ideal, because it defines the chemical potential in terms of other partial molar quantities. The expression with activity coefficients is just another way to express the same quantity. Choice A is incomplete as it doesn't recognize the general validity. Choice B incorrectly suggests this only applies to ideal solutions. Choice C incorrectly limits it to standard state conditions.
For a system containing both neutral and charged species, the chemical potential can be partitioned into ideal and excess contributions. Consider an aqueous electrolyte solution where the chemical potential of the salt AB is $\mu_{AB} = \mu_{AB}^\circ + \nu RT \ln(m \gamma_{\pm})$ , where $\nu = \nu_+ + \nu_-$ is the total number of ions, $m$ is molality, and $\gamma_{\pm}$ is the mean activity coefficient. For a 2:1 electrolyte like CaCl₂ at $m = 0.1$ mol/kg with $\gamma_{\pm} = 0.725$ , what is the chemical potential relative to the standard state?
Answer: $\mu_{CaCl_2} - \mu_{CaCl_2}^\circ = 3RT \ln(0.1 \times 0.725) = 3RT(-2.62) = -19.5 \text{ kJ/mol}$ When working with electrolyte solutions, you need to understand how salts dissociate and how this affects their chemical potential. The key is identifying the stoichiometry correctly. For CaCl₂, the dissociation is: CaCl₂ → Ca²⁺ + 2Cl⁻. This gives us ν₊ = 1 and ν₋ = 2, so the total number of ions is ν = ν₊ + ν₋ = 3. This coefficient appears in front of the RT ln term because it accounts for all ions produced. Using the given equation with ν = 3, m = 0.1 mol/kg, and γ± = 0.725: $\mu_{CaCl_2} - \mu_{CaCl_2}^\circ = 3RT \ln(0.1 \times 0.725) = 3RT \ln(0.0725) = 3RT(-2.62)$ At 298 K, this equals 3 × 8.314 × 298 × (-2.62)/1000 = -19.5 kJ/mol. Option A correctly applies this calculation. Option B reaches the same numerical result but unnecessarily separates the logarithm terms—while mathematically valid due to ln(ab) = ln(a) + ln(b), it's not the most direct approach. Option C uses ν = 2, incorrectly counting only the chloride ions or confusing this with a 1:1 electrolyte. Option D completely omits the activity coefficient, ignoring non-ideal behavior entirely. Remember: always determine the complete dissociation stoichiometry first (ν = total ions per formula unit), then include both concentration and activity coefficient in your calculation. The activity coefficient accounts for ion-ion interactions that make real solutions deviate from ideal behavior.
For a system undergoing a phase transition, the chemical potential serves as the driving force for mass transfer between phases. Consider a pure substance at its triple point where solid, liquid, and vapor phases coexist. If the molar volume changes are $V_l - V_s = 2.1 \times 10^{-6} \text{ m}^3/\text{mol}$ and $V_g - V_l = 3.2 \times 10^{-2} \text{ m}^3/\text{mol}$ , and a small pressure increase of 100 Pa is applied, which statement best describes the resulting changes in chemical potential?
Answer: The vapor phase experiences the largest chemical potential increase, driving condensation to liquid and freezing to solid When you encounter phase equilibrium problems, remember that chemical potential changes drive mass transfer between phases. At equilibrium, all phases have equal chemical potentials, but when conditions change, the phase experiencing the largest chemical potential increase becomes thermodynamically unfavorable. For small pressure changes, the change in chemical potential is given by $\Delta \mu = V_m \Delta P$ , where $V_m$ is the molar volume. Since pressure increases by 100 Pa, each phase experiences a chemical potential increase proportional to its molar volume. Given the molar volume differences, we can determine relative volumes. The vapor phase has the largest molar volume ( $V_g = V_l + 3.2 \times 10^{-2}$ ), followed by liquid ( $V_l = V_s + 2.1 \times 10^{-6}$ ), then solid. Therefore, vapor experiences the largest chemical potential increase, making it thermodynamically unstable relative to the denser phases. This drives condensation from vapor to liquid and deposition from vapor to solid. Answer A is wrong because phases with different molar volumes cannot experience identical chemical potential changes under pressure. Answer C incorrectly suggests solid has the smallest increase drives mass transfer away from it—actually, having the smallest increase makes solid more favorable. Answer D incorrectly focuses on liquid disappearing, but liquid has an intermediate molar volume and would be moderately affected compared to the highly unstable vapor phase. Remember: under pressure increases, phases with larger molar volumes become less stable, driving mass transfer toward denser phases. Always compare molar volumes to predict the direction of phase transitions.
