Free PE Chemical Practice Test

Answer: D. 17.5 wt%

At steady state with no reaction, ethanol in equals ethanol out.

Ethanol in = 100 × 0.40 + 300 × 0.10 = 40 + 30 = 70 kg/h

Total out = 100 + 300 = 400 kg/h

Mass fraction = 70 ÷ 400 = 17.5 wt%

Why not the others:

• A (32.5 wt%): Weights the fractions with the flows swapped (100 kg/h × 10% + 300 kg/h × 40%).
• B (25.0 wt%): A simple average of 40% and 10%. It ignores that Stream 2 is three times larger.
• C (23.3 wt%): Divides the 70 kg/h of ethanol by the 300 kg/h stream instead of the 400 kg/h total.

If you missed it: Write a component balance before you average anything. Fractions never add; masses do.

Source: Fogler, Elements of Chemical Reaction Engineering, 7th ed., companion site, Chapter 1 summary, General Mole Balance (In − Out + Generation = Accumulation). With no generation and no accumulation, In = Out for each component.

Answer: A. 99.4 kg/h

Overall solids balance (solids only leave in the product):

P = (1,000 × 0.12) ÷ 0.42 = 285.7 kg/h

Let B be the bypass flow, so the evaporator gets (1,000 − B). Solids are conserved through the evaporator, so its outlet mass is 0.12(1,000 − B) ÷ 0.58.

Mixing point: B + 0.12(1,000 − B) ÷ 0.58 = 285.7

B + 206.9 − 0.2069B = 285.7 → 0.7931B = 78.8 → B = 99.4 kg/h

Why not the others:

• B (285.7 kg/h): This is the product flow from the overall balance, not the bypass.
• C (347.8 kg/h): Applies the lever-rule fraction (0.58 − 0.42) ÷ (0.58 − 0.12) = 0.348 to the 1,000 kg/h feed. That fraction is the bypass share of the product (0.348 × 285.7 = 99.4 kg/h), not of the feed.
• D (714.3 kg/h): This is the water evaporated, 1,000 − 285.7.

If you missed it: Draw the box around the whole process first, then around the mixing point. The overall balance usually hands you one stream for free.

Source: Fogler companion site, Chapter 1 summary, General Mole Balance, applied here as a steady-state mass balance with no generation.

Answer: C. 40.2 wt%

Take 1 mol of gas as the basis and convert each component to mass:

CO₂: 0.30 × 44.01 = 13.20 g; N₂: 0.70 × 28.01 = 19.61 g

Mass fraction CO₂ = 13.20 ÷ (13.20 + 19.61) = 40.2 wt%

CO₂ is heavier than N₂, so its mass fraction is higher than its mole fraction. That's a quick direction check.

Why not the others:

• A (21.4 wt%): Swaps the molar masses, pairing CO₂ with 28.01 and N₂ with 44.01.
• B (30.0 wt%): Reports the mole fraction unchanged.
• D (47.1 wt%): Divides the CO₂ mass (13.20 g) by N₂'s molar mass instead of by the total mass.

If you missed it: Pick a basis (1 mol or 100 mol), convert every component, then add. Never convert a fraction directly.

Source: OpenStax Chemistry 2e, §3.1 Formula Mass and the Mole Concept, "The Mole" (mass = moles × molar mass).

Answer: C. 555.6 kg/h

Dry solids are the tie component: 1,000 × 0.40 = 400 kg/h in, and the same 400 kg/h out.

Product = 400 ÷ 0.90 = 444.4 kg/h

Water removed = 1,000 − 444.4 = 555.6 kg/h

Why not the others:

• A (444.4 kg/h): The dried product flow, not the water removed.
• B (500.0 kg/h): Subtracts the percentages (60% − 10% = 50%) and applies that to the feed. Percentages on different total flows can't be subtracted.
• D (666.7 kg/h): Divides the solids by 0.60, which mixes up the water and solids fractions.

If you missed it: Find the component that passes through unchanged (here, dry solids) and use it to link the inlet and outlet totals.

Source: Fogler companion site, Chapter 1 summary, General Mole Balance (In − Out + Generation = Accumulation), applied as a steady-state mass balance with no generation.

Answer: B. 3.8%

Basis: 1 mol CH₄. CH₄ + 2 O₂ → CO₂ + 2 H₂O, so the theoretical O₂ is 2 mol.

O₂ supplied = 2 × 1.20 = 2.40 mol; N₂ supplied = 2.40 × 79/21 = 9.03 mol

Flue gas: CO₂ 1, H₂O 2, O₂ 0.40, N₂ 9.03 mol

Dry basis drops the water: total = 1 + 0.40 + 9.03 = 10.43 mol

y(O₂) = 0.40 ÷ 10.43 = 3.8%

Why not the others:

• A (3.2%): The wet-basis fraction, 0.40 ÷ 12.43. The question asks for dry basis.
• C (4.5%): Counts only the N₂ that came with the theoretical O₂ (7.52 mol) and forgets the N₂ carried in with the excess air.
• D (20.0%): The percent excess air, not a flue-gas composition.

If you missed it: Every mole of excess O₂ drags 3.76 mol of N₂ with it. And always check whether the analysis is wet or dry.

Source: OpenStax Chemistry 2e, §4.4 Reaction Yields, "Limiting Reactant" (excess reactant left after the limiting reactant is consumed). Air composition is a stated input.

Answer: B. 88.4%

At steady state, the inert can leave only in the purge:

Purge = 1 mol/h I ÷ 0.08 = 12.5 mol/h

A lost in the purge = 0.92 × 12.5 = 11.5 mol/h

Overall conversion = (99 − 11.5) ÷ 99 = 88.4%

Why not the others:

• A (87.5%): Computes 1 − 12.5/100, mixing the purge total with fresh-feed A.
• C (92.0%): Uses the A fraction in the recycle as if it were the conversion.
• D (98.9%): Assumes only the inert is purged, so almost no A is lost.

If you missed it: For purge problems, balance the inert first. It has one exit, so it sets the purge rate.

Source: Fogler companion site, Chapter 1 summary, General Mole Balance; overall conversion is computed from the overall balance around the whole process.

Answer: B. 12.0%

O₂ needed to react all the NH₃: 100 × 5/4 = 125 mol/h. NH₃ is the limiting reactant.

Percent excess = (supplied − theoretical) ÷ theoretical = (140 − 125) ÷ 125 = 12.0%

Why not the others:

• A (10.7%): Divides the 15 mol/h excess by the 140 mol/h supplied instead of the 125 mol/h theoretical.
• C (25.0%): The stoichiometric ratio 5/4 = 1.25 read as an excess. It ignores the actual feed.
• D (40.0%): Compares O₂ to NH₃ mole for mole (140 vs. 100), ignoring the 5:4 stoichiometry.

If you missed it: Percent excess always divides by the theoretical requirement of the limiting reactant.

Source: OpenStax Chemistry 2e, §4.4 Reaction Yields, "Limiting Reactant" (limiting and excess reactants).

Answer: C. 714 kW

Two steps: heat the liquid, then boil it.

Sensible: 4.18 × (100 − 25) = 313.5 kJ/kg

Latent: 2,257 kJ/kg

Q = 1,000 kg/h × (313.5 + 2,257) kJ/kg ÷ 3,600 s/h = 714 kW

Why not the others:

• A (87.1 kW): Sensible heat only. It never boils the water.
• B (627 kW): Latent heat only. It skips heating from 25°C to 100°C.
• D (743 kW): Heats from 0°C instead of 25°C.

If you missed it: Sketch the heating curve and add one term per segment: liquid, phase change, vapor if any.

Source: OpenStax Chemistry 2e, §5.1 Energy Basics, "Thermal Energy, Temperature, and Heat" (q = c × m × ΔT); DOE-HDBK-1012/1-92, Thermodynamics, HT-01, First Law of Thermodynamics, Eq. 1-22 (steady-flow energy balance in enthalpy terms).

Answer: 357 kg/h (accepted range: 355 to 359 kg/h)

Duty on the process side: Q = 5,000 × 2.5 × (80 − 20) = 750,000 kJ/h

Steam condensed = 750,000 ÷ 2,100 = 357 kg/h

Common wrong answers:

• 476: Uses the outlet temperature (80°C) as the temperature change.
• 750,000: Stops at the heat duty in kJ/h and never converts to steam flow.

If you missed it: Write "heat lost by steam = heat gained by liquid," then solve for the unknown flow.

Source: OpenStax Chemistry 2e, §5.1 (q = c × m × ΔT); DOE-HDBK-1012/1-92, HT-01, First Law, Eq. 1-22.

Answer: C. 54.9°C

Heat lost by Stream 1 = heat gained by Stream 2:

2.0 × 4.18 × (80 − T) = 3.0 × 2.0 × (T − 20)

668.8 − 8.36T = 6.0T − 120 → 14.36T = 788.8 → T = 54.9°C

The hot stream carries more heat capacity per second (8.36 vs. 6.0 kW/K), so the blend lands closer to its temperature.

