Physical Chemistry 1 Quiz: Gibbs Energy Of Mixing

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Gibbs Energy Of MixingQuestion 1 of 20

The molar Gibbs energy of mixing for an ideal ternary solution is given by $\Delta G_{mix} = RT\sum_{i=1}^{3} x_i \ln x_i$ . If this expression is differentiated with respect to the number of moles of component 2 at constant temperature, pressure, and amounts of components 1 and 3, which quantity is obtained?

The partial molar Gibbs energy of component 2 in the solution

The chemical potential difference between component 2 and its pure state

The partial molar Gibbs energy of mixing for component 2

The activity coefficient of component 2 multiplied by RT

The excess partial molar Gibbs energy of component 2

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Physical Chemistry 1 Quiz

Physical Chemistry 1 Quiz: Gibbs Energy Of Mixing

Practice Gibbs Energy Of Mixing 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 Gibbs Energy Of Mixing, 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

The partial molar Gibbs energy of component 2 in the solution
The chemical potential difference between component 2 and its pure state
The partial molar Gibbs energy of mixing for component 2 (correct answer)
The activity coefficient of component 2 multiplied by RT
The excess partial molar Gibbs energy of component 2

Explanation: When you encounter differentiation of thermodynamic mixing functions, you're dealing with partial molar quantities. The key insight is recognizing what happens when you differentiate a total mixing property with respect to the number of moles of one component. Starting with

\Delta G_{mix} = RT\sum_{i=1}^{3} x_i \ln x_i

, when you differentiate with respect to

n_2

at constant T, P,

n_1

, and

n_3

, you obtain

\left(\frac{\partial \Delta G_{mix}}{\partial n_2}\right)_{T,P,n_1,n_3}

. This is the definition of the partial molar Gibbs energy of mixing for component 2, denoted as

\overline{\Delta G_{mix,2}}

. The correct answer is C because this differentiation gives you how the total mixing Gibbs energy changes when you add an infinitesimal amount of component 2 to the solution - precisely the partial molar mixing quantity. Let's examine why the other options are wrong: A describes

\overline{G_2}

, the partial molar Gibbs energy of component 2 in solution, which would be

\left(\frac{\partial G_{total}}{\partial n_2}\right)

, not the mixing property. B refers to

\mu_2 - \mu_2^*

, which is related but not what this specific differentiation yields. D involves activity coefficients, which don't appear in ideal solution expressions since all activity coefficients equal unity. Remember this pattern: differentiating any extensive mixing property with respect to moles gives you the corresponding partial molar mixing property. The word "mixing" in both the original function and the answer is your key clue.

Question 2

An ideal binary solution exhibits a minimum in its Gibbs energy of mixing at a certain composition. If the pure components have identical molar volumes and the solution is prepared at constant temperature and pressure, which statement best describes the relationship between the composition at minimum $\Delta G_{mix}$ and the individual component properties?

The minimum occurs at the composition where the chemical potential derivatives are equal for both components
The minimum occurs at the composition where the partial molar Gibbs energies are minimized simultaneously
The minimum occurs at $x_1 = x_2 = 0.5$ regardless of component properties for ideal solutions (correct answer)
The minimum occurs where the activity coefficients approach unity most rapidly with composition change
The minimum occurs at the composition where the excess Gibbs energy equals the ideal mixing contribution

Explanation: When you encounter questions about Gibbs energy of mixing in ideal solutions, remember that ideal behavior creates specific, predictable patterns that don't depend on the individual component identities. For an ideal binary solution, the Gibbs energy of mixing is given by

\Delta G_{mix} = RT(x_1 \ln x_1 + x_2 \ln x_2)

. To find the minimum, you take the derivative with respect to composition and set it equal to zero. This mathematical analysis reveals that the minimum always occurs at

x_1 = x_2 = 0.5

for any ideal binary system, regardless of what the components actually are. This is a fundamental property of ideal mixing. Option A is incorrect because equal chemical potential derivatives don't define the minimum condition—you need the first derivative of

\Delta G_{mix}

to equal zero. Option B misunderstands the concept; partial molar Gibbs energies aren't simultaneously minimized at the

\Delta G_{mix}

minimum. Option D refers to activity coefficients, but in ideal solutions, activity coefficients are always unity by definition—they don't "approach" unity or change with composition. The key insight is that ideal solutions follow mathematical rules that override individual component properties. When the problem states "ideal binary solution" and mentions identical molar volumes, it's emphasizing that component-specific properties don't matter. Study tip: For ideal solution problems, always start with the fundamental equations for ideal behavior. The math will lead you to composition-independent results that depend only on the system being ideal, not on what the actual components are.

Question 3

Two ideal solutions are prepared at 298 K: Solution I contains equal moles of components A and B, while Solution II contains equal moles of components C and D. The molar Gibbs energy of mixing for Solution I is measured as $-1729 \text{ J/mol}$ . If Solution II exhibits a molar Gibbs energy of mixing of $-1453 \text{ J/mol}$ , what can be concluded about the preparation conditions?