In a binary liquid mixture, the excess chemical potential of component 1 is defined as $\mu_1^E = \mu_1 - \mu_1^{ideal} = RT \ln(\gamma_1)$ , where $\gamma_1$ is the activity coefficient. If experimental data shows that $\frac{\partial \mu_1^E}{\partial x_1} = 1250$ J/mol at $x_1 = 0.30$ and 298 K, what does this indicate about the local solution behavior?
Answer: The derivative $\frac{\partial \ln(\gamma_1)}{\partial x_1} = \frac{1250}{RT} = 0.504$ indicates moderate positive deviations with $\gamma_1$ increasing as $x_1$ increases From $\mu_1^E = RT \ln(\gamma_1)$ , we get $\frac{\partial \mu_1^E}{\partial x_1} = RT \frac{\partial \ln(\gamma_1)}{\partial x_1}$ . Therefore, $\frac{\partial \ln(\gamma_1)}{\partial x_1} = \frac{1250}{(8.314)(298)} = 0.504$ . This indicates $\gamma_1$ increases with $x_1$ , showing positive deviations. Choice A is qualitatively correct but less quantitative. Choice B incorrectly relates the sign to negative deviations. Choice D incorrectly assumes the system is unstable based on limited information.
The chemical potential of a component in an ideal gas mixture differs from that in a real gas mixture due to intermolecular interactions. For a real gas, $\mu_i = \mu_i^\circ(T) + RT \ln(f_i/p^\circ)$ where $f_i$ is the fugacity. If the fugacity coefficient $\phi_i = f_i/(x_i P)$ for component i in a binary mixture is given by $\ln \phi_1 = \frac{P}{RT}(B_{11} + x_2^2 \delta)$ where $\delta = 2B_{12} - B_{11} - B_{22}$ , how does the chemical potential of component 1 in the real gas compare to its value in an ideal gas mixture at the same $T$ , $P$ , and composition?
Answer: $\mu_1^{real} - \mu_1^{ideal} = RT \ln \phi_1 = P(B_{11} + x_2^2 \delta)$ This question tests your understanding of how chemical potentials differ between real and ideal gases due to intermolecular interactions, specifically through the concept of fugacity coefficients. To find how the chemical potential differs between real and ideal gases, you need to recognize that both expressions use the same reference state $\mu_i^\circ(T)$ . For an ideal gas, the chemical potential is $\mu_i^{ideal} = \mu_i^\circ(T) + RT \ln(x_i P/p^\circ)$ , while for a real gas it's $\mu_i^{real} = \mu_i^\circ(T) + RT \ln(f_i/p^\circ)$ . The difference becomes: $\mu_1^{real} - \mu_1^{ideal} = RT \ln(f_1) - RT \ln(x_1 P) = RT \ln(f_1/x_1 P) = RT \ln \phi_1$ Since $\ln \phi_1 = \frac{P}{RT}(B_{11} + x_2^2 \delta)$ , you get: $\mu_1^{real} - \mu_1^{ideal} = RT \cdot \frac{P}{RT}(B_{11} + x_2^2 \delta) = P(B_{11} + x_2^2 \delta)$ Answer A correctly shows both relationships: the general form $RT \ln \phi_1$ and the specific result $P(B_{11} + x_2^2 \delta)$ . Answer B incorrectly includes an extra factor of $P/RT$ , giving $P^2/(RT)$ instead of just $P$ . Answer C omits the crucial $RT$ factor entirely. Answer D incorrectly writes $P/T$ instead of $P$ , missing the gas constant $R$ . Remember: when comparing real vs. ideal gas chemical potentials, the key relationship is always $\mu^{real} - \mu^{ideal} = RT \ln \phi$ , where the fugacity coefficient captures all the non-ideal behavior.
