Thermodynamics carries 8–12 questions, and on this exam it is the chemist’s version: mixtures, vapor-liquid equilibrium and reaction equilibrium matter more than engine cycles.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Chemical
questions on the exam8–12
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What the specification covers
The outline lists eight subtopics. In our words:
Properties of pure components and mixtures: specific volume, internal energy, enthalpy, entropy, free energy, ideal gas.
Reading property data: steam tables, psychrometric charts and T-s, P-h, x-y and T-x-y diagrams.
First and second laws for closed and open systems.
Chemical equilibrium and its temperature dependence.
Heats of reaction and of mixing.
16 thermodynamics problems
0 / 16 answered · 0 correct
Question 1 of 16 · Thermodynamics · 8–12 items
A binary liquid mixture of components A and B exhibits a significant positive deviation from Raoult's law, characterized by activity coefficients strictly greater than unity (γi > 1). At a constant pressure, which of the following best describes the resulting phase equilibrium behavior of this mixture?
Solution
γi > 1 → partial pressures above Raoult's law → total-pressure maximum at constant T → boiling-point minimum at constant P (minimum-boiling azeotrope, e.g. ethanol-water). Negative deviation gives the maximum-boiling kind.
With activity coefficients that change with composition, α = γ1P1sat/(γ2P2sat) changes along the curve; it is constant only for near-ideal systems.
An azeotrope pinches the x-y curve onto the diagonal, so ordinary distillation cannot pass it to give two pure products.
Positive deviation raises the total vapor pressure above Raoult's line; when it is strong enough the pressure shows a maximum, which at constant pressure is a minimum-boiling azeotrope.
A maximum-boiling azeotrope comes from negative deviation (γi < 1), where unlike molecules attract each other more strongly.
γi > 1 is by definition a departure from ideal-solution behavior, so the mixture cannot be ideal.
Question 2 of 16 · Thermodynamics · 8–12 items
A chemical engineer is scaling up a pilot gas-phase reactor. The engineer calculates the standard free energy change (ΔG°) and uses it to find the equilibrium constant (K). To determine if the reaction will proceed spontaneously forward under the actual operating conditions, the engineer evaluates the expression ΔG = ΔG° + RT ln K and finds ΔG = 0, concluding the system is already at equilibrium and will not proceed. However, when the pilot reactor is started, the reaction proceeds vigorously in the forward direction. Which of the following best explains the fundamental error in the engineer's thermodynamic analysis?
Solution
ΔG = ΔG° + RT·ln Q, with ΔG° = −RT·ln K → ΔG = RT·ln(Q/K)
Substituting K for Q makes ΔG = 0 identically. With fresh feed (few products) Q < K, so ΔG < 0 and the reaction runs forward.
Activation energy governs the rate, not the direction; the analysis error is in the thermodynamics, since kinetics cannot make a reaction at equilibrium run forward.
A wrong temperature would shift K, but ΔG would still come out as exactly zero once K replaces Q; the zero result is caused by the substitution, not by the temperature.
ΔG = ΔG° + RT ln Q; since ΔG° = −RT ln K, putting K in place of Q always gives ΔG = 0, whatever the actual feed composition.
Fugacity corrections refine the activities in Q, but they cannot turn a reaction quotient into the equilibrium constant.
Question 3 of 16 · Thermodynamics · 8–12 items
Consider the following gas-phase reaction at equilibrium: N2(g) + 3H2(g) ⇌ 2NH3(g). If the total system pressure is isothermally doubled, which of the following statements correctly describes the subsequent effect on the equilibrium constant (Kp) and the equilibrium mole fraction of ammonia (yNH3)?
Solution
Kp = yNH32/(yN2·yH23)·P−2 = f(T) only
Δn = 2 − 4 = −2, so at fixed T a higher P must be offset by a larger mole-fraction ratio: the equilibrium shifts toward NH3 (Le Chatelier).
Kp is a function of temperature only; with Kp = (y-ratio)·P−2 fixed, doubling P makes the ratio four times larger, so yNH3 increases.
Higher pressure favors the side with fewer gas moles (2 versus 4), which is ammonia, so its fraction rises rather than falls.
Kp depends on temperature alone, and an isothermal compression does not change it; the mole fraction does respond to pressure.
The ammonia fraction does rise, but Kp is fixed at constant temperature; only the equilibrium composition shifts.