Solution II was prepared at a lower temperature than Solution I under identical pressure conditions (correct answer)
Solution II contains components with different molar volumes compared to Solution I components
Solution II was prepared at a higher pressure than Solution I under identical temperature conditions
The measurement for Solution II contains systematic error since both should have identical values
Solution II exhibits slight positive deviation from ideal behavior due to component interactions

Explanation: When you encounter questions about the Gibbs energy of mixing for ideal solutions, remember that for equal molar mixtures, this value depends solely on temperature through the entropy term. For an ideal solution with equal moles of two components, the Gibbs energy of mixing is:

\Delta G_{mix} = RT \ln(0.5) + RT \ln(0.5) = 2RT \ln(0.5) = -RT \ln(4)

Since both solutions contain equal moles of their respective components, they should have identical mixing behavior if prepared under the same conditions. However, Solution II shows a less negative value (-1453 J/mol vs -1729 J/mol), indicating less favorable mixing. The correct answer is A because temperature directly affects

\Delta G_{mix}

. At lower temperatures, the RT term decreases, making

\Delta G_{mix}

less negative. You can verify this: if Solution I was at 298 K and Solution II at a lower temperature, the magnitude of mixing would be reduced for Solution II. Option B is incorrect because molar volumes don't affect the Gibbs energy of mixing for ideal solutions - only the entropy of mixing matters. Option C is wrong because pressure has negligible effect on liquid solution mixing at normal conditions. Option D represents a common misconception - while the solutions should behave identically under identical conditions, different preparation temperatures would legitimately produce different values. Remember: For ideal solution mixing problems, temperature is the primary variable affecting

\Delta G_{mix}

when composition is held constant. Always check if experimental conditions match before assuming measurement error.

Question 4

An ideal solution is formed by mixing $n_1$ moles of component 1 with $n_2$ moles of component 2. If the total Gibbs energy of mixing is $\Delta G_{mix}^{total} = -2500 \text{ J}$ when $n_1 = 2.0$ mol and $n_2 = 3.0$ mol at 298 K, what would be the total Gibbs energy of mixing if 1.0 mol of component 1 is added to the existing solution?

$-3245 \text{ J}$ (proportional increase based on molar addition and composition change) (correct answer)
$-3750 \text{ J}$ (simple scaling based on total mole increase)
$-3128 \text{ J}$ (accounting for new composition and total mole change)
$-2890 \text{ J}$ (modest increase due to composition moving away from optimum)
$-4167 \text{ J}$ (linear extrapolation based on component 1 contribution)

Explanation: When dealing with ideal solution mixing problems, you need to understand that the Gibbs energy of mixing depends on both the total number of moles and the mole fractions of components. For ideal solutions,

\Delta G_{mix} = nRT(x_1 \ln x_1 + x_2 \ln x_2)

, where

n

is total moles and

x_i

are mole fractions. Initially, you have

n_1 = 2.0

mol and

n_2 = 3.0

mol, giving

n_{total} = 5.0

mol with

x_1 = 0.4

and

x_2 = 0.6

. After adding 1.0 mol of component 1, you have

n_1 = 3.0

mol and

n_2 = 3.0

mol, giving

n_{total} = 6.0

mol with new mole fractions

x_1 = x_2 = 0.5

. Using the initial condition to find the mixing term:

-2500 = 5.0 \times 8.314 \times 298 \times (0.4 \ln 0.4 + 0.6 \ln 0.6)

. This gives the logarithmic term as approximately

-0.201

. For the new composition:

\Delta G_{mix}^{new} = 6.0 \times 8.314 \times 298 \times (0.5 \ln 0.5 + 0.5 \ln 0.5) = 6.0 \times 8.314 \times 298 \times (-0.693) = -3245

J. Answer A (-3245 J) correctly accounts for both the increased total moles and the composition change toward equal fractions. Answer B (-3750 J) incorrectly assumes simple proportional scaling. Answer C (-3128 J) uses an incorrect calculation method. Answer D (-2890 J) wrongly assumes the composition change reduces mixing favorability. Remember: ideal solution mixing problems require calculating both the new total moles and new mole fractions—never just scale the original value proportionally.

Question 5

Consider the partial molar Gibbs energy of component 1 in an ideal binary solution: $\bar{G}_1 = G_1^* + RT\ln x_1$ . If a solution initially containing equal moles of components 1 and 2 is diluted by adding pure component 2 until $x_1 = 0.25$ , what is the change in the partial molar Gibbs energy of component 1 at 300 K?

$+3458 \text{ J/mol}$ (increase due to dilution effect and logarithmic dependence)
$-1729 \text{ J/mol}$ (decrease due to enhanced mixing entropy) (correct answer)
$+2744 \text{ J/mol}$ (moderate increase accounting for mole fraction change)
$-3458 \text{ J/mol}$ (decrease due to increased solution entropy)
$+4315 \text{ J/mol}$ (maximum increase due to significant dilution)

Explanation: When you encounter partial molar properties in ideal solutions, remember that these quantities describe how the chemical potential of each component changes with composition. The partial molar Gibbs energy directly relates to the thermodynamic driving force for processes involving that component. To find the change in

\bar{G}_1

, you need to calculate the difference between final and initial states. Initially, with equal moles of both components,

x_1 = 0.50

. After dilution,

x_1 = 0.25

. The change is:

