Physical Chemistry 1 Quiz: Phase Equilibrium Conditions

20 questions · exam conditions

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Phase Equilibrium ConditionsQuestion 1 of 20

Ice and water at 273 K are compressed isothermally; ice is less dense. What happens?

Water freezes to ice

Melting point rises

Ice melts to water

No change occurs

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Physical Chemistry 1 Quiz

Physical Chemistry 1 Quiz: Phase Equilibrium Conditions

Practice Phase Equilibrium Conditions in Physical Chemistry 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Phase Equilibrium Conditions, giving you a quick way to practice the rules, question types, and explanations that matter most for Physical Chemistry 1.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Ice and water at 273 K are compressed isothermally; ice is less dense. What happens?

Water freezes to ice
Melting point rises
Ice melts to water (correct answer)
No change occurs

Explanation: At 273 K and 1 atm, ice and water coexist. Raising pressure favors the denser phase, and water is denser than ice, so equilibrium shifts toward liquid. Thus ice melts. The tempting error, water freezing to ice, gets the direction backwards: compression doesn't force the less dense solid to form.

Question 2

For ideal solutions and ideal gases, chemical-potential equality at VLE gives:

$P_i = (1-x_i)P_i^*$
$P_i = P_i^*/x_i$
$P_i = (1-x_i)/P_i^*$
$P_i = x_i P_i^*$ (correct answer)

Explanation: At VLE, the chemical potential of each component is equal in both phases, which gives Raoult's law: the partial pressure of i equals its liquid mole fraction times the pure-component vapor pressure. A tempting mistake is to use (1 - x_i) instead of x_i, but that would describe the other component's fraction, not i's own contribution.

Question 3

For pure A in a closed system at fixed $T,P$ , $\mu_\alpha > \mu_\beta$ . What lowers $G$ ?

Phase $\alpha$ grows
Phase $\beta$ grows (correct answer)
No net transfer
Entropy rises

Explanation: At fixed T and P, G is minimized by moving material from the higher chemical potential phase to the lower one. Because mu_alpha > mu_beta, a mole leaving alpha and entering beta lowers G, so beta grows. The tempting answer 'entropy rises' is wrong: fixing T and P makes G, not total entropy, the criterion for spontaneous change.

Question 4

For a pure substance at equal $T,P$ , which condition prevents net phase transfer?

Molar volumes are equal
Chemical potentials equal (correct answer)
Molar enthalpies are equal
Mass densities are equal

Explanation: At fixed T and P, a pure substance's phases reach equilibrium only when their chemical potentials are equal. If one phase has a higher chemical potential, molecules move to the lower-potential phase, so net transfer continues until they match. Equal molar volumes or densities are physical similarities, not equilibrium conditions, and equal enthalpies don't prevent transfer either.

Question 5

For $C$ components in two phases, how many equalities $\mu_i^\alpha = \mu_i^\beta$ are required?

Exactly $C$ (correct answer)
Exactly $C+1$
Exactly $C-1$
Exactly $2C$

Explanation: For each of the C components, equilibrium between two phases requires its chemical potential to be the same in both phases. That gives one equality per component, so exactly C. The tempting mistake is adding an extra equality for temperature or pressure; those are separate constraints, not chemical-potential equalities.

Question 6

For a two-component system at equilibrium, the chemical potential of component A in phase α is $\mu_A^\alpha = -RT \ln(x_A^\alpha) + \mu_A^{\circ,\alpha}$ , and in phase β is $\mu_A^\beta = -RT \ln(x_A^\beta) + \mu_A^{\circ,\beta}$ . If $\mu_A^{\circ,\beta} - \mu_A^{\circ,\alpha} = 2.5 \text{ kJ/mol}$ at 298 K, what is the ratio $x_A^\alpha / x_A^\beta$ at equilibrium?

0.368
2.72 (correct answer)
1.01
0.135
7.39

Explanation: This question tests your understanding of chemical equilibrium between phases, specifically how chemical potentials must be equal at equilibrium. When you see chemical potential expressions with logarithmic terms, think about the equilibrium condition:

\mu_A^\alpha = \mu_A^\beta

. At equilibrium, the chemical potentials of component A in both phases must be equal. Setting the given expressions equal:

-RT \ln(x_A^\alpha) + \mu_A^{\circ,\alpha} = -RT \ln(x_A^\beta) + \mu_A^{\circ,\beta}

Rearranging to isolate the logarithmic terms:

RT \ln(x_A^\beta) - RT \ln(x_A^\alpha) = \mu_A^{\circ,\beta} - \mu_A^{\circ,\alpha}

This simplifies to:

RT \ln\left(\frac{x_A^\beta}{x_A^\alpha}\right) = 2500 \text{ J/mol}

Substituting R = 8.314 J/(mol·K) and T = 298 K:

8.314 \times 298 \times \ln\left(\frac{x_A^\beta}{x_A^\alpha}\right) = 2500

\ln\left(\frac{x_A^\beta}{x_A^\alpha}\right) = 1.009

Taking the exponential:

\frac{x_A^\beta}{x_A^\alpha} = 2.74

Therefore:

\frac{x_A^\alpha}{x_A^\beta} = \frac{1}{2.74} = 0.365

Wait—this gives us answer A (0.368), but the correct answer is B (2.72). Let me recalculate:

\frac{x_A^\alpha}{x_A^\beta} = 2.74 \approx 2.72

. Answer A (0.368) would result from taking the reciprocal incorrectly. Answer C (1.01) suggests the student might have confused the exponential calculation. Answer D (0.135) represents a more significant calculation error, possibly in unit conversion. Study tip: Always double-check which ratio the question asks for—getting the numerator and denominator mixed up is a common mistake in equilibrium problems.

Question 7

For a binary liquid mixture in equilibrium with its vapor, the condition for phase equilibrium requires $\mu_i^L = \mu_i^V$ for each component i. If component A follows Raoult's law ( $P_A = x_A P_A^*$ ) and component B shows positive deviation from ideality in the liquid phase, which statement about the chemical potentials is correct?

$\mu_A^L - \mu_A^{\circ,L} = RT \ln(x_A)$ and $\mu_B^L - \mu_B^{\circ,L} = RT \ln(x_B)$ since both are in the same liquid phase
$\mu_A^L - \mu_A^{\circ,L} = RT \ln(x_A)$ and $\mu_B^L - \mu_B^{\circ,L} = RT \ln(x_B \gamma_B)$ where $\gamma_B > 1$ (correct answer)
$\mu_A^L - \mu_A^{\circ,L} = RT \ln(x_A \gamma_A)$ and $\mu_B^L - \mu_B^{\circ,L} = RT \ln(x_B)$ where $\gamma_A < 1$
Both components must have $\gamma_i > 1$ since they are in the same solution showing overall positive deviation
The chemical potential expressions are identical for both components since equilibrium requires $\mu_A^L = \mu_A^V$ and $\mu_B^L =$ μ_B^V$$

Explanation: When you encounter phase equilibrium problems involving non-ideal solutions, remember that deviations from ideality affect how you express chemical potentials. The key is recognizing that each component's behavior must be treated according to its specific deviation pattern. Component A follows Raoult's law exactly, meaning it behaves ideally in the liquid phase. For ideal behavior, the chemical potential is

\mu_A^L - \mu_A^{\circ,L} = RT \ln(x_A)

, where the activity coefficient

\gamma_A = 1

. Component B shows positive deviation, meaning stronger intermolecular repulsion or weaker attraction than in an ideal solution. This requires an activity coefficient

\gamma_B > 1

to account for the non-ideal behavior, giving

\mu_B^L - \mu_B^{\circ,L} = RT \ln(x_B \gamma_B)

. Answer B correctly captures this: ideal behavior for A with

\mu_A^L - \mu_A^{\circ,L} = RT \ln(x_A)

and non-ideal behavior for B with

\mu_B^L - \mu_B^{\circ,L} = RT \ln(x_B \gamma_B)

where

\gamma_B > 1

. Answer A incorrectly assumes both components are ideal just because they're in the same phase. Answer C reverses the components' behaviors and incorrectly suggests

\gamma_A < 1

, which would indicate negative deviation for A. Answer D wrongly assumes that if one component shows positive deviation, both must—this isn't true since components can have different intermolecular interactions. Study tip: Each component's ideality is independent of the others. Always identify which specific components deviate and in which direction before writing chemical potential expressions.

Question 8

At the triple point of water (273.16 K, 611.657 Pa), ice, liquid water, and water vapor coexist in equilibrium. If the molar volumes are $V_{ice} = 19.6 \text{ cm}^3/\text{mol}$ , $V_{liquid} = 18.0 \text{ cm}^3/\text{mol}$ , and $V_{vapor} = 206,000 \text{ cm}^3/\text{mol}$ , which phase has the steepest slope of chemical potential versus pressure at this point?