Kp is unchanged, but the composition is not: Kp = yNH32/(yN2yH23P2), so doubling P forces the mole-fraction ratio up.
Question 4 of 16 · Thermodynamics · 8–12 items
Using the Clapeyron equation, what is the slope dP/dT of the phase boundary on a pressure-temperature diagram for a pure substance, where L is the latent heat and vg − vf is the change in specific volume?
Solution
dP/dT = Δs/Δv = L/(T·(vg − vf))
From equal Gibbs energies of the two phases along the boundary (dgf = dgg); T is absolute temperature.
This is the reciprocal of the Clapeyron slope; it would give a slope that shrinks as the latent heat grows.
This drops the temperature: L/T is the entropy of the phase change, and the slope is Δs/Δv.
Temperature belongs in the denominator with the volume change, not in the numerator.
Along a phase boundary dP/dT = Δs/Δv = L/(TΔv), with L the latent heat and Δv the change in specific volume between the phases.
Question 5 of 16 · Thermodynamics · 8–12 items
Parameter
Component 1
Component 2
Psat at 70 °C (kPa)
80.0
40.0
Margules constants
A12 = 1.2
A21 = 0.8
Calculate the bubble-point pressure of a nonideal binary mixture of Component 1 and Component 2 at 70 °C. The liquid-phase mole fraction of Component 1 is x1 = 0.4. The activity coefficients follow the Margules equations shown.
ln γ1 = x22[A12 + 2(A21 − A12)x1]
ln γ2 = x12[A21 + 2(A12 − A21)x2]
56.0 kPa is the ideal Raoult's-law bubble point (γ1 = γ2 = 1); it ignores the positive deviation the Margules constants describe.
67.9 kPa corrects component 1 only and leaves γ2 = 1, although γ2 = 1.23 here.
70.2 kPa swaps x1 and x2 inside the Margules brackets, which gives the wrong activity coefficients for this composition.
γ1 = e0.36(1.2 − 0.32) = 1.373 and γ2 = e0.16(0.8 + 0.48) = 1.227, so P = 0.4(1.373)(80.0) + 0.6(1.227)(40.0) = 73.4 kPa.
80.9 kPa pairs each activity coefficient with the other component's saturation pressure.
Question 6 of 16 · Thermodynamics · 8–12 items
Component
ΔHf° (kJ/mol)
ΔGf° (kJ/mol)
A (g)
−100
−80
B (g)
−40
−30
C (g)
−20
−20
Consider the gas-phase reaction A ⇌ B + C. The standard enthalpies and Gibbs energies of formation at 298.15 K are given in the table. Assuming the standard heat of reaction is constant, what is the equilibrium constant K at 500 K? Use ln(K2/K1) = −(ΔH°/R)(1/T2 − 1/T1) with R = 8.314 J/(mol·K).
FE Reference Handbook, Thermodynamics: chemical reaction equilibrium (van't Hoff equation).
675 is the ratio K500/K298 from the van't Hoff equation; it still has to multiply K at 298 K.
ΔH° = +40 kJ/mol and ΔG° = +30 kJ/mol give K298 = 5.55 × 10−6; van't Hoff multiplies it by e6.51 = 675, so K500 = 3.74 × 10−3.
7.34 × 10−4 plugs the 298 K value of ΔG° into K = e−ΔG°/RT at 500 K, as if ΔG° did not change with temperature.
5.55 × 10−6 is K at 298.15 K from ΔG° = +30 kJ/mol; it never corrects to 500 K.
8.22 × 10−9 flips the sign in the van't Hoff equation; for an endothermic reaction (ΔH° = +40 kJ/mol) K must grow with temperature, not shrink.
Question 7 of 16 · Thermodynamics · 8–12 items
What does the fugacity coefficient measure in gas-phase thermodynamics?
Solution
φi = fi/(yiP); φ = 1 for an ideal gas
φ < 1 when attractive forces dominate, φ > 1 when repulsive forces dominate (high pressure).
Partial pressure is yiP; the fugacity coefficient is the correction that multiplies it to give the fugacity of a non-ideal component.
Cv is a thermal property and has no connection to the fugacity coefficient.
φ = f/P compares a real gas's fugacity with its pressure; φ = 1 for an ideal gas, so it measures non-ideality.
Liquid compressibility is a volumetric property (isothermal compressibility), not the fugacity coefficient of a gas.