\Delta\bar{G}_1 = RT\ln(0.25) - RT\ln(0.50) = RT\ln\left(\frac{0.25}{0.50}\right) = RT\ln(0.50)

At 300 K:

\Delta\bar{G}_1 = (8.314)(300)\ln(0.50) = -1729 \text{ J/mol}

The negative value makes physical sense because diluting component 1 decreases its chemical potential, making it less "eager" to escape the solution. Answer A gives the correct magnitude but wrong sign—this represents the common error of calculating

\ln(2)

instead of

\ln(0.5)

, essentially reversing the initial and final states. Answer C shows another magnitude error, possibly from incorrectly using natural log properties. Answer D has the right sign but wrong magnitude, suggesting a calculation error in the logarithm evaluation. Study tip: Always check that your sign makes chemical sense. Diluting a component (decreasing its mole fraction) should decrease its partial molar Gibbs energy, giving a negative

\Delta\bar{G}

. The logarithmic relationship means small composition changes can create significant thermodynamic effects.

Question 6

A researcher measures the Gibbs energy of mixing for an ideal ternary solution at two different temperatures and finds that $\Delta G_{mix}(350\text{ K})/\Delta G_{mix}(280\text{ K}) = 1.25$ for the same composition. However, when the calculation is repeated using the ideal solution equation, the theoretical ratio is found to be different. What is the most likely explanation for this discrepancy?

The solution exhibits slight positive deviations from ideality at higher temperature due to increased molecular motion
Experimental error in temperature measurement, as the theoretical ratio should be exactly $350/280 = 1.25$ (correct answer)
The solution composition changed between measurements due to preferential evaporation of volatile components
Temperature-dependent activity coefficients are affecting the mixing behavior despite assumed ideality
Pressure effects on the liquid phase become significant at elevated temperature

Explanation: When analyzing mixing behavior in ideal solutions, remember that the Gibbs energy of mixing depends directly on temperature through the fundamental equation:

\Delta G_{mix} = RT \sum_i n_i \ln x_i

. Since this relationship is linear in temperature, the ratio of mixing energies at two temperatures should equal the ratio of those temperatures. Let's verify the theoretical calculation. For an ideal ternary solution at constant composition, the ratio should be

\frac{\Delta G_{mix}(350\text{ K})}{\Delta G_{mix}(280\text{ K})} = \frac{350}{280} = 1.25

. The experimental value of 1.25 matches this theoretical prediction exactly, confirming that B is correct—this appears to be experimental error in temperature measurement, though the "error" actually produced the theoretically expected result. A is incorrect because increased molecular motion at higher temperatures doesn't cause positive deviations from ideality in the thermodynamic sense being tested here. The fundamental mixing equation already accounts for temperature effects. C is wrong because the problem states measurements were taken "for the same composition." If preferential evaporation had occurred, the composition would have changed, violating this condition. D misses the point entirely. By definition, ideal solutions have activity coefficients equal to 1 at all temperatures. Temperature-dependent activity coefficients would indicate non-ideal behavior, contradicting the premise. Study tip: For ideal solution problems, always check if experimental ratios match the simple temperature ratio

T_1/T_2

. Perfect agreement often indicates the system is truly behaving ideally, making apparent "discrepancies" actually confirmations of theory.

Question 7

Two researchers measure the Gibbs energy of mixing for the same ideal binary solution at 300 K but report different values: Researcher A reports $\Delta G_{mix} = -1800 \text{ J/mol}$ , while Researcher B reports $\Delta G_{mix} = -900 \text{ J}$ . Both claim their measurements are correct. What is the most likely explanation for this discrepancy?

Researcher A used a different temperature scale (Celsius vs. Kelvin) in their calculations
Researcher B measured a sample containing 0.5 mol of solution while A reported per mole (correct answer)
Researcher A included enthalpy contributions while B measured only entropy effects
Researcher B used a different standard state definition for the pure components
Researcher A corrected for non-ideal behavior while B assumed perfect ideality

Explanation: When you encounter Gibbs energy problems with seemingly contradictory results, always check the units and basis for the measurements—this reveals whether researchers are reporting intensive vs. extensive properties. The key insight is recognizing that

\Delta G_{mix}

can be reported either as an intensive property (per mole of solution) or as an extensive property (total for the actual sample). Researcher A reports -1800 J/mol, which is the standard intensive form. Researcher B reports -900 J total for their specific sample. If Researcher B's sample contained 0.5 mol of solution, then their intensive value would be -900 J ÷ 0.5 mol = -1800 J/mol—identical to Researcher A's result. Both measurements are correct; they're just expressing the same thermodynamic result on different bases. Option A is incorrect because temperature scale errors would affect the calculated values themselves, not just the units, and both researchers used the same temperature (300 K). Option C is wrong because for ideal solutions,

\Delta H_{mix} = 0

, so including enthalpy contributions wouldn't change the Gibbs energy. Option D is incorrect because different standard states would fundamentally alter the calculated mixing values, not just scale them proportionally. Study tip: Always examine units carefully in thermodynamics problems. When values differ by simple numerical factors (like 2×), suspect that one researcher reported per mole while another reported for their actual sample size. This distinction between intensive and extensive properties is frequently tested.

