Free FE Chemical Practice Test
Forty original FE Chemical practice questions with a worked solution under every one — no signup. The set samples all 17 knowledge areas in the current NCEES specifications (effective July 2020); it is unofficial, shorter than the real exam, and doesn't predict a pass.
Question 1 — Diluting caustic
FEC-001 · Material/Energy Balances · Choose one answer
A 20 wt% NaOH solution flows at 100 lbm/h into a steady-state mixer, where it is diluted with pure water to make a 5 wt% NaOH product. What water flow rate is required?
- A. 300 lbm/h
- B. 20 lbm/h
- C. 75 lbm/h
- D. 400 lbm/h
Show answer and worked solution
Answer: A. 300 lbm/h
NaOH is a tie component: it enters only with the feed and leaves only in the product.
NaOH in = 0.20 × 100 = 20 lbm/h. Product flow = 20 / 0.05 = 400 lbm/h. Water added = 400 − 100 = 300 lbm/h.
- D (400) is the total product flow, not the water you add.
- C (75) is 100(1 − 0.05/0.20), which applies a dilution fraction to the feed instead of balancing NaOH mass.
- B (20) is the NaOH flow itself.
Relation: Component balance: in = out at steady state, no reaction. Handbook search words to try: mass balance.
Source: LearnChemE, Material Balances (Important Equations).
Question 2 — Reynolds number in a water line
FEC-002 · Fluid Mechanics/Dynamics · Choose one answer
Water (ρ = 998 kg/m³, μ = 1.0 × 10⁻³ Pa·s) flows at an average velocity of 2.0 m/s through a pipe with an inside diameter of 5.0 cm. Which best describes the flow?
- A. Re ≈ 1.0 × 10², laminar
- B. Re ≈ 1.0 × 10⁵, turbulent
- C. Re ≈ 1.0 × 10⁵, laminar
- D. Re ≈ 2.0 × 10³, transitional
Show answer and worked solution
Answer: B. Re ≈ 1.0 × 10⁵, turbulent
Re = ρVD/μ = (998)(2.0)(0.050) / (1.0 × 10⁻³) = 99,800.
Pipe flow is generally laminar below about 2,000 and turbulent above roughly 3,000 (references draw the boundary slightly differently). At about 100,000, this flow is clearly turbulent.
- A comes from entering viscosity in centipoise (1.0) instead of Pa·s.
- C has the right number with the wrong regime.
- D is a guess near the transition range, not a calculation.
Relation: Re = ρVD/μ (dimensionless). Handbook search words to try: Reynolds number.
Source: LearnChemE, Pressure Drop and Power in Pipe Flow (Important Equations: Reynolds number, head loss, pump power); OpenStax College Physics 2e, §12.5 The Onset of Turbulence (Eq. 12.53 and approximate Reynolds-number flow regimes).
Question 3 — First bubble from a benzene–toluene liquid
FEC-003 · Thermodynamics · Choose one answer
A liquid mixture of benzene (B) and toluene (T) with x_B = 0.40 is at its bubble point. At this temperature, use P_sat,B = 180 kPa and P_sat,T = 72 kPa, and assume ideal liquid and vapor behavior. What is the benzene mole fraction in the first vapor bubble, y_B?
- A. 0.400
- B. 0.500
- C. 0.625
- D. 0.714
Show answer and worked solution
Answer: C. 0.625
Bubble pressure: P = x_B·P_sat,B + x_T·P_sat,T = (0.40)(180) + (0.60)(72) = 72 + 43.2 = 115.2 kPa.
Raoult's law for benzene: y_B = x_B·P_sat,B / P = 72 / 115.2 = 0.625. The vapor is richer in the more volatile component, as it should be.
- A assumes the vapor has the same composition as the liquid.
- D is 180/(180 + 72), which ignores the liquid composition.
- B has no basis in the data.
Relation: Raoult's law, yᵢP = xᵢPᵢ^sat; bubble pressure P = Σ xᵢPᵢ^sat. Handbook search words to try: Raoult, vapor-liquid equilibrium.
Source: LearnChemE, Raoult's Law and VLE (Important Equations).
Question 4 — CSTR space time
FEC-004 · Chemical Reaction Engineering · Choose one answer
A liquid-phase, first-order reaction A → products has k = 0.20 min⁻¹. What space time is needed to reach 90% conversion in a single ideal CSTR at steady state, constant temperature, and constant density?
- A. 45 min
- B. 4.5 min
- C. 11.5 min
- D. 50 min
Show answer and worked solution
Answer: A. 45 min
Start from the CSTR design equation, V = (F_A0 − F_A)/(−r_A). For a first-order reaction at constant density this becomes τ = X / [k(1 − X)].
τ = 0.90 / [(0.20)(0.10)] = 45 min.
- C (11.5 min) is the plug-flow reactor answer, τ = −ln(1 − X)/k. For first-order kinetics at high conversion, a PFR needs far less volume than a CSTR, which is worth remembering.
- B (4.5 min) drops the (1 − X) term.
- D (50 min) drops the X in the numerator.
Relation: CSTR design equation; rate evaluated at outlet (well-mixed) conditions. Handbook search words to try: CSTR, continuous stirred tank reactor.
Source: Fogler, Elements of Chemical Reaction Engineering 7e, Ch. 1 summary (Univ. of Michigan) (general mole balance; CSTR and PFR design equations).
Question 5 — Counterflow LMTD
FEC-005 · Heat Transfer · Choose one answer
In a counterflow double-pipe exchanger, oil cools from 150 °C to 90 °C while water heats from 30 °C to 70 °C. What is the log-mean temperature difference?
- A. 55.8 °C
- B. 60.0 °C
- C. 70.0 °C
- D. 69.5 °C
Show answer and worked solution
Answer: D. 69.5 °C
Pair the temperatures at each physical end. Counterflow: ΔT₁ = 150 − 70 = 80 °C; ΔT₂ = 90 − 30 = 60 °C.
LMTD = (80 − 60) / ln(80/60) = 20 / 0.2877 = 69.5 °C.
- A (55.8) is the parallel-flow LMTD (end differences of 120 and 20 °C). Check the flow arrangement before pairing end temperatures.
- C (70.0) is the arithmetic mean of the end differences. It is close here only because the two ends are similar.
- B (60.0) is just the smaller end difference.
Relation: LMTD = (ΔT₁ − ΔT₂)/ln(ΔT₁/ΔT₂). Handbook search words to try: log mean temperature difference.
Source: MIT Unified Engineering, Thermodynamics notes §18.5.2 (Eq. 18.35, counterflow LMTD).
Question 6 — One Newton-Raphson step
FEC-006 · Mathematics · Choose one answer
Use one Newton-Raphson iteration on f(x) = x³ − 2x − 5, starting from x₀ = 2. What is x₁?
- A. 1.900
- B. 2.050
- C. 2.095
- D. 2.100
Show answer and worked solution
Answer: D. 2.100
f(2) = 8 − 4 − 5 = −1. f′(x) = 3x² − 2, so f′(2) = 10.
x₁ = x₀ − f(x₀)/f′(x₀) = 2 − (−1/10) = 2.100.
- A (1.900) is a sign error: subtracting +0.1 instead of −0.1.
- C (2.095) is close to the true root (about 2.0946). More iterations approach it, but one step gives 2.100.