Question 8 of 16 · Thermodynamics · 8–12 items
In a binary vapor-liquid equilibrium, what does the relative volatility α measure?
Solution
αAB = KA/KB = (yA/xA)/(yB/xB) = γAPAsat/(γBPBsat)
FE Reference Handbook, Chemical Engineering: vapor-liquid equilibrium and relative volatility.
αAB = KA/KB = (yA/xA)/(yB/xB); the larger it is, the easier the separation by distillation.
γA/γB is only one factor of α in modified Raoult's law; α also contains the ratio of saturation pressures.
Boiling-point difference hints at volatility but is not α; α is a ratio of equilibrium K-values at the same T and P.
The slope dy/dx of the x-y curve changes along the curve and equals α only at x → 0.
Question 9 of 16 · Thermodynamics · 8–12 items
For an ideal binary mixture obeying Raoult's law, the total pressure is written as Ptotal = x1P1sat + x2P2sat. What underlying assumption is built into this formulation?
Solution
Raoult's law: yiP = xiPisat. It assumes an ideal liquid solution (γi = 1, like and unlike interactions equal) and an ideal-gas vapor; non-ideal liquids need γi, non-ideal vapors need fugacity coefficients.
Raoult's law treats the liquid as an ideal solution (A-B interactions equal to A-A and B-B, γi = 1), with an ideal-gas vapor phase.
Raoult's law assumes the vapor is an ideal gas (yiP = partial pressure); a non-ideal vapor would need fugacity coefficients.
Each partial pressure xiPisat is directly proportional to the liquid mole fraction, so it depends on composition by construction.
The equation uses each component's own saturation pressure; if they were equal, there would be nothing to separate.
Question 10 of 16 · Thermodynamics · 8–12 items
What is the standard enthalpy of formation of a compound?
Solution
ΔHf° = enthalpy change for forming 1 mol of the compound from its elements in their standard states (25 °C, 1 bar); elements in their standard states have ΔHf° = 0.
That is the heat of solution; formation refers to building the compound from its elements.
ΔHf° is defined for one mole formed from its elements in their standard states (25 °C, 1 bar), with elements set to zero.
That is the heat of combustion, a different reference reaction, though formation enthalpies can be derived from it by Hess's law.
That is a latent heat (fusion, vaporization), not a heat of formation.
Question 11 of 16 · Thermodynamics · 8–12 items
Using the Clausius-Clapeyron equation, if the heat of vaporization of a liquid is 40 kJ/mol and its vapor pressure is 1 bar at 373 K, what is its vapor pressure at 393 K? (R = 8.314 J/mol·K)
Given
Heat of vaporization ΔHvap = 40 kJ/mol
Vapor pressure P1 = 1 bar at T1 = 373 K
Temperature T2 = 393 K
Gas constant R = 8.314 J/(mol·K)
Find
Vapor pressure at 393 K P2 (bar)
Solution
ln(P2/P1) = (ΔHvap/R)(1/T1 − 1/T2)
ln(P2/1 bar) = (40,000 J/mol ÷ 8.314 J/(mol·K))(1/373 K − 1/393 K) = 0.656 → P2 = 1.93 bar
FE Reference Handbook, Thermodynamics: Clausius-Clapeyron equation.
0.519 bar reverses the temperature terms (1/T2 − 1/T1); vapor pressure must rise when the liquid is heated.
1.00 bar puts ΔHvap in kJ/mol with R in J/(mol·K), which makes the exponent a thousand times too small.
1.66 bar adds the logarithm to P1 (1 + 0.656) instead of raising e to it.
ln(P2/P1) = (40,000/8.314)(1/373 − 1/393) = 0.656, so P2 = 1 bar × e0.656 = 1.93 bar.
Question 12 of 16 · Thermodynamics · 8–12 items
Component
Vapor Mole Fraction (yi)
Saturation Pressure (Pisat)
Activity Coefficient (γi)
1
0.40
100 kPa
1.20
2
0.60
200 kPa
1.50
A binary nonideal vapor mixture is at phase equilibrium with its liquid at a constant temperature. Using the modified Raoult's law equation and the phase data provided in the table, calculate the dew point pressure of this mixture.
Which of the following statements best describes why a spontaneous gas-phase reaction with a highly negative standard Gibbs free energy of reaction (ΔG° ≪ 0) typically does not proceed to 100% completion in a closed system?
