PE Chemical Exam Prep
Start your PE Chemical exam prep with the original, unofficial problem below. All 36 problems, their worked solutions, and a 12-week study plan are free on this page.
Problem 1 · Mass balances with reaction (recycle)
CHE-MEB-01 · Mass/Energy Balances · Select one
A reactor converts A → B. Fresh feed is 100 mol/h of pure A. Single-pass conversion in the reactor is 25%. A separator after the reactor recovers all unreacted A and recycles it to mix with the fresh feed; the product stream contains no A. At steady state, what is the recycle flow of A?
- A. 75 mol/h
- B. 133 mol/h
- C. 300 mol/h
- D. 400 mol/h
Show answer and solution
Answer: C. 300 mol/h
No A leaves in the product, so overall conversion is 100%: all 100 mol/h of fresh A ends up as B. The reactor converts only 25% of what enters it, so the reactor feed F satisfies 0.25F = 100 mol/h, giving F = 400 mol/h. Recycle = 400 − 100 = 300 mol/h.
Why the other choices are wrong:
- A (75 mol/h) is the unreacted A if the fresh feed passed through the reactor once with no recycle.
- B (133 mol/h) is 100 ÷ 0.75, which mixes up the converted and unconverted fractions.
- D (400 mol/h) is the reactor feed (fresh feed plus recycle), not the recycle itself.
Tip: Draw two boundaries: the overall process (overall conversion) and the reactor alone (single-pass conversion). Recycle problems usually need both.
Source: DOE-HDBK-1012/3-92, Fluid Flow, conservation of mass (Eq. 3-4), p. 10
By Castleport Test Prep Editorial Team · Exam facts last verified October 8, 2026
On this page: Topic weights · Your only reference · 36 problems · 12-week plan · Exam-day rules · Pass rates · Quick answers
The PE Chemical exam at a glance
- 80 questions on a computer, offered year-round at Pearson VUE test centers.
- 9-hour appointment: a 2-minute nondisclosure agreement, an 8-minute tutorial, the 8-hour exam, and a 50-minute scheduled break.
- $400 paid to NCEES. Your licensing board may charge its own fee too.
- One reference: the electronic NCEES PE Chemical Reference Handbook, shown on screen. No design standards, no books, and no notes of your own.
- Units: both SI and U.S. customary.
- Question types: multiple choice plus alternative item types (select all that apply, point and click, drag and drop, and fill in the blank). Each question is scored right or wrong, with no partial credit.
- Results: pass or fail, typically 7–10 business days after the exam.
- First-time pass rate: 55% of 225 first-time takers, January–June 2026.
What to study first: the 7 areas by published question range
NCEES lists seven knowledge areas, each with a range of questions. The current NCEES Chemical page still links the specification effective January 1, 2020, as verified October 8, 2026.
| Rank | Knowledge area | Questions (of 80) | What it covers | Practice here |
|---|---|---|---|---|
| 1 | Plant Design and Operation | 15–23 | Safety, health, and environment; design, economics, materials, and process control; operation and maintenance | Problems 2–9 |
| 2 | Mass/Energy Balances | 12–18 | Mass balances with and without reaction (recycle, bypass, purge); energy balances with and without reaction | Problems 1, 10–14 |
| 3 | Thermodynamics | 11–17 | Properties and state functions, cycles, chemical equilibrium, phase equilibrium | Problems 15–19 |
| 4 | Fluids | 10–16 | Mechanical-energy balance, pipe flow, compressible flow, pumps, compressors, flow measurement | Problems 20–24 |
| 5 | Heat Transfer | 9–14 | Conduction, convection, radiation, phase change, heat exchanger design and rating | Problems 25–28 |
| 6 | Mass Transfer | 7–11 | Diffusion, staged separations such as distillation, absorption, extraction, and other separations | Problems 29–32 |
| 7 | Chemical Reaction Engineering | 6–10 | Rate laws, kinetics, reactor types, conversion, yield, and selectivity | Problems 33–36 |
The question ranges come from NCEES. The ranking is ours, based on the middle of each range, and it says nothing about difficulty. The ranges overlap, so study all seven areas. Two things stand out:
- Plant Design and Operation is the biggest area. It mixes process safety, economics, control valves, and operations. Don't treat it as the "soft" section.
- Mass/energy balances and thermodynamics together carry 23–35 questions. They also feed every other area, so weak balances cost you points everywhere.
Your only reference on exam day: the handbook
On the PE Chemical exam, the only reference is the electronic NCEES PE Chemical Reference Handbook. It appears on screen next to the questions. Unlike some other PE exams, PE Chemical supplies no design standards. You can't bring your own books or notes, so older advice about tabbing a stack of reference books is out of date.
You can download the same handbook for free from your MyNCEES account. Practice with it from day one. On exam day, you search it with a search box on the left side of the screen. Ctrl+F doesn't work. A handbook you've used for 12 weeks is much faster than one you're seeing for the first time.
A handbook lookup drill
Do this drill in week 1 and again in week 11. For each item, open your downloaded handbook, find the relationship, and note where it is, what you searched, and how long it took. If you can't find one, write that down too. Knowing what isn't in the handbook tells you what you need to understand well enough to set up yourself.
- The mechanical-energy balance (Bernoulli's equation with friction and pump work)
- The Reynolds number, and where flow turns turbulent
- The friction factor, and whether the chart or equation uses Darcy or Fanning
- Net positive suction head (NPSH) available
- The log mean temperature difference and its correction factor
- The overall heat transfer coefficient, including fouling
- The compressibility factor or an equation of state for real gases
- The Arrhenius equation
- Design equations for a CSTR and a plug-flow reactor
- The Fenske equation for minimum stages
Keep a lookup log as you study. Here's one way to set it up:
| Topic | Where it is in your handbook | Search term that worked | Time to find | What slowed you down |
|---|---|---|---|---|
| Example: LMTD with correction factor | (record the section you found) | (record the words that worked) | (minutes) | (for example, searched "LMTD" but the handbook spells it out) |
36 original practice problems with worked solutions
These 36 problems are original and unofficial. They are not NCEES questions, and they are not a full-length exam. Problems are spread across the seven areas roughly in line with the NCEES question ranges. Most are select-one questions. Problem 18 is select all that apply, and Problem 14 asks you to enter a number. The real exam also uses point-and-click and drag-and-drop questions, which don't fit on a text page.
Work each problem before you open its solution. Each solution explains the right answer, why each wrong choice is tempting, and where to read more.
Plant Design and Operation (Problems 2–9)
Problem 2 · Hazards identification and management (exposure limits)
CHE-PD-01 · Plant Design and Operation · Select one
During an 8-hour shift, an operator is exposed to a solvent at 120 ppm for 4 hours, 60 ppm for 2 hours, and 0 ppm for 2 hours. The solvent's 8-hour time-weighted average (TWA) limit is 100 ppm. Using the OSHA computation in 29 CFR 1910.1000(d)(1)(i), what is the equivalent 8-hour exposure, and is it within the limit?
- A. 75 ppm; within the limit
- B. 90 ppm; within the limit
- C. 100 ppm; at the limit
- D. 120 ppm; over the limit
Show answer and solution
Answer: A. 75 ppm; within the limit
E = (CaTa + CbTb + …) ÷ 8 = (120 × 4 + 60 × 2 + 0 × 2) ÷ 8 = 600 ÷ 8 = 75 ppm, which is below 100 ppm.
Why the other choices are wrong:
- B (90 ppm; within the limit) averages the two nonzero concentrations and ignores how long each lasted.
- C (100 ppm; at the limit) divides by the 6 exposed hours. OSHA's formula divides by the 8-hour shift.
- D (120 ppm; over the limit) compares the peak concentration to a time-weighted limit.
Tip: A TWA is concentration × hours summed, then divided by 8. Check separately whether the substance also has a ceiling or short-term limit.