- B (2.050) halves the correct Newton correction of 0.100.
Relation: Newton-Raphson: xₙ₊₁ = xₙ − f(xₙ)/f′(xₙ). Handbook search words to try: Newton's method.
Source: OpenStax Calculus Vol. 1, §4.9 Newton’s Method (Newton iteration formula and algorithm).
Question 7 — Minimum stages at total reflux
FEC-007 · Mass Transfer and Separation · Choose one answer
A binary distillation must produce a distillate with x_D = 0.95 and bottoms with x_B = 0.05 (light-component mole fractions). The relative volatility is constant at α = 2.5. What is the Fenske minimum-stage result at total reflux, before rounding up to a whole number of stages?
- A. 2.5
- B. 3.2
- C. 6.4
- D. 12.9
Show answer and worked solution
Answer: C. 6.4
Fenske equation: N_min = ln{[x_D/(1 − x_D)]·[(1 − x_B)/x_B]} / ln α.
N_min = ln[(0.95/0.05)(0.95/0.05)] / ln 2.5 = ln(361) / 0.9163 = 6.4 stages.
The unrounded result is about 6.427; rounding up gives 7 whole theoretical stages. The answer choices ask for the Fenske calculation before that rounding.
- B (3.2) uses only the distillate term, not the separation between both ends.
- D (12.9) doubles the answer.
- A (2.5) is just α.
Relation: Fenske equation (total reflux, constant relative volatility). Handbook search words to try: Fenske, total reflux.
Source: Skogestad, NTNU distillation lecture slides (pp. 13–16: Fenske equation, stage convention, minimum and total reflux).
Question 8 — Acetate buffer pH
FEC-008 · Chemistry and Biology · Choose one answer
A buffer contains 0.20 M sodium acetate and 0.10 M acetic acid. Use pKa = 4.76 for acetic acid. Using concentrations in place of activities, what is the approximate pH?
- A. 2.88
- B. 4.46
- C. 4.76
- D. 5.06
Show answer and worked solution
Answer: D. 5.06
Henderson–Hasselbalch: pH = pKa + log([A⁻]/[HA]) = 4.76 + log(0.20/0.10) = 4.76 + 0.30 = 5.06.
- B (4.46) inverts the ratio.
- C (4.76) is the pH only when [A⁻] = [HA].
- A (2.88) is roughly the pH of 0.10 M acetic acid alone, ignoring the acetate.
Relation: pH = pKa + log([A⁻]/[HA]). Handbook search words to try: pH, buffer.
Source: OpenStax Chemistry 2e, §14.6 Buffers (The Henderson–Hasselbalch Equation).
Question 9 — Scaling an exchanger cost
FEC-009 · Process Design · Choose one answer
A heat exchanger with 100 m² of area cost $80,000 when the cost index was 600. Estimate the cost of a 250 m² exchanger at a new cost index of 800. For this hypothetical estimate, use C₂ = C₁(A₂/A₁)^0.6(I₂/I₁), where A is exchanger area and I is the cost index.
- A. $185,000
- B. $107,000
- C. $139,000
- D. $267,000
Show answer and worked solution
Answer: A. $185,000
Scale for size, then update for time:
Cost = 80,000 × (250/100)^0.6 × (800/600) = 80,000 × 1.733 × 1.333 = ≈ $185,000.
- D scales linearly with size (exponent 1.0), ignoring economy of scale.
- C applies the size exponent but forgets the cost index.
- B applies the cost index but forgets the size change.
Relation: C₂ = C₁(S₂/S₁)ⁿ × (I₂/I₁), with n given in the problem. Handbook search words to try: cost index, capacity.
Source: Model supplied in this question (hypothetical inputs; calculated directly).
Question 10 — Choosing a catalyst by expected value
FEC-010 · Probability and Statistics · Choose one answer
A plant is choosing among three catalyst options (profit over the evaluation period). Option A: 60% chance of +$120,000 and 40% chance of −$20,000. Option B: certain +$55,000. Option C: 90% chance of +$70,000 and 10% chance of −$40,000. On an expected-value basis, which option should be chosen?
- A. Option A, EV = $50,000
- B. Option A, EV = $64,000
- C. Option B, EV = $55,000
- D. Option C, EV = $59,000
Show answer and worked solution
Answer: B. Option A, EV = $64,000
EV(A) = 0.6(120,000) + 0.4(−20,000) = $64,000. EV(B) = $55,000. EV(C) = 0.9(70,000) + 0.1(−40,000) = $59,000.
Option A has the highest expected value. Expected value ignores risk tolerance, so a risk-averse owner might still prefer the certain option. Answer the question asked: here, expected value.
- A ($50,000) averages the two outcomes without weighting by probability.
- C and D compute their own options correctly, but neither is the highest.
Relation: E[X] = Σ pᵢxᵢ. Handbook search words to try: expected value.
Source: OpenStax Introductory Statistics 2e, §4.2 Expected Value (μ = ΣxP(x)).
Question 11 — Recycle around a low-conversion reactor
FEC-011 · Material/Energy Balances · Choose one answer
At steady state, fresh feed of 100 mol/h of pure A is combined with a recycle stream and fed to a reactor where the single-pass conversion of A is 25%. A separator recovers all unreacted A and recycles it, so the overall conversion is 100%. What is the recycle flow rate?
- A. 75 mol/h
- B. 133 mol/h
- C. 400 mol/h
- D. 300 mol/h
Show answer and worked solution
Answer: D. 300 mol/h
Let R be the recycle flow. Reactor feed = 100 + R. Unreacted A leaving the reactor = 0.75(100 + R), and all of it is recycled:
0.75(100 + R) = R → 75 = 0.25R → R = 300 mol/h.
Check: reactor feed = 400 mol/h, and 25% of 400 = 100 mol/h reacts, matching the fresh feed.
- C (400) is the reactor feed, not the recycle.
- A (75) is the unreacted A from one pass of fresh feed alone.
- B (133) is 100/0.75, a misapplied ratio.
Relation: Mole balance on A (in − out + generation = accumulation) around the reactor and the overall process. Handbook search words to try: recycle, conversion.
Source: Fogler, Elements of Chemical Reaction Engineering 7e, Ch. 1 summary (Univ. of Michigan) (general mole balance; CSTR and PFR design equations).
Question 12 — LFL of a fuel blend
FEC-012 · Safety, Health, and Environment · Choose one answer
A fuel gas is 60 mol% methane and 40 mol% propane on a fuel-only basis. Use lower flammability limits in air of 5.0 vol% (methane) and 2.1 vol% (propane). Using Le Chatelier's mixing rule, LFL_mix = 1 / Σ(yᵢ/LFLᵢ), where yᵢ is each fuel's fraction on a fuel-only basis, what is the approximate LFL of the mixture?
- A. 2.1 vol%
- B. 3.2 vol%
- C. 3.8 vol%
- D. 5.0 vol%
Show answer and worked solution
Answer: B. 3.2 vol%
LFL_mix = 1 / (0.60/5.0 + 0.40/2.1) = 1 / (0.120 + 0.190) = 3.2 vol%.
The rule is a weighted harmonic average, which pulls the result toward the component with the lower limit. This is a mixture-rule estimate using the supplied limits, not a safe operating limit.
- C (3.8) is a simple weighted average, which is not how the rule combines limits.