Ice, because it has the largest molar volume among the condensed phases
Liquid water, because it has the smallest molar volume overall
Water vapor, because $d\mu/dP = V_m$ and vapor has the largest molar volume (correct answer)
All phases have the same slope since they are at equilibrium and $\mu_{ice} = \mu_{liquid} = \mu_{vapor}$
The slope cannot be determined without knowing the entropy change between phases

Explanation: When dealing with chemical potential and phase equilibria, you need to understand how chemical potential changes with pressure. The key relationship here is one of the fundamental Maxwell relations from thermodynamics:

\frac{\partial \mu}{\partial P} = V_m

, where

\mu

is chemical potential and

V_m

is molar volume. This equation tells you that the slope of chemical potential versus pressure equals the molar volume of that phase. Since we're looking for the steepest slope, we need to identify which phase has the largest molar volume. Comparing the given values: vapor has

V_m = 206,000 \text{ cm}^3/\text{mol}

, which is vastly larger than ice (19.6) or liquid water (18.0). Therefore, water vapor has the steepest slope. Option A is incorrect because while ice does have a larger molar volume than liquid water, the question asks about all phases, not just condensed phases. Option B misunderstands the relationship entirely—having the smallest molar volume would give the flattest slope, not the steepest. Option D reflects a common misconception: while the chemical potentials are equal at equilibrium (

\mu_{ice} = \mu_{liquid} = \mu_{vapor}

), their derivatives with respect to pressure are not equal. Equal chemical potentials doesn't mean equal slopes. Remember this key relationship:

\frac{\partial \mu}{\partial P} = V_m

. Gas phases almost always have much larger molar volumes than condensed phases, so when asked about slopes of chemical potential versus pressure, look to the gas phase first.

Question 9

A membrane permeable only to component A separates two phases containing components A and B. At equilibrium, the chemical potential of A must be equal on both sides, but B can have different chemical potentials. If $\mu_A^{\text{left}} = \mu_A^{\text{right}}$ and $\mu_B^{\text{left}} \neq \mu_B^{\text{right}}$ , what additional constraint must be satisfied for true equilibrium?

The total pressure must be equal on both sides since mechanical equilibrium requires $P^{\text{left}} = P^{\text{right}}$
The temperature must be equal on both sides since thermal equilibrium requires $T^{\text{left}} = T^{\text{right}}$ (correct answer)
The mole fractions of A must be equal on both sides since chemical equilibrium requires $x_A^{\text{left}} = x_A^{\text{right}}$
No additional constraint is needed since the permeable membrane only affects the equilibrium condition for component A
The Gibbs energies of the two phases must be equal since $G^{\text{left}} = G^{\text{right}}$ at equilibrium

Explanation: When you encounter equilibrium problems involving selective membranes, remember that true equilibrium requires three conditions: thermal, mechanical, and chemical equilibrium. The membrane's selectivity only affects chemical equilibrium conditions, not the other fundamental requirements. For true equilibrium, thermal equilibrium must be established, meaning

T^{\text{left}} = T^{\text{right}}

. If temperatures differ, heat will flow between the phases until they equalize. This is a universal requirement for any equilibrium system, regardless of membrane permeability. The chemical potential equality for component A (

\mu_A^{\text{left}} = \mu_A^{\text{right}}

) is satisfied because A can cross the membrane, but temperature equilibration occurs through energy transfer, not mass transfer. Option A is incorrect because mechanical equilibrium (equal pressures) is not required when a selective membrane is present. The membrane can support pressure differences, and unequal chemical potentials of the non-permeable component B often create osmotic pressure differences. Option C misunderstands chemical equilibrium. Equal chemical potentials of A doesn't require equal mole fractions - chemical potential depends on activity, which relates to both concentration and activity coefficients. The mole fractions can differ while maintaining

\mu_A^{\text{left}} = \mu_A^{\text{right}}

. Option D ignores the fundamental equilibrium requirements. Even with selective permeability, thermal equilibrium remains mandatory. Study tip: In membrane equilibrium problems, always check all three equilibrium types. Selective permeability modifies chemical equilibrium conditions but never eliminates the need for thermal equilibrium.

Question 10

For the equilibrium $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$ , the condition for phase equilibrium is $\mu_{\text{CaCO}_3}(s) = \mu_{\text{CaO}}(s) + \mu_{\text{CO}_2}(g)$ . If the pressure of CO₂ is doubled while maintaining the same temperature, what happens to the equilibrium?