Source: OSHA, 29 CFR 1910.1000(d)(1), 8-hour time-weighted average computation
Problem 3 · Hazards identification and management (PSM applicability)
CHE-PD-02 · Plant Design and Operation · Select one
Consider only 29 CFR 1910.119(a)(1)(ii), the flammables provision of OSHA's Process Safety Management (PSM) standard. Each site is an occupied manufacturing facility (not retail, not well drilling or servicing, not a remote unoccupied facility). Which one is covered by PSM?
- A. 15,000 lb of a flammable liquid (flashpoint below 100 °F) kept below its normal boiling point in an atmospheric tank used only for storage or transfer to storage and not connected to a process vessel
- B. 8,000 lb of propane in a process
- C. 20,000 lb of propane used only as fuel for comfort heating and not part of a process with another covered chemical
- D. 12,000 lb of propane, a Category 1 flammable gas, held in interconnected process vessels
Show answer and solution
Answer: D. 12,000 lb of propane, a Category 1 flammable gas, held in interconnected process vessels
Paragraph (a)(1)(ii) covers a process involving 10,000 lb or more of a Category 1 flammable gas, or a flammable liquid with a flashpoint below 100 °F, on site in one location, unless an exception applies. Interconnected vessels count as one process, and 12,000 lb exceeds the threshold.
Why the other choices are wrong:
- A (15,000 lb of a flammable liquid in an atmospheric tank used only for storage or transfer to storage) fits the flammable-liquid storage exception in (a)(1)(ii)(B); OSHA's enforcement guidance distinguishes storage/transfer-only tanks from tanks connected to a process vessel.
- B (8,000 lb of propane in a process) is below the 10,000-lb threshold.
- C (20,000 lb of propane used only as fuel for comfort heating and not part of a process with another covered chemical) matches exception (a)(1)(ii)(A) for hydrocarbon fuels used solely for workplace consumption.
Tip: For applicability questions, check the threshold first, then each exception, then the facility exclusions in (a)(2).
Sources: OSHA, 29 CFR 1910.119(a)(1)(ii)(A)–(B), PSM application to flammables; OSHA PSM compliance guidance, flammable-liquid atmospheric storage tanks and process connections
Problem 4 · Chemical hazards (flammability of mixtures)
CHE-PD-03 · Plant Design and Operation · Select one
A fuel gas is 50 mol% methane and 50 mol% propane on a fuel-only basis. Use lower flammable limits (LFL) of 5.0 vol% for methane and 2.0 vol% for propane in air. Using Le Chatelier's mixing rule, estimate the LFL of the mixture.
- A. 2.0 vol%
- B. 2.9 vol%
- C. 3.5 vol%
- D. 7.0 vol%
Show answer and solution
Answer: B. 2.9 vol%
1/LFL_mix = Σ(y_i/LFL_i) = 0.5/5.0 + 0.5/2.0 = 0.10 + 0.25 = 0.35, so LFL_mix = 1/0.35 = 2.86 ≈ 2.9 vol%.
Why the other choices are wrong:
- A (2.0 vol%) takes the lowest component limit, which overstates how easily the mixture ignites.
- C (3.5 vol%) is a simple average. The rule weights by reciprocals, so the more flammable component pulls the result down.
- D (7.0 vol%) adds the two limits.
Tip: Le Chatelier's rule is an estimate. Mixtures with very different components, or conditions far from ambient, need measured data.
Problem 5 · Process design (scale-up and economics)
CHE-PD-04 · Plant Design and Operation · Select one
A 100,000 t/yr plant cost $40 million. Using the six-tenths capacity exponent and ignoring time and location adjustments, estimate the cost of a 200,000 t/yr plant of the same design.
- A. $56.6 million
- B. $67.3 million
- C. $60.6 million
- D. $80.0 million
Show answer and solution
Answer: C. $60.6 million
Cost₂ = Cost₁ × (Capacity₂/Capacity₁)^0.6 = 40 × 2^0.6 = 40 × 1.516 = $60.6 million.
Why the other choices are wrong:
- A ($56.6 million) uses an exponent of 0.5.
- B ($67.3 million) uses an exponent of 0.75.
- D ($80.0 million) scales cost linearly, with no economy of scale.
Tip: The 0.6 exponent is a rule of thumb. Real exponents vary by equipment type, and cost-index (time) and location corrections are separate steps.
Source: NASA TM X-73397, Handbook of Estimating Data, Factors, and Procedures (1977), Part II.A.2 “The Six-Tenths Factor in Estimating”; DOE G 413.3-21, Cost Estimating Guide (2011), §5.2.4 Capacity Factor Method, p. 21
Problem 6 · Process design (economics)
CHE-PD-05 · Plant Design and Operation · Select one
A heat-recovery project costs $1,000,000 now and saves $300,000 at the end of each year for 5 years, with no salvage value. At a 10% discount rate, the net present value is closest to:
- A. −$137,000
- B. +$500,000
- C. +$1,137,000
- D. +$137,000
Show answer and solution
Answer: D. +$137,000
Uniform present value factor = [(1 + d)^N − 1] ÷ [d(1 + d)^N] = (1.1^5 − 1) ÷ (0.1 × 1.1^5) = 3.7908. Present value of savings = 300,000 × 3.7908 = $1,137,236. NPV = 1,137,236 − 1,000,000 ≈ +$137,000.
Why the other choices are wrong:
- A (−$137,000) has the sign reversed. The discounted savings exceed the investment.
- B (+$500,000) adds up the savings without discounting (5 × 300,000 − 1,000,000).
- C (+$1,137,000) is the present value of the savings before subtracting the investment.
Tip: Write the cash-flow timeline first: what happens at time zero, and what happens at the end of each year.
Problem 7 · Instrumentation and process control (control valve sizing)
CHE-PD-06 · Plant Design and Operation · Select one
A liquid control valve must pass 200 gpm of water (specific gravity 1.0) with a 16-psi pressure drop across the valve. Assume no attached fittings and nonchoked flow. What valve flow coefficient Cv is required?
- A. 50
- B. 12.5
- C. 200
- D. 800
Show answer and solution
Answer: A. 50
For liquids in U.S. units, Q (gpm) = Cv × √(ΔP/SG) with ΔP in psi. So Cv = 200 ÷ √(16/1.0) = 200 ÷ 4 = 50.
Why the other choices are wrong:
- B (12.5) divides by ΔP instead of by its square root.
- C (200) ignores the pressure drop entirely.
- D (800) multiplies by √ΔP instead of dividing.
Tip: Check that a bigger pressure drop gives a smaller required Cv for the same flow. If it doesn't, the formula is upside down.
Source: Fisher (Emerson) Catalog 12, Sec. 2, Sizing Valves for Liquids, Step 6, p. 2-3
Problem 8 · Instrumentation and process control (valve fail position)
CHE-PD-07 · Plant Design and Operation · Select one
A pneumatic valve controls cooling water to the jacket of a reactor running an exothermic reaction. The hazard analysis identifies loss of cooling as the dominant risk; overcooling has no safety consequence. On loss of instrument air, how should the valve behave?
- A. Fail closed, using an air-to-open actuator
- B. Fail open, using an air-to-close actuator
- C. Fail in place, because the controller will recover
- D. Fail closed, because a safety instrumented system makes the valve's fail position irrelevant
Show answer and solution
Answer: B. Fail open, using an air-to-close actuator
A fail-open valve opens when its actuating energy is lost. An air-to-close valve is held closed by air pressure, so it opens when the air fails. That keeps cooling water flowing during an instrument-air failure, which is the safe direction for this hazard.
Why the other choices are wrong:
- A (Fail closed, using an air-to-open actuator) would cut off cooling at exactly the moment the reaction needs it.
- C (Fail in place, because the controller will recover) relies on a control loop whose final element may have just lost power.
- D (Fail closed, because a safety instrumented system makes the valve's fail position irrelevant) is wrong because added protection layers supplement a safe fail position. They don't replace it.