- A and D are the individual components' limits.
Relation: Le Chatelier's mixing rule (given in the question). Handbook search words to try: flammability, lower flammable limit.
Source: Zlochower and Green, The limiting oxygen concentration and flammability limits of gases and gas mixtures (PDF p. 5, Eq. 1 (Le Chatelier mixture rule)).
Question 13 — Friction loss in a horizontal pipe
FEC-013 · Fluid Mechanics/Dynamics · Choose one answer
Water (ρ = 998 kg/m³) flows at 2.0 m/s through 100 m of horizontal pipe with an inside diameter of 5.0 cm. The Darcy friction factor is 0.020. Ignoring minor losses, what is the frictional pressure drop?
- A. 8.2 kPa
- B. 39.9 kPa
- C. 160 kPa
- D. 79.8 kPa
Show answer and worked solution
Answer: D. 79.8 kPa
Darcy–Weisbach head loss: h_f = f(L/D)(V²/2g) = 0.020 × (100/0.050) × (2.0²/(2 × 9.81)) = 8.15 m.
For a horizontal pipe with no velocity change, ΔP = ρgh_f = 998 × 9.81 × 8.15 = 79.8 kPa.
Always check whether a friction factor is Darcy or Fanning. The Fanning factor is one-fourth of the Darcy factor.
- C (160) drops the 2 in V²/2g.
- B (39.9) uses V instead of V².
- A (8.2) reports the head in meters as if it were kPa.
Relation: Darcy–Weisbach: h_f = f(L/D)(V²/2g); ΔP = ρgh_f. Handbook search words to try: Darcy, head loss, friction factor.
Source: LearnChemE, Pressure Drop and Power in Pipe Flow (Important Equations: Reynolds number, head loss, pump power).
Question 14 — Net present worth of heat recovery
FEC-014 · Economics · Choose one answer
A heat-recovery project costs $120,000 now and saves $25,000 at the end of each year for 8 years, with no salvage value. At an interest rate of 10% per year, what is the net present worth, and should the project be accepted on a net-present-worth basis?
- A. +$13,400; accept
- B. +$133,400; accept
- C. +$80,000; accept
- D. −$26,700; reject
Show answer and worked solution
Answer: A. +$13,400; accept
(P/A, 10%, 8) = [(1.10)⁸ − 1] / [0.10(1.10)⁸] = 5.335.
Present worth of savings ≈ $133,373. NPW ≈ 133,373 − 120,000 = +$13,373, closest to +$13,400, so accept on this basis.
- B is the present worth of the savings alone; it forgets the investment.
- C ignores the time value of money (8 × 25,000 − 120,000).
- D discounts the 8-year total as one payment at year 8, as if all the savings arrived at the end.
Relation: Uniform-series present worth factor (P/A, i, n). Handbook search words to try: present worth, uniform series.
Source: OpenStax Principles of Finance 2e, §8.2 Annuities (ordinary annuities; Eq. 8.14 present value).
Question 15 — Heat flux through an insulated furnace wall
FEC-015 · Heat Transfer · Choose one answer
A furnace wall has 0.20 m of firebrick (k = 1.0 W/m·K) covered by 0.05 m of insulation (k = 0.05 W/m·K). The inner surface is at 600 °C and the outer surface is at 40 °C. Assume one-dimensional conduction through a plane wall and negligible contact resistance. What is the steady heat flux through the wall?
- A. 467 W/m²
- B. 2,240 W/m²
- C. 2,800 W/m²
- D. 3,360 W/m²
Show answer and worked solution
Answer: A. 467 W/m²
Conduction resistances in series, per unit area: R = L₁/k₁ + L₂/k₂ = 0.20/1.0 + 0.05/0.05 = 0.20 + 1.00 = 1.20 m²·K/W.
q″ = ΔT / R = 560 / 1.20 = 467 W/m². The thin insulation carries most of the resistance, which is the point of insulation.
- C (2,800) ignores the insulation.
- D (3,360) adds conductances (k/L) instead of resistances.
- B (2,240) adds the thicknesses and uses one conductivity.
Relation: Fourier's law; series resistances R = L/(kA) add. Handbook search words to try: conduction, thermal resistance.
Source: MIT Unified Engineering, Thermodynamics notes, composite wall (thermal resistances in series, R = L/kA); OpenStax University Physics Vol. 2, §1.6 Mechanisms of Heat Transfer (Eq. 1.9 (conduction) and Eq. 1.10 (net radiation with emissivity)).
Question 16 — Nitrogen in a tank (U.S. units)
FEC-016 · Thermodynamics · Choose one answer
A 100 ft³ tank holds nitrogen (MW = 28.01 lbm/lbmol) at 150 psia and 80 °F. Use R = 10.73 psia·ft³/(lbmol·°R) and treat the gas as ideal. What is the mass of nitrogen in the tank?
- A. 2.59 lbm
- B. 36.3 lbm
- C. 72.6 lbm
- D. 489 lbm
Show answer and worked solution
Answer: C. 72.6 lbm
Convert to absolute temperature: T = 80 + 459.67 = 539.67 °R.
n = PV/RT = (150)(100) / (10.73 × 539.67) = 2.590 lbmol. Mass = 2.590 × 28.01 = 72.6 lbm.
- D (489) uses 80 °F instead of °R. Gas-law problems always need absolute temperature.
- A (2.59) is the answer in lbmol, not lbm.
- B (36.3) uses the atomic mass of N (14) instead of N₂.
Relation: PV = nRT with absolute temperature and pressure. Handbook search words to try: ideal gas, gas constant.
Source: OpenStax Chemistry 2e, §9.2 The Ideal Gas Law (The Ideal Gas Law; kelvin requirement).
Question 17 — Temperature sensitivity of a rate constant
FEC-017 · Chemical Reaction Engineering · Choose one answer
A reaction has an activation energy of 80 kJ/mol. Using R = 8.314 J/(mol·K), by what factor does the rate constant increase when the temperature rises from 300 K to 320 K?
- A. 1.07
- B. 2.0
- C. 4.5
- D. 7.4
Show answer and worked solution
Answer: D. 7.4
k₂/k₁ = exp[(Ea/R)(1/T₁ − 1/T₂)] = exp[(80,000/8.314)(1/300 − 1/320)] = exp(9,622 × 2.083 × 10⁻⁴) = exp(2.005) = 7.4.
- A (1.07) is 320/300, a linear temperature ratio.
- B (2.0) is approximately ln(k₂/k₁), before taking the exponential.
- C has no basis.
Relation: Arrhenius: k = A·exp(−Ea/RT). Handbook search words to try: Arrhenius, activation energy.
Source: OpenStax Chemistry 2e, §12.5 Collision Theory (Arrhenius equation, two-temperature form).
Question 18 — Steel bolts in a copper plate
FEC-018 · Materials Science · Choose one answer
Carbon-steel bolts are used to fasten a large copper plate that will be immersed in seawater; the bolts and plate are in electrical contact. What is the most likely result?
- A. The copper plate corrodes because its exposed area is larger.
- B. The steel bolts corrode rapidly, because steel is anodic to copper and the small anode is coupled to a large cathode.
- C. Neither corrodes significantly, because the metals protect each other.
- D. Both corrode at about the same rate as they would if they weren't in contact.