Why not the others:

• A (44.0°C): Weights by mass flow only and ignores the different heat capacities.
• B (50.0°C): A simple average of the two temperatures.
• D (60.6°C): Weights by heat capacity only and ignores the flow rates.

If you missed it: The weighting factor in a mixing energy balance is ṁ·cp, not ṁ alone.

Source: OpenStax Chemistry 2e, §5.1 Energy Basics, "Thermal Energy, Temperature, and Heat" (q = c × m × ΔT; heat lost by one substance equals heat gained by the other).

Answer: B. 69.0 kg/h

Energy in = energy out, and the steam ends up as part of the 60°C liquid:

1,000 × 83.9 + m × 2,676 = (1,000 + m) × 251.1

m × (2,676 − 251.1) = 1,000 × (251.1 − 83.9)

m = 167,200 ÷ 2,424.9 = 69.0 kg/h

Why not the others:

• A (62.5 kg/h): Divides by the steam enthalpy alone, as if the condensed steam left at 0 kJ/kg rather than at 60°C.
• C (74.1 kg/h): Credits the steam with only its latent heat (2,257 kJ/kg) and forgets that the condensate then cools from 100°C to 60°C.
• D (167.2 kg/h): The water's enthalpy gain in kJ/kg times 1,000, read as a steam flow.

If you missed it: With tabulated enthalpies, write in = out for every stream at its own state. Don't split it into latent and sensible pieces unless you must.

Source: DOE-HDBK-1012/1-92, Thermodynamics, HT-01, First Law of Thermodynamics, Eq. 1-22 (steady-flow energy balance in enthalpy terms). Enthalpies are supplied as inputs.

Answer: A. Remove about 4.58 MW

ΔH°rxn = Σ ΔH°f(products) − Σ ΔH°f(reactants) = (−393.5 + 0) − (−110.5 − 241.8) = −41.2 kJ/mol

The reaction is exothermic. Heat released = 41.2 kJ/mol × 400,000 mol/h ÷ 3,600 s/h = 4,578 kW, so about 4.58 MW must be removed.

Why not the others:

• B (Add about 4.58 MW): Right magnitude, wrong sign. A negative ΔH means heat is released.
• C (Add about 0.31 MW): Uses liquid water's heat of formation (−285.8 kJ/mol). The stem says H₂O(g).
• D (Remove about 43.7 MW): Uses only the CO₂ heat of formation and never subtracts the reactants.

If you missed it: Check the phase of every species in the equation before you pull a heat of formation.

Source: OpenStax Chemistry 2e, §5.3 Enthalpy, "Standard Enthalpy of Formation" and "Hess's Law."

Answer: A. Remove 13.3 kJ

Choose a path: react at 25°C, then heat the products to 400°C. Enthalpy is a state function, so any path gives the same total.

React at 25°C: −41.2 kJ

Heat products: (45.0 + 29.5) × (400 − 25) = 27,940 J = +27.9 kJ

Q = −41.2 + 27.9 = −13.3 kJ, so 13.3 kJ must be removed per mole of CO

Why not the others:

• B (Remove 41.2 kJ): Uses only the heat of reaction, as if the products left at 25°C.
• C (Add 27.9 kJ): Counts only the sensible heat of the products.
• D (Remove 69.1 kJ): Subtracts the sensible heat instead of adding it. Heating the products absorbs part of the heat released.

If you missed it: Draw the path (reactants at T₁ → react at reference → products to T₂) and give each leg its own sign.

Source: OpenStax Chemistry 2e, §5.3 Enthalpy, "Hess's Law" (enthalpy change of a total process equals the sum of the steps); §5.1 (q = c × m × ΔT, used here per mole).

Answer: B. 802.5 kJ

ΔH°c = [−393.5 + 2(−241.8)] − [−74.6] = −877.1 + 74.6 = −802.5 kJ/mol

So 802.5 kJ is released per mole of methane with water vapor in the products.

Why not the others:

• A (560.7 kJ): Counts only one mole of water.
• C (877.1 kJ): Leaves out the methane's own heat of formation.
• D (890.5 kJ): Uses liquid water (−285.8 kJ/mol). The difference is the heat released when the water vapor condenses.

If you missed it: Multiply each heat of formation by its coefficient, and check the phase of the water.

Source: OpenStax Chemistry 2e, §5.3 Enthalpy, "Standard Enthalpy of Formation" and "Hess's Law."

Answer: C. 582 m³/h

V̇ = ṅRT ÷ P = (100,000 mol/h × 8.314 × 350) ÷ 500,000 Pa = 582 m³/h

Why not the others:

• A (58.2 m³/h): Uses 5 MPa instead of 5 bar (500 kPa).
• B (128 m³/h): Uses the temperature in °C (76.85) instead of kelvin.
• D (5,820 m³/h): Uses 50 kPa for 5 bar.

If you missed it: Convert to absolute T and P first: 1 bar = 100 kPa.

Source: OpenStax Chemistry 2e, §9.2, "The Ideal Gas Law" (PV = nRT); DOE-HDBK-1012/1-92, HT-01, Ideal Gas Law, Eq. 1-43 (absolute P and T).

Answer: A. 44.9 m³/h

The compressibility factor corrects the ideal gas law: PV = ZnRT.

V̇ = ZṅRT ÷ P = 0.90 × 100,000 × 8.314 × 300 ÷ 5,000,000 = 44.9 m³/h

Z below 1 means the real gas takes up less volume than an ideal gas would at the same conditions.

Why not the others:

• B (49.9 m³/h): The ideal-gas volume; it ignores Z.
• C (55.4 m³/h): Divides by Z instead of multiplying.
• D (449 m³/h): Uses 5 bar instead of 50 bar.

If you missed it: Remember the meaning of Z: actual molar volume divided by ideal molar volume at the same T and P. Then the direction is obvious.

Source: OpenStax Chemistry 2e, §9.6 Non-Ideal Gas Behavior (compressibility factor Z = PV_m/RT, the measured molar volume relative to the ideal molar volume).

Answer: D. Not possible, because the Carnot limit for these temperatures is 50%

No engine operating between two reservoirs can beat a reversible (Carnot) engine:

η_max = 1 − T_C/T_H = 1 − 300/600 = 50%

The claimed 55% exceeds that, so the claim violates the second law.

Why not the others:

• A: The first law allows anything up to 100%; the second law caps it lower.
• B: Converts the kelvin temperatures to °C. The Carnot equation needs absolute temperatures.
• C: There's no universal 33% cap. The limit depends on the reservoir temperatures.

If you missed it: For any cycle claim, compute the Carnot limit first. It takes ten seconds and catches impossible numbers.

Source: DOE-HDBK-1012/1-92, HT-01, Second Law of Thermodynamics, Carnot efficiency η = 1 − T_C/T_H (Eq. 1-23), with absolute temperatures.

Answer: D. 5.82 MW

Ideal work = 3,230 − 2,520 = 710 kJ/kg

Turbine efficiency = actual work ÷ ideal work, so actual work = 0.82 × 710 = 582.2 kJ/kg

Power = 10 kg/s × 582.2 kJ/kg = 5.82 MW

Why not the others:

• A (26.5 MW): Multiplies the flow by the actual exit enthalpy (2,647.8 kJ/kg) instead of by the enthalpy drop.
• B (8.66 MW): Divides by efficiency. That's the rule for pumps and compressors, which need more work than ideal.
• C (7.10 MW): The isentropic (ideal) power.

If you missed it: Ask whether the machine produces or consumes work. Producers deliver less than ideal; consumers need more.

Source: DOE-HDBK-1012/1-92, HT-01, Second Law, Power Plant Components, Eqs. 1-26 and 1-27 (turbine efficiency); Eq. 1-34 (pump efficiency).

Answer: 446 K (accepted range: 445 to 447 K)

For an isentropic process of an ideal gas, T₂/T₁ = (P₂/P₁)^[(k − 1)/k]:

T₂ = 300 × 4^(0.4/1.4) = 300 × 4^0.2857 = 300 × 1.486 = 446 K

Common wrong answers:

• 808: Uses the exponent 1/k instead of (k − 1)/k.
• 38,400: Uses the exponent k/(k − 1), which is the inverse relation (pressure ratio from temperature ratio).

If you missed it: The exponent on the pressure ratio is small, (k − 1)/k ≈ 0.29 for air. If the answer more than doubles the temperature for a 4:1 ratio, the exponent is upside down.

Source: NASA Glenn Research Center, "Isentropic Flow Equations", Eq. #4 (p/pt = (T/Tt)^[γ/(γ−1)]), rearranged for temperature.

Answer: C. 24.3%

Basis 1 mol A, extent ξ: A = 1 − ξ, B = 2ξ, total = 1 + ξ.

