FE Chemical Exam Prep: 25 Free Practice Questions
Start your FE Chemical exam prep with 25 original, unofficial questions spanning all 17 NCEES topic areas, then use the topic map and eight-week plan. This is a starter set, not a full-length exam or a prediction of passing.
25 FE Chemical practice questions
Question 1. Evaporator solute balance
8. Material/Energy Balances · Single best answer
An evaporator receives 500 kg/h of an aqueous solution containing 12% nonvolatile solute by mass. At steady state the liquid product contains 30% solute by mass. No reaction occurs, all solute leaves in the liquid product, and the vapor is pure water. What is the vapor flow rate?
- A. 200 kg/h
- B. 300 kg/h
- C. 440 kg/h
- D. 500 kg/h
Show answer and explanation
Answer: B — 300 kg/h
Balance the component that only has one way out. Solute in = 500 × 0.12 = 60 kg/h, so the liquid product is 60 ÷ 0.30 = 200 kg/h. The total balance then gives vapor = 500 − 200 = 300 kg/h.
Why the others miss: A is the liquid product, not the vapor. C is all the water that came in, but some water stays in the product. D sends the whole feed overhead, which the nonvolatile solute can't do.
Work them untimed first, with your copy of the FE Reference Handbook open. Try each question before revealing its explanation.
Jump to: Topic map · Eight-week study plan · Exam rules · Answer key
Question 2. Excess air for methane
8. Material/Energy Balances · Single best answer
Methane burns completely: CH₄ + 2O₂ → CO₂ + 2H₂O. One mole of CH₄ is fed with 20% excess air. Take air as 21 mol% O₂ and 79 mol% N₂. How many moles of air are fed?
- A. 9.52 mol
- B. 2.40 mol
- C. 11.43 mol
- D. 9.03 mol
Show answer and explanation
Answer: C — 11.43 mol
Theoretical O₂ is 2 mol. With 20% excess, O₂ fed = 2 × 1.20 = 2.40 mol. Air carries that oxygen at 21%, so air fed = 2.40 ÷ 0.21 = 11.43 mol.
Why the others miss: A is theoretical air (0% excess). B is the oxygen fed, not the air. D is the nitrogen that comes along with the fed air (11.43 × 0.79).
Source: LearnChemE: Combustion Reactions (percent excess air)
Question 3. Sensible heat duty
8. Material/Energy Balances · Numeric entry · answer in kW
Liquid water flows steadily at 2.0 kg/s and is heated from 20 °C to 80 °C with no phase change. Use constant cp = 4.18 kJ/(kg·K). Neglect shaft work, kinetic and potential energy changes, and pressure effects on enthalpy. What is the heat duty, to the nearest kW?
Show answer and explanation
Answer: 502 kW (501.6 kW unrounded)
With the stated assumptions, the steady-flow energy balance gives Q̇ = ṁ·cp·ΔT = 2.0 × 4.18 × 60 = 501.6 kJ/s ≈ 502 kW. A temperature difference of 60 °C equals 60 K, so no offset is needed.
Common mistakes: Adding 273 to each temperature before subtracting changes nothing; adding it to only one temperature is the classic mistake. kJ/s is the same unit as kW.
Source: OpenStax University Physics Vol. 2, §1.4 (Q = mcΔT); LearnChemE: First Law – Open Systems (energy balance)
Question 4. Reynolds number in a small pipe
6. Fluid Mechanics/Dynamics · Single best answer
A Newtonian liquid with density 950 kg/m³ and dynamic viscosity 2.0 × 10⁻³ Pa·s flows at a mean velocity of 0.80 m/s in a pipe with a 25 mm inside diameter. Using Re = ρvD/μ, what is the Reynolds number?
- A. 95
- B. 950
- C. 95,000
- D. 9,500
Show answer and explanation
Answer: D — 9,500
Convert the diameter first: 25 mm = 0.025 m. Then Re = (950)(0.80)(0.025) ÷ 0.0020 = 9,500. The units cancel, so Re is dimensionless.
Why the others miss: A and B are 100× and 10× too small, and C is 10× too large — each is a power-of-ten slip in the diameter or viscosity. Check the exponent before you use Re to choose a friction factor.
Source: LearnChemE: Introduction to Pipe Flow (Reynolds number)
Question 5. Hydrostatic gauge pressure
6. Fluid Mechanics/Dynamics · Single best answer
An open tank holds water (ρ = 1,000 kg/m³) to a depth of 10.0 m. Using g = 9.81 m/s², what is the gauge pressure at the bottom?
- A. 9.81 kPa
- B. 98.1 kPa
- C. 981 kPa
- D. 199.4 kPa
Show answer and explanation
Answer: B — 98.1 kPa
For a static fluid open to the atmosphere, gauge pressure at depth h is ρgh = 1,000 × 9.81 × 10.0 = 98,100 Pa = 98.1 kPa.