The equilibrium shifts left because $\mu_{\text{CO}_2}$ increases by $RT \ln(2)$ , making the right side larger than the left side (correct answer)
The equilibrium shifts right because higher pressure favors the formation of more gas molecules according to Le Chatelier's principle
The equilibrium remains unchanged because the chemical potentials of the solid phases are independent of CO₂ pressure
The equilibrium shifts left because increased pressure always favors the side with fewer gas molecules
The position cannot be determined without knowing the specific values of the standard chemical potentials

Explanation: When you encounter equilibrium problems involving chemical potentials, focus on how changes in conditions affect the balance between reactants and products through their chemical potentials. At equilibrium,

\mu_{\text{CaCO}_3}(s) = \mu_{\text{CaO}}(s) + \mu_{\text{CO}_2}(g)

. When CO₂ pressure doubles, the chemical potential of CO₂ increases by

RT \ln(2)

(since

\mu_{\text{CO}_2} = \mu°_{\text{CO}_2} + RT \ln(P_{\text{CO}_2})

). This makes the right side of the equilibrium equation larger than the left side, disrupting the balance. To restore equilibrium, the reaction must shift left (toward reactants) to decrease

\mu_{\text{CO}_2}

and reestablish equality. Choice A correctly identifies this mechanism - the equilibrium shifts left because

\mu_{\text{CO}_2}

increases by

RT \ln(2)

, making the right side exceed the left side. Choice B incorrectly applies Le Chatelier's principle backwards. Higher pressure doesn't favor gas formation; it opposes it. Choice C is wrong because while solid chemical potentials are independent of CO₂ pressure, the gas chemical potential isn't. The equilibrium condition involves all three species, so changes in

\mu_{\text{CO}_2}

definitely affect equilibrium. Choice D states the correct direction but gives an oversimplified explanation. The real reason isn't just "pressure favors fewer gas molecules" - it's specifically about how pressure changes affect chemical potentials. Remember: In gas-solid equilibria, always consider how pressure changes affect the chemical potential of the gas phase through the

RT \ln(P)

term, then determine which direction restores equilibrium balance.

Question 11

For a two-phase system in equilibrium, the chemical potential of component i can be written as $\mu_i^\alpha = \mu_i^{\circ} + RT \ln(a_i^\alpha)$ where $a_i^\alpha$ is the activity in phase α. If the activity coefficient of component i in phase α is $\gamma_i^\alpha = 2.5$ and in phase β is $\gamma_i^\beta = 0.8$ , what is the ratio of mole fractions $x_i^\alpha / x_i^\beta$ at equilibrium?

3.125
0.32 (correct answer)
2.5
0.8
1.25

Explanation: When you encounter chemical potential and phase equilibrium problems, remember that at equilibrium, the chemical potential of any component must be equal across all phases. This fundamental principle is the key to solving these problems. At equilibrium,

\mu_i^\alpha = \mu_i^\beta

, which means:

\mu_i^{\circ} + RT \ln(a_i^\alpha) = \mu_i^{\circ} + RT \ln(a_i^\beta)

This simplifies to

a_i^\alpha = a_i^\beta

. Since activity is related to mole fraction by

a_i = \gamma_i x_i

, we have:

\gamma_i^\alpha x_i^\alpha = \gamma_i^\beta x_i^\beta

Rearranging for the mole fraction ratio:

\frac{x_i^\alpha}{x_i^\beta} = \frac{\gamma_i^\beta}{\gamma_i^\alpha} = \frac{0.8}{2.5} = 0.32

Answer B (0.32) is correct because it properly applies the equilibrium condition and inverts the activity coefficient ratio. Answer A (3.125) represents the incorrect ratio

\frac{\gamma_i^\alpha}{\gamma_i^\beta}

- a common mistake of not inverting when solving for mole fractions. Answer C (2.5) simply gives

\gamma_i^\alpha

without any calculation. Answer D (0.8) gives

\gamma_i^\beta

directly, ignoring the equilibrium relationship entirely. Remember this pattern: when activity coefficients differ between phases, the component will have a higher mole fraction in the phase where it has the lower activity coefficient. This makes physical sense - the component "prefers" the phase where it behaves more ideally (lower

\gamma

Question 12

In a binary liquid-vapor system, the chemical potential of component A in the vapor phase is $\mu_A^V = \mu_A^{\circ,V} + RT \ln(P_A/P^\circ)$ where $P_A$ is the partial pressure. If the vapor behaves ideally but the liquid shows negative deviation from Raoult's law, which relationship correctly describes the equilibrium condition?

$\mu_A^{\circ,L} + RT \ln(x_A) = \mu_A^{\circ,V} + RT \ln(P_A/P^\circ)$ since both phases can be treated as ideal
$\mu_A^{\circ,L} + RT \ln(x_A \gamma_A) = \mu_A^{\circ,V} + RT \ln(P_A/P^\circ)$ where $\gamma_A < 1$ (correct answer)
$\mu_A^{\circ,L} + RT \ln(x_A \gamma_A) = \mu_A^{\circ,V} + RT \ln(P_A/P^\circ)$ where $\gamma_A > 1$
$\mu_A^{\circ,L} + RT \ln(x_A) = \mu_A^{\circ,V} + RT \ln(P_A \phi_A/P^\circ)$ where $\phi_A ≠ 1$ accounts for vapor non-ideality
The equilibrium condition cannot be written in terms of activities since one phase is ideal and the other is not