Question 8

The partial molar Gibbs energy of component i in an ideal solution is $\bar{G}_i = G_i^* + RT\ln x_i$ . For a binary solution where $x_1 = 0.3$ , if the partial molar Gibbs energies of both components are equal ( $\bar{G}_1 = \bar{G}_2$ ), what is the relationship between the standard state Gibbs energies?

$G_2^* - G_1^* = RT\ln(2.33)$ (component 2 standard state is significantly higher)
$G_1^* - G_2^* = RT\ln(2.33)$ (component 1 standard state is higher by factor of 2.33)
$G_2^* - G_1^* = RT\ln(0.429)$ (component 2 standard state is lower to compensate for mole fraction difference) (correct answer)
$G_1^* = G_2^*$ (standard states must be equal for equal partial molar values in ideal solutions)
$G_2^* - G_1^* = RT\ln(0.7/0.3)$ (difference equals the direct logarithmic ratio of mole fractions)

Explanation: When dealing with partial molar Gibbs energies in ideal solutions, you're examining how the chemical potential of each component depends on both its standard state and its mole fraction. The key insight is that when partial molar Gibbs energies are equal, the differences in mole fractions must be compensated by differences in standard states. Given that

\bar{G}_1 = \bar{G}_2

, we can set up the equation:

G_1^* + RT\ln x_1 = G_2^* + RT\ln x_2

With

x_1 = 0.3

, we have

x_2 = 0.7

. Rearranging:

G_2^* - G_1^* = RT\ln x_1 - RT\ln x_2 = RT\ln\left(\frac{x_1}{x_2}\right) = RT\ln\left(\frac{0.3}{0.7}\right) = RT\ln(0.429)

This confirms answer C is correct. Component 2's standard state must be lower to offset its higher mole fraction. Option A incorrectly uses

\ln(2.33)

, which would come from calculating

\frac{x_2}{x_1} = \frac{0.7}{0.3} = 2.33

but with the wrong sign relationship. Option B makes the same numerical error while also reversing which component has the higher standard state. Option D assumes the standard states must be equal, ignoring that different mole fractions require compensating differences in standard states to achieve equal partial molar Gibbs energies. Remember: in ideal solutions, when partial molar properties are equal despite different compositions, the standard states must differ to compensate. The component with the smaller mole fraction needs a higher standard state Gibbs energy.

Question 9

A student calculates the Gibbs energy of mixing for an ideal binary solution and obtains $\Delta G_{mix} = -RT[x_1\ln x_1 + x_2\ln x_2]$ . When checking the units, they find that if R is in $\text{J mol}^{-1} \text{K}^{-1}$ and T is in K, the expression gives units of $\text{J mol}^{-1}$ . However, their experimental data is for 0.5 mol of solution. What correction must be applied to compare theory with experiment?

Multiply the theoretical value by 0.5 to get total Gibbs energy of mixing for the sample (correct answer)
Divide the experimental value by 0.5 to get molar Gibbs energy of mixing
Convert experimental units to $\text{J mol}^{-1}$ by dividing by molar mass
Apply activity coefficient corrections since the sample size affects ideality
Use the partial molar quantities instead of total molar quantities for comparison

Explanation: When dealing with thermodynamic mixing properties, you need to distinguish between molar quantities (per mole of solution) and total quantities (for your entire sample). The theoretical expression

\Delta G_{mix} = -RT[x_1\ln x_1 + x_2\ln x_2]

gives the molar Gibbs energy of mixing in units of

\text{J mol}^{-1}

. Since your experimental data represents 0.5 mol of solution, you're measuring the total Gibbs energy change for that specific amount. To compare theory with experiment, you must convert the molar theoretical value to match your sample size by multiplying by 0.5 mol. This gives you the total

\Delta G_{mix}

for your 0.5 mol sample in units of J. Looking at the incorrect options: Option B suggests converting experimental data to molar quantities, but this approach is unnecessarily complicated since you can easily scale the theoretical prediction to match your sample size. Option C misunderstands the unit issue entirely—this isn't about converting between mass and moles using molar mass, but rather about intensive vs. extensive properties. Option D incorrectly assumes that sample size affects solution ideality, but ideal solution behavior is independent of the total amount of solution present. Study tip: Always check whether thermodynamic quantities are expressed as molar (intensive) or total (extensive) properties. When comparing theory to experiment, make sure both values represent the same basis—either both per mole or both for the total sample amount.

Question 10

Two volatile liquids A and B form an ideal solution. At 298 K, the molar Gibbs energy of mixing per mole of solution is $-1247 \text{ J/mol}$ when the mole fraction of A is 0.30. If the temperature is increased to 350 K while maintaining the same composition, what is the new molar Gibbs energy of mixing?