Tip: The fail position comes from the process hazard, not a fixed rule. If a different hazard dominated (for example, freezing), the answer could reverse.
Source: Fisher (Emerson) Control Valve Handbook (Aug. 2023), Ch. 1 glossary: fail-open, fail-closed, p. 26; Kuphaldt, Lessons in Industrial Automation, “Valve Failure Modes”
Problem 9 · Operation (management of change)
CHE-PD-08 · Plant Design and Operation · Select one
At a PSM-covered process, which change can proceed without the written management-of-change (MOC) procedures required by 29 CFR 1910.119(l)?
- A. Raising the maximum operating temperature limit by 10 °C
- B. Switching to a different supplier's catalyst formulation
- C. Replacing a failed pump with one that satisfies the original design specification
- D. Reordering the steps in the startup procedure
Show answer and solution
Answer: C. Replacing a failed pump with one that satisfies the original design specification
Paragraph (l)(1) excludes “replacements in kind,” which paragraph (b) defines as a replacement that satisfies the design specification.
Why the other choices are wrong:
- A (Raising the maximum operating temperature limit by 10 °C) changes the process technology and its safe operating limits.
- B (Switching to a different supplier's catalyst formulation) changes process chemicals.
- D (Reordering the steps in the startup procedure) changes procedures. Process chemicals, technology, equipment, and procedures are all named in (l)(1).
Tip: When a change touches chemicals, technology, equipment, or procedures, assume MOC applies unless it truly meets the design specification.
Source: OSHA, 29 CFR 1910.119(b) “replacement in kind” and (l)(1) management of change
Mass/Energy Balances (Problems 10–14)
Problem 1, at the top of this page, is also a mass balance.
Problem 10 · Mass balances with reaction (combustion, excess air)
CHE-MEB-02 · Mass/Energy Balances · Select one
Methane burns completely: CH₄ + 2O₂ → CO₂ + 2H₂O. Air is 21 mol% O₂. With 20% excess air, how many moles of air are fed per mole of CH₄?
- A. 2.4 mol
- B. 9.5 mol
- C. 13.7 mol
- D. 11.4 mol
Show answer and solution
Answer: D. 11.4 mol
Theoretical O₂ is 2 mol. With 20% excess, O₂ fed = 2 × 1.20 = 2.4 mol. Air fed = 2.4 ÷ 0.21 = 11.4 mol.
Why the other choices are wrong:
- A (2.4 mol) is the oxygen fed, not the air.
- B (9.5 mol) is the theoretical air, with no excess (2 ÷ 0.21).
- C (13.7 mol) applies the 20% excess twice (11.4 × 1.2).
Tip: Percent excess is always measured against the theoretical (stoichiometric) amount.
Sources: OpenStax Chemistry 2e, §4.3 Reaction Stoichiometry; DOE-HDBK-1012/3-92, Fluid Flow, conservation of mass (Eq. 3-4), p. 10
Problem 11 · Mass balances with no reaction (composition)
CHE-MEB-03 · Mass/Energy Balances · Select one
An evaporator concentrates 1,000 kg/h of a 10 wt% sugar solution to 40 wt% sugar. No sugar leaves with the vapor. How much water is evaporated?
- A. 750 kg/h
- B. 250 kg/h
- C. 300 kg/h
- D. 900 kg/h
Show answer and solution
Answer: A. 750 kg/h
Sugar balance: 0.10 × 1,000 = 100 kg/h sugar = 0.40 × P, so P = 250 kg/h of product. Water evaporated = 1,000 − 250 = 750 kg/h.
Why the other choices are wrong:
- B (250 kg/h) is the concentrated product rate.
- C (300 kg/h) subtracts the percentages (40 − 10 = 30%) and applies that to the feed.
- D (900 kg/h) removes all the water in the feed, which would leave pure sugar.
Tip: Find a component that passes through unchanged (a tie component) and balance it first.
Source: DOE-HDBK-1012/3-92, Fluid Flow, conservation of mass (Eq. 3-4), p. 10
Problem 12 · Mass balances with no reaction (purge)
CHE-MEB-04 · Mass/Energy Balances · Select one
A recycle loop receives 1,000 mol/h of fresh feed containing 1.0 mol% of an inert, I. The inert doesn't react and leaves only through a purge stream taken from the recycle, which has the recycle composition. To hold the recycle at 10 mol% inert, what purge flow is needed at steady state?
- A. 0 mol/h
- B. 100 mol/h
- C. 10 mol/h
- D. 1,000 mol/h
Show answer and solution
Answer: B. 100 mol/h
At steady state, inert in = inert out: 0.010 × 1,000 = 10 mol/h = 0.10 × P, so P = 100 mol/h.
Why the other choices are wrong:
- A (0 mol/h) would let the inert accumulate in the loop without limit.
- C (10 mol/h) treats the purge as pure inert. The purge carries the recycle composition, so it is 90% useful material.
- D (1,000 mol/h) purges the entire fresh feed.
Tip: Purge exists so that something that enters can also leave. Balance that component over the whole process.
Source: DOE-HDBK-1012/3-92, Fluid Flow, conservation of mass (Eq. 3-4), p. 10
Problem 13 · Energy balances with reaction (heat of reaction)
CHE-MEB-05 · Mass/Energy Balances · Select one
Use standard enthalpies of formation (kJ/mol): CH₄(g) −74.8, CO₂(g) −393.5, H₂O(g) −241.8, H₂O(l) −285.8. What is ΔH°rxn for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)?
- A. −802.3 kJ/mol
- B. −877.1 kJ/mol
- C. −890.3 kJ/mol
- D. +802.3 kJ/mol
Show answer and solution
Answer: A. −802.3 kJ/mol
ΔH°rxn = Σ(products) − Σ(reactants) = [−393.5 + 2(−241.8)] − [−74.8 + 2(0)] = −877.1 + 74.8 = −802.3 kJ/mol. O₂ is an element in its standard state, so its ΔH°f is zero.
Why the other choices are wrong:
- B (−877.1 kJ/mol) forgets to subtract the formation enthalpy of methane.
- C (−890.3 kJ/mol) uses liquid water, the higher-heating-value case. The reaction as written makes water vapor.
- D (+802.3 kJ/mol) subtracts products from reactants.
Tip: Check the phase of every species. Liquid versus vapor water changes the answer by about 88 kJ per mole of methane.
Source: OpenStax Chemistry 2e, §5.3 Enthalpy (formation enthalpies, Hess's law, Table 5.2)
Problem 14 · Energy balances with no reaction (sensible and latent heat)
CHE-MEB-06 · Mass/Energy Balances · Enter a number
How much heat is needed to take 10.0 kg of liquid water at 25 °C to saturated vapor at 100 °C and 1 atm? Use cp = 4.18 kJ/(kg·K) for the liquid and a latent heat of vaporization of 2,257 kJ/kg. Enter the answer in kJ, to the nearest 100 kJ.
Work it out and write your number down before you open the solution.
Show answer and solution
Answer: 25,700 kJ
Sensible heat: 10.0 × 4.18 × (100 − 25) = 3,135 kJ. Latent heat: 10.0 × 2,257 = 22,570 kJ. Total = 25,705 kJ, or 25,700 kJ to the nearest 100.
Common misses: 22,570 kJ (latent heat only) or 3,135 kJ (sensible heat only).
Tip: Split a heating path at every phase change: sensible to the boiling point, then latent.
Source: DOE-HDBK-1012/2-92, Heat Transfer, Q = ṁ·cp·ΔT (Eq. 2-15), p. 45; latent heat of condensation, p. 37
Thermodynamics (Problems 15–19)
Problem 15 · State functions (nonideal gas, compressibility)
CHE-TH-01 · Thermodynamics · Select one
A 2.00 m³ tank holds a gas at 5.00 MPa and 300 K. The compressibility factor is Z = 0.90. How many moles are in the tank? Use R = 8.314 J/(mol·K).