Kp = (y_B P)² ÷ (y_A P) = 4ξ²P ÷ [(1 − ξ)(1 + ξ)] = 4ξ²P ÷ (1 − ξ²)

With P = 2: 8ξ² = 0.5(1 − ξ²) → 8.5ξ² = 0.5 → ξ = 0.243, or 24.3% conversion

Why not the others:

• A (11.1%): Drops the square on the B term.
• B (22.1%): Forgets that total moles grow to 1 + ξ.
• D (33.3%): Uses 1 bar instead of 2 bar. Because moles increase, higher pressure pushes conversion down.

If you missed it: Build an extent table (start, change, end, mole fraction) before writing K.

Source: OpenStax Chemistry 2e, §13.2 Equilibrium Constants, Qp and Kp for gas-phase reactions; Kp expression confirmed on the LibreTexts mirror of §13.2.

Answer: A. Reverse, making more N₂O₄

Compute the reaction quotient with the current partial pressures:

Qp = (P_NO₂)² ÷ P_N₂O₄ = 0.50² ÷ 1.0 = 0.25

Qp (0.25) > Kp (0.15): there are too many products, so the reaction runs in reverse until Q falls to K.

Why not the others:

• B: Forward would make Q even larger.
• C: Equilibrium requires Q = K.
• D: The partial pressures are all Q needs.

If you missed it: Q < K, go forward. Q > K, go back. Q = K, stay.

Source: OpenStax Chemistry 2e, §13.2 Equilibrium Constants (Qp for gases; Q < K shifts forward, Q > K shifts reverse).

Answer: A and D

A: Four moles of gas become two. Compression shifts equilibrium toward fewer gas moles, toward NH₃.

D: For an exothermic reaction, lowering temperature raises K, favoring products.

Why not the others:

• B: Higher temperature lowers K for an exothermic reaction.
• C: A catalyst speeds the approach to equilibrium without changing the equilibrium constant.
• E: Expansion lowers pressure and shifts toward more gas moles, away from NH₃.

If you missed it: Separate "where equilibrium lies" (K, composition) from "how fast you get there" (rate, catalyst).

Source: OpenStax Chemistry 2e, §13.3 Shifting Equilibria: Le Châtelier's Principle, "Effect of Pressure/Volume Changes," "Effect of a Change in Temperature," and "Effect of a Catalyst."

Answer: C. 0.627

Partial pressures: benzene = 0.40 × 136 = 54.4 kPa; toluene = 0.60 × 54 = 32.4 kPa

Bubble-point pressure = 54.4 + 32.4 = 86.8 kPa

y(benzene) = 54.4 ÷ 86.8 = 0.627

The vapor is richer in the more volatile component, which is why distillation works.

Why not the others:

• A (0.373): The toluene fraction in the vapor.
• B (0.400): The liquid composition. Vapor and liquid differ at equilibrium.
• D (0.716): Ratios the pure vapor pressures (136 ÷ 190) and ignores the liquid composition.

If you missed it: Raoult's law gives each partial pressure; Dalton's law turns partial pressures into vapor mole fractions.

Source: OpenStax Chemistry 2e, §11.4 Colligative Properties, "Vapor Pressure Lowering" (P_A = X_A P°_A) and "Distillation of Solutions"; §9.3, "The Pressure of a Mixture of Gases: Dalton's Law" (P_A = X_A P_total).

Answer: B. 0.431 atm

ln(P₂/P₁) = (ΔHvap/R)(1/T₁ − 1/T₂)

= (38,600 ÷ 8.314) × (1/351 − 1/330) = 4,643 × (−1.813 × 10⁻⁴) = −0.842

P₂ = 1 atm × e^(−0.842) = 0.431 atm

Below the boiling point, the vapor pressure must be below 1 atm, which is a quick sanity check.

Why not the others:

• A (0.144 atm): Takes 10 to the power −0.842 instead of e.
• C (0.999 atm): Leaves ΔHvap in kJ/mol while R is in J/(mol·K).
• D (2.32 atm): Flips the sign of the temperature term. Above 1 atm is impossible below the normal boiling point.

If you missed it: Check units on ΔH and R together, and check the direction of the answer before you calculate.

Source: OpenStax Chemistry 2e, §10.3 Phase Transitions, "Boiling Points" (two-point Clausius–Clapeyron equation).

Answer: C. 8.7 mg/L

Henry's law uses the gas's partial pressure: P_O₂ = 0.21 × 1.00 = 0.21 atm.

C = kP = 1.3 × 10⁻³ × 0.21 = 2.73 × 10⁻⁴ mol/L

× 32.0 g/mol = 8.7 × 10⁻³ g/L = 8.7 mg/L

Why not the others:

• A (0.27 mg/L): Reports 0.273 mmol/L as if it were mg/L.
• B (7.6 mg/L): Uses N₂'s molar mass (28.0 g/mol) by mistake.
• D (41.6 mg/L): Uses the total pressure, as if the water were exposed to pure O₂.

If you missed it: In Henry's law, P is always the partial pressure of the dissolving gas.

Source: OpenStax Chemistry 2e, §11.3 Solubility, "Solutions of Gases in Liquids" (C_g = kP_g). The constant is supplied as an input.

Answer: A. 0.284

At the dew point, the liquid mole fractions must sum to 1, with x_i = y_i P ÷ P_i*:

1/P = y_B/P_B + y_T/P_T = 0.5/136 + 0.5/54 → P = 77.3 kPa

x_B = 0.5 × 77.3 ÷ 136 = 0.284

The first liquid is rich in the heavier component, toluene, the mirror image of the bubble point.

Why not the others:

• B (0.358): Ratios the pure vapor pressures with equal weights (0.5 × 136 ÷ 190). It never finds the dew-point pressure.
• C (0.500): Assumes the liquid matches the vapor. At equilibrium they differ.
• D (0.716): The toluene fraction in the drop.

If you missed it: Bubble point: sum of y = 1. Dew point: sum of x = 1. Write whichever sum the question fixes.

Source: OpenStax Chemistry 2e, §11.4 Colligative Properties, "Vapor Pressure Lowering" (Raoult's law); §9.3, "Dalton's Law." The dew-point relation is derived from the two laws.

Answer: D. 1,008 W/m²

The layers are thermal resistances in series:

R = 0.20/1.2 + 0.10/0.15 = 0.167 + 0.667 = 0.833 m²·K/W

q = ΔT ÷ R = (900 − 60) ÷ 0.833 = 1,008 W/m²

The firebrick–insulation interface sits at 900 − 1,008 × 0.167 = 732°C, a check worth making when the insulation has a temperature limit.

Why not the others:

• A (6,300 W/m²): Adds the layer conductances (k/Δx) as if the layers were in parallel.
• B (5,040 W/m²): Uses only the firebrick's resistance.
• C (1,260 W/m²): Uses only the insulating layer's resistance.

If you missed it: Layers the heat must pass through one after another add as resistances, not conductances.

Source: DOE-HDBK-1012/2-92, Heat Transfer, HT-02, Conduction Heat Transfer, "Equivalent Resistance Method" (Q̇ = ΔT/R_th, R_th = Δx/k, Eq. 2-6), p. 9; composite-wall example, p. 10.

Answer: B. 63.5 W/m

For radial conduction through a cylinder:

Q/L = 2πk(T_i − T_o) ÷ ln(r_o/r_i) = 2π × 0.05 × 140 ÷ ln 2 = 43.98 ÷ 0.693 = 63.5 W/m

Why not the others:

• A (44.0 W/m): A plane-wall calculation using the inner surface area. The heat-flow area grows with radius, so a flat-wall formula is wrong here.
• C (88.0 W/m): The same plane-wall error using the outer surface area.
• D (146 W/m): Uses log₁₀ 2 instead of ln 2.

If you missed it: When insulation thickness is comparable to the pipe radius, use the log form. The plane-wall formula only works for thin layers.

Source: DOE-HDBK-1012/2-92, Heat Transfer, HT-02, Conduction—Cylindrical Coordinates, Q̇ = 2πkL(ΔT)/ln(r_o/r_i), Eq. 2-8, p. 13.

Answer: B. 9.6 kW

Q = hAΔT = 8 × 30 × (60 − 20) = 9,600 W = 9.6 kW

A real heat-loss check would add radiation too; the question leaves it out to isolate convection.

Why not the others:

• A (0.32 kW): Leaves out the area.
• C (14.4 kW): Uses the surface temperature (60) as the temperature difference.
• D (80.0 kW): Uses the surface temperature in kelvin (333 K) as the difference. Convection uses a temperature difference, in which °C and K steps are equal.

If you missed it: Convection and conduction use temperature differences; radiation uses absolute temperatures to the fourth power.

Source: DOE-HDBK-1012/2-92, HT-02, Convection Heat Transfer, Q̇ = hAΔT, Eq. 2-9, p. 19.