Show answer and worked solution
Answer: B. The steel bolts corrode rapidly, because steel is anodic to copper and the small anode is coupled to a large cathode.
In seawater, steel is less noble than copper in the galvanic series, so it becomes the anode and corrodes while the copper is protected.
A small anode coupled to a large cathode concentrates the galvanic attack on the anode. Steel fasteners in copper are a classic example of this area effect.
Fixes include matching fastener material to the plate, electrically isolating the joint, or coating the cathode.
- A reverses the roles. Area changes the rate, not which metal is the anode.
- C overlooks the steel–copper galvanic couple; the steel bolts are not protected by being attached to copper.
- D ignores the galvanic couple.
Relation: Galvanic series; cathode-to-anode area effect. Handbook search words to try: galvanic series, corrosion.
Source: WJE, Galvanic Corrosion primer (pp. 2–3: galvanic series in seawater, small-anode/large-cathode effect and mitigation examples).
Question 19 — First-order step response
FEC-019 · Process Control · Choose one answer
A first-order process has a steady-state gain K = 2 (output units per input unit) and a time constant τ = 5 min, with no dead time. Starting at steady state, the input steps up by 3 units at t = 0. How much has the output changed at t = 5 min?
- A. 1.90 units
- B. 2.21 units
- C. 3.79 units
- D. 6.00 units
Show answer and worked solution
Answer: C. 3.79 units
Step response: Δy(t) = KΔu(1 − e^(−t/τ)) = (2)(3)(1 − e⁻¹) = 6 × 0.632 = 3.79 units.
At one time constant, a first-order process has completed 63.2% of its final change.
- D (6.00) is the final change as t → ∞.
- B (2.21) uses e⁻¹ instead of (1 − e⁻¹); that is the change still remaining.
- A (1.90) forgets the gain.
Relation: Δy(t) = KΔu(1 − e^(−t/τ)). Handbook search words to try: first-order, time constant.
Source: APMonitor, First Order Systems (Process time constant: 63.2% of final change at t = τ).
Question 20 — Minimum reflux statements
FEC-020 · Mass Transfer and Separation · Select all that apply
For a conventional binary distillation with specified, achievable product compositions, which statements about the minimum reflux limit are true? Select all that apply.
- A. It would require an infinite number of equilibrium stages.
- B. On a McCabe-Thiele diagram, an operating line touches the equilibrium curve (a pinch point).
- C. It gives the minimum number of theoretical stages.
- D. No distillate product is withdrawn.
- E. Practical columns operate at a reflux ratio above the minimum.
Choose all true statements and no false statements. No partial credit.
Show answer and worked solution
Answer: A, B, E
At minimum reflux the operating line pinches the equilibrium curve. Near the pinch, each stage changes the composition by almost nothing, so an infinite number of stages would be needed. Real columns run above the minimum to keep the stage count finite.
- C and D describe total reflux, the opposite limit. At total reflux no distillate is withdrawn and the stage count is at its minimum (the Fenske result in Question 7).
Relation: Minimum reflux ↔ pinch and infinite stages; total reflux ↔ minimum stages. Handbook search words to try: McCabe-Thiele, minimum reflux.
Source: Skogestad, NTNU distillation lecture slides (pp. 13–16: Fenske equation, stage convention, minimum and total reflux).
End of Section 1. This is a convenient study-break point; this page does not lock earlier questions. On the real exam, after about half the questions you review and submit the section before the scheduled break, and you cannot return to that submitted section (NCEES Examinee Guide, pp. 11–12).
Section 2: Questions 21–40
Question 21 — Air for methane combustion
FEC-021 · Material/Energy Balances · Choose one answer
Methane is burned at 10 mol/h with 20% excess air: CH₄ + 2 O₂ → CO₂ + 2 H₂O. Air is 21 mol% O₂. What is the air feed rate?
- A. 24.0 mol/h
- B. 57.1 mol/h
- C. 95.2 mol/h
- D. 114 mol/h
Show answer and worked solution
Answer: D. 114 mol/h
Theoretical O₂ = 2 × 10 = 20 mol/h. With 20% excess, O₂ fed = 1.20 × 20 = 24 mol/h.
Air = 24 / 0.21 = 114 mol/h.
Percent excess is always based on the theoretical requirement for complete combustion, even if combustion turns out incomplete.
- C (95.2) is the theoretical air, with no excess.
- A (24.0) is the O₂ flow, not the air flow.
- B (57.1) uses 1 mol O₂ per mol CH₄, an unbalanced equation.
Relation: Excess air relative to theoretical air for complete combustion. Handbook search words to try: combustion, excess air.
Source: LearnChemE, Combustion Reactions (Important Equations: complete-combustion basis and percent excess air).
Question 22 — Chemostat substrate level
FEC-022 · Chemistry and Biology · Choose one answer
A chemostat grows cells on one limiting substrate following Monod kinetics, μ = μ_max·S/(K_s + S), with μ_max = 0.50 h⁻¹ and K_s = 2.0 g/L. The sterile feed contains 20 g/L substrate, and the dilution rate is D = 0.30 h⁻¹. Assuming no cell death or maintenance, what is the steady-state substrate concentration with positive biomass (no washout) in the vessel?
- A. 1.2 g/L
- B. 3.0 g/L
- C. 5.0 g/L
- D. 20 g/L
Show answer and worked solution
Answer: B. 3.0 g/L
At steady state with positive biomass in a chemostat, the specific growth rate equals the dilution rate: μ = D.
0.30 = 0.50·S/(2.0 + S) → S = K_s·D/(μ_max − D) = (2.0)(0.30)/(0.50 − 0.30) = 3.0 g/L.
Under these assumptions the positive-biomass substrate result doesn't depend on the feed concentration, provided the feed can sustain that state. Here the washout threshold is μ(S_feed) = 0.50 × 20/(2 + 20) = 0.455 h⁻¹, above D = 0.30 h⁻¹; the calculated 3.0 g/L is also below the 20 g/L feed concentration. D < μ_max alone is not sufficient. The feed concentration sets how much biomass you get.
- A (1.2) uses μ_max in the denominator instead of (μ_max − D).
- C (5.0) swaps D and μ_max in the numerator.
- D (20) is the feed concentration, which you would see only at washout.
Relation: Chemostat steady state μ = D with Monod kinetics. Handbook search words to try: Monod, chemostat.
Source: MIT OCW 10.37, Lecture 13 (chemostats) (pp. 2–3: positive-biomass μ = D, Monod substrate result, finite-feed washout limit).
Question 23 — Pump shaft power
FEC-023 · Fluid Mechanics/Dynamics · Numerical entry
A pump moves 0.010 m³/s of liquid against a pressure rise of 300 kPa, with no elevation or velocity change. The pump efficiency (useful fluid power ÷ shaft power) is 75%. What shaft power is required? Enter kW to one decimal place.
Your answer: ______ kW
Show answer and worked solution
Answer: 4.0 kW
Fluid power = Q·ΔP = 0.010 × 300,000 = 3,000 W.
Shaft power = 3,000 / 0.75 = 4,000 W = 4.0 kW.
- Multiplying by the efficiency (2.25 kW) goes the wrong way: the shaft must supply more than the fluid receives.
- Stopping at fluid power (3.0 kW) leaves out the efficiency.
Relation: Fluid power = QΔP (= h_P·γ·Q); shaft power = fluid power / η. Handbook search words to try: pump power, efficiency.