- A. 4.45 mol
- B. 3,608 mol
- C. 4,009 mol
- D. 4,455 mol
Show answer and solution
Answer: D. 4,455 mol
PV = ZnRT, so n = PV ÷ (ZRT) = (5.00 × 10⁶ Pa × 2.00 m³) ÷ (0.90 × 8.314 × 300) = 4,455 mol.
Why the other choices are wrong:
- A (4.45 mol) uses 5,000 (kPa) in an equation that needs pascals.
- B (3,608 mol) multiplies by Z instead of dividing.
- C (4,009 mol) is the ideal-gas answer (Z = 1). With Z below 1, more moles fit in the same tank.
Tip: Keep R and pressure in matching units. With R = 8.314 J/(mol·K), pressure must be in Pa and volume in m³.
Source: OpenStax Chemistry 2e, §9.6 Non-Ideal Gas Behavior (compressibility factor Z)
Problem 16 · Power cycles (compressors)
CHE-TH-02 · Thermodynamics · Select one
Air (treat as an ideal gas with γ = 1.4) at 300 K is compressed reversibly and adiabatically through a pressure ratio of 4. What is the outlet temperature?
- A. 300 K
- B. 446 K
- C. 807 K
- D. 1,200 K
Show answer and solution
Answer: B. 446 K
For an isentropic ideal-gas process, T₂/T₁ = (P₂/P₁)^((γ − 1)/γ) = 4^0.2857 = 1.486, so T₂ = 300 × 1.486 = 446 K.
Why the other choices are wrong:
- A (300 K) is the isothermal case, not adiabatic compression.
- C (807 K) uses an exponent of 1/γ instead of (γ − 1)/γ.
- D (1,200 K) assumes temperature rises in direct proportion to pressure.
Tip: A real compressor with less than 100% isentropic efficiency discharges hotter than this ideal temperature.
Source: NASA Glenn Research Center, Isentropic Flow Equations (Eq. 4 and Eq. 6)
Problem 17 · Power cycles (refrigeration)
CHE-TH-03 · Thermodynamics · Select one
An ideal (Carnot) refrigerator keeps a space at −10 °C and rejects heat at 35 °C. What is its coefficient of performance (COP)?
- A. −0.22
- B. 0.17
- C. 5.85
- D. 6.85
Show answer and solution
Answer: C. 5.85
COP_R = T_c ÷ (T_h − T_c), using absolute temperatures: 263.15 ÷ (308.15 − 263.15) = 263.15 ÷ 45 = 5.85.
Why the other choices are wrong:
- A (−0.22) uses Celsius temperatures (−10 ÷ 45).
- B (0.17) inverts the ratio.
- D (6.85) is the Carnot heat-pump COP, T_h ÷ (T_h − T_c), which is always the refrigerator COP plus 1.
Tip: Temperature ratios always need kelvins (or degrees Rankine). Differences can stay in Celsius.
Source: OpenStax University Physics Vol. 2, §4.5 The Carnot Cycle (Carnot coefficients of performance)
Problem 18 · Chemical equilibria (temperature and pressure dependence)
CHE-TH-04 · Thermodynamics · Select all that apply
For the gas-phase reaction A(g) + B(g) ⇌ C(g), ΔH < 0, which changes increase the equilibrium conversion of A? Select all that apply.
- A. Raise the temperature
- B. Compress the mixture to a smaller volume at constant temperature
- C. Add a catalyst
- D. Remove C continuously as it forms
- E. Lower the temperature
Show answer and solution
Answer: B, D, E
Compressing the mixture favors the side with fewer moles of gas (one mole of C versus two of reactants). Removing product shifts the equilibrium toward C. For an exothermic reaction, heat acts like a product, so lowering the temperature increases K.
Why the others are wrong:
- A (Raise the temperature) Raising the temperature lowers K for an exothermic reaction.
- C (Add a catalyst) A catalyst speeds up both directions equally. Equilibrium is reached faster, but K doesn't change.
On the real exam, a select-all question counts only if you pick every correct option and nothing else. There's no partial credit.
Tip: Plants often run exothermic reactions warmer than equilibrium alone would favor, because the rate at low temperature is too slow. That's a design trade-off, not a contradiction.
Source: OpenStax Chemistry 2e, §13.3 Shifting Equilibria: Le Châtelier's Principle
Problem 19 · Phase equilibria (Raoult's law, bubble point)
CHE-TH-05 · Thermodynamics · Select one
An ideal liquid mixture is 40 mol% benzene and 60 mol% toluene. At the temperature of interest, the pure-component vapor pressures are 180 kPa for benzene and 74 kPa for toluene. What is the benzene mole fraction in the first bubble of vapor that forms?
- A. 0.29
- B. 0.40
- C. 0.71
- D. 0.62
Show answer and solution
Answer: D. 0.62
Raoult's law: p_B = 0.40 × 180 = 72.0 kPa and p_T = 0.60 × 74 = 44.4 kPa. Bubble-point pressure P = 72.0 + 44.4 = 116.4 kPa. y_B = 72.0 ÷ 116.4 = 0.62.
Why the other choices are wrong:
- A (0.29) is 1 − 0.71 and has no physical basis here.
- B (0.40) assumes the vapor has the same composition as the liquid.
- C (0.71) is 180 ÷ (180 + 74), which ignores the liquid composition.
Tip: The vapor is always richer in the more volatile component. If your answer isn't, recheck the setup.
Source: OpenStax Chemistry 2e, §11.4 Colligative Properties (Raoult's law, total vapor pressure)
Fluids (Problems 20–24)
Problem 20 · Mechanical-energy balance (Reynolds number)
CHE-FL-01 · Fluids · Select one
Water (ρ = 1,000 kg/m³, μ = 0.001 Pa·s) flows at 2.0 m/s in a pipe with an inside diameter of 0.050 m. Which statement is correct?
- A. Re = 1.0 × 10⁵; turbulent
- B. Re = 100; laminar
- C. Re = 1.0 × 10⁵; laminar
- D. Re = 2,500; transitional
Show answer and solution
Answer: A. Re = 1.0 × 10⁵; turbulent
Re = ρvD ÷ μ = (1,000 × 2.0 × 0.050) ÷ 0.001 = 100,000. That is far above the roughly 3,500 where flow is reliably turbulent.
Why the other choices are wrong:
- B (Re = 100; laminar) uses μ = 1, treating 1 cP as if it were 1 Pa·s.
- C (Re = 1.0 × 10⁵; laminar) has the right number but the wrong regime.
- D (Re = 2,500; transitional) doesn't follow from the given data.
Tip: Water's viscosity near room temperature is about 1 cP = 0.001 Pa·s. Unit slips with viscosity are common.
Source: DOE-HDBK-1012/3-92, Fluid Flow, Reynolds number (Eq. 3-7) and flow regimes, pp. 19–20
Problem 21 · Incompressible flow (piping systems)
CHE-FL-02 · Fluids · Select one
Water (ρ = 1,000 kg/m³) flows at 2.0 m/s through 100 m of pipe with a 0.10 m inside diameter. The Darcy friction factor is 0.020. What is the frictional pressure drop?
- A. 20 kPa
- B. 40 kPa
- C. 80 kPa
- D. 160 kPa
Show answer and solution
Answer: B. 40 kPa
ΔP = f (L/D)(ρv²/2) = 0.020 × (100/0.10) × (1,000 × 2.0² ÷ 2) = 0.020 × 1,000 × 2,000 = 40,000 Pa = 40 kPa.
Why the other choices are wrong:
- A (20 kPa) drops a factor of 2 somewhere in the length-to-diameter ratio or velocity head.
- C (80 kPa) leaves out the ½ in the velocity head.
- D (160 kPa) treats 0.020 as a Fanning friction factor and multiplies by 4.
Tip: Darcy f = 4 × Fanning f. Check which one your chart, correlation, or handbook equation uses before plugging in.