Answer: C. 3,084 W

The surface emits σAT₁⁴ and absorbs σAT₂⁴ from its surroundings, so the net loss is:

Q = σA(T₁⁴ − T₂⁴) = 5.67 × 10⁻⁸ × 1.0 × (500⁴ − 300⁴) = 5.67 × 10⁻⁸ × 5.44 × 10¹⁰ = 3,084 W

Why not the others:

• A (90.7 W): Raises the temperature difference to the fourth power, (200)⁴.
• B (151 W): Uses °C instead of kelvin.
• D (3,544 W): Total emission at 500 K. It ignores radiation absorbed from the surroundings.

If you missed it: Radiation always uses absolute temperature, and the fourth power applies to each temperature, not the difference.

Source: DOE-HDBK-1012/2-92, HT-02, Radiant Heat Transfer, "Black Body Radiation" (Q̇ = σAT⁴, Eq. 2-12), p. 26. The net-exchange form for a black surface in black surroundings is derived from that law.

Answer: C. 19,100 kg/h

Heat removed from the vapor: Q = 2,000 × 400 = 800,000 kJ/h. The solvent stays at its saturation temperature; only its phase changes.

Water flow = Q ÷ (cp ΔT) = 800,000 ÷ (4.18 × 10) = 19,100 kg/h

Why not the others:

• A (5,470 kg/h): Uses the outlet temperature (35) as ΔT.
• B (7,660 kg/h): Uses the inlet temperature (25) as ΔT.
• D (191,000 kg/h): Uses ΔT = 1 K, dropping the temperature rise altogether.

If you missed it: On a condenser, the condensing side's duty is latent heat; the coolant's duty is sensible heat. Set them equal.

Source: OpenStax Chemistry 2e, §5.1 (q = c × m × ΔT); OpenStax §10.3 (heat of vaporization); DOE-HDBK-1012/1-92, HT-01, First Law, Eq. 1-22.

Answer: 1 → a, 2 → b, 3 → c, 4 → d

Nucleate boiling is the efficient regime: bubbles stir the liquid at the surface. As heat flux rises, bubbles can merge into patches of vapor film (partial film boiling), and at still higher surface temperature a stable vapor blanket forms (film boiling). Vapor conducts heat poorly, so the surface temperature can climb sharply as film coverage develops. Departure from nucleate boiling (DNB) occurs at the critical heat flux (CHF), where nucleate boiling breaks down and the surface moves into the transition/partial-film region.

Why not the others:

• e: Heating a liquid with no vapor formation is single-phase convection, not a boiling regime.

If you missed it: Picture the surface: bubbles leaving (nucleate), patches of film (partial film), then a full blanket (film). DNB/CHF is the limit where nucleate boiling breaks down—not a claim that stable film boiling appears instantaneously.

Source: DOE-HDBK-1012/2-92, HT-02, Boiling Heat Transfer summary (nucleate boiling, film boiling, DNB, and critical heat flux).

Answer: 20.6 m² (accepted range: 20.4 to 20.8 m²)

Counterflow end differences: ΔT₁ = 160 − 50 = 110 K; ΔT₂ = 60 − 20 = 40 K

LMTD = (110 − 40) ÷ ln(110/40) = 70 ÷ 1.012 = 69.2 K

A = Q ÷ (U × LMTD) = 500,000 ÷ (350 × 69.2) = 20.6 m²

Common wrong answers:

• 19.0: Uses the arithmetic mean, (110 + 40)/2 = 75 K, which overstates the driving force when the end differences are unequal.
• 29.0: Pairs the ends as if the flow were parallel (140 K and 10 K), giving an LMTD of 49.3 K.

If you missed it: Sketch both streams along the exchanger and label which ends face each other before computing ΔT₁ and ΔT₂.

Source: DOE-HDBK-1012/2-92, HT-02, Heat Transfer Terminology, "Log Mean Temperature Difference" (Eq. 2-2), p. 3; Heat Exchangers, Q̇ = U₀A₀ΔT_lm, pp. 37 and 39.

Answer: C. 67.1°C

Duty from the oil side: Q = 4.0 × 2.2 × (150 − 90) = 528 kW

Water temperature rise = 528 ÷ (3.0 × 4.18) = 42.1 K

Outlet = 25 + 42.1 = 67.1°C

Why not the others:

• A (42.1°C): The temperature rise, never added to the 25°C inlet.
• B (56.6°C): Uses the oil's mass flow (4.0 kg/s) for the water.
• D (105.0°C): Uses the oil's heat capacity for the water.

If you missed it: Work one stream completely (duty), then the other. Keep each stream's own flow and cp together.

Source: OpenStax Chemistry 2e, §5.1 (q = c × m × ΔT); DOE-HDBK-1012/1-92, HT-01, First Law, Eq. 1-22.

Answer: D. Double the outside film coefficient

1/U = 1/h_i + Δx/k + 1/h_o = 0.000667 + 0.000040 + 0.002000 = 0.002707 → U = 369 W/(m²·K)

The outside film is about three-quarters of the total resistance, so it controls.

Why not the others:

• A: Helps, but the inside film was only about a quarter of the resistance.
• B and C: The wall is about 1.5% of the resistance, so even removing it barely moves U.

If you missed it: Before improving an exchanger, list each resistance and attack the largest one.

Source: DOE-HDBK-1012/2-92, HT-02, Convection Heat Transfer, overall coefficient for thin-walled tubes, U₀ = 1/(1/h₁ + Δr/k + 1/h₂), Eq. 2-11.

Answer: A.

Rearrange Q = UA·ΔT_lm:

U = 400,000 ÷ (25 × 40) = 400 W/(m²·K)

That's about 27% below the 550 W/(m²·K) clean value. Deposits on a heat-transfer surface act like an insulating layer, so a falling U over time is the classic sign of fouling.

Why not the others:

• B: The comparison that matters is with the clean value, and 400 is lower, not higher.
• C: Leaves the duty in kW instead of W.
• D: Leaves out the LMTD (400 ÷ 25).

If you missed it: For performance checks, back-calculate U from operating data and trend it against the clean or design value.

Source: DOE-HDBK-1012/2-92, HT-02, Heat Exchangers, Q̇ = U₀A₀ΔT_lm, pp. 37 and 39; deposits acting as an insulating blanket that reduces heat transfer, p. 50.

Answer: B. rate = k[A]²

For a rate law of the form rate = k[A]^m[B]^n:

• Trials 1 → 2: [A] doubles, [B] fixed, rate ×4. 2^m = 4, so m = 2.
• Trials 2 → 3: [B] triples, [A] fixed, rate unchanged. 3^n = 1, so n = 0.

Why not the others:

• A: Would double the rate when [A] doubles and triple it when [B] triples.
• C: Gets the A order right but wrongly includes B.
• D: First order in A would only double the rate.

If you missed it: Change one concentration at a time and ask: by what power of the concentration ratio did the rate change?

Source: OpenStax Chemistry 2e, §12.3 Rate Laws (rate law form and the method of initial rates). The orders are derived from the supplied data.

Answer: D. 53.6 kJ/mol

ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂)

ln 2 = (Ea ÷ 8.314) × (1/300 − 1/310) = (Ea ÷ 8.314) × 1.075 × 10⁻⁴

Ea = 0.693 × 8.314 ÷ 1.075 × 10⁻⁴ = 53,600 J/mol = 53.6 kJ/mol

Why not the others:

• A (0.58 kJ/mol): Uses °C (27 and 37) instead of kelvin.
• B (6.45 kJ/mol): Leaves out R.
• C (23.3 kJ/mol): Uses log₁₀ 2 instead of ln 2.

If you missed it: Arrhenius is a natural-log, absolute-temperature equation. Both slips are easy to make under time pressure.

Source: OpenStax Chemistry 2e, §12.5 Collision Theory, "Activation Energy and the Arrhenius Equation" (two-temperature form, written there as ln(k₁/k₂) = (Ea/R)(1/T₂ − 1/T₁), which is algebraically the same).

Answer: B. 0.75

A reacted = 100 − 20 = 80 mol/h

Yield = F_D ÷ (F_A0 − F_A) = 60 ÷ 80 = 0.75

Why not the others:

• A (0.60): Divides by A fed (100) instead of A reacted. Some texts call this a yield too, which is why reading the stated definition matters.
• C (0.80): The conversion of A.
• D (3.0): The overall selectivity of D to U (60 ÷ 20).

If you missed it: Write the definition from the question in symbols before plugging in. Yield and selectivity have several conventions.

Source: Fogler companion site, Chapter 8 summary, item 1, "Types of Multiple Reactions, Selectivity and Yield" (overall yield Ỹ_D = F_D/(F_A0 − F_A); overall selectivity S̃_D/U = F_D/F_U).

Answer: D. 72.0 m³

For a first-order CSTR: τk = X/(1 − X)

τ = 0.90 ÷ [0.25 × (1 − 0.90)] = 36.0 min

V = τ × v₀ = 36.0 × 2.0 = 72.0 m³

A plug-flow reactor would need only τ = ln(10)/0.25 = 9.2 min, or 18.4 m³. The CSTR runs entirely at the low outlet concentration, so it needs about four times the volume at 90% conversion.