Why the others miss: D adds an assumed atmospheric pressure of 101.3 kPa to the gauge pressure. Gauge pressure is measured relative to the atmosphere, so you don't add it. A and C are factor-of-10 errors.
Source: OpenStax University Physics Vol. 1, §14.1 (pressure at depth, Eq. 14.4); OpenStax University Physics Vol. 1, §14.2 Measuring Pressure
Question 6. Ideal-gas volume
7. Thermodynamics · Numeric entry · answer in m³
A gas is modeled as ideal. A 2.00 kmol sample is at 200 kPa (absolute) and 300 K. Use R = 8.314 kPa·m³/(kmol·K). What is its volume, rounded to two decimal places?
Show answer and explanation
Answer: 24.94 m³ (24.942 m³ unrounded)
V = nRT/P = (2.00)(8.314)(300) ÷ 200 = 24.942 m³ ≈ 24.94 m³. The pressure is already absolute and the temperature is already in kelvin, and the kmol in n cancels the kmol in R.
Common mistakes: Using mol with a gas constant expressed per kmol, or the reverse, introduces a factor-of-1,000 error.
Question 7. Raoult's law vapor composition
7. Thermodynamics · Single best answer
An ideal liquid solution is 50 mol% A and 50 mol% B. At the system temperature, the pure-component vapor pressures are 100 kPa for A and 40 kPa for B. Assume the equilibrium vapor is an ideal gas mixture containing only A and B. What is the mole fraction of A in the equilibrium vapor?
- A. 0.50
- B. 0.29
- C. 0.71
- D. 0.40
Show answer and explanation
Answer: C — 0.71
Raoult's law gives each partial pressure: p_A = 0.5 × 100 = 50 kPa and p_B = 0.5 × 40 = 20 kPa, so the total is 70 kPa. In the vapor, mole fraction = partial pressure ÷ total pressure, so y_A = 50 ÷ 70 = 0.71.
Why the others miss: A is the liquid composition. The vapor is richer in the more volatile component. B is y_B. D is the vapor-pressure ratio 40/100, which isn't a composition.
Source: OpenStax Chemistry 2e, §11.4 (Raoult's law, Vapor Pressure Lowering); OpenStax Chemistry 2e, §9.3 (partial pressure and mole fraction)
Question 8. Plane-wall conduction in U.S. units
9. Heat Transfer · Single best answer
Heat flows steadily in one dimension through a uniform plane wall with no internal heat generation. The wall area is 12 ft², the thickness is 3.0 in., and the thermal conductivity is 0.50 Btu/(h·ft·°F). The faces are held at 120 °F and 80 °F. Neglecting edge effects, what is the heat-transfer rate?
- A. 80 Btu/h
- B. 480 Btu/h
- C. 960 Btu/h
- D. 11,520 Btu/h
Show answer and explanation
Answer: C — 960 Btu/h
Put the thickness in the length unit used by k: 3.0 in. = 0.25 ft. With ΔT = 40 °F, q = kAΔT/L = (0.50)(12)(40) ÷ 0.25 = 960 Btu/h.
Why the others miss: A treats 3.0 as feet. B corresponds to double the thickness or half the ΔT. D multiplies by 12 again after the conversion. Also note that the question asks for the rate, not the flux per square foot.
Source: OpenStax University Physics Vol. 2, §1.6, Eq. 1.9 (conduction)
Question 9. Log-mean temperature difference
9. Heat Transfer · Single best answer
In a counterflow heat exchanger, the temperature difference between the hot and cold streams is 80 K at one end and 60 K at the other. What is the log-mean temperature difference (LMTD)?
- A. 69.5 K
- B. 70.0 K
- C. 20.0 K
- D. 140 K
Show answer and explanation
Answer: A — 69.5 K
LMTD = (ΔT₁ − ΔT₂) ÷ ln(ΔT₁/ΔT₂) = (80 − 60) ÷ ln(80/60) = 20 ÷ 0.2877 = 69.5 K.
Why the others miss: B is the arithmetic mean. It's close here because the end differences are similar, but it overstates the true mean and the gap grows as the ends diverge. C is just the numerator, and D is the sum.
Source: Virtual Labs (NITK): Heat Exchangers — Theory (LMTD)
Question 10. Binary flash vapor flow
10. Mass Transfer and Separation · Single best answer
A steady, nonreacting flash drum receives 100 kmol/h of a binary feed that is 0.40 mole fraction A. The equilibrium vapor is 0.70 mole fraction A and the liquid is 0.20 mole fraction A. There are no other streams. What is the vapor flow rate?
- A. 40 kmol/h
- B. 60 kmol/h
- C. 20 kmol/h
- D. 80 kmol/h
Show answer and explanation
Answer: A — 40 kmol/h
Write both balances. Total: 100 = V + L. Component A: 100(0.40) = 0.70V + 0.20L. Substituting L = 100 − V gives 40 = 0.50V + 20, so V = 40 kmol/h and L = 60 kmol/h.