The Gibbs-Duhem equation for a binary system states that $n_1 d\mu_1 + n_2 d\mu_2 = 0$ at constant temperature and pressure. In an experiment, the chemical potential of component 1 increases by $d\mu_1 = +150$ J/mol when the system composition changes slightly. If $n_1 = 0.75$ mol and $n_2 = 1.25$ mol, what constraint does this place on the chemical potential change of component 2?
Answer: $d\mu_2 = -150 \times \frac{0.75}{1.25} = -90$ J/mol, maintaining thermodynamic equilibrium through the Gibbs-Duhem relationship The Gibbs-Duhem equation is a fundamental thermodynamic relationship that constrains how chemical potentials can change in a mixture. When you encounter this equation, remember that it enforces a balance: changes in chemical potentials of different components cannot be independent—they must be coupled in a specific way. To find $d\mu_2$ , you simply rearrange the Gibbs-Duhem equation: $n_1 d\mu_1 + n_2 d\mu_2 = 0$ , which gives $d\mu_2 = -\frac{n_1}{n_2} d\mu_1$ . Substituting the values: $d\mu_2 = -\frac{0.75}{1.25} \times 150 = -90$ J/mol. This confirms answer B is correct. Answer A incorrectly inverts the mole fraction ratio. The constraint depends on $\frac{n_1}{n_2}$ , not $\frac{n_2}{n_1}$ . This gives the wrong magnitude and represents a common algebraic error when rearranging the equation. Answer C assumes the changes must be equal and opposite, ignoring the crucial role of the mole numbers. The Gibbs-Duhem equation weighs the changes by the amount of each component present—equal changes would only occur if $n_1 = n_2$ . Answer D suggests you need additional information about activity coefficients. However, the Gibbs-Duhem equation is exact and applies regardless of solution non-ideality. The constraint it imposes depends only on composition (mole numbers) and the observed change in $\mu_1$ . Remember: The Gibbs-Duhem equation provides a direct constraint that's always calculable from mole numbers alone—no additional thermodynamic data required.
At phase equilibrium between two phases (α and β) containing the same components, the chemical potentials must be equal: $\mu_i^{\alpha} = \mu_i^{\beta}$ for each component i. For a binary system where component 1 has $\mu_1^{\alpha} = -15.2$ kJ/mol and component 2 has $\mu_2^{\alpha} = -8.7$ kJ/mol in phase α, which statement correctly describes the constraints on the β phase?
Answer: The chemical potentials $\mu_1^{\beta} = -15.2$ and $\mu_2^{\beta} = -8.7$ kJ/mol determine the composition and other intensive properties of phase β through fundamental equations At equilibrium, $\mu_i^{\alpha} = \mu_i^{\beta}$ for each component. These chemical potential values, combined with the fundamental thermodynamic equations for phase β, determine its composition and other intensive properties. The chemical potentials are intensive properties that must match exactly at equilibrium, not just their differences. Choice A misses that these values constrain other properties. Choice B incorrectly suggests composition is independently determined. Choice C incorrectly focuses only on differences rather than absolute values.
The chemical potential of a component in a polymer blend can be described using the Flory-Huggins theory. For a blend of two polymers with degrees of polymerization $N_1 = 500$ and $N_2 = 1000$ , and interaction parameter $\chi_{12} = 0.02$ , the chemical potential of polymer 1 is $\mu_1 = \mu_1^\circ + RT[\ln \phi_1 + (1 - \frac{N_1}{N_2})\phi_2 + N_1 \chi_{12} \phi_2^2]$ . If $\phi_1 = 0.4$ , what is the excess chemical potential (deviation from ideal mixing) of polymer 1?