Source: DOE-HDBK-1012/3-92, Fluid Flow, Darcy's equation (Eq. 3-14), p. 32
Problem 22 · Pumps (net positive suction head)
CHE-FL-03 · Fluids · Select one
A pump draws 60 °C water (ρ = 983 kg/m³, vapor pressure 19.9 kPa) from an open tank at 101.3 kPa. The liquid surface is 2.0 m below the pump centerline, and suction-line friction loss is 1.0 m. The pump's required NPSH is 4.0 m. What is the available NPSH, and is it adequate?
- A. 5.4 m; adequate
- B. 6.4 m; adequate
- C. 7.5 m; adequate
- D. 9.4 m; adequate
Show answer and solution
Answer: A. 5.4 m; adequate
NPSHa = (P_atm − P_vapor) ÷ (ρg) − suction lift − friction = (101,300 − 19,900) ÷ (983 × 9.81) − 2.0 − 1.0 = 8.44 − 3.0 = 5.4 m. That exceeds the 4.0 m required.
Why the other choices are wrong:
- B (6.4 m; adequate) forgets the suction-line friction.
- C (7.5 m; adequate) forgets to subtract the vapor pressure.
- D (9.4 m; adequate) adds the 2.0 m as if the tank surface were above the pump.
Tip: Hot liquid, a suction lift, and a long suction line all reduce available NPSH. Each one is a fair exam variable.
Source: DOE-HDBK-1012/3-92, Fluid Flow, net positive suction head (Eq. 3-19), p. 49
Problem 23 · Compressible flow (choked flow)
CHE-FL-04 · Fluids · Select one
Air (γ = 1.4) flows isentropically through a converging nozzle from a large reservoir at 500 kPa absolute. Below about what back pressure does the mass flow stop increasing?
- A. 250 kPa absolute
- B. 417 kPa absolute
- C. 264 kPa absolute
- D. 500 kPa absolute; the flow is always choked
Show answer and solution
Answer: C. 264 kPa absolute
Mass flow reaches its maximum when the throat reaches Mach 1 (choked flow). At M = 1, p/p_t = (1 + 0.2)^(−3.5) = 0.528, so the critical pressure is 0.528 × 500 = 264 kPa. Lowering the back pressure further doesn't increase the flow.
Why the other choices are wrong:
- A (250 kPa absolute) rounds the critical ratio to one-half.
- B (417 kPa absolute) uses the temperature ratio at Mach 1 (0.833) instead of the pressure ratio.
- D (500 kPa absolute; the flow is always choked) is wrong because at back pressures above the critical value the nozzle is unchoked and the mass flow still responds to back pressure.
Tip: For air, remember the critical pressure ratio is about 0.53. For other gases, compute it from γ.
Source: NASA Glenn Research Center, Mass Flow Choking (maximum flow at Mach 1); NASA Glenn Research Center, Isentropic Flow Equations (Eq. 4 and Eq. 6)
Problem 24 · Flow measurement
CHE-FL-05 · Fluids · Select one
An orifice meter reads 100 m³/h at a differential pressure of 25 kPa. With fluid properties and the discharge coefficient unchanged, what flow does a differential of 9 kPa indicate?
- A. 13 m³/h
- B. 36 m³/h
- C. 77.5 m³/h
- D. 60 m³/h
Show answer and solution
Answer: D. 60 m³/h
For differential-pressure meters, flow is proportional to the square root of the differential: Q = 100 × √(9/25) = 100 × 0.6 = 60 m³/h.
Why the other choices are wrong:
- A (13 m³/h) assumes flow varies with the differential squared.
- B (36 m³/h) assumes flow varies directly with the differential.
- C (77.5 m³/h) averages the readings rather than applying the square-root relationship.
Tip: The square-root relationship is also why these meters lose accuracy at low flow: a small differential becomes hard to read.
Source: DOE-HDBK-1012/3-92, Fluid Flow, venturi flow and differential pressure, pp. 27–29
Heat Transfer (Problems 25–28)
Problem 25 · Heat transfer with no phase change (conduction)
CHE-HT-01 · Heat Transfer · Select one
A furnace wall has 0.20 m of brick (k = 0.80 W/(m·K)) and 0.050 m of insulation (k = 0.040 W/(m·K)). The inner and outer surfaces differ by 50 K. What is the steady heat flux through the wall?
- A. 40.0 W/m²
- B. 33.3 W/m²
- C. 200 W/m²
- D. 240 W/m²
Show answer and solution
Answer: B. 33.3 W/m²
Resistances in series add. Per square meter: R = 0.20/0.80 + 0.050/0.040 = 0.25 + 1.25 = 1.50 m²·K/W. Flux q = ΔT ÷ R = 50 ÷ 1.50 = 33.3 W/m².
Why the other choices are wrong:
- A (40.0 W/m²) uses the insulation layer alone.
- C (200 W/m²) uses the brick layer alone.
- D (240 W/m²) adds the fluxes each layer would carry on its own. Resistances add, not fluxes.
Tip: The layer with the biggest resistance controls the flux. Here the thin insulation does most of the work.
Source: DOE-HDBK-1012/2-92, Heat Transfer, equivalent resistance method (Eq. 2-6), pp. 9–11
Problem 26 · Heat exchange equipment design (LMTD)
CHE-HT-02 · Heat Transfer · Select one
In a counterflow heat exchanger, the hot stream cools from 150 °C to 90 °C and the cold stream heats from 30 °C to 70 °C. The log mean temperature difference (LMTD) is closest to:
- A. 55.8 °C
- B. 60.0 °C
- C. 69.5 °C
- D. 70.0 °C
Show answer and solution
Answer: C. 69.5 °C
Counterflow end differences: 150 − 70 = 80 °C at one end and 90 − 30 = 60 °C at the other. LMTD = (80 − 60) ÷ ln(80/60) = 20 ÷ 0.2877 = 69.5 °C.
Why the other choices are wrong:
- A (55.8 °C) pairs the ends as if the flow were parallel (120 and 20 °C).
- B (60.0 °C) takes the smaller end difference.
- D (70.0 °C) takes the arithmetic mean. It's close here only because the two end differences are similar.
Tip: Sketch the two temperature profiles before pairing the ends. Counterflow pairs hot-in with cold-out.
Source: DOE-HDBK-1012/2-92, Heat Transfer, log mean temperature difference (Eq. 2-2), p. 3
Problem 27 · Heat exchange equipment design (area, F-factor)
CHE-HT-03 · Heat Transfer · Select one
For the counterflow exchanger in Problem 26 (LMTD = 69.5 °C), the duty is 500 kW, U = 500 W/(m²·K), and the multipass correction factor is F = 0.90. What heat transfer area is required?
- A. 12.9 m²
- B. 14.4 m²
- C. 17.9 m²
- D. 16.0 m²
Show answer and solution
Answer: D. 16.0 m²
Q = U × A × F × ΔT_lm, so A = 500,000 ÷ (500 × 0.90 × 69.5) = 16.0 m².
Why the other choices are wrong:
- A (12.9 m²) multiplies by F instead of dividing.
- B (14.4 m²) ignores the F correction.
- C (17.9 m²) uses the parallel-flow LMTD of 55.8 °C and also omits the 0.90 correction factor.
Tip: F is never greater than 1. A correction factor that shrinks the required area is applied the wrong way.
Sources: DOE-HDBK-1012/2-92, Heat Transfer, Q = U·A·ΔT_lm, p. 37; Carnegie Mellon University, Chemical Unit Operations, corrected heat-exchanger duty Q = UA·F·LMTD
Problem 28 · Overall heat transfer coefficient
CHE-HT-04 · Heat Transfer · Select one
A thin flat metal wall separates two fluids. The film coefficients are 1,000 W/(m²·K) on one side and 2,500 W/(m²·K) on the other. The wall is 2.0 mm thick with k = 50 W/(m·K). Neglect fouling. What is the overall coefficient U?