Why the others miss: B is the liquid flow. C and D fail the component balance. As a quick check, the feed composition (0.40) sits between the liquid (0.20) and vapor (0.70) compositions, so both phases exist.
Source: LearnChemE: Flash Separations (Important Equations); LearnChemE: Material Balances (Important Equations)
Question 11. McCabe–Thiele rectifying line
10. Mass Transfer and Separation · Single best answer
A distillation column operates with constant molal overflow, a total condenser, and a reflux ratio R = L/D = 3. What is the slope of the rectifying-section operating line on a McCabe–Thiele diagram?
- A. 3.00
- B. 0.25
- C. 1.33
- D. 0.75
Show answer and explanation
Answer: D — 0.75
The rectifying operating line is y = [R/(R + 1)]x + x_D/(R + 1). Its slope is L/V = R/(R + 1) = 3/4 = 0.75.
Why the others miss: A uses R itself as the slope. B is 1/(R + 1), the coefficient on x_D in the intercept term. C inverts the slope. For the finite reflux ratio and total condenser here, V = L + D > L, so the slope is less than 1.
Source: LearnChemE: McCabe–Thiele Diagrams (rectifying operating line and reflux ratio)
Question 12. First-order batch reaction time
12. Chemical Reaction Engineering · Single best answer
A liquid-phase, first-order reaction A → B runs in an isothermal, constant-volume batch reactor with k = 0.10 min⁻¹. How long does it take to reach 90% conversion of A?
- A. 9.0 min
- B. 23.0 min
- C. 90 min
- D. 1.05 min
Show answer and explanation
Answer: B — 23.0 min
For first-order kinetics, ln(C_A0/C_A) = kt. At 90% conversion, C_A = 0.10·C_A0, so t = ln(10) ÷ 0.10 = 23.0 min.
Why the others miss: A assumes the concentration falls linearly with time. C is the CSTR answer in the next question. D uses 10% conversion instead of 90%.
Question 13. First-order CSTR space time
12. Chemical Reaction Engineering · Single best answer
A liquid-phase, first-order reaction A → B with k = 0.10 min⁻¹ and constant density runs in an ideal isothermal, steady-state continuous stirred-tank reactor (CSTR). What space time τ gives 90% conversion?
- A. 9.0 min
- B. 23.0 min
- C. 90 min
- D. 10.0 min
Show answer and explanation
Answer: C — 90 min
For a first-order CSTR, τk = X/(1 − X), so τ = 0.90 ÷ (0.10 × 0.10) = 90 min. A perfectly mixed tank runs the whole reaction at the low outlet concentration, which is why it needs about 4× the batch time from the previous question.
Why the others miss: A evaluates the rate at the feed concentration instead of the outlet concentration. B is the batch answer. D is just 1/k.
Source: Fogler, Elements of Chemical Reaction Engineering (7th ed. companion), Ch. 5 summary
Question 14. Arrhenius temperature effect
12. Chemical Reaction Engineering · Single best answer
A reaction follows the Arrhenius equation with a constant pre-exponential factor and an activation energy of 50 kJ/mol. By what factor does its rate constant increase when the temperature rises from 300 K to 310 K? Use R = 8.314 J/(mol·K).
- A. 1.91
- B. 1.03
- C. 0.52
- D. 2.00
Show answer and explanation
Answer: A — 1.91
ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂) = (50,000 ÷ 8.314)(1/300 − 1/310) = 0.647, so k₂/k₁ = e^0.647 = 1.91.
Why the others miss: B is the temperature ratio 310/300, which treats the temperature dependence as linear. C is the inverse (k₁/k₂). D is the 'rate doubles every 10 K' rule of thumb, which is not a calculation.
Question 15. Molarity of a NaOH solution
5. Chemistry and Biology · Single best answer
4.0 g of NaOH (molar mass 40 g/mol) is dissolved in water to make 500 mL of solution. What is the molarity?
- A. 0.10 M
- B. 0.050 M
- C. 8.0 M
- D. 0.20 M
Show answer and explanation
Answer: D — 0.20 M
Moles of NaOH = 4.0 ÷ 40 = 0.10 mol. Molarity = moles of solute per liter of solution = 0.10 ÷ 0.500 L = 0.20 M.
Why the others miss: A is the moles, not divided by the volume. C is grams per liter (8.0 g/L). B multiplies by the 0.500 L volume instead of dividing by it.
Question 16. Six-tenths rule cost scaling
14. Process Design · Single best answer
A process unit costs $100,000. Estimate the cost of a similar unit with twice the capacity, built at the same cost-index date, using a capacity exponent of 0.6.
- A. $151,600
- B. $200,000
- C. $66,000
- D. $141,400
Show answer and explanation
Answer: A — $151,600
Cost scales as C₂ = C₁(S₂/S₁)^n = 100,000 × 2^0.6 = $151,572 ≈ $151,600. The exponent below 1 captures economy of scale.
Why the others miss: B scales cost linearly with capacity. C applies the ratio upside down (0.5^0.6), which estimates a unit half the size. D uses an exponent of 0.5 instead of 0.6.