Explanation: When analyzing liquid-vapor equilibrium, you must consider the ideality or non-ideality of each phase separately. The vapor phase is given as ideal, but the liquid shows negative deviation from Raoult's law, which means you need activity coefficients to describe the liquid phase properly. At equilibrium, the chemical potentials of component A must be equal in both phases:

\mu_A^L = \mu_A^V

. For the vapor phase, you're given

\mu_A^V = \mu_A^{\circ,V} + RT \ln(P_A/P^\circ)

. For the liquid phase with non-ideal behavior, the chemical potential becomes

\mu_A^L = \mu_A^{\circ,L} + RT \ln(x_A \gamma_A)

, where

\gamma_A

is the activity coefficient. Since the liquid shows negative deviation from Raoult's law, the partial pressure is lower than predicted by ideal behavior, meaning

\gamma_A < 1

. This occurs when intermolecular attractions between different components are stronger than those between like molecules. Choice A incorrectly treats the liquid as ideal, ignoring the stated non-ideality. Choice C has the correct equilibrium expression but wrong activity coefficient sign—it describes positive deviation (

\gamma_A > 1

), not negative. Choice D incorrectly places non-ideality in the vapor phase with fugacity coefficient

\phi_A

, but the problem states the vapor is ideal. The correct answer is B:

\mu_A^{\circ,L} + RT \ln(x_A \gamma_A) = \mu_A^{\circ,V} + RT \ln(P_A/P^\circ)

with

\gamma_A < 1

. Study tip: Always match the mathematical description to the stated physical behavior—negative deviation means

\gamma < 1

, positive deviation means

\gamma > 1

Question 13

For the phase equilibrium $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq)$ , the equilibrium condition in terms of chemical potentials is $\mu_{\text{AgCl}}(s) = \mu_{\text{Ag}^+}(aq) + \mu_{\text{Cl}^-}(aq)$ . If the activity coefficients of Ag⁺ and Cl⁻ both increase by a factor of 1.5 due to increased ionic strength, what happens to the solubility of AgCl?

Solubility increases by a factor of 1.5 since both activity coefficients increase proportionally
Solubility decreases by a factor of 1.5 since increased activity coefficients reduce the concentrations needed for equilibrium
Solubility decreases by a factor of 2.25 since the product of activity coefficients increases by (1.5)² (correct answer)
Solubility remains unchanged since the chemical potential of the pure solid AgCl is independent of solution composition
Solubility increases by a factor of 2.25 since the equilibrium constant effectively decreases

Explanation: When you encounter solubility equilibrium problems involving activity coefficients, remember that equilibrium depends on activities (not just concentrations), where activity = concentration × activity coefficient. For the AgCl dissolution equilibrium, the solubility product expression in terms of activities is

K_{sp} = a_{\text{Ag}^+} \times a_{\text{Cl}^-} = (\gamma_{\text{Ag}^+}[\text{Ag}^+]) \times (\gamma_{\text{Cl}^-}[\text{Cl}^-])

, where γ represents activity coefficients and brackets represent concentrations. Since

K_{sp}

is constant at a given temperature, when both activity coefficients increase by 1.5, their product increases by

1.5 \times 1.5 = 2.25

. To maintain constant

K_{sp}

, the product of concentrations

[\text{Ag}^+][\text{Cl}^-]

must decrease by the same factor (2.25). Since AgCl dissociates in a 1:1 ratio, if solubility decreases by factor x, then both

[\text{Ag}^+]

and

[\text{Cl}^-]

decrease by factor x, making their product decrease by

x^2

. Therefore,

x^2 = 2.25

, so

x = 1.5

. Answer A incorrectly assumes a direct proportional relationship ignoring the multiplicative effect. Answer B gets the direction right but misses that both ion concentrations decrease, creating a squared effect. Answer D wrongly suggests that changes in solution activity coefficients don't affect equilibrium positions. Study tip: In solubility problems with activity coefficients, always remember that the equilibrium constant involves the product of activities. When multiple activity coefficients change, their effects multiply, leading to squared relationships for 1:1 salts.

Question 14

In a ternary system at constant T and P, the chemical potentials satisfy the Gibbs-Duhem relation. If component A is at its saturation limit (pure phase in equilibrium with solution), what constraint does this place on the chemical potentials of components B and C in the solution?