- A. 694 W/(m²·K)
- B. 714 W/(m²·K)
- C. 1,750 W/(m²·K)
- D. 3,500 W/(m²·K)
Show answer and solution
Answer: A. 694 W/(m²·K)
1/U = 1/h₁ + Δx/k + 1/h₂ = 1/1,000 + 0.002/50 + 1/2,500 = 0.00100 + 0.00004 + 0.00040 = 0.00144, so U = 694 W/(m²·K).
Why the other choices are wrong:
- B (714 W/(m²·K)) leaves out the wall resistance. It's small here but not zero.
- C (1,750 W/(m²·K)) averages the two film coefficients.
- D (3,500 W/(m²·K)) adds the film coefficients instead of their resistances.
Tip: U is always lower than the smallest individual coefficient. Fouling resistances, when given, add to the same sum.
Source: DOE-HDBK-1012/2-92, Heat Transfer, overall heat transfer coefficient (Eq. 2-11), pp. 23 and 37
Mass Transfer (Problems 29–32)
Problem 29 · Staged separations (minimum stages)
CHE-MT-01 · Mass Transfer · Select one
A binary distillation column at total reflux produces a distillate with 95 mol% light key and bottoms with 5 mol% light key. The relative volatility is constant at 2.5. Counting the partial reboiler as a stage, what value does the Fenske equation give for the minimum number of theoretical stages, N_min?
- A. 3.2
- B. 6.4
- C. 7.4
- D. 12.9
Show answer and solution
Answer: B. 6.4
Fenske: separation factor S = (0.95/0.05) × (0.95/0.05) = 361. N_min = ln S ÷ ln α = ln 361 ÷ ln 2.5 = 5.889 ÷ 0.916 = 6.4 theoretical stages, including the reboiler. The Fenske result may be fractional; a physical design that uses whole equilibrium stages must meet or exceed the required separation after applying the chosen stage-counting convention.
Why the other choices are wrong:
- A (3.2) uses only the distillate ratio (0.95/0.05) in S.
- C (7.4) adds the reboiler a second time.
- D (12.9) doubles the correct answer.
Tip: Read the problem's stage-counting convention carefully: total stages, stages inside the column, or trays.
Problem 30 · Staged separations (reflux and operating lines)
CHE-MT-02 · Mass Transfer · Select one
A distillation column with a total condenser runs at reflux ratio R = L/D = 3, with distillate composition x_D = 0.96. Assuming constant molar overflow, which equation is the rectifying-section operating line?
- A. y = 0.75x + 0.32
- B. y = 0.75x + 0.96
- C. y = 0.75x + 0.24
- D. y = 3x − 1.92
Show answer and solution
Answer: C. y = 0.75x + 0.24
Slope = R ÷ (R + 1) = 3/4 = 0.75. Intercept = x_D ÷ (R + 1) = 0.96/4 = 0.24. Check: at x = 0.96, y = 0.72 + 0.24 = 0.96, so the line passes through (x_D, x_D) as it must.
Why the other choices are wrong:
- A (y = 0.75x + 0.32) uses x_D ÷ R for the intercept.
- B (y = 0.75x + 0.96) forgets to divide the intercept by R + 1.
- D (y = 3x − 1.92) uses R itself as the slope.
Tip: The (x_D, x_D) check catches most operating-line mistakes in seconds.
Source: Skogestad, NTNU lecture notes “Distillation (multistage with reflux),” operating lines, slides 8–10
Problem 31 · Gas-liquid operations (absorption)
CHE-MT-03 · Mass Transfer · Select one
A dilute solute is absorbed from a gas into solute-free water. Equilibrium is y = 1.2x (mole fractions). The design removes 95% of the solute. What is the minimum liquid-to-gas molar ratio, (L/G)min?
- A. 0.06
- B. 1.14
- C. 1.20
- D. 1.37
Show answer and solution
Answer: B. 1.14
At minimum liquid flow, the liquid leaving the bottom is in equilibrium with the entering gas, so x_out = y_in ÷ 1.2. With solute-free entering liquid and dilute flows, (L/G)min = (y_in − 0.05y_in) ÷ (y_in/1.2 − 0) = 0.95 × 1.2 = 1.14.
Why the other choices are wrong:
- A (0.06) uses only the unabsorbed 5% fraction.
- C (1.20) would correspond to 100% removal.
- D (1.37) is 1.2 times the minimum, a typical design liquid rate rather than the minimum.
Tip: Real absorbers run above the minimum ratio, because at the minimum the column would need infinite height.
Problem 32 · Other separations (liquid-liquid extraction)
CHE-MT-04 · Mass Transfer · Select one
A single equilibrium stage contacts 100 kg of water containing a solute with 50 kg of an immiscible solvent. The distribution ratio is K = (kg solute/kg solvent) ÷ (kg solute/kg water) = 4, constant. What fraction of the solute is extracted?
- A. 50%
- B. 80%
- C. 100%
- D. 67%
Show answer and solution
Answer: D. 67%
Let x = kg solute per kg water at equilibrium. Solute in water = 100x; solute in solvent = 50 × 4x = 200x. Fraction extracted = 200x ÷ (100x + 200x) = 2/3 = 67%.
Why the other choices are wrong:
- A (50%) assumes the solute splits evenly between the phases.
- B (80%) is K ÷ (1 + K), which ignores the solvent-to-water ratio.
- C (100%) assumes complete extraction in one equilibrium stage.
Tip: Splitting the same 50 kg of solvent into two 25-kg crosscurrent stages would extract 75%. More stages beat one big stage.
Source: DOE-HDBK-1012/3-92, Fluid Flow, conservation of mass (Eq. 3-4), p. 10
Chemical Reaction Engineering (Problems 33–36)
Problem 33 · Rate equation (Arrhenius)
CHE-CRE-01 · Chemical Reaction Engineering · Select one
A rate constant doubles when the temperature rises from 300 K to 310 K. Estimate the activation energy.
- A. 0.58 kJ/mol
- B. 6.4 kJ/mol
- C. 23.3 kJ/mol
- D. 53.6 kJ/mol
Show answer and solution
Answer: D. 53.6 kJ/mol
Ea = R ln(k₂/k₁) ÷ (1/T₁ − 1/T₂) = 8.314 × ln 2 ÷ (1/300 − 1/310) = 5.763 ÷ 1.075 × 10⁻⁴ = 53,600 J/mol = 53.6 kJ/mol.
Why the other choices are wrong:
- A (0.58 kJ/mol) uses Celsius temperatures.
- B (6.4 kJ/mol) leaves out R.
- C (23.3 kJ/mol) uses log base 10 instead of the natural log.
Tip: The old rule that “rates double every 10 °C” corresponds to an activation energy of about 50 kJ/mol near room temperature.
Source: OpenStax Chemistry 2e, §12.5 Collision Theory (two-point Arrhenius equation)
Problem 34 · Conversion in reactors (PFR)
CHE-CRE-02 · Chemical Reaction Engineering · Select one
An irreversible first-order liquid-phase reaction with k = 0.20 min⁻¹ runs at constant density in an isothermal plug-flow reactor (PFR) with space time τ = 10 min. What is the conversion?
- A. 20%
- B. 66.7%
- C. 86.5%
- D. 100%
Show answer and solution
Answer: C. 86.5%
For a first-order reaction in a PFR, X = 1 − e^(−kτ) = 1 − e^(−2.0) = 0.865.
Why the other choices are wrong:
- A (20%) uses k alone.
- B (66.7%) is the CSTR result, kτ ÷ (1 + kτ).
- D (100%) assumes kτ greater than 1 means complete conversion.
Tip: For the same volume and positive-order kinetics, a PFR beats a single CSTR because it never operates at the low outlet concentration everywhere.
Problem 35 · Conversion in reactors (reactors in series)
CHE-CRE-03 · Chemical Reaction Engineering · Select one
The same reaction (k = 0.20 min⁻¹, first order) runs in two equal continuous stirred-tank reactors (CSTRs) in series, each with τ = 5 min. What is the overall conversion?