Source: LearnChemE, Pressure Drop and Power in Pipe Flow (Important Equations: Reynolds number, head loss, pump power).
Question 24 — Best possible refrigerator
FEC-024 · Thermodynamics · Choose one answer
A refrigerator removes heat from a space at 260 K and rejects heat to surroundings at 300 K. What is the maximum possible coefficient of performance?
- A. 6.5
- B. 0.133
- C. 0.867
- D. 7.5
Show answer and worked solution
Answer: A. 6.5
Carnot refrigerator: COP_R = T_L / (T_H − T_L) = 260 / 40 = 6.5.
- D (7.5) is the Carnot heat pump COP, T_H/(T_H − T_L). Read which device the problem describes.
- B (0.133) is the Carnot engine efficiency between these temperatures.
- C (0.867) is T_L/T_H.
Relation: Carnot refrigerator COP = T_L/(T_H − T_L). Handbook search words to try: coefficient of performance, refrigeration.
Source: OpenStax University Physics Vol. 2, §4.5 The Carnot Cycle (Eqs. 4.6 and 4.7, refrigerator and heat-pump coefficients of performance).
Question 25 — Reading P&ID instrument tags
FEC-025 · Process Design · Matching
Using the conventions in the linked P&ID legend, match each instrument tag on a piping and instrumentation diagram (P&ID) to its meaning.
| Tag | Meaning |
|---|---|
| FIC | ? |
| TT | ? |
| LAH | ? |
| PSV | ? |
Meanings to place: Pressure safety valve · Level alarm, high · Flow indicating controller · Temperature transmitter
Counts as correct only if all four matches are right.
Show answer and worked solution
Answer:
FIC → flow indicating controller · TT → temperature transmitter · LAH → level alarm, high · PSV → pressure safety valve.
The first letter is the measured variable (F = flow, T = temperature, L = level, P = pressure). In these tags, I means indicate, C means control, T means transmit, A means alarm, and V means valve; H marks high. In PSV, S modifies pressure to identify the safety function. S can instead mean switch when used as a succeeding function letter, so read the letter position and the diagram legend.
- A common slip is reading the second T in “TT” as temperature again. Only the first letter names the variable.
Relation: Instrument-identification letters and combinations in the cited P&ID legend. Handbook search words to try: P&ID, instrumentation symbols.
Source: WorleyParsons P&ID legend hosted by MIMOSA (p. 1, Tables 1 and 2: identification letters and common letter combinations).
Question 26 — Diffusion across a thin layer
FEC-026 · Mass Transfer and Separation · Choose one answer
Species A diffuses at steady state across a 2.0 mm layer by equimolar counterdiffusion, with a linear concentration profile. The diffusivity is 1.0 × 10⁻⁵ m²/s, and the concentration difference across the layer is 0.40 mol/m³. What is the magnitude of the molar flux of A?
- A. 8.0 × 10⁻⁹ mol/(m²·s)
- B. 2.0 × 10⁻⁶ mol/(m²·s)
- C. 2.0 × 10⁻³ mol/(m²·s)
- D. 5.0 × 10⁻³ mol/(m²·s)
Show answer and worked solution
Answer: C. 2.0 × 10⁻³ mol/(m²·s)
Fick's first law, J = −D(dC/dx). With a linear profile, the gradient is ΔC/L, so the flux magnitude is D·ΔC/L.
J = (1.0 × 10⁻⁵)(0.40)/(0.0020) = 2.0 × 10⁻³ mol/(m²·s).
- B leaves L as 2 (millimeters not converted).
- A multiplies by the thickness instead of dividing.
- D leaves out ΔC.
Relation: Fick's first law. Handbook search words to try: Fick, diffusion.
Source: MIT 3.091, Lecture Notes No. 9: Diffusion (pp. 2–3: linear concentration gradient and Fick’s first law).
Question 27 — Radiation from a hot pipe
FEC-027 · Heat Transfer · Choose one answer
A small pipe surface (area 0.50 m², emissivity 0.80) at 500 K sits in a large room whose walls are at 300 K. Using σ = 5.67 × 10⁻⁸ W/(m²·K⁴), what is the net radiation heat loss from the pipe?
- A. 36 W
- B. 1,230 W
- C. 1,540 W
- D. 2,470 W
Show answer and worked solution
Answer: B. 1,230 W
Small body in large surroundings: q = εσA(T_s⁴ − T_surr⁴) = 0.80 × 5.67 × 10⁻⁸ × 0.50 × (500⁴ − 300⁴) = ≈ 1,230 W.
- A (36 W) uses (T_s − T_surr)⁴. Raise each temperature to the fourth power first, then subtract.
- C (1,540 W) treats the surface as a blackbody (ε = 1).
- D (2,470 W) leaves out the area (it is the flux in W/m²).
Relation: Net radiation: q = εσA(T_s⁴ − T_surr⁴), absolute temperatures. Handbook search words to try: radiation, Stefan-Boltzmann.
Source: OpenStax University Physics Vol. 2, §1.6 Mechanisms of Heat Transfer (Eq. 1.9 (conduction) and Eq. 1.10 (net radiation with emissivity)).
Question 28 — Flushing salt from a tank
FEC-028 · Mathematics · Choose one answer
A well-mixed 2.0 m³ tank initially contains brine at 40 kg/m³ salt. Pure water flows in at 0.50 m³/min, and mixed solution flows out at the same rate. What is the salt concentration in the tank after 6.0 min?
- A. 2.0 kg/m³
- B. 8.9 kg/m³
- C. 20.0 kg/m³
- D. 31.1 kg/m³
Show answer and worked solution
Answer: B. 8.9 kg/m³
Unsteady salt balance: V(dC/dt) = −QC, which integrates to C(t) = C₀·e^(−Qt/V).
Q/V = 0.50/2.0 = 0.25 min⁻¹, so C(6) = 40·e^(−1.5) = 40 × 0.2231 = 8.9 kg/m³.
- A (2.0) drops V and uses e^(−Qt) = e^(−3).
- D (31.1) is the decrease in concentration, 40(1 − e^(−1.5)), not the concentration remaining.
- C (20.0) assumes linear dilution.
Relation: Accumulation = in − out; first-order linear ODE. Handbook search words to try: differential equation, first-order.
Source: Fogler, Elements of Chemical Reaction Engineering 7e, Ch. 1 summary (Univ. of Michigan) (general mole balance; CSTR and PFR design equations).
Question 29 — Immersion heater energy
FEC-029 · Engineering Sciences · Choose one answer
A DC electric immersion heater draws 12 A at 240 V. If all the electrical energy becomes heat, how much heat is delivered in 30 minutes?
- A. 86.4 kJ
- B. 2,880 kJ
- C. 5,184 kJ
- D. 10,368 kJ
Show answer and worked solution
Answer: C. 5,184 kJ
P = IV = 12 × 240 = 2,880 W. Energy = P·t = 2,880 W × 1,800 s = 5,184,000 J = 5,184 kJ.
- A uses 30 (minutes) instead of 1,800 seconds.
- B reports the power in watts as if it were kilojoules.
- D uses one hour instead of 30 minutes.
Relation: P = IV; energy = power × time. Handbook search words to try: electrical power, Ohm's law.