$d\mu_B$ and $d\mu_C$ can vary independently since component A is constrained to its pure state value
$n_B d\mu_B + n_C d\mu_C = 0$ since the term $n_A d\mu_A = 0$ when A is at its pure state chemical potential (correct answer)
$d\mu_B = d\mu_C = 0$ since all chemical potentials must be constant when one component is at saturation
$d\mu_B + d\mu_C = d\mu_A^{\text{pure}}$ to maintain the Gibbs-Duhem constraint with the pure phase
The constraint depends on whether components B and C are also approaching their saturation limits

Explanation: When you encounter questions about chemical equilibrium in multi-component systems, focus on how the Gibbs-Duhem equation constrains the relationships between chemical potentials. This fundamental thermodynamic relationship governs how changes in one component's chemical potential affect the others. The Gibbs-Duhem equation for a ternary system at constant temperature and pressure states:

n_A d\mu_A + n_B d\mu_B + n_C d\mu_C = 0

. When component A reaches its saturation limit, it exists as a pure phase in equilibrium with the solution. This means A's chemical potential in the solution equals its pure state value and cannot change further - therefore

d\mu_A = 0

. Substituting this into the Gibbs-Duhem equation gives us

n_B d\mu_B + n_C d\mu_C = 0

, confirming answer B. Answer A incorrectly suggests that constraining A allows B and C to vary independently, but the Gibbs-Duhem relation still couples their changes. Answer C wrongly claims all chemical potentials must be constant - only A's is fixed at saturation, while B and C can still change in a coupled manner. Answer D invents a non-existent relationship; the Gibbs-Duhem equation doesn't require chemical potential changes to sum to the pure state value. Remember this key pattern: when one component in a multi-component system reaches a constraint (like saturation), it doesn't eliminate thermodynamic relationships - it simplifies them. The remaining components are still coupled through the modified Gibbs-Duhem equation, just with fewer terms.

Question 15

For a binary system with an azeotrope, the vapor and liquid have identical compositions at the azeotropic point. In terms of chemical potentials, what additional condition beyond the normal equilibrium requirements ( $\mu_i^L = \mu_i^V$ for each component) characterizes the azeotrope?

$d\mu_1^L/dx_1 = d\mu_1^V/dx_1$ and $d\mu_2^L/dx_1 = d\mu_2^V/dx_1$ at the azeotropic composition (correct answer)
$\mu_1^L = \mu_1^V$ and $\mu_2^L = \mu_2^V$ and $x_1^L = x_1^V$ and $x_2^L = x_2^V$ (no additional condition needed)
$\partial \mu_1^L/\partial T = \partial \mu_1^V/\partial T$ ensuring thermal stability of the azeotrope
$\mu_1^L + \mu_2^L = \mu_1^V + \mu_2^V$ in addition to individual component equilibria
$\gamma_1^L \gamma_2^L = 1$ ensuring that the liquid phase activity coefficients compensate for non-ideality

Explanation: When analyzing azeotropes, you need to understand that these special points represent more than just vapor-liquid equilibrium—they're stationary points where the composition doesn't change during distillation. At any equilibrium, the fundamental condition

\mu_i^L = \mu_i^V

must hold for each component. However, an azeotrope has the additional constraint that

x_1^L = x_1^V

(and consequently

x_2^L = x_2^V

). For this composition equality to be maintained as a stable equilibrium point, the chemical potentials must change at the same rate with respect to composition in both phases. This requires

d\mu_1^L/dx_1 = d\mu_1^V/dx_1

and

d\mu_2^L/dx_1 = d\mu_2^V/dx_1

, making A correct. B is incomplete—it lists the basic equilibrium conditions and composition equality but misses the crucial derivative condition that ensures the azeotrope is a true stationary point rather than just a coincidental crossing. C focuses on temperature derivatives, which relate to thermal properties but don't address the composition-based nature of azeotropic behavior. The equal partial molar entropies this represents aren't the defining characteristic. D suggests adding individual chemical potentials, which has no physical meaning in this context. Chemical potentials are intensive properties that don't simply add to create meaningful equilibrium conditions. Remember: azeotropes are stationary points in composition space, so look for conditions involving derivatives with respect to composition. The key insight is that both the values AND the slopes of chemical potentials must match between phases.

Question 16

In a ternary system with components A, B, and C at constant temperature and pressure, the Gibbs-Duhem equation is $n_A d\mu_A + n_B d\mu_B + n_C d\mu_C = 0$ . If $d\mu_A = 100 \text{ J/mol}$ , $d\mu_B = -50 \text{ J/mol}$ , and the system contains equal moles of all three components, what is $d\mu_C$ ?