- A. 75%
- B. 50%
- C. 66.7%
- D. 86.5%
Show answer and solution
Answer: A. 75%
For n equal first-order CSTRs in series, X = 1 − 1/(1 + kτ)ⁿ = 1 − 1/(1 + 1.0)² = 1 − 0.25 = 0.75.
Why the other choices are wrong:
- B (50%) is the conversion of one 5-minute tank.
- C (66.7%) is a single CSTR with τ = 10 min.
- D (86.5%) is the PFR limit, which more tanks in series approach.
Tip: This formula holds only for first-order kinetics.
Problem 36 · Yield and selectivity
CHE-CRE-04 · Chemical Reaction Engineering · Select one
Parallel reactions A → B (desired) and A → C both have 1:1 stoichiometry. Feeding 100 mol of A gives 20 mol A, 60 mol B, and 20 mol C at the outlet. Defining yield as moles of B formed per mole of A reacted, what is the yield?
- A. 0.60
- B. 0.75
- C. 0.80
- D. 3.0
Show answer and solution
Answer: B. 0.75
A reacted = 100 − 20 = 80 mol. B formed = 60 mol. Yield = 60 ÷ 80 = 0.75.
Why the other choices are wrong:
- A (0.60) is B formed per mole of A fed, a different yield definition.
- C (0.80) is the conversion of A.
- D (3.0) is the selectivity ratio of B to C.
Tip: Yield and selectivity have several textbook definitions. Use the one the problem states.
How to read your results
Count how many you got right in each area. Use that count to decide what to study next, not to predict your exam result. NCEES scales scores and doesn't publish the passing score. Thirty-six problems also can't cover every topic in the specification. For example, these problems don't include radiation, crystallization, or materials of construction.
For each miss, write down the problem number and the type of mistake (see the error table in the study plan). Your misses are the first page of your study plan.
Your 12-week PE Chemical study plan
This plan is an example, not an official requirement. It assumes about 10 focused hours a week (about 120 hours total), a full-time job, and a test date booked about 13 weeks out. No number of hours guarantees a pass. Time per week roughly follows the NCEES question ranges, and every week ends with a check you can actually see.
| Week | Focus | Learn and look up | Practice and rework | Done when |
|---|---|---|---|---|
| 1 | Setup and baseline | Confirm your board's approval steps and pick a testing window. Download the handbook from MyNCEES. Run the lookup drill above. | Work all 36 problems untimed before reading any solutions. Log every miss. | Your error log is started and your three weakest areas are named. |
| 2 | Mass balances | Recycle, bypass, purge, reaction stoichiometry, limiting and excess reactants | Rework Problems 1, 10, 11, and 12, changing one input each time. | You draw overall and unit boundaries before writing any equation. |
| 3 | Energy balances and basic thermodynamics | Sensible and latent heat, heats of reaction, real-gas behavior, first and second laws, compressors, refrigeration | Problems 13–17 | You check units, and kelvins versus degrees Celsius, on every line. |
| 4 | Chemical and phase equilibrium | How K changes with temperature, Le Châtelier's principle, Raoult's and Henry's laws, bubble and dew points, flash, activity coefficients | Problems 18–19 | You can find each relationship in the handbook in under a minute. |
| 5 | Fluids | Mechanical-energy balance, Darcy versus Fanning friction factors, compressible flow, pumps and NPSH, compressors, flow meters | Problems 20–24 | Your lookup log has an entry for every fluids equation you used. |
| 6 | Heat transfer | Conduction, convection, radiation, phase change, overall coefficients, fouling, LMTD with correction factor, effectiveness-NTU | Problems 25–28 | You can size an exchanger (find the area) and rate one (find the duty). |
| 7 | Mass transfer | Staged separations, McCabe-Thiele, Fenske, reflux, absorption and stripping, extraction, drying, adsorption, membranes | Problems 29–32 | Every operating line you draw passes the (x_D, x_D) check. |
| 8 | Reaction engineering, plus safety, health, and environment | Rate laws, Arrhenius, batch, CSTR, and plug-flow reactors, series and parallel reactions, heat effects; hazard analysis, relief, flammability, exposure limits, process safety management | Problems 33–36 and 2–4 | You know which design equation goes with which reactor and reaction order. |
| 9 | Plant design, operation, and maintenance | Economics (present worth, scale-up), materials and corrosion, equipment selection, P&IDs, control valves, fail positions, safety instrumented systems, management of change, troubleshooting | Problems 5–9 | You have a one-page list of handbook locations for design and economics topics. |
| 10 | Full rehearsal | Sit one full-length, 80-question practice exam in a single day, using only the handbook. Split it into two halves with a break, and treat the first half as final once you move on. | Log every miss by area and by error type. | You can name your two weakest areas from real timed work. |
| 11 | Fix the leaks | Study the areas that cost you the most in the rehearsal. | Work fresh problems in those areas. Then run Block A and Block B below. | You have a short repair list based on work you actually did. |
| 12 | Consolidate and rest | Rerun the lookup drill and practice your five slowest lookups. Confirm your ID, calculator, and route. | Rework every logged miss. Learn nothing new the day before. | Your exam-day plan is done. |
Two timed mixed blocks
Each block has 18 problems from this page. Give yourself 108 minutes, which is the exam's average of 6 minutes per question. Use only the handbook and an approved calculator. Because you've seen these problems before, the blocks train pacing and switching between topics. They don't replace a fresh full-length exam.
- Block A: Problems 1, 2, 15, 20, 25, 29, 33, 3, 10, 16, 21, 26, 30, 4, 11, 17, 22, 5
- Block B: Problems 34, 27, 31, 6, 12, 18, 23, 35, 28, 32, 7, 13, 19, 24, 36, 8, 14, 9
Adjust the plan to your life
- About 6 hours a week: stretch the same tasks over about 20 weeks. Keep the full rehearsal.
- About 15 hours a week: compress the plan to about 8 weeks. Keep the full rehearsal.
- A bad week: do one problem a day for 20 minutes. Small sessions keep the handbook familiar.
Turn each miss into your next study task
| Error type | What it looks like | Fix |
|---|---|---|
| Wrong model | You used a CSTR equation for a plug-flow reactor, or isothermal for adiabatic. | Before solving, write one line naming the system, boundary, and assumptions. |
| Wrong handbook relationship | You found an equation that looked right but had different variables or conditions. | Read the variable definitions and limits before plugging in. Add the correct location to your lookup log. |
| Units | Gauge versus absolute pressure, °C versus K, Pa versus kPa, Darcy versus Fanning | Carry units on every line. Convert first. |
| Arithmetic | Right setup, wrong number | Estimate the answer's size before you calculate. Recheck exponents and logs. |
| Misread question | You missed "minimum," "per mole of A reacted," or "not." | Underline what's being asked and its units before you start. |
| Time | You ran out of time or rushed the end. | Practice timed sets. Make a first pass that skips anything over about 8 minutes. |
If your exam is next week
Don't try to learn a new area now. Work through Block A or Block B under time limits, rework your misses, and run the handbook lookup drill once more. Then confirm your ID, calculator, and route to the test center, and get some sleep.
If you're preparing for a retake
If you didn't pass, NCEES gives you a diagnostic report. It shows your performance in each knowledge area on a 0–15 scale, next to the average performance of examinees who passed. Start there. Put the most time where a big gap meets a big question range. A small gap in Plant Design and Operation can be worth more than a large one in Reaction Engineering. You can take the exam once per testing window and up to three times in 12 months, but your board may be stricter. See NCEES exam results and the NCEES exam retake policy.
How to know you're ready
There's no official readiness test, but these signs mean your preparation is working:
- You can finish mixed, timed sets without giving up on any area.
- Most of your misses have a named error type, and you've reworked them.