Question 17. Integrating a changing flow rate
1. Mathematics · Numeric entry · answer in L
Liquid enters a vessel at q(t) = 2.0 + 0.50t L/min for 0 ≤ t ≤ 4.0 min, where t is in minutes (so the 0.50 coefficient has units of L/min²). How many liters enter during this interval?
Show answer and explanation
Answer: 12 L
The volume is the integral of the rate: ∫₀⁴ (2.0 + 0.50t) dt = [2.0t + 0.25t²]₀⁴ = 8 + 4 = 12 L.
Common mistakes: Multiplying the final rate (4.0 L/min) by 4 min gives 16 L, which assumes the highest flow for the whole interval. The integral accounts for the flow rising over time.
Source: OpenStax Calculus Vol. 1, §5.3 The Fundamental Theorem of Calculus
Question 18. Expected downtime
2. Probability and Statistics · Single best answer
For a changeover, the downtime D is 2 min with probability 0.25, 5 min with probability 0.50, and 9 min with probability 0.25. What is the expected value E[D]?
- A. 3.00 min
- B. 5.00 min
- C. 5.25 min
- D. 5.33 min
Show answer and explanation
Answer: C — 5.25 min
The expected value is the probability-weighted sum: E[D] = 2(0.25) + 5(0.50) + 9(0.25) = 0.50 + 2.50 + 2.25 = 5.25 min. First confirm that the probabilities add to 1.
Why the others miss: B is the most likely single outcome, not the expected value. D is the unweighted average of 2, 5, and 9. A omits the 9-minute outcome and its probability.
Source: OpenStax Introductory Statistics 2e, §4.2 (expected value)
Question 19. Resistor power
3. Engineering Sciences · Single best answer
A 60 Ω resistive heater is connected across a 120 V DC supply. How much power does it dissipate?
- A. 2.0 W
- B. 240 W
- C. 7,200 W
- D. 120 W
Show answer and explanation
Answer: B — 240 W
For a resistor, P = V²/R = 120² ÷ 60 = 14,400 ÷ 60 = 240 W. (The current is 120 ÷ 60 = 2.0 A, and P = IV = 2.0 × 120 gives the same 240 W.)
Why the others miss: A is the current in amperes, not a power. C multiplies V by R. D just repeats the voltage.
Source: OpenStax University Physics Vol. 2, §9.5 Electrical Energy and Power
Question 20. Galvanic corrosion: zinc and steel
4. Materials Science · Single best answer
Zinc is electrically connected to carbon steel in an electrolyte under conditions where zinc is the more readily oxidized metal. Which statement describes the intended galvanic protection mechanism?
- A. The steel becomes the anode and corrodes faster.
- B. The zinc acts as the anode and corrodes preferentially, protecting the steel.
- C. Neither metal corrodes while the two are in electrical contact.
- D. Both metals corrode at the same rate.
Show answer and explanation
Answer: B — The zinc acts as the anode and corrodes preferentially, protecting the steel.
Zinc is oxidized at the anode, while the steel is made the cathode. This is the principle behind sacrificial-anode cathodic protection; the protective zinc is consumed rather than making corrosion stop everywhere.
Why the others miss: A reverses the roles. C overlooks consumption of the zinc anode. D does not describe the preferential oxidation on which this protection mechanism depends.
Source: OpenStax Chemistry 2e, §17.6 Corrosion (cathodic protection)
Question 21. Bulk density of a packed solid
11. Solids Handling · Single best answer
Dry, nonporous particles with a density of 2,500 kg/m³ settle into a bed with an interparticle void fraction (porosity) of 0.40. Neglect the mass of air in the voids. What is the bulk density of the bed?
- A. 1,000 kg/m³
- B. 2,500 kg/m³
- C. 4,167 kg/m³
- D. 1,500 kg/m³
Show answer and explanation
Answer: D — 1,500 kg/m³
Porosity ε = 1 − ρ_bulk/ρ_particle, so ρ_bulk = ρ_particle(1 − ε) = 2,500 × 0.60 = 1,500 kg/m³. Only 60% of the bed volume is solid.
Why the others miss: A multiplies by the void fraction instead of the solid fraction. B is the particle density. C divides when it should multiply, giving a bed denser than its own particles, which is impossible.
Question 22. Present value of one future payment
13. Economics · Single best answer
A project receives a single payment of $12,100 exactly two years from today. At an effective annual interest rate of 10%, what is the present value? Ignore taxes and all other cash flows.
- A. About $10,083
- B. $10,000
- C. $11,000
- D. $14,641
Show answer and explanation
Answer: B — $10,000
Rearranging FV = PV(1 + r)ⁿ gives PV = 12,100 ÷ (1.10)² = 12,100 ÷ 1.21 = $10,000.
Why the others miss: A uses simple interest, dividing by 1 + 2(0.10). C discounts for only one year. D compounds forward instead of discounting back to today.