$-1247 \text{ J/mol}$ (unchanged because composition is constant)
$-1466 \text{ J/mol}$ (proportional increase with temperature) (correct answer)
$-1060 \text{ J/mol}$ (decreased magnitude due to higher temperature)
$-1434 \text{ J/mol}$ (using entropy of mixing at new temperature)
$-1092 \text{ J/mol}$ (accounting for temperature dependence of activity)

Explanation: When you encounter problems about ideal solution mixing at different temperatures, focus on the fundamental thermodynamic relationship for Gibbs energy of mixing:

\Delta G_{mix} = \Delta H_{mix} - T\Delta S_{mix}

. For ideal solutions,

\Delta H_{mix} = 0

because there are no intermolecular interactions between different components. This means

\Delta G_{mix} = -T\Delta S_{mix}

, where the entropy of mixing depends only on composition:

\Delta S_{mix} = -R(x_A \ln x_A + x_B \ln x_B)

. First, calculate the entropy of mixing at the given composition (

x_A = 0.30

x_B = 0.70

\Delta S_{mix} = -8.314[0.30 \ln(0.30) + 0.70 \ln(0.70)] = 4.19 \text{ J/(mol·K)}

Since

\Delta G_{mix} = -T\Delta S_{mix}

, at 350 K:

\Delta G_{mix} = -350 × 4.19 = -1466 \text{ J/mol}

This confirms answer B is correct. Answer A incorrectly assumes Gibbs energy of mixing is temperature-independent, ignoring the

-T\Delta S_{mix}

term. Answer C suggests decreased magnitude, but since

\Delta G_{mix}

is proportional to temperature for ideal solutions, the magnitude actually increases. Answer D gives an arbitrary value that doesn't follow the correct proportional relationship. Study tip: For ideal solution problems, remember that

\Delta G_{mix} = -T\Delta S_{mix}

(no enthalpy term), so the Gibbs energy of mixing is directly proportional to temperature at constant composition. This makes temperature scaling straightforward: just multiply by the temperature ratio.

Question 11

Consider the function $f(x) = x\ln x + (1-x)\ln(1-x)$ which appears in the Gibbs energy of mixing expression for ideal binary solutions. At what value of x does this function have its minimum value, and what is that minimum value?

Minimum at $x = 0.5$ with value $-0.693$ (symmetric composition gives maximum mixing) (correct answer)
Minimum at $x = 0.368$ with value $-0.632$ (related to natural exponential function properties)
Minimum at $x = 0.5$ with value $-0.500$ (equimolar mixture with moderate mixing entropy)
Minimum at $x = 0.632$ with value $-0.693$ (asymmetric optimum due to logarithmic weighting)
Minimum at $x = 0.5$ with value $-1.000$ (maximum entropy configuration for binary system)

Explanation: When you encounter the Gibbs energy of mixing function for ideal binary solutions, you're looking at how entropy changes when two components mix. This function

f(x) = x\ln x + (1-x)\ln(1-x)

represents the configurational entropy term, and finding its minimum tells us the composition that maximizes mixing entropy (since

\Delta S_{mix} = -R \cdot f(x)

). To find the minimum, take the derivative and set it equal to zero:

f'(x) = \ln x + 1 - \ln(1-x) - 1 = \ln x - \ln(1-x) = \ln\left(\frac{x}{1-x}\right) = 0

. This gives us

\frac{x}{1-x} = 1

, so

x = 1-x

, which means

x = 0.5

. At

x = 0.5

f(0.5) = 0.5\ln(0.5) + 0.5\ln(0.5) = 2 \times 0.5 \times (-0.693) = -0.693

. The second derivative

f''(x) = \frac{1}{x} + \frac{1}{1-x}

is always positive for

0 < x < 1

, confirming this is indeed a minimum. Answer A is correct - the symmetric composition (

x = 0.5

) gives maximum mixing entropy, corresponding to the minimum of this function at

-0.693

. Answer B incorrectly places the minimum at

x = 0.368

(which is

1/e

), confusing this with exponential function properties that don't apply here. Answer C has the right composition but wrong minimum value. Answer D incorrectly suggests an asymmetric optimum at

x = 0.632

, missing that logarithmic functions are symmetric around

x = 0.5

for this particular form. Remember: for ideal mixing problems, maximum entropy (and minimum Gibbs energy of mixing) always occurs at equimolar composition due to the inherent symmetry of the mixing process.

Question 12

An ideal solution exhibits a Gibbs energy of mixing of $\Delta G_{mix} = -1500 \text{ J/mol}$ at 298 K. The same solution at 298 K undergoes isothermal expansion from 1 bar to 0.5 bar. Assuming the solution behaves as an incompressible liquid, what is the new Gibbs energy of mixing?

$-1500 \text{ J/mol}$ (unchanged because mixing is independent of pressure for liquids) (correct answer)
$-1650 \text{ J/mol}$ (decreased due to volume expansion enhancing mixing)
$-1350 \text{ J/mol}$ (increased due to reduced molecular interactions at lower pressure)
$-750 \text{ J/mol}$ (proportional to pressure change for ideal gas behavior)
$-3000 \text{ J/mol}$ (doubled due to volume doubling at half pressure)

Explanation: This question tests your understanding of how pressure affects the thermodynamic properties of incompressible liquids, particularly the Gibbs energy of mixing for ideal solutions. For an incompressible liquid, the key insight is that volume remains constant regardless of pressure changes. The Gibbs energy of mixing for an ideal solution depends only on the entropy of mixing (which relates to the random distribution of molecules) and any enthalpy effects. Since we're dealing with an ideal solution,