- You can find the relationships you need in the handbook quickly.
- You've completed at least one full-length rehearsal with the handbook only.
- Your last weeks are mostly rework, not new material.
Exam-day rules that change how you practice
- Pacing: 480 minutes for 80 questions is an average of 6 minutes each. Practice at that pace.
- Two halves, one clock: after about half the questions, you review and submit that section. After you submit, you can't go back to it. The scheduled break comes after that first section. Practice making decisions you won't revisit.
- Breaks: an unused break doesn't add time to the exam. Unscheduled breaks are allowed, but the exam clock keeps running.
- Guessing: your score is based on the number of questions you answer correctly. Wrong answers don't cost extra points, so answer every question.
- Unscored questions: some questions are pretest items that don't count. You can't tell which ones, so treat every question as real.
- Calculator: bring one NCEES-approved calculator. For 2026 exams, the approved models are the HP 33s, HP 35s, Casio fx-115 and fx-991 models, and TI-30X and TI-36X models. An on-screen TI-30XS is also available. If you test in 2027, check NCEES's current calculator policy. NCEES was still listing the 2026 approved models when this page was checked on October 8, 2026.
- Scratch paper: the test center gives you two reusable booklets and three markers. Practice with your own markers and a small writing area.
- Arrival and ID: arrive 30 minutes early. The name on your ID must match the name on your appointment confirmation.
PE Chemical pass rates
| Group | Takers | Pass rate |
|---|---|---|
| First-time takers | 225 | 55% |
| Repeat takers | 81 | 32% |
These figures come from NCEES and cover January–June 2026 (updated July 2026). In that table, PE Chemical had the lowest first-time pass rate among the year-round PE exams, which ranged from 60% to 73% for the others. Pass rates describe a group of past test takers, not your personal odds. NCEES says first-time and repeat takers are graded to the same standard.
Study resources
These are the most useful places to go next.
- NCEES PE Chemical exam page: the official format, fees, and pass rates.
- PE Chemical exam specification (PDF): the full topic list behind the seven areas.
- MyNCEES: your account, registration, and the free download of the PE Chemical Reference Handbook.
- NCEES PE Chemical practice exam: NCEES's own practice exam, an e-book with 80 questions and solutions, including alternative item types. It's based on the specification that took effect January 1, 2020. As of October 8, 2026, NCEES listed it at $59.95, nonrefundable. It's a good choice for the week 10 rehearsal.
- LearnChemE PE Chemical Exam Review: screencasts, simulations, and self-study modules from the University of Colorado Boulder, organized by the seven exam topics.
Official next steps
- Check your licensing board's requirements. Your board decides whether you're eligible. Find it in the NCEES member board directory.
- Register for the exam in MyNCEES. Follow your board's approval process. See how NCEES exam registration works.
- Schedule your appointment with Pearson VUE. You can reschedule or cancel at least 48 hours before your appointment for a $50 fee paid to Pearson VUE.
- Request accommodations, if you need them, when you register. Pearson VUE manages approved accommodations, and you can't schedule your appointment until a decision is made. See NCEES ADA accommodations and our guide to NCEES exam accommodations.
Quick answers
Do I register with NCEES or my state board?
Both are involved. Your licensing board sets eligibility and approval rules. You register and pay the exam fee in MyNCEES, then schedule with Pearson VUE. Read more about NCEES exam registration.
Do I need to pass the FE exam or have four years of experience first?
Your board decides. NCEES designed the PE exam for engineers with at least four years of experience after college. If you haven't taken the FE yet, see FE Chemical exam prep.
Is the PE Chemical exam open book?
No. You can't bring any books or notes. The NCEES PE Chemical Reference Handbook is provided on screen, and it's the only reference you get.
How long should I study?
NCEES doesn't set a number. The plan on this page uses about 120 hours over 12 weeks as an example. Shift time toward your weakest areas, especially heavily weighted ones.
How is the exam scored?
NCEES counts your correct answers, with no penalty for wrong ones, and converts that count to a scaled score. You get a pass or fail result. NCEES doesn't publish the passing score.
Can I retake it if I fail?
Yes. You can take it once per testing window (January–March, April–June, July–September, October–December) and up to three times in 12 months. Your board may set stricter limits. See the NCEES exam retake policy.
Does passing the exam give me a license?
No. Your licensing board issues the license after you meet all of its requirements.
What other PE exams does Castleport cover?
See PE exam prep for the other PE disciplines.
Sources
Official NCEES sources
- NCEES, PE Chemical exam page: format, appointment length, fee, question types, results timing, and pass rates (January–June 2026, updated July 2026).
- NCEES, PE Chemical exam specification, effective January 1, 2020 (PDF), pp. 1–3: knowledge areas, question ranges, and units.
- NCEES, Examinee Guide, May 2026 (PDF): eligibility and accommodations (pp. 2–3), registration period and retake limits (pp. 4–5), scheduling fees and test-day rules (pp. 6–9), the handbook and search tools (p. 10), sections, breaks, and scoring of all item types (pp. 11–12), scoring and results (p. 14), exam table showing no design standards for PE Chemical (p. 16), and the diagnostic report (p. 20).
- NCEES, memo to member boards on 2026 exam changes, October 20, 2025 (PDF), p. 1: the 2026 approved calculator list.
- NCEES, PE Chemical practice exam listing: format, specification basis, and price as of October 8, 2026.
Teaching sources for the practice problems
These sources explain the methods used in the problems. They are not references you'll have during the exam.
- U.S. Department of Energy, DOE-HDBK-1012/2-92, Thermodynamics, Heat Transfer, and Fluid Flow, Volume 2 (PDF)
- U.S. Department of Energy, DOE-HDBK-1012/3-92, Thermodynamics, Heat Transfer, and Fluid Flow, Volume 3 (PDF)
- OpenStax, Chemistry 2e, sections 4.3, 5.3, 9.6, 11.4, 12.5, and 13.3
- OpenStax, University Physics Volume 2, section 4.5, The Carnot Cycle
- NASA Glenn Research Center, Isentropic Flow Equations and Mass Flow Choking
- Fogler, Elements of Chemical Reaction Engineering, 7th ed., Chapter 5 FAQ
- Skogestad, NTNU, "Distillation (multistage with reflux)" lecture notes (PDF)
- U.S. EPA, Air Pollution Control Cost Manual, Section 5.2, Chapter 1 (PDF)
- OSHA, 29 CFR 1910.1000, 29 CFR 1910.119, and PSM compliance guidance on atmospheric storage tanks and process connections
- Zlochower and Green (NIOSH), "The limiting oxygen concentration and flammability limits of gases and gas mixtures" (2009)
- NASA, TM X-73397, Handbook of Estimating Data, Factors, and Procedures (PDF)
- U.S. Department of Energy, DOE G 413.3-21, Cost Estimating Guide (PDF)
- NIST, IR 85-3273-40, Energy Price Indices and Discount Factors for Life-Cycle Cost Analysis, 2025 (PDF)
- Emerson, Fisher Catalog 12, Section 2, Sizing Valves for Liquids (PDF) and Fisher Control Valve Handbook (PDF)
- Kuphaldt, Lessons in Industrial Automation, "Valve Failure Modes"
- Carnegie Mellon University, Chemical Unit Operations — corrected heat-exchanger duty using F·LMTD
How this page was checked
Exam facts on this page were last verified October 8, 2026, against the NCEES sources above. Each practice problem was recalculated and checked against its teaching source. The problems have not been reviewed by NCEES or by a licensed professional engineer. This guide was developed with AI-assisted research and editing. Source checking is not professional engineering review. Read our methodology.
Castleport Test Prep is an independent exam prep publisher. It is not affiliated with, endorsed by, or approved by the National Council of Examiners for Engineering and Surveying (NCEES). Exam and credential names identify their subjects; trademarks belong to their respective owners. The practice problems are original and unofficial, and nothing on this page guarantees a passing result or a license.