Why not the others:

• A (7.2 m³): Uses τ = X/k, leaving out the (1 − X) term.
• B (18.4 m³): The PFR volume.
• C (36.0 m³): The space time in minutes, never multiplied by the flow rate.

If you missed it: Write the design equation for the reactor type named, solve for τ, then convert τ to volume.

Source: University of Michigan ChE 344, Lecture 6, "Isothermal reactor design" (CSTR: τk = X/(1−X); PFR: ln[1/(1−X)] = τk; τ = V/v₀).

Answer: B. 96.7%

Each tank: τ = 36 ÷ 2.0 = 18 min, so τk = 4.5.

For a first-order CSTR, the fraction of A remaining per tank is 1 ÷ (1 + τk) = 1/5.5.

After two tanks: (1/5.5)² = 0.033, so X = 96.7%

Splitting the same volume into two tanks in series beats one big tank (90%), because the first tank runs at a higher concentration of A, so it reacts faster.

Why not the others:

• A (99.99%): A 72 m³ plug-flow reactor (τk = 9), not two CSTRs.
• C (90.0%): The single 72 m³ CSTR from Question 40.
• D (81.8%): Stops after the first tank.

If you missed it: For CSTRs in series, apply the single-tank equation tank by tank; the outlet of one is the feed to the next.

Source: University of Michigan ChE 344, Lecture 6, "Isothermal reactor design" (first-order CSTR, X = τk/(1 + τk)), applied to each tank in turn.

Answer: C. 13.4 h

The first-order integrated rate law: ln([A]₀/[A]) = kt. At 80% conversion, [A]₀/[A] = 5.

t = ln 5 ÷ 0.12 = 1.609 ÷ 0.12 = 13.4 h

Why not the others:

• A (5.8 h): Uses log₁₀ 5.
• B (6.7 h): Divides the conversion by k, as if the rate stayed at its initial value.
• D (33.3 h): Uses the CSTR design equation, X/[k(1 − X)].

If you missed it: A batch reactor follows the same time path as a plug-flow reactor: ln form, not the CSTR form.

Source: OpenStax Chemistry 2e, §12.4 Integrated Rate Laws, "The Integrated Rate Law for First-Order Reactions" (ln([A]₀/[A]ₜ) = kt).

Answer: 160°C (accepted range: 159 to 161°C)

For an adiabatic reactor with ΔCp = 0:

T = T₀ + (−ΔH_rx)X ÷ ΣΘᵢCpᵢ = 40 + (60,000 × 0.80) ÷ 400 = 40 + 120 = 160°C

Common wrong answers:

• 190: Uses complete conversion instead of 80%.
• 120: Reports the temperature rise and forgets to add the feed temperature.

If you missed it: In an adiabatic reactor, temperature and conversion move together along a straight line. Know its slope and intercept.

Source: Fogler companion site, Chapter 11 summary, item 1, "Adiabatic CSTR, PFR, Batch, PBR," Eq. 1.B, valid when Ẇs = 0 and ΔCp = 0.

Answer: D. 1,800; laminar

87 cP = 0.087 Pa·s

Re = ρvD ÷ μ = 870 × 1.8 × 0.10 ÷ 0.087 = 1,800

The DOE handbook calls flow laminar below Re = 2,000, transitional from 2,000 to 3,500, and turbulent above 3,500. Other references draw these lines slightly differently, so use the thresholds your reference states.

Why not the others:

• A (180,000): Converts 87 cP to 0.00087 Pa·s, treating centipoise as if 1 cP were 10⁻⁵ Pa·s.
• B (18,000): Converts 87 cP to 0.0087 Pa·s, a factor-of-10 slip.
• C: 1,800 is below the 2,000 transition threshold.

If you missed it: Memorize 1 cP = 0.001 Pa·s. Viscosity conversions cause more Reynolds number errors than the formula does.

Source: DOE-HDBK-1012/3-92, Fluid Flow, HT-03, Laminar and Turbulent Flow, "Reynolds Number," Eq. 3-7, p. 19; regime thresholds, p. 20.

Answer: B. 16.8 kPa

Continuity: area scales with diameter squared, so halving the diameter quadruples the velocity: v₂ = 1.5 × 4 = 6.0 m/s.

Bernoulli, horizontal and frictionless: P₁ − P₂ = ρ(v₂² − v₁²)/2 = 998 × (36 − 2.25) ÷ 2 = 16.8 kPa

Why not the others:

• A (3.4 kPa): Scales velocity with diameter instead of diameter squared (v₂ = 3.0 m/s).
• C (18.0 kPa): Ignores the upstream velocity head.
• D (33.7 kPa): Drops the ½ in the kinetic-energy term.

If you missed it: Do continuity first (velocities), then Bernoulli (pressures).

Source: DOE-HDBK-1012/3-92, Fluid Flow, HT-03, Continuity Equation, Eq. 3-5, p. 11; Simplified Bernoulli Equation, Eq. 3-11, p. 22.

Answer: D. 40.4 m

Extended Bernoulli between the two liquid surfaces: pump head = elevation gain + pressure-head gain + friction loss.

Pressure head = 200,000 ÷ (998 × 9.81) = 20.4 m

H_p = 15 + 20.4 + 5 = 40.4 m

Why not the others:

• A (20.0 m): Elevation plus friction only; ignores the vessel pressure.
• B (30.4 m): Subtracts friction instead of adding it.
• C (35.4 m): Leaves out friction.

If you missed it: List every term the pump must overcome (lift, pressure, friction, velocity) and add them in consistent units of head.

Source: DOE-HDBK-1012/3-92, HT-03, Extended Bernoulli, Eq. 3-12 (pump head and friction head), p. 25.

Answer: C. 59.9 kPa

h_f = f (L/D)(v²/2g) = 0.020 × (150/0.10) × (2.0² ÷ 19.62) = 6.12 m of water

ΔP = ρg h_f = 998 × 9.81 × 6.12 = 59.9 kPa

Why not the others:

• A (15.0 kPa): Divides the friction factor by 4, mixing up the Darcy and Fanning conventions.
• B (29.9 kPa): Uses v instead of v² (with v = 2.0, that halves the answer).
• D (240 kPa): Multiplies by 4, converting a Darcy factor as if it were Fanning.

If you missed it: Check which friction factor your chart or equation uses. Darcy = 4 × Fanning.

Source: DOE-HDBK-1012/3-92, HT-03, Head Loss, "Darcy's Equation" (H_f = f(L/D)(v²/2g), Eq. 3-14).

Answer: C. 2.25

With f, L and D fixed, Darcy's equation makes head loss proportional to velocity squared:

h_f ∝ v², and 1.5² = 2.25

That's why a "small" throughput increase can overload a pump: half again as much flow more than doubles the friction loss.

Why not the others:

• A (1.22): Uses √1.5, as if loss scaled with the square root of flow.
• B (1.50): Assumes loss is proportional to flow. In Darcy's equation it goes with velocity squared.
• D (3.38): Uses 1.5³, the power scaling of the pump laws, not friction loss.

If you missed it: Tie each relationship to its exponent: turbulent friction loss ∝ Q², pump power ∝ N³, orifice flow ∝ √ΔP.

Source: DOE-HDBK-1012/3-92, HT-03, Head Loss, Darcy's Equation, H_f = f(L/D)(v²/2g), Eq. 3-14.

Answer: A.

Setting Mach = 1 in the isentropic pressure relation gives the critical pressure ratio:

P*/P₀ = [2/(k + 1)]^[k/(k − 1)] = (2/2.30)^(4.333) = 0.546

P = 0.546 × 800 = 437 kPa*

The header (350 kPa) is below 437 kPa, so the throat reaches Mach 1 and the flow chokes. Further lowering the header pressure can't increase the mass flow.

Why not the others:

• B: The pressure ratio 350/800 = 0.44 is below the critical ratio, so the flow is choked.
• C: A choked throat stays at P*, not at the downstream pressure. The remaining expansion happens after the throat.
• D: Uses 0.528, the critical ratio for k = 1.4. This gas has k = 1.30.

If you missed it: Compute P* from the gas's own k every time. The familiar 0.528 is for air-like gases only.

Source: NASA Glenn Research Center, "Isentropic Flow Equations", Eq. 6 (p/pt as a function of Mach number), and "Mass Flow Choking" (flow limited at Mach 1 at the throat). The critical ratio shown is Eq. 6 evaluated at M = 1.

Answer: B. 9.6 m

NPSHa = (P_surface − P_vapor) ÷ ρg + static head − friction loss

= (101,300 − 19,900) ÷ (983 × 9.81) + 2.0 − 0.8 = 8.44 + 2.0 − 0.8 = 9.6 m

If the pump's required NPSH is lower than this with a margin, it shouldn't cavitate.