\Delta H_{mix} = 0

, and the mixing depends solely on the configurational entropy:

\Delta G_{mix} = \Delta H_{mix} - T\Delta S_{mix} = -T\Delta S_{mix}

. When pressure changes from 1 bar to 0.5 bar, the molecular composition and arrangement in the solution remain identical because the liquid is incompressible. The entropy of mixing, which depends only on the mole fractions of components, stays constant. Therefore,

\Delta G_{mix}

remains

-1500 \text{ J/mol}

. Answer A is correct because mixing thermodynamics in incompressible liquids are indeed independent of pressure changes. Answer B incorrectly assumes volume expansion occurs, but incompressible liquids don't expand. Answer C wrongly suggests that pressure affects molecular interactions in ideal solutions, but ideal solutions have no interaction energy changes upon mixing. Answer D incorrectly applies ideal gas behavior to a liquid system, where pressure-volume relationships are fundamentally different. Study tip: Remember that "incompressible" means volume is truly constant with pressure, making most pressure-dependent thermodynamic effects negligible for liquids compared to gases.

Question 13

Two separate ideal solutions are prepared at 298 K: Solution X contains 0.6 mol fraction of component A and 0.4 mol fraction of component B, while Solution Y contains 0.4 mol fraction of component A and 0.6 mol fraction of component B. If equal volumes of these solutions are mixed to form Solution Z, assuming additive volumes and identical molar volumes for all components, what is the ratio $\Delta G_{mix}^Z / \Delta G_{mix}^X$ ?

$1.00$ (identical mixing due to symmetrical composition inversion)
$0.97$ (slightly less negative due to averaging effect reducing mixing entropy)
$1.03$ (slightly more negative due to enhanced compositional diversity) (correct answer)
$2.00$ (doubled mixing energy due to combination of two solutions)
$0.50$ (reduced mixing due to partial cancellation of compositional effects)

Explanation: When dealing with ideal solution mixing problems, you need to calculate the Gibbs free energy of mixing using

\Delta G_{mix} = RT \sum n_i \ln x_i

, where the sum goes over all components. For Solution X:

\Delta G_{mix}^X = RT[0.6 \ln(0.6) + 0.4 \ln(0.4)]

For Solution Y:

\Delta G_{mix}^Y = RT[0.4 \ln(0.4) + 0.6 \ln(0.6)]

Notice these are identical due to symmetry - just swapped compositions. When equal volumes mix to form Solution Z, assuming additive volumes and identical molar volumes, you get equal molar amounts from each solution. This creates a final composition of 0.5 mol fraction for each component. For Solution Z:

\Delta G_{mix}^Z = RT[0.5 \ln(0.5) + 0.5 \ln(0.5)] = RT \ln(0.5)

Calculating the numerical values:

$\Delta G_{mix}^X = RT[0.6(-0.511) + 0.4(-0.916)] = -0.673RT$
$\Delta G_{mix}^Z = RT \ln(0.5) = -0.693RT$

Therefore:

\Delta G_{mix}^Z / \Delta G_{mix}^X = -0.693RT / (-0.673RT) = 1.03

Answer A incorrectly assumes symmetrical compositions lead to identical mixing energies. Answer B suggests reduced entropy, but moving toward equal composition actually maximizes mixing entropy. Answer D wrongly treats this as additive energies rather than a new equilibrium state. Study tip: Remember that mixing entropy is maximized at equal mole fractions (0.5, 0.5), making the mixing process more thermodynamically favorable and

\Delta G_{mix}

more negative.

Question 14

A ternary ideal solution contains components X, Y, and Z with mole fractions 0.50, 0.30, and 0.20, respectively, at 300 K. If component Y is selectively removed until its mole fraction becomes 0.15, while maintaining the ratio of X to Z constant, what is the change in molar Gibbs energy of mixing?

$+392 \text{ J/mol}$ (less negative mixing due to composition change toward pure components) (correct answer)
$-156 \text{ J/mol}$ (more negative mixing due to increased entropy from redistribution)
$+278 \text{ J/mol}$ (decreased mixing entropy due to removal of component Y)
$-298 \text{ J/mol}$ (enhanced mixing from new composition with maintained X:Z ratio)
$+445 \text{ J/mol}$ (positive change due to reduced total mixing entropy)

Explanation: When you encounter problems involving changes to ideal solution compositions, focus on how the molar Gibbs energy of mixing responds to entropy changes. For ideal solutions,

\Delta G_{mix} = RT \sum x_i \ln x_i

, where the mixing becomes less favorable (less negative) as you move toward pure components. Let's calculate both states. Initially:

x_X = 0.50

x_Y = 0.30

x_Z = 0.20

. The initial

\Delta G_{mix} = RT(0.50 \ln 0.50 + 0.30 \ln 0.30 + 0.20 \ln 0.20) = -2714 \text{ J/mol}

at 300 K. After removing Y while keeping the X:Z ratio constant (initially 0.50:0.20 = 2.5:1), and setting

x_Y = 0.15

, we have 0.85 moles left for X and Z. To maintain the ratio:

x_X = 0.85 \times \frac{2.5}{3.5} = 0.607

and

x_Z = 0.85 \times \frac{1}{3.5} = 0.243

. The final

\Delta G_{mix} = RT(0.607 \ln 0.607 + 0.15 \ln 0.15 + 0.243 \ln 0.243) = -2322 \text{ J/mol}

. The change is:

\Delta(\Delta G_{mix}) = -2322 - (-2714) = +392 \text{ J/mol}

, confirming answer A. Answer B incorrectly suggests more negative mixing, but removing a component reduces mixing entropy. Answer C gets the sign right but calculates the wrong magnitude. Answer D misunderstands that maintaining ratios doesn't enhance mixing when you're removing material overall. Study tip: Remember that removing components from ideal solutions always makes mixing less favorable (more positive

\Delta G_{mix}

) because you're reducing the system's entropy of mixing.