Source: OpenStax University Physics Vol. 2, §9.5 Electrical Energy and Power (Eq. 9.12, P = IV).
Question 30 — Settling velocity of a fine particle
FEC-030 · Solids Handling · Choose one answer
A spherical particle (diameter 50 μm, density 2,500 kg/m³) settles in water (density 1,000 kg/m³, viscosity 1.0 × 10⁻³ Pa·s). Assuming Stokes' law applies, what is its terminal settling velocity?
- A. 0.51 mm/s
- B. 2.0 mm/s
- C. 3.4 mm/s
- D. 37 mm/s
Show answer and worked solution
Answer: B. 2.0 mm/s
Stokes' law: v_t = g·d²·(ρ_p − ρ_f) / (18μ) = 9.81 × (50 × 10⁻⁶)² × 1,500 / (18 × 1.0 × 10⁻³) = 2.04 × 10⁻³ m/s = 2.0 mm/s.
Check the assumption: particle Re = ρ_f·v·d/μ ≈ 0.10, inside the range where Stokes' law holds (up to about 0.2).
- A (0.51) uses the radius instead of the diameter.
- C (3.4) uses the particle density instead of the density difference.
- D (37) leaves out the 18.
Relation: Stokes' law, valid at low particle Reynolds number. Handbook search words to try: Stokes, terminal velocity.
Source: Richard G. Holdich, Fundamentals of Particle Technology, Ch. 5 (printed p. 46 (PDF p. 2), Eqs. 5.5–5.6: Stokes settling and particle Reynolds number).
Question 31 — Yield in parallel reactions
FEC-031 · Chemical Reaction Engineering · Choose one answer
Parallel reactions A → B (desired) and A → C occur in a reactor fed 100 mol of A. The product contains 20 mol A, 60 mol B, and 20 mol C. Define the fractional yield of B as mol B formed per mol A reacted. What is it?
- A. 0.60
- B. 0.75
- C. 0.80
- D. 3.0
Show answer and worked solution
Answer: B. 0.75
A reacted = 100 − 20 = 80 mol. Fractional yield of B = 60/80 = 0.75.
- A (0.60) divides by A fed. That is also a valid yield definition somewhere, which is exactly why problems state which one they mean. Use the definition given.
- C (0.80) is the conversion of A.
- D (3.0) is the selectivity of B relative to C (60/20).
Relation: Yield based on reactant consumed; selectivity = desired/undesired. Handbook search words to try: yield, selectivity.
Source: Fogler, Elements of Chemical Reaction Engineering 7e, Ch. 8 summary (Univ. of Michigan) (yield and selectivity definitions).
Question 32 — Steam for a water heater
FEC-032 · Material/Energy Balances · Numerical entry
Water flowing at 2.0 kg/s is heated from 20 °C to 80 °C (c_p = 4.18 kJ/kg·K) in an indirect heat exchanger by saturated steam that condenses completely, with no subcooling. Use a latent heat of 2,100 kJ/kg for the steam at its operating pressure. Ignoring heat losses, what steam flow is required? Enter kg/s to three decimal places.
Your answer: ______ kg/s
Show answer and worked solution
Answer: 0.239 kg/s
Heat to the water: Q = ṁc_pΔT = 2.0 × 4.18 × 60 = 501.6 kW.
Steam: ṁ_steam = Q/λ = 501.6 / 2,100 = 0.239 kg/s.
The steam gives up only its latent heat because it leaves as saturated liquid. If the condensate were subcooled, you would add that sensible heat too. When latent heat is not supplied, use the vapor-to-liquid enthalpy difference at the operating pressure.
- Dividing by c_p or using the water's temperature change for the steam mixes up the two streams. Balance heat lost by the steam against heat gained by the water.
Relation: Steady-flow energy balance: Q = ṁc_pΔT (water) = ṁλ (steam). Handbook search words to try: steam tables, latent heat.
Source: OpenStax University Physics Vol. 2, §1.4 Heat Transfer, Specific Heat, and Calorimetry (Eq. 1.5, Q = mcΔT); OpenStax University Physics Vol. 2, §1.5 Phase Changes (Eq. 1.8, Q = mL_v for vaporization/condensation).
Question 33 — Spotting an out-of-control subgroup
FEC-033 · Probability and Statistics · Choose one answer
An X̄ control chart tracks the mean of subgroups of n = 9 independent measurements. The process mean is 250, and the standard deviation of individual measurements is known to be 6. With 3-sigma limits, which subgroup mean signals an out-of-control condition?
- A. 255.1
- B. 251.0
- C. 243.5
- D. None of these; all are within limits
Show answer and worked solution
Answer: C. 243.5
Standard error of a subgroup mean: σ/√n = 6/3 = 2.
Limits: 250 ± 3(2) = 244 to 256. Only 243.5 falls outside, below the lower limit.
- D is what you conclude if you use σ = 6 instead of σ/√n (limits of 232 to 268).
- A (255.1) is close to the upper limit but still inside it.
- B (251.0) is near the center line.
Relation: X̄-chart limits: mean ± 3σ/√n. Handbook search words to try: control chart, control limits.
Source: NIST/SEMATECH e-Handbook, §6.3.2.1 Shewhart X-bar and R charts (control limits for subgroup means).
Question 34 — Stress in a loaded rod
FEC-034 · Materials Science · Choose one answer
A solid round stainless-steel rod with a diameter of 10 mm carries an axial tensile load of 15 kN. What is the normal stress in the rod?
- A. 191 MPa
- B. 19.1 MPa
- C. 47.7 MPa
- D. 764 MPa
Show answer and worked solution
Answer: A. 191 MPa
A = πd²/4 = π(0.010)²/4 = 7.854 × 10⁻⁵ m².
σ = F/A = 15,000 / (7.854 × 10⁻⁵) = 1.91 × 10⁸ Pa = 191 MPa.
- C (47.7) uses πd², putting the diameter where the radius belongs in πr².
- D (764) uses πr²/4, dividing by 4 twice.
- B (19.1) is a factor-of-10 slip in the unit conversion.
Relation: Axial stress σ = F/A. Handbook search words to try: stress, strain.
Source: OpenStax University Physics Vol. 1, §12.3 Stress, Strain, and Elastic Modulus (Eq. 12.34, tensile stress = F/A).
Question 35 — Fail position for reactor cooling
FEC-035 · Process Control · Choose one answer
A spring-return pneumatic control valve manipulates cooling-water flow to the jacket of a reactor running a strongly exothermic reaction. A completed hazard assessment specifies that loss of instrument air must leave this valve fully open: cooling-water pressure remains available, and full cooling does not introduce another hazard in this scenario. Which valve action meets that requirement?
- A. Fail open (air-to-close)
- B. Fail closed (air-to-open)
- C. Fail in last position
- D. Fail position doesn't matter if the reactor has a relief valve
Show answer and worked solution
Answer: A. Fail open (air-to-close)
The stated requirement is full cooling-water flow after loss of air, so this valve must fail open. A spring-to-open valve needs air to close it, which is why it is called air-to-close. The required fail position comes from the application-specific hazard assessment, not from a universal rule about exothermic reactors.
- B cuts off cooling and does not meet the stated fail-open requirement.
- C leaves the valve at its previous position rather than ensuring it is fully open.
- D ignores the fail-open requirement; the presence of a relief valve does not satisfy it.