Answer: $\mu_1^{excess} = RT[(1 - 0.5)(0.6) + 500(0.02)(0.36)] = RT(0.3 + 3.6) = 3.9 RT$ When you encounter Flory-Huggins theory problems, you're dealing with polymer solution thermodynamics where the excess chemical potential represents deviations from ideal mixing behavior. The key is identifying which terms in the chemical potential expression represent non-ideal contributions. The given chemical potential expression has three terms after $\mu_1°$ : $\ln \phi_1$ (configurational entropy), $(1 - \frac{N_1}{N_2})\phi_2$ (size asymmetry effect), and $N_1 \chi_{12} \phi_2^2$ (interaction energy). For excess chemical potential, you exclude the ideal entropy term $\ln \phi_1$ and keep only the non-ideal contributions. With $N_1 = 500$ , $N_2 = 1000$ , $\chi_{12} = 0.02$ , and $\phi_1 = 0.4$ (so $\phi_2 = 0.6$ ): $\mu_1^{excess} = RT[(1 - \frac{500}{1000})(0.6) + 500(0.02)(0.6)^2]$ $= RT[(0.5)(0.6) + 500(0.02)(0.36)]$ $= RT[0.3 + 3.6] = 3.9RT$ Choice A correctly includes both non-ideal terms. Choice B omits the size asymmetry term, capturing only the interaction contribution. Choice C includes only the size effect while ignoring interactions. Choice D incorrectly includes the ideal entropy term $\ln \phi_1$ , which isn't part of the excess quantity. Remember: "excess" properties always exclude ideal mixing contributions. In Flory-Huggins theory, the $\ln \phi_i$ terms represent ideal configurational entropy, while size asymmetry and interaction terms represent deviations from ideality.
For a component in a multicomponent system, the chemical potential can be written as $\mu_i = \left(\frac{\partial G}{\partial n_i}\right)_{T,P,n_{j \neq i}}$ . In a system where the total Gibbs energy is $G = n_1(-50.0) + n_2(-30.0) + n_1 n_2(8.0) + n_1^2(2.0) + n_2^2(1.5)$ kJ/mol, what is the chemical potential of component 1 when $n_1 = 2.0$ mol and $n_2 = 1.5$ mol?
Answer: $\mu_1 = -50.0 + 1.5(8.0) + 4.0(2.0) = -30.0$ kJ/mol, applying the partial derivative correctly to all terms Taking the partial derivative: $\mu_1 = \frac{\partial G}{\partial n_1} = -50.0 + n_2(8.0) + 2n_1(2.0) + 0 = -50.0 + 8.0n_2 + 4.0n_1$ . Substituting values: $\mu_1 = -50.0 + 8.0(1.5) + 4.0(2.0) = -50.0 + 12.0 + 8.0 = -30.0$ kJ/mol. Choice A incorrectly calculates the derivative of the $n_1^2$ term. Choice C uses wrong coefficients. Choice D incorrectly treats the quadratic term.
The chemical potential of water in a biological cell can be expressed as $\mu_w = \mu_w^* + RT \ln a_w + V_w \Pi$ , where $V_w$ is the molar volume of water, $\Pi$ is the osmotic pressure, and $a_w$ is the water activity. For a cell in equilibrium with an external solution, both having the same temperature, if the internal osmotic pressure is 0.8 MPa and the external osmotic pressure is 0.3 MPa, and $V_w = 18 \times 10^{-6} \text{ m}^3/\text{mol}$ , what must be the ratio of water activities ( $a_w^{internal}/a_w^{external}$ ) for equilibrium?