Source: OpenStax Principles of Finance, §7.2 (FV = PV(1 + r)ⁿ)
Question 23. First-order process step response
15. Process Control · Single best answer
A first-order process with no dead time has a gain K = 2 (output units per input unit) and time constant τ. It starts at steady state, and the input steps up by 3 units at t = 0. By how much has the output changed at t = τ?
- A. 3.79 output units
- B. 6.00 output units
- C. 1.90 output units
- D. 2.21 output units
Show answer and explanation
Answer: A — 3.79 output units
The final change is K·Δu = 2 × 3 = 6. A first-order response reaches 63.2% of its final change after one time constant: Δy(τ) = 6(1 − e⁻¹) = 6 × 0.632 = 3.79.
Why the others miss: B is the final steady-state change, reached only as t → ∞. C applies 63.2% to the input step and forgets the gain. D is the 36.8% that remains to go at t = τ.
Source: APMonitor: Graphical Method, FOPDT to Step Test (Kp = Δy/Δu; 0.632Δy)
Question 24. Finding the right safety data sheet (SDS) sections
16. Safety, Health, and Environment · Select exactly two
Under OSHA's Safety Data Sheet format, which two sections cover (1) precautions for safe handling and storage and (2) exposure controls and personal protection? Select exactly two.
- A. Section 1 — Identification
- B. Section 7 — Handling and storage
- C. Section 4 — First-aid measures
- D. Section 8 — Exposure controls/personal protection
Show answer and explanation
Answer: B and D (both required)
Section 7 covers safe-handling precautions and storage conditions, including incompatibilities. Section 8 covers exposure limits (such as OSHA permissible exposure limits), engineering controls, and personal protective equipment.
Why the others miss: Section 1 identifies the product and supplier, and Section 4 covers first aid after an exposure. Knowing the section numbers tells you where to look, not which PPE suits a specific chemical or task.
Question 25. Competence under the NSPE Code
17. Ethics and Professional Practice · Single best answer
An engineer experienced in process calculations is asked to take independent technical responsibility for a specialized structural design. The engineer has no education or experience in structural engineering. Under the NSPE Code of Ethics, what is the appropriate response?
- A. Accept, because competence in one engineering field carries over to others.
- B. Sign the design after checking its arithmetic.
- C. Accept, but add a note disclosing the lack of structural experience.
- D. Decline that technical responsibility and arrange for a qualified engineer to take it on.
Show answer and explanation
Answer: D — Decline that technical responsibility and arrange for a qualified engineer to take it on.
The NSPE Code says engineers perform services only in areas of their competence (I.2). They take assignments only when qualified in the specific technical field (II.2.a), and they don't sign documents in areas where they lack competence (II.2.b). Bringing in a qualified engineer fixes the actual gap.
Why the others miss: A treats expertise as transferable across all fields. B mistakes an arithmetic check for technical competence. C discloses the problem but doesn't solve it. Actual licensing and signing rules also depend on the jurisdiction.
Source: NSPE Code of Ethics for Engineers (July 2019), I.2 and II.2.a–c
Answer key
Show all answers
| # | Topic area | Answer |
|---|---|---|
| 1 | 8. Material/Energy Balances | B — 300 kg/h |
| 2 | 8. Material/Energy Balances | C — 11.43 mol |
| 3 | 8. Material/Energy Balances | 502 kW (501.6 kW unrounded) |
| 4 | 6. Fluid Mechanics/Dynamics | D — 9,500 |
| 5 | 6. Fluid Mechanics/Dynamics | B — 98.1 kPa |
| 6 | 7. Thermodynamics | 24.94 m³ (24.942 m³ unrounded) |
| 7 | 7. Thermodynamics | C — 0.71 |
| 8 | 9. Heat Transfer | C — 960 Btu/h |
| 9 | 9. Heat Transfer | A — 69.5 K |
| 10 | 10. Mass Transfer and Separation | A — 40 kmol/h |
| 11 | 10. Mass Transfer and Separation | D — 0.75 |
| 12 | 12. Chemical Reaction Engineering | B — 23.0 min |
| 13 | 12. Chemical Reaction Engineering | C — 90 min |
| 14 | 12. Chemical Reaction Engineering | A — 1.91 |
| 15 | 5. Chemistry and Biology | D — 0.20 M |
| 16 | 14. Process Design | A — $151,600 |
| 17 | 1. Mathematics | 12 L |
| 18 | 2. Probability and Statistics | C — 5.25 min |
| 19 | 3. Engineering Sciences | B — 240 W |
| 20 | 4. Materials Science | B — The zinc acts as the anode and corrodes preferentially, protecting the steel. |
| 21 | 11. Solids Handling | D — 1,500 kg/m³ |
| 22 | 13. Economics | B — $10,000 |
| 23 | 15. Process Control | A — 3.79 output units |
| 24 | 16. Safety, Health, and Environment | B and D (both required) |
| 25 | 17. Ethics and Professional Practice | D — Decline that technical responsibility and arrange for a qualified engineer to take it on. |
Scoring: Give yourself one point per correct item, with both answers required for Question 24 and no partial credit. For numeric answers, use the stated unit and rounding.