Why not the others:

• A (5.6 m): Treats the liquid level as below the pump (a suction lift).
• C (11.2 m): Adds the friction loss instead of subtracting it.
• D (11.7 m): Ignores the vapor pressure. At 60°C, that's a big mistake.

If you missed it: NPSH is the margin above boiling at the pump inlet. Anything that lowers inlet pressure or raises vapor pressure eats it.

Source: DOE-HDBK-1012/3-92, HT-03, Centrifugal Pumps, "Cavitation," p. 48, and "Net Positive Suction Head" (NPSH = P_suction − P_saturation, Eq. 3-19; keep available NPSH above required), p. 49.

Answer: 89.4 m³/h (accepted range: 89.0 to 89.8 m³/h)

The operating point is where the pump curve meets the system curve:

60 − 0.0020Q² = 20 + 0.0030Q² → 40 = 0.0050Q² → Q² = 8,000 → Q = 89.4 m³/h

The head there is 60 − 0.0020 × 8,000 = 44 m.

Common wrong answers:

• 109.5: Ignores the system's 20 m static head.
• 200.0: Subtracts the curve coefficients (0.0030 − 0.0020) instead of adding them.

If you missed it: Set the two curves equal and collect the Q² terms on one side.

Source: DOE-HDBK-1012/3-92, HT-03, Centrifugal Pumps, "System Characteristic Curve" and "System Operating Point" (operating point at the intersection of the pump and system curves), p. 52.

Answer: D. Series, because the heads of pumps in series add at the same flow

In series, the same flow passes through both pumps, and their heads add. In parallel, each pump develops the same head and the flows add. So series is the choice for a high-head service; parallel is the choice for more flow.

Where the system actually operates still depends on the system curve.

Why not the others:

• A: Reverses the rule. In parallel, each pump produces the same head.
• B: In parallel, flows add, but a centrifugal pump's head falls, not rises, as flow increases.
• C: Only the series arrangement adds heads.

If you missed it: Series: add heads at the same flow. Parallel: add flows at the same head.

Source: DOE-HDBK-1012/3-92, HT-03, "System Use of Multiple Centrifugal Pumps": Centrifugal Pumps in Parallel, p. 53; Centrifugal Pumps in Series, p. 54.

Answer: A. 8.5 kW

Power scales with speed cubed: P₂ = 15 × (1,450/1,750)³ = 15 × 0.569 = 8.5 kW

At the new speed, flow is 82.9 m³/h and head 27.5 m.

Why not the others:

• B (10.3 kW): Scales power with speed squared (that's head).
• C (12.4 kW): Scales power linearly (that's flow).
• D (26.4 kW): Inverts the speed ratio.

If you missed it: Remember the 1-2-3 pattern: flow ∝ N, head ∝ N², power ∝ N³.

Source: DOE-HDBK-1012/3-92, HT-03, Centrifugal Pumps, "Pump Laws," Eqs. 3-20 to 3-22, p. 49.

Answer: C. 60.0 m³/h

For a head-type meter, flow is proportional to the square root of the differential pressure:

Q₂ = 100 × √(9/25) = 100 × 0.60 = 60.0 m³/h

Why not the others:

• A (13.0 m³/h): Squares the pressure ratio.
• B (36.0 m³/h): Treats flow as proportional to ΔP.
• D (167 m³/h): Inverts the ratio. Lower ΔP must mean lower flow.

If you missed it: A head meter's square-root response means poor resolution at low flow: a quarter of the ΔP is still half the flow.

Source: DOE-HDBK-1013/1-92, Instrumentation and Control, IC-04, Flow Detectors, "Head Flow Meters," p. 1 (square-root extraction) and Eq. 4-3 (V̇ = K√ΔP), p. 5.

Answer: C. 2.52

Relative volatility compares how readily the two components vaporize: α = (y_B/x_B) ÷ (y_T/x_T). With Raoult's law, y_i/x_i = P_i*/P, so P cancels:

α = P_B ÷ P_T = 136 ÷ 54 = 2.52

That's why benzene–toluene distillation is fairly easy: α well above 1. As α approaches 1, the number of stages needed climbs sharply.

Why not the others:

• A (0.40): The inverse, toluene relative to benzene.
• B (1.59): The square root of the ratio.
• D (82 kPa): The difference of the vapor pressures. Relative volatility is a dimensionless ratio.

If you missed it: α is always the more volatile component over the less volatile one, so it's greater than 1.

Source: NPTEL, Mass Transfer Operation I, Module 5, Lecture 1, §5.1.5 Relative volatility (definition, Eq. 5.3), p. 4; OpenStax Chemistry 2e, §11.4 (Raoult's law).

Answer: C. 6.4

N_min = ln{[x_D/(1 − x_D)] × [(1 − x_B)/x_B]} ÷ ln α

= ln[(0.95/0.05) × (0.95/0.05)] ÷ ln 2.5 = ln 361 ÷ 0.916 = 6.4

That is the equation result for the minimum number of equilibrium stages. If a design problem asks for an integer physical stage count, apply the problem’s stage-count convention and round up as required; a total condenser is not normally counted as an equilibrium stage, while a reboiler may be.

Why not the others:

• A (2.8): Mixes log₁₀ in the numerator with ln in the denominator.
• B (3.2): Uses only the distillate ratio (19) and forgets the bottoms term.
• D (12.9): Double-counts by squaring the separation factor again.

If you missed it: The separation factor is the product of the top and bottom composition ratios. Keep the same log base in numerator and denominator, and distinguish the calculated N_min value from an integer equipment count.

Source: FAMU-FSU College of Engineering, "Distillation" lecture, slide p. 10 (Fenske equation, total reflux, equilibrium stages); NTNU separations exam solution, 2015, p. 2 (S = αᴺ, N_min = ln S / ln α).

Answer: 1 → a, 2 → b, 3 → c, 4 → d

q is the fraction of the feed that joins the liquid flowing down the column.

• Saturated liquid: q = 1, so the slope's denominator is zero and the line is vertical.
• Saturated vapor: q = 0, so the slope is zero: horizontal.
• Partly vaporized: 0 < q < 1, giving a negative slope.
• Subcooled liquid: q > 1 (the cold feed also condenses some vapor), giving a positive slope.

Why not the others:

• e: The 45° diagonal is where the operating lines lie at total reflux. It isn't a q-line.

If you missed it: Plug each q into −q/(1 − q) once and sketch the result. The four cases stick after one pass.

Source: NPTEL, Mass Transfer Operation I, Module 5, Lecture 4, Table 5.3 (q values by feed condition), p. 6; q-line slope, Eq. 5.28, pp. 8–9.

Answer: D. 38.9 kmol/h

Total balance: F = D + B. Light-key balance: F·z_F = D·x_D + B·x_B.

Combining: D = F(z_F − x_B) ÷ (x_D − x_B) = 100 × (0.40 − 0.05) ÷ (0.95 − 0.05) = 38.9 kmol/h

Why not the others:

• A (61.1 kmol/h): The bottoms flow.
• B (42.1 kmol/h): Divides the light key fed by the distillate purity, ignoring what's lost in the bottoms.
• C (40.0 kmol/h): Assumes every mole of light key goes overhead as pure product.

If you missed it: Two unknowns (D and B) need two balances: total and one component.

Source: NPTEL, Mass Transfer Operation I, Module 5, Lecture 5, Solution 5.2, p. 10 (F = D + B and F·x_F = D·x_D + B·x_B).

Answer: C. 1.44

Saturated liquid gives a vertical q-line at x = 0.40. The pinch is on the equilibrium curve there:

y* = 2.5 × 0.40 ÷ (1 + 1.5 × 0.40) = 1.0 ÷ 1.6 = 0.625

At minimum reflux, the rectifying line runs from (0.95, 0.95) to (0.40, 0.625):

Slope = Rmin/(Rmin + 1) = (0.95 − 0.625) ÷ (0.95 − 0.40) = 0.591

Rmin = 0.591 ÷ (1 − 0.591) = 1.44

A real column runs at a reflux ratio above this minimum.

Why not the others:

• A (0.59): Reports the operating-line slope L/V, not the reflux ratio L/D.
• B (0.69): Inverts the reflux ratio (1/1.44).
• D (2.17): Multiplies Rmin by 1.5. That's an operating reflux above the minimum, not the minimum itself.

If you missed it: Locate the pinch first (q-line meets equilibrium curve), then read Rmin from the slope of the line through the distillate point.

Source: NPTEL, Mass Transfer Operation I, Module 5, Lecture 5, minimum reflux at the intersection of the q-line and the equilibrium curve, pp. 5–6; Lecture 1, Eq. 5.4 (x–y relation from relative volatility), p. 4.