Relation: Spring-return pneumatic actuator: spring-to-open / air-to-close. Handbook search words to try: control valve, fail-safe.
Source: Spirax Sarco, Control Valve Actuators and Positioners (Pneumatic actuators: spring-return action and selection on loss of air).
Question 36 — Where an SDS lists flash point
FEC-036 · Safety, Health, and Environment · Choose one answer
You need a solvent's flash point and its upper and lower flammability limits. Under OSHA's safety data sheet section numbering, which section should you go to?
- A. Section 2, Hazard(s) Identification
- B. Section 7, Handling and Storage
- C. Section 9, Physical and Chemical Properties
- D. Section 11, Toxicological Information
Show answer and worked solution
Answer: C. Section 9, Physical and Chemical Properties
OSHA's Appendix D lists flammability, lower and upper explosion/flammability limits, and flash point among the required Section 9 items.
Appendix D also requires the sheet to say when no applicable information was found for a required subheading; a Section 9 heading does not guarantee that every property has a numerical value.
- A gives the hazard classification (for example, a flammable-liquid category), not the underlying property data.
- B covers safe handling and storage practices.
- D covers health effects such as LD50 values.
Relation: 29 CFR 1910.1200 Appendix D, Table D.1. Handbook search words to try: safety data sheet.
Source: OSHA, 29 CFR 1910.1200 Appendix D (Appendix D introductory paragraph; Table D.1, Section 9(f)–(h)).
Question 37 — Straight-line book value
FEC-037 · Economics · Choose one answer
Equipment costs $500,000, has a 10-year life, and a $50,000 salvage value. Using straight-line depreciation, what is its book value at the end of year 4?
- A. $180,000
- B. $300,000
- C. $455,000
- D. $320,000
Show answer and worked solution
Answer: D. $320,000
Annual depreciation = (500,000 − 50,000)/10 = $45,000.
Book value after 4 years = 500,000 − 4(45,000) = $320,000.
- B ignores the salvage value ($50,000 per year).
- A is the accumulated depreciation, not the book value.
- C is the book value after only one year.
Relation: Straight-line depreciation = (cost − salvage)/life. Handbook search words to try: depreciation, straight line.
Source: OpenStax Principles of Financial Accounting, §11.3 (straight-line method; book value).
Question 38 — Which change is inherently safer?
FEC-038 · Process Design · Choose one answer
A plant stores and reacts a toxic intermediate in a large batch reactor. Which proposed change is an inherently safer design measure rather than an added safeguard?
- A. Replace the large batch reactor with a small continuous reactor so much less intermediate is on hand at any time
- B. Add a high-temperature interlock that shuts off the feed
- C. Install a larger relief valve with a scrubber
- D. Write a new operating procedure and retrain operators
Show answer and worked solution
Answer: A. Replace the large batch reactor with a small continuous reactor so much less intermediate is on hand at any time
Inherently safer design reduces or removes the hazard itself. The Center for Chemical Process Safety groups the approaches as substitute, minimize, moderate, and simplify. Cutting the inventory of a hazardous material is minimize.
- B is an active safeguard: a control system that manages the hazard without reducing it.
- C is an added mitigation device.
- D is a procedural safeguard and depends on people following it.
- All four can be worthwhile. Only A directly reduces the hazardous inventory in this comparison; that does not establish that the modified process is safer for every hazard.
Relation: CCPS inherently safer design strategies; inherent vs. passive, active, and procedural layers. Handbook search words to try: inherently safer, process safety.
Source: Hendershot, “Inherently Safer Design: The Fundamentals,” AIChE CEP (Jan. 2012) (CCPS strategies: substitute, minimize, moderate, simplify; inherent/passive/active/procedural).
Question 39 — Normality of sulfuric acid
FEC-039 · Chemistry and Biology · Choose one answer
9.8 g of sulfuric acid (H₂SO₄, MW = 98.08 g/mol) is dissolved in water to make 500 mL of solution. What is its normality for complete acid-base neutralization?
- A. 0.10 N
- B. 0.20 N
- C. 0.40 N
- D. 0.80 N
Show answer and worked solution
Answer: C. 0.40 N
Moles = 9.8/98.08 = 0.0999 mol. Molarity = 0.0999/0.500 L = 0.200 M.
H₂SO₄ donates 2 protons per mole in full neutralization, so normality = 2 × 0.200 = 0.40 N.
- B (0.20) is the molarity.
- A (0.10) is the number of moles, not a concentration.
- D (0.80) applies the factor of 2 twice.
Relation: N = n × M, where n is equivalents per mole for the reaction. Handbook search words to try: normality, molarity.
Source: Harvey, Analytical Chemistry 2.1, Appendix 16.1 Normality (LibreTexts) (N = n × M; n = 2 for H₂SO₄).
Question 40 — Pressured to approve an undersized relief valve
FEC-040 · Ethics and Professional Practice · Choose one answer
An engineer finds that a pressure-relief valve on a new reactor is undersized for a credible runaway scenario. Her supervisor tells her to approve the design anyway to keep the project on schedule. Under the NSPE Code of Ethics, what should she do?
- A. Approve it, because responsibility shifts to the supervisor who gave the instruction
- B. Approve it but keep a private note of her concerns in case of an incident
- C. Resign immediately without telling anyone, to avoid being associated with the project
- D. Decline to approve the nonconforming design, document her concerns to her employer, and, if her judgment is overruled where life or property is endangered, notify her employer or client and other appropriate authority
Show answer and worked solution
Answer: D. Decline to approve the nonconforming design, document her concerns to her employer, and, if her judgment is overruled where life or property is endangered, notify her employer or client and other appropriate authority
The NSPE Code says engineers shall hold paramount the safety, health, and welfare of the public; shall approve only engineering documents that conform to applicable standards; and, if their judgment is overruled where life or property is endangered, shall notify their employer or client and such other authority as may be appropriate.
- A and B both approve a document she knows is unsafe.
- C walks away without notifying anyone, which leaves the hazard in place.
Relation: NSPE Code, Canon I.1; Rules II.1.a–b. Handbook search words to try: ethics, code of ethics.
Source: NSPE Code of Ethics for Engineers (rev. July 2019) (Canon I.1; Rules II.1.a and II.1.b).