How to read your result: Count the questions you got right in the core five areas (Questions 1–11) separately from the rest. Use the explanations for any missed core questions to choose which of weeks 1–5 needs more time. If you missed the reaction questions (12–14), move week 6's reaction block earlier. This set has only one to three questions per topic, too few to measure a topic precisely. Treat a miss as a pointer to your next study session, not a verdict.
What's on the FE Chemical exam: the 17-topic priority map
NCEES publishes a range of questions for each of the 17 knowledge areas in the FE Chemical exam specifications (effective beginning with the July 2020 exams). The ranges come straight from that document. The tiers are our recommended study order, not NCEES weights or difficulty ratings.
| # | NCEES knowledge area | Questions on the exam | Our study tier | Practice first |
|---|---|---|---|---|
| 8 | Material/Energy Balances | 10–15 | Core | Draw a system boundary and label every stream. Then work steady and unsteady balances, recycle and bypass, and combustion. |
| 6 | Fluid Mechanics/Dynamics | 8–12 | Core | Reynolds number with the right viscosity, the mechanical energy balance and pressure losses, flow meters, and pumps |
| 7 | Thermodynamics | 8–12 | Core | Property tables and phase diagrams, first and second law, Raoult's law and fugacity, chemical equilibrium, and heats of reaction |
| 9 | Heat Transfer | 8–12 | Core | Conduction, convection coefficients, overall U and fouling, LMTD and NTU, and radiation basics |
| 10 | Mass Transfer and Separation | 8–12 | Core | Flash and distillation balances, McCabe–Thiele, absorption and extraction, HTU/NTU, diffusion, and drying |
| 5 | Chemistry and Biology | 7–11 | Next | Molarity, normality, pH, redox, and solubility product; organic basics; cell biology and bioprocessing |
| 12 | Chemical Reaction Engineering | 7–11 | Next | Rate laws, Arrhenius, conversion/yield/selectivity, and batch vs. CSTR vs. PFR design equations |
| 14 | Process Design | 7–11 | Next | Reading PFDs and P&IDs, equipment sizing, cost indices and capacity scaling, and design standards |
| 1 | Mathematics | 6–9 | Next | Calculus, differential equations, matrices, numerical methods, and significant figures |
| 16 | Safety, Health, and Environment | 5–8 | Keep in rotation | SDS sections, HAZOP and LOPA, relief and overpressure protection, and waste regulations |
| 2 | Probability and Statistics | 4–6 | Keep in rotation | Expected value, distributions, confidence intervals, t-tests, regression, and control limits |
| 3 | Engineering Sciences | 4–6 | Keep in rotation | Force and momentum, work/energy/power, and Ohm's and Kirchhoff's laws |
| 4 | Materials Science | 4–6 | Keep in rotation | Property effects of temperature and stress, corrosion, and polymers, ceramics, and composites |
| 13 | Economics | 4–6 | Keep in rotation | Time value of money, break-even and benefit-cost, and comparing projects with unequal lives |
| 15 | Process Control | 4–6 | Keep in rotation | First- and second-order dynamics, gain and time constant, PID, and feedback vs. feedforward vs. cascade |
| 11 | Solids Handling | 3–5 | Keep in rotation | Particle size and bulk properties, crushing and grinding, and conveying and storage |
| 17 | Ethics and Professional Practice | 3–5 | Keep in rotation | Codes of ethics, contracts, public safety, and intellectual property |
Abbreviations in the map: U = overall heat-transfer coefficient; NTU = number of transfer units; HTU = height of a transfer unit; CSTR = continuous stirred-tank reactor; PFR = plug-flow reactor; PFD = process flow diagram; P&ID = piping and instrumentation diagram; HAZOP = hazard and operability study; LOPA = layer of protection analysis; PID = proportional–integral–derivative control.
Why the core five come first: adding their published question-range bounds gives 42 to 63 of the 110 questions. The sum runs from 10 + 4 × 8 at the low end to 15 + 4 × 12 at the high end. Their balance and transport methods also support reactor and separation calculations. Don't skip the smaller areas, though.
The range endpoints are not a single 110-question exam form; do not add all lower or all upper bounds and treat that sum as the exam length. Counting how many subtopics a knowledge area lists won't tell you how heavily it's tested; use the published question ranges.