Answer: $\frac{a_w^{internal}}{a_w^{external}} = \exp\left(\frac{-0.5 \times 10^6 \times 18 \times 10^{-6}}{RT}\right) = \exp(-3.63)$ When dealing with chemical potential equilibrium across cell membranes, you need to understand that at equilibrium, the chemical potentials must be equal on both sides. This means $\mu_w^{internal} = \mu_w^{external}$ . Setting up the equilibrium condition using the given equation: $\mu_w^* + RT \ln a_w^{internal} + V_w \Pi_{internal} = \mu_w^* + RT \ln a_w^{external} + V_w \Pi_{external}$ The standard chemical potential terms cancel out, leaving: $RT \ln a_w^{internal} + V_w \Pi_{internal} = RT \ln a_w^{external} + V_w \Pi_{external}$ Rearranging to isolate the activity ratio: $RT \ln\left(\frac{a_w^{internal}}{a_w^{external}}\right) = V_w(\Pi_{external} - \Pi_{internal})$ Substituting the values: $\Pi_{external} - \Pi_{internal} = 0.3 - 0.8 = -0.5$ MPa $\ln\left(\frac{a_w^{internal}}{a_w^{external}}\right) = \frac{V_w(-0.5 \times 10^6)}{RT} = \frac{-0.5 \times 10^6 \times 18 \times 10^{-6}}{RT}$ Therefore: $\frac{a_w^{internal}}{a_w^{external}} = \exp(-3.63)$ , which is answer B. Answer A uses the wrong sign (positive instead of negative). Answer C incorrectly uses 0.8 MPa instead of the pressure difference, and with the wrong sign. Answer D uses 0.8 MPa instead of the pressure difference but has the correct negative sign. Remember: in equilibrium problems, always set chemical potentials equal and focus on the difference between conditions, not absolute values. The sign matters crucially for determining which direction the equilibrium favors.
The Gibbs-Duhem equation relates changes in chemical potentials in a multicomponent system: $\sum_{i} n_i d\mu_i = 0$ at constant temperature and pressure. For a binary solution where component 1 follows Raoult's law ( $\mu_1 = \mu_1^* + RT \ln x_1$ ) and component 2 shows positive deviations with $\ln \gamma_2 = \beta x_1^2$ , what constraint does the Gibbs-Duhem equation impose on the parameter $\beta$ ?
Answer: $\beta$ must satisfy $\ln \gamma_1 = \beta x_2^2$ to maintain symmetry The Gibbs-Duhem equation is a fundamental constraint that ensures thermodynamic consistency in multicomponent systems. When you encounter problems involving activity coefficients in binary solutions, this equation forces the behavior of both components to be mathematically linked—you can't arbitrarily specify the deviation from ideality for just one component. Starting with the Gibbs-Duhem equation at constant T and P: $x_1 d\ln a_1 + x_2 d\ln a_2 = 0$ . Since component 1 follows Raoult's law, $a_1 = x_1$ and $\gamma_1 = 1$ initially. For component 2, $\ln \gamma_2 = \beta x_1^2$ . To find the constraint, we need $d\ln \gamma_2 = 2\beta x_1 dx_1$ . Since $dx_1 = -dx_2$ in a binary system, this becomes $d\ln \gamma_2 = -2\beta x_1 dx_2$ . Substituting into Gibbs-Duhem: $x_1 d\ln \gamma_1 + x_2(-2\beta x_1) dx_2/dx_2 = 0$ , which gives us $x_1 d\ln \gamma_1 = 2\beta x_1 x_2 dx_2$ . Integrating and using $x_2 = 1-x_1$ , we find that $\ln \gamma_1 = \beta x_2^2$ , confirming answer C. Option A provides a specific numerical relationship that isn't derived from thermodynamic constraints. Option B incorrectly suggests $\beta$ can be arbitrary—the Gibbs-Duhem equation prevents this. Option D states the general differential relationship but doesn't identify the specific constraint on $\beta$ . Remember: In binary solution problems, always check that proposed activity coefficient expressions satisfy the Gibbs-Duhem equation—both components' deviations from ideality must be thermodynamically consistent.
The chemical potential of an ideal gas component in a mixture is $\mu_i = \mu_i^\circ(T) + RT \ln\left(\frac{P_i}{P^\circ}\right)$ , where $P_i$ is the partial pressure and $P^\circ = 1$ bar. For a gas mixture at 400 K where component A has $\mu_A = -25.6$ kJ/mol and $P_A = 0.50$ bar, what would be the chemical potential of pure component A at the same temperature and 2.0 bar pressure?