Score yourself
Count one point for each question you got completely right: every correct box on Question 20 and all four matches on Question 25, and nothing extra. That is how NCEES scores its own select-all and matching items — right or wrong, no partial credit. For this practice tally, a skipped question is unanswered; a solution opened before answering is reviewed, not correct. Keep correct, incorrect, unanswered, and reviewed separate. Use a score out of 40 only after answering all 40 without opening solutions first; otherwise record the four counts and use the set for review.
| NCEES knowledge area | Questions on the real exam | Questions here | Your score |
|---|---|---|---|
| 1. Mathematics | 6–9 | Q6, Q28 | ___ / 2 |
| 2. Probability and Statistics | 4–6 | Q10, Q33 | ___ / 2 |
| 3. Engineering Sciences | 4–6 | Q29 | ___ / 1 |
| 4. Materials Science | 4–6 | Q18, Q34 | ___ / 2 |
| 5. Chemistry and Biology | 7–11 | Q8, Q22, Q39 | ___ / 3 |
| 6. Fluid Mechanics/Dynamics | 8–12 | Q2, Q13, Q23 | ___ / 3 |
| 7. Thermodynamics | 8–12 | Q3, Q16, Q24 | ___ / 3 |
| 8. Material/Energy Balances | 10–15 | Q1, Q11, Q21, Q32 | ___ / 4 |
| 9. Heat Transfer | 8–12 | Q5, Q15, Q27 | ___ / 3 |
| 10. Mass Transfer and Separation | 8–12 | Q7, Q20, Q26 | ___ / 3 |
| 11. Solids Handling | 3–5 | Q30 | ___ / 1 |
| 12. Chemical Reaction Engineering | 7–11 | Q4, Q17, Q31 | ___ / 3 |
| 13. Economics | 4–6 | Q14, Q37 | ___ / 2 |
| 14. Process Design | 7–11 | Q9, Q25, Q38 | ___ / 3 |
| 15. Process Control | 4–6 | Q19, Q35 | ___ / 2 |
| 16. Safety, Health, and Environment | 5–8 | Q12, Q36 | ___ / 2 |
| 17. Ethics and Professional Practice | 3–5 | Q40 | ___ / 1 |
| Total | 110 | 40 | ___ / 40 |
The official ranges come from the FE Chemical specifications, pp. 1–4. They are ranges for individual areas, not percentages to add together; their endpoints need not sum to 110. Use the “Your score” cells for completed, unassisted items only, or record how many in that area you reviewed or left unanswered.
What your score means. It's your result on these 40 original questions, nothing more. NCEES converts the number of correct responses on scored items to a scaled score and reports the outcome as pass or fail; it doesn't publish a passing percentage (Examinee Guide, p. 14), so a percentage here can't be turned into a pass or fail. And with one to four questions per area, the area scores point you somewhere to look. They aren't a measurement of your strengths.
Answer key
Show answer key
Turn misses into a review list
For every question you missed, guessed, or opened the solution on before answering, fill in one row:
| Question | What went wrong | Corrected setup (equation + units) | Next action |
|---|---|---|---|
| e.g., Q13 | Setup: used V instead of V² | h_f = f(L/D)(V²/2g) | Rework two pipe-loss problems from scratch |
| Concept · Setup · Units · Arithmetic · Definition · Handbook lookup |
Then work in this order:
- Start with the heavy areas. Material/Energy Balances carries 10–15 questions on the real exam; Solids Handling carries 3–5. When your review needs are otherwise comparable, give the higher-count area priority; a tiny topic sample cannot measure which area will cost you more points.
- Rebuild each missed setup in the handbook. Find the governing equation with the search words under the question. If a word finds nothing, try another one. On exam day you search the handbook with a search box, and Ctrl+F doesn't work, so this habit is part of the skill (Examinee Guide, p. 10).
- Watch for repeat patterns. Check repeated unit and definition mistakes first. In this set those are Q13 (Darcy vs. Fanning), Q16 (absolute temperature), Q24 (refrigerator vs. heat pump), and Q31 (yield on A fed vs. A reacted).
- For topic review, the University of Colorado Boulder's free LearnChemE FE Chemical Exam Review organizes screencasts and modules by exam topic.
How close is this to the real FE Chemical exam?
It samples the same knowledge areas, but it's shorter and it isn't official.
- Topics. The real exam has 110 questions across the 17 knowledge areas in the NCEES FE Chemical specifications, in effect since the July 2020 exams. This set gives each area a share of 40 questions roughly in proportion to the middle of its official range. It samples every area but can't cover every subtopic, and NCEES's difficulty mix is its own.
- Question types. Most real questions are single-answer multiple choice. NCEES also uses multiple-correct, point-and-click, drag-and-drop, and fill-in-the-blank items (Examinee Guide, p. 11). This set includes a select-all item (Q20), a matching item (Q25), and two fill-in-the-blank items (Q23 and Q32). It has no point-and-click item.
- Units. The exam uses both SI and U.S. Customary units. Questions 1 and 16 use U.S. Customary units; most other calculations here use SI units.
- Pretest items. NCEES includes a limited number of unscored pretest items that are not identified to candidates (Examinee Guide, p. 11). Every question here is available for your practice tally.
Exam-day facts that change how you practice
| What | What NCEES says |
|---|---|
| Length | 110 questions in 5 hours 20 minutes of exam time. The appointment adds a 2-minute nondisclosure agreement, an 8-minute tutorial, and an optional 25-minute scheduled break (Examinee Guide, p. 16). |
| Two sections | All the exam time is given up front. After about half the questions you review and submit them, and you can't return to them afterward (p. 11). |
| Reference | The current FE Reference Handbook appears onscreen as a searchable PDF. Search with the box on the left; Ctrl+F isn't available (p. 10). You can download the same handbook free from MyNCEES to practice with. |
| Calculator | One approved model. The Examinee Guide, p. 8, explains the calculator rule and links to the current approved-model policy. A TI-30XS is also available onscreen. |
| Scratch work | The test center gives you 2 reusable booklets and 3 markers (p. 9). |
Fees, scheduling, retakes, and results are covered in the NCEES Examinee Guide. Use the instructions for your licensing board when checking eligibility or applying for Engineer Intern status; completing this practice set does not establish either.
How to get the most from this set
Have a calculator, scratch paper, and the free FE Reference Handbook from your MyNCEES account open. On exam day that handbook is your only reference, so look things up there instead of in your notes (Examinee Guide, p. 10).
- First pass: untimed. Work each question, then read the whole solution, including on questions you got right. Wrong options illustrate setup, unit, and definition mistakes.
- Second pass: timed. The real exam averages about 2.9 minutes per question (320 minutes ÷ 110). At that pace, 40 questions take about 1 hour 56 minutes. That's our pacing suggestion, not an NCEES time limit, and a second attempt after reading the solutions measures review, not a fresh start.
- Practice the handbook, not your notes. If you solved something from memory, find the equation in the handbook anyway.
Use original practice questions, not shared or recalled exam material. NCEES treats disclosure of protected exam material as an exam irregularity, with consequences that can include invalidated results (Examinee Guide, p. 13).
Sources
Exam facts on this page were checked against NCEES sources on October 6, 2026. The solutions link their underlying principles or identify the supplied hypothetical model.
- NCEES, FE Chemical CBT Exam Specifications, effective beginning with the July 2020 examinations.
- NCEES, Examinee Guide (May 2026): items allowed and calculator (pp. 8–9), reference materials (p. 10), exam format and item types (p. 11), irregularities (p. 13), scoring (p. 14), FE timing (p. 16).
- NCEES, FE exam overview, computer-based testing, and reference handbook access.
All 40 questions are original practice items prepared for this page. This version received an AI-assisted source and calculation check, including independent solution calculations and checks of the answer choices. That is not a review by a licensed engineer. Question 9 uses the hypothetical cost model supplied in its stem; source links in the other solutions support the underlying principles, not the invented practice values. See our methodology and editorial standards.
Castleport Test Prep is an independent exam prep publisher and is not affiliated with, endorsed by, or approved by NCEES (National Council of Examiners for Engineering and Surveying). These practice questions are original and unofficial; they are not real or recalled exam items. Exam and credential names identify the subjects discussed, and trademarks belong to their respective owners. Practice results don't guarantee passing the exam or licensure.
By the Castleport Test Prep Editorial Team · Last verified October 6, 2026