Your eight-week FE Chemical study plan
Assumptions: about 8 hours a week (for example, 1 hour on four weeknights plus 2 hours each weekend day). This plan is aimed at a senior or a graduate up to a few years out of school. It's our editorial plan, not an NCEES requirement, and no number of hours guarantees a result.
| Week | Where the 8 hours go | What to do |
|---|---|---|
| 1 | Starter set 1.5 h · handbook setup 0.5 h · Material/Energy Balances 5 h · error log 1 h | Download the current FE Reference Handbook from your MyNCEES account. Work the 25 questions above untimed and log every miss. Drill balances: mixing, evaporation, recycle and bypass, combustion, and sensible and latent heat. |
| 2 | Thermodynamics 4 h · Chemistry and Biology 3 h · review 1 h | Steam tables and phase diagrams, first and second law, Raoult's law, and heats of reaction. Then molarity and normality, pH and pKa, redox, and the biology list in the topic map. |
| 3 | Fluid Mechanics 5 h · Mathematics 2 h · review 1 h | Reynolds number, friction factor and pressure drop, the mechanical energy balance, pumps and flow meters. Then calculus, ordinary differential equations (ODEs), and Newton–Raphson. |
| 4 | Heat Transfer 5 h · Probability and Statistics 2 h · review 1 h | Conduction, convection correlations, overall U, LMTD and NTU, and radiation. Then expected value, distributions, confidence intervals, and control limits. |
| 5 | Mass Transfer and Separation 5 h · Solids Handling 2 h · review 1 h | Flash, McCabe–Thiele, absorption and extraction, HTU/NTU, humidification and drying. Then particle properties, size reduction, and conveying. |
| 6 | Reaction Engineering 4 h · Process Control 2 h · Engineering Sciences 1 h · review 1 h | Batch vs. CSTR vs. PFR for the same rate law, Arrhenius, and selectivity. First-order response, PID, and control strategies. Basic dynamics and circuits. |
| 7 | Process Design 2 h · Safety 2 h · Economics 1.5 h · Materials 1 h · Ethics 0.5 h · review 1 h | PFD/P&ID reading and cost scaling, SDS, HAZOP and relief devices, present worth and project comparison, corrosion, and a named code of ethics |
| 8 | Timed starter set 1.5 h · targeted drills 4 h · review misses 2 h · exam-day check 0.5 h | Allow about 73 minutes for the 25 questions above, then use the remainder of the first block to record where time went. Use the four-hour block to rebuild missed setups and repeat the handbook lookup drill below. Rework every miss from a blank page. Confirm your ID, calculator, and appointment details. |
That's 64 planned hours, and every one of the 17 topics has a dedicated block. The 73-minute exercise is scaled from 320 ÷ 110 × 25 ≈ 72.7 minutes, not a simulation of the full exam. Repeating this familiar set checks your method and pacing on these items; it does not establish readiness for unseen questions. Additional free topic practice is linked below to extend this starter set; all 25 questions and explanations are available on this page.
Adjust the plan to your situation
- Out of school three or more years: stretch weeks 1–4 into six weeks, spending the extra time on balances, thermodynamics, and fluids, for a 10-week plan. This is an editorial adjustment for rebuilding fundamentals, not a study-hour requirement established by pass-rate data.
- Only 4 hours a week: run each week over two calendar weeks, for about 16 weeks.
- Exam in four weeks: combine weeks 1 and 2, then 3 and 4, then 5 and 6, then 7 and 8. That preserves the 64-hour plan and requires about 16 hours a week. With less time available, prioritize the errors you actually find rather than pretending the same workload fits.
- Retaking after a fail: start from your NCEES diagnostic report (see Exam rules) and give your lowest-scoring areas the first weeks. Keep a lighter pass over everything else.
Graduation-timing context: NCEES's 2014–25 data for ABET first-time takers at member-board sites, across all FE disciplines, reports 71.34% before graduation, 70.62% within 12 months after graduation, and 63.43% at 12 or more months after graduation. These are the source's cohort labels and aggregate pass rates, not Chemical-only results or evidence that a particular study schedule causes a pass (NCEES Squared 2025, p. 15).
What one study hour looks like
10 minutes reviewing one concept → 10 minutes finding it in the handbook → 25 minutes solving problems → 10 minutes checking the solutions → 5 minutes logging what went wrong. Shift the minutes as needed; it's a starting routine, not a rule.
Turn misses into your next session
Write down why an answer went wrong, not just that it did. A units slip and a wrong model need different fixes. After reading an explanation, rework the setup from a blank page, because remembering the answer letter isn't the same as solving the problem.
| Question / topic | What went wrong | Correct setup | Error type | Next action |
|---|---|---|---|---|
| Example: Q13, CSTR | Used the feed concentration in the rate | Rate is evaluated at the outlet: C_A = C_A0(1 − X), so τk = X/(1 − X) | Model / setup | Rework Q12 and Q13 side by side; compare batch and CSTR for one more rate law |
Useful error types: concept, model/setup, units, handbook lookup, arithmetic/calculator, time.
Free help for a weak topic: LearnChemE's FE Chemical review organizes free screencasts, simulations, and self-study modules by FE topic. Use it after your error log shows you which topic to work on.
Using the FE Reference Handbook on exam day
You can't bring notes or your own copy. NCEES supplies the current FE Reference Handbook onscreen as a searchable PDF, on a 24-inch monitor next to the questions. You search it with the search box in the exam software; Ctrl+F doesn't work (NCEES Examinee Guide, May 2026, p. 10). Download the current version from your MyNCEES account and study with that same file. NCEES lets you print it for personal use but not share or repost it (NCEES help: reference handbooks).