Answer: $\mu_A^{pure} = \mu_A^\circ + (8.314 \times 10^{-3})(400)\ln(2.0)$ where $\mu_A^\circ = -25.6 + 2.31 = -23.3$ kJ/mol, giving $\mu_A^{pure} = -21.0$ kJ/mol First, find $\mu_A^\circ$ from the mixture data: $-25.6 = \mu_A^\circ + RT\ln(0.50/1.0)$ , so $\mu_A^\circ = -25.6 - (3.326)(-0.693) = -23.3$ kJ/mol. Then for pure A at 2.0 bar: $\mu_A^{pure} = -23.3 + (3.326)\ln(2.0) = -21.0$ kJ/mol. Choice A skips finding the standard chemical potential. Choice B incorrectly assumes the reference state changes. Choice D incorrectly applies direct proportionality.
For a component undergoing a chemical reaction, its chemical potential appears in the reaction Gibbs energy: $\Delta_r G = \sum \nu_i \mu_i$ . Consider the reaction A + B ⇌ C where the chemical potentials are $\mu_A = \mu_A^\circ - 2000 \text{ J/mol}$ , $\mu_B = \mu_B^\circ + 1500 \text{ J/mol}$ , and $\mu_C = \mu_C^\circ + 800 \text{ J/mol}$ . If the standard reaction Gibbs energy is $\Delta_r G^\circ = -5000 \text{ J/mol}$ , what is the driving force for the forward reaction?
Answer: $\Delta_r G = -5000 + 800 - (-2000) - 1500 = -3700 \text{ J/mol}$ When you encounter chemical potential problems involving reaction Gibbs energy, you need to carefully apply the formula $\Delta_r G = \sum \nu_i \mu_i$ where $\nu_i$ represents the stoichiometric coefficients (positive for products, negative for reactants). For the reaction A + B ⇌ C, the stoichiometric coefficients are: $\nu_A = -1$ , $\nu_B = -1$ , and $\nu_C = +1$ . The reaction Gibbs energy becomes: $\Delta_r G = (-1)\mu_A + (-1)\mu_B + (+1)\mu_C$ Substituting the given chemical potentials: $\Delta_r G = -(\mu_A^\circ - 2000) - (\mu_B^\circ + 1500) + (\mu_C^\circ + 800)$ This simplifies to: $\Delta_r G = (\mu_C^\circ - \mu_A^\circ - \mu_B^\circ) + (800 - (-2000) - 1500)$ Since $\Delta_r G^\circ = \mu_C^\circ - \mu_A^\circ - \mu_B^\circ = -5000$ J/mol: $\Delta_r G = -5000 + 800 + 2000 - 1500 = -3700$ J/mol Answer A correctly applies the stoichiometric coefficients and handles the signs properly. Answer B incorrectly adds all the deviation terms without considering stoichiometry. Answer C makes the same error as B, treating all deviations as positive contributions. Answer D incorrectly subtracts the reactant deviations when they should be added due to the negative stoichiometric coefficients. Remember: always assign negative stoichiometric coefficients to reactants and positive to products, then carefully track the algebra when substituting chemical potential expressions. The signs of the deviation terms matter!
The chemical potential can be used to derive colligative property relationships. For a dilute solution where the solvent obeys Raoult's law, $\mu_1 = \mu_1^* + RT \ln x_1$ , and the vapor pressure is $P_1 = x_1 P_1^*$ . If the freezing point depression is given by $\Delta T_f = K_f m_2$ where $K_f$ is the cryoscopic constant and $m_2$ is molality, and we know that $K_f = \frac{RT_{f,1}^{*2} M_1}{1000 \Delta_{fus} H_1}$ where $M_1$ is the molar mass of solvent in g/mol, what is the relationship between the chemical potential change and the freezing point depression for small $m_2$ ?