Question 15

A researcher measures the Gibbs energy of mixing for what is believed to be an ideal binary solution and finds $\Delta G_{mix} = -1850 \text{ J/mol}$ at 298 K when $x_A = 0.60$ . If this solution were truly ideal, what should the measured value be, and what does the discrepancy suggest?

Expected: $-1730 \text{ J/mol}$ ; the solution shows slight positive deviation from ideality
Expected: $-1670 \text{ J/mol}$ ; the solution shows negative deviation indicating stronger intermolecular forces (correct answer)
Expected: $-1670 \text{ J/mol}$ ; the solution shows positive deviation indicating weaker intermolecular forces
Expected: $-1730 \text{ J/mol}$ ; the solution shows negative deviation due to association effects

Explanation: For an ideal solution with

x_A = 0.60, x_B = 0.40

\Delta G_{mix,ideal} = RT(0.60 \ln 0.60 + 0.40 \ln 0.40) = (8.314)(298)(0.60(-0.511) + 0.40(-0.916)) = (8.314)(298)(-0.673) = -1670 \text{ J/mol}

. The measured value (-1850 J/mol) is more negative than ideal, indicating negative deviation from Raoult's law, which occurs when intermolecular forces between unlike molecules are stronger than those between like molecules. Choice A has wrong expected value and deviation type. Choice C has correct expected value but wrong deviation interpretation. Choice D has wrong expected value.

Question 16

An ideal binary solution is formed by gradually adding component B to pure component A at constant temperature T. At what mole fraction of B does the rate of change of molar Gibbs energy of mixing with respect to $x_B$ equal $-2RT$ ?

$x_B = e^{-2} - e^{-1} \approx 0.232$
$x_B = \frac{1}{e^2 + 1} \approx 0.119$ (correct answer)
$x_B = \frac{e^2}{1 + e^2} \approx 0.881$
$x_B = 1 - \frac{1}{e^2 + 1} \approx 0.881$

Explanation: For

\Delta G_{mix} = RT((1-x_B) \ln(1-x_B) + x_B \ln x_B)

, taking the derivative:

\frac{d\Delta G_{mix}}{dx_B} = RT(-\ln(1-x_B) + \ln x_B) = RT \ln\left(\frac{x_B}{1-x_B}\right)

. Setting this equal to

-2RT

\ln\left(\frac{x_B}{1-x_B}\right) = -2

, so

\frac{x_B}{1-x_B} = e^{-2}

. Solving:

x_B = e^{-2}(1-x_B)

, giving

x_B(1+e^{-2}) = e^{-2}

, so

x_B = \frac{e^{-2}}{1+e^{-2}} = \frac{1}{e^2+1} \approx 0.119

. Choice A uses incorrect algebra. Choices C and D give the complement value.

Question 17

Two ideal solutions are prepared at the same temperature: Solution I contains equal moles of components A and B, while Solution II contains components A and B in a 3:1 molar ratio. How does the magnitude of the molar Gibbs energy of mixing compare between these solutions?

Solution I has a larger magnitude because maximum mixing occurs at equal mole fractions (correct answer)
Solution II has a larger magnitude because the entropy contribution is proportional to the mole ratio
Both solutions have equal magnitudes since they contain the same components
Solution II has a larger magnitude because the asymmetric composition increases the driving force

Explanation: For ideal solutions,

\Delta G_{mix} = RT(x_A \ln x_A + x_B \ln x_B)

. Solution I:

x_A = x_B = 0.5

, giving

\Delta G_{mix} = RT(2 × 0.5 \ln 0.5) = RT(-0.693)

. Solution II:

x_A = 0.75, x_B = 0.25

, giving

\Delta G_{mix} = RT(0.75 \ln 0.75 + 0.25 \ln 0.25) = RT(-0.562)

. The equal mole fraction case maximizes the magnitude. Choice B incorrectly relates entropy to mole ratios. Choice C ignores composition effects. Choice D incorrectly suggests asymmetry increases magnitude.

Question 18

For an ideal solution, the relationship between the Gibbs energy of mixing and temperature can be used to determine the entropy of mixing. If $\Delta G_{mix} = -RT \ln 2$ for an equimolar binary mixture, what is the molar entropy of mixing at any temperature for this composition?