Lookup drill (15 minutes): find the relation you'd need for a Reynolds number, an ideal-gas volume, plane-wall conduction, an LMTD, a first-order CSTR, and a present-worth factor. For each one, note the search term that found it fastest, the units of every input, and one assumption that limits it. Record search terms that work in your version rather than page numbers.
Also note which kinds of questions the handbook won't solve for you. Ethics, contracts, many safety questions, and reading a P&ID depend on understanding, not on finding a formula.
FE Chemical exam rules that affect how you prepare
| Rule | What NCEES says | What it means for you |
|---|---|---|
| Length | 110 questions in 5 h 20 min of exam time (FE exam page; Examinee Guide, p. 16) | That's about 2.9 minutes (≈2 min 55 s) per question on average. It's an average, not a per-question limit. |
| Appointment | The FE page calls the appointment 6 hours. The guide's table, p. 16 lists 5 h 55 min: a 2-min nondisclosure agreement, an 8-min tutorial, the exam, and a 25-min scheduled break. | Plan for about 6 hours and follow your appointment confirmation. |
| Two halves | After about half the questions you review and submit, and you can't reopen them. Then you can take the scheduled break. Unscheduled breaks use exam time. (Guide, pp. 11–12) | Review your flagged questions before submitting the first half. |
| Question types | Multiple choice and alternative item types, including multiple-correct, point-and-click, drag-and-drop, and fill-in-the-blank. All are scored right or wrong with no partial credit. Some unscored pretest items are mixed in. (NCEES CBT page; Guide p. 11) | Practice numeric-entry and select-all questions, like Questions 3, 6, 17, and 24 above. |
| Scoring | Pass/fail on a scaled score. There's no deduction for wrong answers, and NCEES doesn't publish the passing score. (Guide, p. 14) | Answer every question, even if you have to guess. |
| Calculator | One NCEES-approved calculator. A TI-30XS is also available onscreen. (Guide, p. 8) | Confirm your handheld model with NCEES through your MyNCEES account before exam day. The onscreen calculator is an alternative, not permission to bring an unapproved model. |
| Scratch work | Two reusable booklets and three markers, supplied at the test center. (Guide, p. 9) | Practice working without pencil and paper. |
| Fee | $225, paid to NCEES. Your licensing board may charge its own application fee. (Guide, p. 3) | Check your board's process before you register. |
| Results | Usually 7–10 days after the exam, reported as pass/fail. If you fail, you get a diagnostic report showing your performance in each knowledge area. (Guide, p. 14) | Use the diagnostic report to reorder your plan. |
| Retakes | One attempt per testing window (Jan–Mar, Apr–Jun, Jul–Sep, Oct–Dec), and no more than three in any 12 months. Some boards are stricter. (Guide, p. 5) | A failed attempt costs you at least the rest of that testing window. |
| Eligibility | Set by each licensing board, not by NCEES. (Guide, p. 2) | Use the jurisdiction information on the NCEES FE exam page to confirm your board's registration process. |
Historical pass rate: in NCEES's fiscal year from October 1, 2024, to September 30, 2025, 68% of 1,873 first-time FE Chemical takers passed. Among 316 repeat takers, 35% passed. Among the 1,654 first-time takers with an EAC/ABET-accredited bachelor's degree, the rate was 69% (NCEES Squared 2025, pp. 10–11). These are group statistics for that year, not anyone's individual odds.
FE results and the path to licensure
Do I need the FE to become a PE? Passing the FE is the usual first exam step toward a PE license, but the rules for engineer-intern (EI/EIT) certification and licensure belong to your state board. Passing the FE doesn't by itself let you practice as a professional engineer. NCEES explains the FE's place in the licensure process.
Sources and independence
By Castleport Test Prep Editorial Team · Last verified: October 6, 2026
On that date we checked the exam specifications, format, timing, item types, handbook access, fees, retake rules, and pass rates against the NCEES sources below. We also checked each practice question's underlying principle against the source cited with it and independently checked every answer, including recalculating numerical results. AI tools assisted with drafting, source checking, and calculation checks. These are source and arithmetic checks, not a professional engineering review. Editorial standards.
- NCEES, FE Chemical CBT Exam Specifications, effective beginning with the July 2020 exams, pp. 1–4
- NCEES, FE exam page
- NCEES, Examinee Guide, May 2026 (current guide: ncees.org/examinee-guide)
- NCEES, Computer-based testing: alternative item types
- NCEES Help, Exam reference handbooks
- NCEES, Squared 2025, FY October 1, 2024–September 30, 2025
- Teaching sources for each question are linked inside its explanation.
Castleport Test Prep is an independent exam prep publisher and is not affiliated with, endorsed by, or approved by NCEES. The practice questions on this page are original and are not taken from any actual exam. Exam and credential names are used only to identify the exam; trademarks belong to their respective owners.