Solution
G(ξ) of the reacting mixture includes the entropy of mixing, so it has a minimum short of full conversion; there ΔG = ΔG° + RT ln Q = 0 and Q = K. A very negative ΔG° only makes K very large.
At equilibrium the forward rate is not zero; it equals the reverse rate.
Mixing reactants and products lowers G, so the total G of the mixture passes through a minimum at partial conversion; that minimum is the equilibrium point.
ΔG° is fixed at a given temperature and does not depend on composition; it is the actual ΔG = ΔG° + RT ln Q that rises toward zero.
K is a constant at fixed temperature; what changes with conversion is the reaction quotient Q.
Formation enthalpies set ΔH° and the temperature dependence of K, but the stopping point is a Gibbs-energy minimum, not an enthalpy limit.
Question 14 of 16 · Thermodynamics · 8–12 items
Consider a multicomponent system with a liquid phase and a vapor phase initially not in equilibrium. For a particular species i, the chemical potential in the liquid phase is μiL and in the vapor phase is μiV. If μiL > μiV at constant temperature and pressure, which of the following describes the spontaneous direction of mass transfer and the thermodynamic driving force?
Species i evaporates until μiL = μiV, the phase-equilibrium condition.
At constant T and P a species moves down its chemical-potential gradient; liquid to vapor lowers total G until μiL = μiV.
Species move from higher to lower chemical potential; here the liquid is higher, so transfer runs liquid to vapor, not the reverse.
Spontaneous change at constant T and P lowers G; equilibrium is a minimum, never a maximum.
The direction is right, but the driving force is a decrease in G; moving species i lowers μi in the liquid rather than raising it.
Question 15 of 16 · Thermodynamics · 8–12 items
The Helmholtz energy is A = U − TS, so dA = −S dT − P dV. Which Maxwell relation follows from this differential?
Solution
dA = −S dT − P dV → (∂A/∂T)V = −S, (∂A/∂V)T = −P
∂2A/∂V∂T = ∂2A/∂T∂V → (∂S/∂V)T = (∂P/∂T)V
FE Reference Handbook, Thermodynamics: Maxwell relations (fundamental property relations).
This is a valid Maxwell relation, but it comes from the Gibbs energy, dG = −S dT + V dP, not from dA.
This pairs the wrong variables: (∂S/∂T)V = Cv/T, which is not a Maxwell relation at all.
From dA = −S dT − P dV, equal mixed second derivatives give (∂S/∂V)T = (∂P/∂T)V.
This is not a valid identity; the internal-energy and enthalpy relations are (∂T/∂V)S = −(∂P/∂S)V and (∂T/∂P)S = (∂V/∂S)P.
Question 16 of 16 · Thermodynamics · 8–12 items
During a plant scale-up from a benchtop reactor to a pilot-scale vessel, an engineer uses standard thermodynamic tables to determine the equilibrium state of a binary mixture. Which of the following sets of properties depends directly on the system mass and cannot be used alone to define the intensive thermodynamic state from standard tables?
Solution
Extensive properties (total H, S, V) scale with mass; intensive ones (T, P, specific or molar properties, μ) do not. Tables are indexed by intensive properties: divide a total by m (or n) before using it.
Molar quantities and chemical potential are intensive; they do not change with system size.
Totals scale with the amount of material (extensive); a property table is entered with intensive values, so H, S and V must first be divided by the mass or moles.
All three are intensive and are exactly what a property table is entered with.
Specific (per-mass) properties are intensive, so any two of them fix the state of a pure substance.
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Common traps
Ideal gas where tables are needed. Steam near saturation and any compressed liquid are far from ideal; take properties from the steam tables.
Sign of the heat of reaction. Exothermic means a negative ΔH; a balance that treats it as heat removed predicts a cooler outlet where it should run hotter.
Celsius in an exponential. Every equilibrium and vapor-pressure equation takes absolute temperature.
Unit systems. The reference’s steam charts include USCS (its Mollier chart is in Btu/lb); check the units before reading a value.
Sources: FE Reference Handbook, Thermodynamics chapter; Smith, Van Ness and Abbott, Introduction to Chemical Engineering Thermodynamics.
The outline lists cycles and efficiencies, so expect a question or two; the core of the area is phase and chemical equilibrium. Our cycle problems are on the FE Mechanical page.
Which property charts deserve practice time?
The ones the specification names: steam tables, psychrometric charts, T-s, P-h, x-y and T-x-y diagrams.
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