$\Delta S_{mix} = 2R \ln 2$ (factor of 2 from the equimolar composition)
$\Delta S_{mix} = -R \ln 2$ (negative due to the negative Gibbs energy change)
$\Delta S_{mix} = R \ln 2$ (independent of temperature as expected for ideal mixing) (correct answer)
$\Delta S_{mix} = RT \ln 2$ (includes the temperature dependence explicitly)

Explanation: This question tests your understanding of fundamental thermodynamic relationships, specifically how Gibbs energy relates to entropy through temperature dependence. When you encounter mixing problems, remember that the key relationship is

\left(\frac{\partial G}{\partial T}\right)_P = -S

. To find the entropy of mixing, you need to take the partial derivative of the given Gibbs energy expression with respect to temperature at constant pressure. Starting with

\Delta G_{mix} = -RT \ln 2

, you differentiate:

\Delta S_{mix} = -\frac{\partial(\Delta G_{mix})}{\partial T} = -\frac{\partial(-RT \ln 2)}{\partial T} = R \ln 2

. Notice that the temperature cancels out completely, leaving only

R \ln 2

. Option A (

\Delta S_{mix} = 2R \ln 2

) incorrectly assumes the equimolar composition introduces an additional factor of 2, but the

\ln 2

term already accounts for the 50:50 mixture. Option B (

\Delta S_{mix} = -R \ln 2

) makes the sign error of thinking entropy should be negative because Gibbs energy is negative, but remember the negative sign in the derivative relationship. Option D (

\Delta S_{mix} = RT \ln 2

) fails to properly differentiate, incorrectly retaining the temperature dependence. The correct answer is C:

\Delta S_{mix} = R \ln 2

, which is positive (mixing increases entropy) and temperature-independent, as expected for ideal solutions. Study tip: Always remember that entropy is the negative temperature derivative of Gibbs energy. For ideal mixing problems, entropy of mixing is always temperature-independent and positive.

Question 19

A student calculates the Gibbs energy of mixing for an ideal binary solution and obtains $\Delta G_{mix} = +850 \text{ J/mol}$ at 300 K with $x_A = 0.70$ . What is the most likely error in the student's calculation?

Used $+RT$ instead of $-RT$ in the temperature factor, making the result positive
Calculated $\ln(x_A + x_B)$ instead of $(x_A \ln x_A + x_B \ln x_B)$ in the composition term
Used natural logarithm instead of base-10 logarithm in the mixing formula
Forgot the negative sign in the standard formula $\Delta G_{mix} = -RT(x_A \ln x_A + x_B \ln x_B)$ (correct answer)

Explanation: The correct value should be

\Delta G_{mix} = RT(0.70 \ln 0.70 + 0.30 \ln 0.30) = (8.314)(300)(-0.610) = -1522 \text{ J/mol}

. Since the student got +850 J/mol, they likely calculated

RT(x_A \ln x_A + x_B \ln x_B)

correctly but forgot that the standard formula includes a negative sign. Choice A would give a larger positive value. Choice B would give

RT \ln 1 = 0

. Choice C would give a different magnitude but still negative value.

Question 20

An ideal binary solution at 298 K has a molar Gibbs energy of mixing of $-1400 \text{ J/mol}$ . If the temperature is increased to 350 K while maintaining the same composition, and assuming the enthalpy of mixing remains zero (ideal behavior), what will be the new molar Gibbs energy of mixing?

$-1192 \text{ J/mol}$ (decreased magnitude due to temperature effects)
$-1400 \text{ J/mol}$ (unchanged because composition is constant)
$-1644 \text{ J/mol}$ (proportional increase with temperature) (correct answer)
$-1750 \text{ J/mol}$ (increased magnitude from enhanced thermal mixing)

Explanation: When you encounter questions about Gibbs energy of mixing at different temperatures, focus on how the entropy term changes while enthalpy remains constant for ideal solutions. For an ideal binary solution, the Gibbs energy of mixing is

\Delta G_{mix} = \Delta H_{mix} - T\Delta S_{mix}

. Since we're told the enthalpy of mixing remains zero (ideal behavior), this simplifies to

\Delta G_{mix} = -T\Delta S_{mix}

. The entropy of mixing depends only on composition, not temperature, so

\Delta S_{mix}

stays constant when composition is unchanged. At 298 K:

-1400 = -T_1 \Delta S_{mix}

, so

\Delta S_{mix} = \frac{1400}{298} \text{ J/(mol·K)}

At 350 K with the same composition:

\Delta G_{mix} = -350 × \frac{1400}{298} = -1644 \text{ J/mol}

This confirms answer C is correct - the magnitude increases proportionally with temperature. Answer A incorrectly assumes the magnitude should decrease with temperature, which contradicts the negative relationship between

\Delta G_{mix}

and temperature for ideal mixing. Answer B wrongly suggests that constant composition means constant

\Delta G_{mix}

, ignoring the temperature dependence of the entropy term. Answer D gives an arbitrary larger negative value without proper proportional scaling. Study tip: For ideal solution problems, remember that

\Delta G_{mix} = -T\Delta S_{mix}

when

\Delta H_{mix} = 0

. The entropy of mixing depends only on composition, so changing temperature while keeping composition constant creates a direct proportional relationship:

\frac{\Delta G_{mix,2}}{\Delta G_{mix,1}} = \frac{T_2}{T_1}