Answer: A. 114 kmol/h

At minimum liquid flow, the operating line touches the equilibrium line at the rich (bottom) end, where the leaving liquid is in equilibrium with the entering gas:

x_out = y_in ÷ m = 0.02 ÷ 1.2 = 0.01667

Solute absorbed = 100 × 0.02 × 0.95 = 1.90 kmol/h

L_min = 1.90 ÷ 0.01667 = 114 kmol/h (that is, (L/G)_min = m × fractional recovery = 1.2 × 0.95)

Why not the others:

• B (120 kmol/h): Uses m alone, which corresponds to 100% recovery.
• C (126 kmol/h): Divides m by the recovery instead of multiplying.
• D (171 kmol/h): A typical design rate at 1.5 × minimum, not the minimum.

If you missed it: Find the pinch first. For absorption with linear equilibrium, it sits at the end where the gas enters.

Source: University at Buffalo CE 407, Lecture 03, slide 13 (minimum liquid flow when the operating line first contacts the equilibrium curve) and slide 2 (absorption factor). The result above is derived from that pinch condition with the stated dilute, linear assumptions.

Answer: B. 75.0%

Fraction left in the water after one stage = V_aq ÷ (K V_org + V_aq)

Each 25 L stage: 100 ÷ (4 × 25 + 100) = 0.50

After two stages: 0.50² = 0.25 left, so 75.0% extracted

A single 50 L extraction leaves 100 ÷ (200 + 100) = 1/3, extracting only 66.7%. Same solvent, better result in two smaller portions.

Why not the others:

• A (66.7%): One 50 L extraction.
• C (80.0%): K/(K + 1), which assumes equal phase volumes in a single stage.
• D (88.9%): Uses two full 50 L stages, which is 100 L of solvent.

If you missed it: For a fixed amount of solvent, more smaller stages beat one large stage. Write the fraction remaining per stage and raise it to the number of stages.

Source: LibreTexts, Harvey, Analytical Chemistry 2.1, §7.7 Liquid–Liquid Extractions, Eqs. 7.6 (one extraction) and 7.7 (n extractions).

Answer: B. 50.0 kg

At 20°C, the 100 kg of water can hold only 30 kg of solute in a saturated solution. The rest crystallizes:

Crystals = 80 − 30 = 50.0 kg

That's a 62.5% recovery of the solute. Evaporating some water or cooling further would push more out.

Why not the others:

• A (80.0 kg): Assumes all the solute crystallizes.
• C (37.1 kg): Reads the solubility as 30 kg per 100 kg of solution, not per 100 kg of water.
• D (30.0 kg): The solute that stays dissolved.

If you missed it: Check the solubility basis (per water or per solution) before you do anything else.

Source: OpenStax Chemistry 2e, §11.3 Solubility (saturated solutions; solubility of solids varies with temperature). The crystal yield is a solute mass balance on the supplied data.

Answer: A. 40 ppm; below the limit

E = (C_a T_a + C_b T_b + …) ÷ 8 = (80 × 3 + 40 × 2 + 0 × 3) ÷ 8 = 320 ÷ 8 = 40 ppm

That's below 50 ppm. Short-term or ceiling limits, if the substance has them, are separate checks.

Why not the others:

• B: Computes correctly but misreads 40 versus 50.
• C (60 ppm): Averages 80 and 40 without weighting by time.
• D (64 ppm): Divides by the 5 exposed hours instead of 8.

If you missed it: The divisor is always 8 for an 8-hour TWA, even when part of the shift has zero exposure.

Source: 29 CFR 1910.1000(d)(1)(i) (cumulative exposure E = (C_aT_a + C_bT_b + … C_nT_n) ÷ 8, not to exceed the 8-hour TWA limit).

Answer: A. Equivalent exposure 1.1; the mixture limit is exceeded

E_m = C₁/L₁ + C₂/L₂ + C₃/L₃ = 200/500 + 40/100 + 15/50 = 0.40 + 0.40 + 0.30 = 1.1

The mixture value must not exceed 1, so this exposure is over the limit even though each vapor alone is under its own.

Why not the others:

• B (0.37): Averages the three ratios instead of adding them.
• C: Calculates correctly but misreads the criterion. The limit is 1.
• D: Checking each substance alone misses the combined effect the mixture formula exists to catch.

If you missed it: For mixtures, add the fractions of each limit. The sum must stay at or below 1.

Source: 29 CFR 1910.1000(d)(2)(i) (E_m = C₁/L₁ + C₂/L₂ + … + C_n/L_n; E_m shall not exceed 1).

Answer: B. 1 × 10⁻⁵ per year

An independent protection layer (IPL) must work independently of the initiating event. The alarm shares the transmitter whose failure may have started the scenario, so it can't be credited.

Mitigated consequence frequency = 0.1 × 0.01 × 0.01 = 1 × 10⁻⁵ per year

Why not the others:

• A (1 × 10⁻³): Credits only one of the two valid IPLs.
• C (1 × 10⁻⁶): Credits the alarm too, which breaks the independence requirement.
• D (0.12): Adds the PFDs to the initiating frequency. Probabilities of failure on demand multiply.

If you missed it: For each safeguard, ask: "If the initiating cause happened, would this still work?" If the answer depends on the same component, it isn't independent.

Source: AIChE/CCPS, LOPA overview (estimate the frequency of the mitigated consequence by combining initiating-event frequency with IPL PFDs; IPLs must be independent); see also HSE Research Report RR716 (2009), §3.5.3 (layers sharing components do not satisfy independence).

Answer: D.

The rule requires secondary containment for the entire capacity of the largest single container, plus sufficient freeboard to contain precipitation. The dike must also be sufficiently impervious to contain discharged oil.

Why not the others:

• A: The rule is based on the largest single container, not the sum. (Designing for more is allowed; it isn't the minimum.)
• B: Averaging could leave the largest tank's contents uncontained.
• C: Leaves out the precipitation freeboard the rule requires.

If you missed it: For containment sizing, ask two questions: which single failure is credible, and what else (rain, firewater) will be in the dike at the same time?

Source: 40 CFR 112.8(c)(2), eCFR (secondary containment for bulk storage tank installations).

Answer: B. 220 to 250 psig

For ASME containers built to the code editions in that row of Table H-26 (which includes the 1968 Division 1 edition), the start-to-discharge setting must be at least 88% and at most 100% of the container's design pressure:

0.88 × 250 = 220 psig; 1.00 × 250 = 250 psig

The table's maximum allows a plus tolerance of up to 10% of the set pressure marked on the valve. That tolerance doesn't move the setting itself above design pressure.

Why not the others:

• A: Uses 80% as the minimum. The table says 88%.
• C and D: Set the valve above design pressure. A relief device is set at or below the pressure the container is designed for, not above it.

If you missed it: Relief settings are anchored to the vessel's design pressure, and they sit at or below it.

Source: 29 CFR 1910.110(b)(10)(iv) and Table H-26 (container safety relief valve start-to-discharge settings as a percentage of design pressure; footnote on plus tolerance).

Answer: A. About −$17,000; reject

Present value of a 5-year annuity at 12%:

P/A = [1 − 1/(1.12)⁵] ÷ 0.12 = 3.6048

PV of inflows = 550,000 × 3.6048 = $1,982,600

NPV = 1,982,600 − 2,000,000 = about −$17,000

A negative NPV means the project doesn't earn the 12% the company requires, so on this test it's rejected, even though it pays back the $2.0 million in nominal dollars.

Why not the others:

• B (+$750,000): Adds up the cash flows without discounting.
• C (+$1.98 million): The present value of the inflows. It forgets to subtract the investment.
• D (+$85,000): Discounts at 10% instead of 12%.

If you missed it: NPV = PV(inflows) − PV(outflows). Put the investment in, and use the rate the question gives.

Source: Principles of Finance (OpenStax), LibreTexts mirror, §8.3 Annuities, Eq. 8.14 (present value of an ordinary annuity); §16.3 Net Present Value (NPV) Method (accept positive NPV, reject negative).

Answer: D. Pump B: about $88,600 vs. $100,400 for Pump A

Present worth factor for a 10-year annuity at 8%: P/A = [1 − 1/(1.08)¹⁰] ÷ 0.08 = 6.7101

Pump A: 40,000 + 9,000 × 6.7101 = $100,400

Pump B: 55,000 + 5,000 × 6.7101 = $88,600

Pump B's extra $15,000 up front buys $4,000 a year in savings worth about $26,800 today, so B wins.

Why not the others:

• A: Compares only the first cost.
• B: Right pump, but adds the cash flows without discounting.
• C: Discounting shrinks the savings but doesn't erase them: $4,000 a year for 10 years is still worth about $26,800 today, more than the $15,000 premium.

If you missed it: Put every cost on the same time basis (present worth) before comparing alternatives.

Source: Principles of Finance (OpenStax), LibreTexts mirror, §8.3 Annuities, Eq. 8.14 (present value of an ordinary annuity).

Answer: A.

This is the pattern of chloride stress corrosion cracking (SCC). It needs three things together: chloride ions, dissolved oxygen and tensile stress (weld residual stress is a common source). Chromium-carbide sensitization near welds makes it worse. The controls are low chloride and oxygen and low-carbon steels.