Answer: $\Delta \mu_1 = -\frac{\Delta_{fus} H_1 \Delta T_f}{T_{f,1}^*}$ This question tests your understanding of how chemical potential relates to colligative properties, specifically freezing point depression. The key insight is recognizing that at equilibrium, the chemical potential of the solvent in the liquid phase must equal that in the solid phase. At the freezing point, we have equilibrium between liquid and solid solvent: $\mu_1^{liquid} = \mu_1^{solid}$ . For the pure solvent at its normal freezing point $T_{f,1}^*$ , this gives us $\mu_1^* = \mu_1^{solid}$ . When solute is added, the chemical potential of the liquid solvent decreases by $\Delta \mu_1 = RT \ln x_1$ , where $x_1 < 1$ . To restore equilibrium at the new (lower) freezing point, we use the Gibbs-Helmholtz relationship. The change in chemical potential with temperature is $\frac{d\mu}{dT} = -S$ , and since $\Delta S_{fus} = \frac{\Delta_{fus} H_1}{T_{f,1}^*}$ , we get: $\Delta \mu_1 = -\Delta S_{fus} \Delta T_f = -\frac{\Delta_{fus} H_1 \Delta T_f}{T_{f,1}^*}$ This confirms answer A is correct. Answer B has $T_{f,1}^{*2}$ in the denominator, which would come from incorrectly applying the $K_f$ expression directly. Answer C starts correctly but incorrectly equates the logarithmic and linear forms. Answer D inappropriately includes the conversion factors from the $K_f$ expression (1000 and $M_1$ ) that don't belong in the fundamental thermodynamic relationship. Study tip: Remember that chemical potential changes are fundamentally linked to entropy changes via temperature derivatives. The $K_f$ expression is useful for calculations but the core thermodynamic relationship comes from equilibrium conditions.
The chemical potential of water in an aqueous solution can be expressed as $\mu_{H_2O} = \mu_{H_2O}^* + RT \ln(a_{H_2O})$ , where $a_{H_2O}$ is the activity of water. If the partial molar volume of water in the solution is $\bar{V}_{H_2O} = 17.8$ mL/mol and the system is compressed at constant temperature and composition, how does the chemical potential change with pressure?
Answer: $\frac{d\mu_{H_2O}}{dP} = \bar{V}_{H_2O} = 17.8 \times 10^{-6}$ m³/mol, representing the direct pressure dependence of the partial molar Gibbs energy The fundamental thermodynamic relationship for chemical potential gives $\left(\frac{\partial \mu_i}{\partial P}\right)_{T,n} = \bar{V}_i$ . This is the definition of partial molar volume and applies regardless of solution ideality. At constant T and composition, $d\mu_{H_2O} = \bar{V}_{H_2O} dP$ . Choice B incorrectly suggests this only applies to ideal solutions. Choice C incorrectly adds an activity term that's already accounted for in the partial molar volume. Choice D incorrectly limits this to standard state conditions.
Consider a ternary system with components A, B, and C at equilibrium. The chemical potential of component B changes according to $d\mu_B = -\bar{S}_B dT + \bar{V}_B dP + \left(\frac{\partial \mu_B}{\partial n_A}\right)_{T,P,n_C} dn_A + \left(\frac{\partial \mu_B}{\partial n_C}\right)_{T,P,n_A} dn_C$ . If the system undergoes a process where $dT = 0$ , $dP = 0$ , but the amounts of A and C change while maintaining constant total moles, what determines the sign of $d\mu_B$ ?
Answer: The relative magnitudes of $\left(\frac{\partial \mu_B}{\partial n_A}\right)_{T,P,n_C}$ and $\left(\frac{\partial \mu_B}{\partial n_C}\right)_{T,P,n_A}$ along with the constraint $dn_A + dn_C = 0$ At constant T, P, and total moles, we have $dn_A + dn_B + dn_C = 0$ . Since $n_B$ is constant (we're looking at changes in $\mu_B$ ), we get $dn_A + dn_C = 0$ , so $dn_C = -dn_A$ . Thus: $d\mu_B = \left(\frac{\partial \mu_B}{\partial n_A}\right)_{T,P,n_C} dn_A - \left(\frac{\partial \mu_B}{\partial n_C}\right)_{T,P,n_A} dn_A$ . The sign depends on the relative magnitudes of these cross-derivatives. Choice B ignores the C component contribution. Choice C makes an unfounded generalization. Choice D misapplies the Gibbs-Duhem equation.

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