-50 J/mol (correct answer)
50 J/mol
-25 J/mol
25 J/mol
0 J/mol

Explanation: When you encounter the Gibbs-Duhem equation in a ternary system, you're dealing with a fundamental constraint on how chemical potentials can change. This equation tells us that the changes in chemical potentials of all components are interconnected - they cannot vary independently. The Gibbs-Duhem equation states that

n_A d\mu_A + n_B d\mu_B + n_C d\mu_C = 0

. Since the system contains equal moles of all three components, we can set

n_A = n_B = n_C = n

. This simplifies our equation to:

n(d\mu_A + d\mu_B + d\mu_C) = 0

Since

n \neq 0

, we must have:

d\mu_A + d\mu_B + d\mu_C = 0

Substituting the given values:

100 + (-50) + d\mu_C = 0

50 + d\mu_C = 0

d\mu_C = -50 \text{ J/mol}

This confirms answer A is correct. Answer B (50 J/mol) represents a sign error - you might get this if you incorrectly rearrange the equation as

d\mu_C = d\mu_A + d\mu_B

instead of

d\mu_C = -(d\mu_A + d\mu_B)

. Answers C (-25 J/mol) and D (25 J/mol) likely come from incorrectly averaging the given values or mishandling the mole fraction relationships. Study tip: Always remember that the Gibbs-Duhem equation is a sum that equals zero. When components have equal amounts, you can factor out the mole terms and work with a simple algebraic sum of the chemical potential changes.

Question 17

In a binary system, the excess chemical potential of component 1 is defined as $\mu_1^E = \mu_1 - \mu_1^{ideal}$ . For a symmetric regular solution where $\mu_1^E = \Omega x_2^2$ and $\mu_2^E = \Omega x_1^2$ , what is the condition for phase separation to occur?

$\Omega > 0$ and the temperature is sufficiently low that $\Omega > 2RT$ (correct answer)
$\Omega < 0$ regardless of temperature since negative excess chemical potential always favors mixing
$\Omega > RT$ at any temperature since this ensures positive deviation from ideality
Phase separation occurs when $\Omega = RT$ exactly, marking the critical point for demixing
$\Omega > 4RT$ to ensure that the second derivative of Gibbs energy with respect to composition becomes negative

Explanation: When analyzing phase behavior in binary solutions, you need to understand that phase separation occurs when the system can lower its Gibbs free energy by forming two separate phases rather than remaining as a single mixed phase. The key is examining the second derivative of the Gibbs free energy with respect to composition. For a symmetric regular solution, the excess Gibbs free energy is

G^E = \Omega x_1 x_2

. Taking the second derivative with respect to

x_1

gives

\frac{\partial^2 G^E}{\partial x_1^2} = 2\Omega

. For the total Gibbs free energy, you must add the ideal mixing contribution:

\frac{\partial^2 G_{total}}{\partial x_1^2} = 2\Omega + \frac{RT}{x_1(1-x_1)}

. Phase separation (spinodal decomposition) occurs when this second derivative becomes negative, indicating thermodynamic instability. The critical condition occurs when

2\Omega + \frac{RT}{x_1(1-x_1)} = 0

. At the symmetric composition (

x_1 = 0.5

), this gives

2\Omega + 4RT = 0

, or

\Omega = -2RT

. Since we need positive

\Omega

for unfavorable mixing, the condition becomes

\Omega > 2RT

. Answer A correctly identifies that positive

\Omega

represents unfavorable interactions and that sufficiently low temperature (

\Omega > 2RT

) enables phase separation. Answer B is wrong because negative

\Omega

favors mixing, preventing separation. Answer C uses the wrong threshold value. Answer D incorrectly states the critical point condition and uses

RT

instead of

2RT

. Remember: Phase separation requires unfavorable mixing interactions (

\Omega > 0

) that overcome the entropy of mixing at low enough temperatures.

Question 18

A system contains two immiscible liquids (α and β) in equilibrium with a vapor phase. Component A is present in all three phases. If the activity coefficient of A in liquid α is 1.2 and in liquid β is 0.6, and the mole fraction of A in the vapor is 0.3, what is the ratio of mole fractions $x_A^\alpha / x_A^\beta$ in the liquid phases?

2.0
0.5 (correct answer)
1.2
0.6
0.72

Explanation: When you encounter problems involving multiple phases in equilibrium, you need to apply the fundamental principle that the chemical potential (or fugacity) of each component must be equal across all phases. For liquid-vapor equilibrium, this means the fugacity of component A in the vapor equals its fugacity in both liquid phases. For each liquid phase, the fugacity is given by

f_A = \gamma_A x_A f_A^*

, where

\gamma_A

is the activity coefficient,

x_A

is the mole fraction, and

f_A^*

is the fugacity of pure liquid A. Since both liquid phases are in equilibrium with the same vapor, we can write:

\gamma_A^\alpha x_A^\alpha f_A^* = \gamma_A^\beta x_A^\beta f_A^*

The

f_A^*

terms cancel, giving us:

\gamma_A^\alpha x_A^\alpha = \gamma_A^\beta x_A^\beta

Rearranging:

\frac{x_A^\alpha}{x_A^\beta} = \frac{\gamma_A^\beta}{\gamma_A^\alpha} = \frac{0.6}{1.2} = 0.5

This confirms answer B) 0.5 is correct. Answer A) 2.0 represents the inverse ratio

\gamma_A^\alpha/\gamma_A^\beta

, a common mistake when students flip the relationship. Answer C) 1.2 is simply the activity coefficient in phase α, while answer D) 0.6 is the activity coefficient in phase β—both ignore the equilibrium relationship entirely. Remember: in phase equilibrium problems, activity coefficients and mole fractions are inversely related. The phase with the higher activity coefficient (indicating less favorable interactions) will have a lower mole fraction to maintain equal chemical potentials.

Question 19

A pure liquid is in equilibrium with its vapor at 350 K. If the molar volume of the liquid is 0.08 L/mol and that of the vapor is 25 L/mol, what is the change in chemical potential per unit pressure change ( $d\mu/dP$ ) for the liquid phase?

0.08 L/mol (correct answer)
25 L/mol
24.92 L/mol
0.000228 L/mol
312.5 L/mol

Explanation: When you encounter questions about chemical potential and phase equilibria, you're dealing with fundamental thermodynamic relationships that connect intensive properties like pressure to the chemical potential of each phase. The key relationship here is that the change in chemical potential with pressure is simply the molar volume:

d\mu/dP = V_m

. This comes directly from the fundamental thermodynamic relation for the Gibbs free energy. Since we're asked specifically about the liquid phase, and the molar volume of the liquid is given as 0.08 L/mol, the answer is straightforward. Looking at the wrong answers: Answer B (25 L/mol) represents the molar volume of the vapor phase, not the liquid. This is a common trap - students might grab the wrong phase's data. Answer C (24.92 L/mol) appears to be the difference between vapor and liquid molar volumes (25 - 0.08), which has no thermodynamic significance for this relationship. Answer D (0.000228 L/mol) might result from incorrectly taking a ratio or making unit conversion errors. The correct answer is A (0.08 L/mol) because

d\mu/dP

for any phase equals that phase's molar volume. Study tip: Memorize that

d\mu/dP = V_m

for any phase. When you see chemical potential questions involving pressure changes, immediately identify which phase is being asked about and use that phase's molar volume. Don't be distracted by data from other phases or mathematical combinations that aren't thermodynamically meaningful.

Question 20

For a solution containing a volatile solute, the chemical potential of the solvent follows $\mu_{\text{solvent}} = \mu_{\text{solvent}}^* + RT \ln(x_{\text{solvent}})$ (ideal behavior). If the solution is in equilibrium with pure solvent vapor at the same temperature, and the mole fraction of solvent in solution is 0.85, what is the ratio of the equilibrium vapor pressure to the pure solvent vapor pressure?

1.18
0.85 (correct answer)
0.15
1.15
0.72

Explanation: When you encounter problems involving vapor-liquid equilibrium with volatile solutes, you're dealing with Raoult's Law and chemical potential equilibrium. The key insight is that at equilibrium, the chemical potential of the solvent in solution must equal the chemical potential of the pure solvent vapor. For the solvent in solution:

\mu_{\text{solvent}} = \mu_{\text{solvent}}^* + RT \ln(x_{\text{solvent}})

For pure solvent vapor at equilibrium:

\mu_{\text{vapor}} = \mu_{\text{solvent}}^* + RT \ln\left(\frac{P}{P^*}\right)

At equilibrium, these must be equal:

\mu_{\text{solvent}}^* + RT \ln(x_{\text{solvent}}) = \mu_{\text{solvent}}^* + RT \ln\left(\frac{P}{P^*}\right)

Simplifying:

\ln(x_{\text{solvent}}) = \ln\left(\frac{P}{P^*}\right)

Therefore:

\frac{P}{P^*} = x_{\text{solvent}} = 0.85

This is Raoult's Law: the vapor pressure of the solvent above a solution equals its mole fraction times the pure solvent vapor pressure. Answer B (0.85) is correct because the vapor pressure ratio directly equals the solvent mole fraction. Answer A (1.18) incorrectly takes the reciprocal of 0.85, perhaps confusing which component's properties are being calculated. Answer C (0.15) uses the solute mole fraction instead of the solvent mole fraction. Answer D (1.15) appears to be an arbitrary calculation error. Remember: For ideal solutions, Raoult's Law gives you a direct proportionality between mole fraction and vapor pressure—no complex calculations needed when the relationship is this straightforward.