Why not the others:

• B: More oxygen supplies one of the conditions that SCC needs.
• C: That's galvanic protection logic. The less noble carbon steel would corrode, but it does nothing about stress, chloride or oxygen at the cracks.
• D: Hotter water doesn't remove chloride, oxygen or tensile stress, so it doesn't touch the cause.

If you missed it: For SCC, name the three legs (environment species, oxygen, tensile stress) and pick the fix that removes at least one of them.

Source: DOE-HDBK-1015/1-93, Chemistry, CH-02, Specialized Corrosion, “Stress Corrosion Cracking,” “Chloride Stress Corrosion Cracking (Stainless Steels),” and “Galvanic Corrosion.”

Answer: D.

This is galvanic corrosion: two dissimilar metals with different potentials, in electrical contact, in an electrolyte (seawater). The less noble, more active metal (carbon steel) becomes the anode and corrodes. The more noble metal (stainless steel) becomes the cathode and is protected.

Why not the others:

• A: Reverses the roles. Stainless is the more noble metal here.
• B: Passivity protects the stainless steel's own surface; it doesn't protect the coupled carbon steel.
• C: Coupling changes the rates. That's the whole point of galvanic corrosion.

If you missed it: For any mixed-metal joint in a wet environment, identify the less noble metal: it's the one at risk. Insulating the joint or matching the metals removes the problem.

Source: DOE-HDBK-1015/1-93, Chemistry, CH-02, "Galvanic Corrosion," p. 23 (the less resistant, active metal becomes the anodic corrosion site; the more noble metal is cathodic and protected).

Answer: B. 5.0 m³

Holdup volume = flow × time = (30 m³/h ÷ 60 min/h) × 5 min = 2.5 m³

That liquid occupies half the drum, so the drum volume is 2.5 ÷ 0.50 = 5.0 m³

Why not the others:

• A (2.5 m³): The liquid holdup alone, without the vapor space above normal level.
• C (75 m³): Treats the 30 m³/h as per minute and applies the 50% factor.
• D (150 m³): Multiplies 30 by 5 with no time conversion.

If you missed it: Residence time is volume divided by volumetric flow; make the time units match.

Source: University of Michigan ChE 344, Lecture 6 (space time τ = V/v₀). The holdup time and fill level are supplied design inputs.

Answer: A.

On loss of air, the actuator spring drives each valve to its fail position. An air-to-close, spring-to-open valve fails open; an air-to-open, spring-to-close valve fails closed.

• Cooling water should keep flowing to remove reaction heat: fail-open, so air-to-close.
• Fuel gas should stop adding heat: fail-closed, so air-to-open.

Why not the others:

• B: Reverses both. Cooling would stop and fuel would flow, the worst combination on an exothermic system.
• C: Fuel is right, but cooling water would shut off.
• D: Cooling is right, but fuel would keep flowing.

If you missed it: Decide the safe state from the process hazard first, then pick the actuator that gets there without air.

Source: DOE-HDBK-1013/2-92, Instrumentation and Control, IC-07, Process Controls, "Valve Actuators" (air-to-close, spring-to-open = fail-open; air-to-open, spring-to-close = fail-closed).

Answer: 45.0 (accepted range: 44.5 to 45.5)

Q = Cv √(ΔP/SG), so Cv = Q ÷ √(ΔP/SG)

Cv = 200 ÷ √(16 ÷ 0.81) = 200 ÷ 4.444 = 45.0

In practice you'd select a valve whose rated Cv covers this point within its good control range, not one sized exactly to it.

Common wrong answers:

• 50.0: Ignores specific gravity (treats the liquid as water).
• 55.6: Multiplies ΔP by SG instead of dividing.
• 10.1: Forgets the square root.

If you missed it: Lighter liquids pass more flow at the same ΔP, so they need a slightly smaller Cv than water. Use that to check the direction of your answer.

Source: Swagelok, "Valve Sizing" Technical Bulletin MS-06-84-E, "Liquid Flow" (q = N₁Cv√(Δp/G_f), with N₁ = 1.0 for gpm and psi).

Answer: A. 2.87 m

The transmitter measures hydrostatic pressure, which equals density × g × height. The hot water is less dense, so the same height produces less pressure, and the transmitter converts it using the calibration density:

Indicated = 3.00 × 958 ÷ 1,000 = 2.87 m

It reads low by about 13 cm. That's why level systems on hot or variable-density liquids need density compensation.

Why not the others:

• B: Assumes the reading doesn't depend on density.
• C: Inverts the density ratio. Less dense liquid gives a lower reading, not a higher one.
• D (2.75 m): Applies the density ratio twice (3.00 × 0.958²).

If you missed it: A DP level transmitter infers height from pressure, so anything that changes density changes the reading.

Source: DOE-HDBK-1013/1-92, Instrumentation and Control, IC-03, Level Detectors: dry reference leg output proportional to hydrostatic head, p. 8; hydrostatic head = density × height (Eq. 3-3) and the need for density compensation as temperature changes, p. 12.

Answer: A and B

Under 1910.119(i)(2), the review confirms, before hazardous chemicals enter the process, that construction and equipment meet design specifications; that safety, operating, maintenance and emergency procedures are in place and adequate; that a process hazard analysis has been done for a new facility (or management-of-change requirements met for a modified one); and that operator training is complete.

Why not the others:

• C: Replacements in kind are excluded from management of change; the rule doesn't require a new HAZOP for them.
• D: Running on the hazardous chemical is exactly what the review must come before.
• E: The review is a pre-startup requirement, not a post-startup one.

If you missed it: Read "pre-startup" literally: everything on the list must be true before the first hazardous chemical arrives.

Source: 29 CFR 1910.119(i)(2), eCFR (pre-startup safety review) and (l)(1) (management of change; replacements in kind excluded).

Answer: D.

The MOC procedure covers changes to process chemicals, technology, equipment and procedures, and changes to facilities that affect a covered process. Raising a documented operating limit changes the process technology. Before it happens, the rule requires addressing the technical basis, the impact on safety and health, procedure changes, the time period, and authorization.

Why not the others:

• A and C: Identical replacements are "replacements in kind," which the rule excludes from MOC.
• B: A scheduled inspection is part of mechanical integrity. Nothing about the process changes.

If you missed it: Ask: "After this, is anything different from the documented design or procedures?" If yes, it's MOC. If it's identical, it's replacement in kind.

Source: 29 CFR 1910.119(l)(1) and (l)(2), eCFR (management of change; replacements in kind excluded; considerations before a change).

Answer: D. 0.99

The service fails only if both pumps fail. For a parallel (redundant) system, the failure probabilities multiply:

F = 0.10 × 0.10 = 0.01, so R = 1 − 0.01 = 0.99

Independence is doing the work here. A shared power supply or suction line would break it.

Why not the others:

• A (0.81): The series result (both must run), R = 0.90 × 0.90.
• B (0.90): One pump, no credit for the spare.
• C (0.95): Averages one and two pumps. Reliabilities don't combine by averaging.

If you missed it: Series: multiply reliabilities. Parallel: multiply failure probabilities, then subtract from 1.

Source: NIST/SEMATECH e-Handbook of Statistical Methods, §8.1.8.2 Series model and §8.1.8.3 Parallel or redundant model (system failure probability is the product of the component failure probabilities).

Answer: D.

The mechanical integrity element requires inspections and tests that follow recognized and generally accepted good engineering practices, at a frequency consistent with manufacturers' recommendations and good engineering practice, and more frequently if prior operating experience shows it's necessary. Each inspection or test must be documented.

Why not the others:

• A: The rule sets no single fixed interval for all equipment.
• B: Waiting for failure defeats the purpose of preventive inspection.
• C: The employer sets and follows the program; it doesn't wait for an inspector.

If you missed it: Mechanical integrity is risk- and experience-based: start from the manufacturer and good practice, then tighten intervals when experience says so.

Source: 29 CFR 1910.119(j)(4)(i)–(iv), eCFR (inspection and testing; frequency; documentation).

Answer: A and B

Cavitation happens when the pressure at the impeller eye falls below the liquid's vapor pressure. NPSHa is the margin between the suction pressure and the vapor pressure.

• A: A higher level adds static head at the suction.
• B: A clogged strainer adds friction loss in the suction line; cleaning it removes that loss.

Why not the others:

• C: Throttling the suction adds pressure drop, lowering NPSHa. Throttle on the discharge side, if at all.
• D: Hotter liquid has a higher vapor pressure, which lowers NPSHa.
• E: A smaller pipe raises velocity and friction loss on the suction side.

If you missed it: For each action, ask whether it raises suction pressure or lowers vapor pressure. Only those help.

Source: DOE-HDBK-1012/3-92, HT-03, Centrifugal Pumps, "Cavitation," p. 48, and "Net Positive Suction Head" (NPSH = P_suction − P_saturation, Eq. 3-19; keep NPSHa above NPSHr), p. 49.