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NCEES Engineering Licensure

FE Chemical Practice Exam

FE Chemical exam practice — the NCEES discipline exam, your first step toward the PE.

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Built for FE Chemical candidates

  • NCEES FE Chemical topic mix: material & energy balances, thermodynamics, fluid mechanics, heat & mass transfer, reaction engineering, and process control
  • Answerable from the NCEES FE Reference Handbook + first principles
  • Mix of calculation and conceptual items, in SI and US customary units
  • Step-by-step solutions on every question (first 2 free; all with Premium)
  • Follow-up study chat to dig into any concept that trips you up

The FE Chemical exam is one of the seven NCEES Fundamentals of Engineering discipline exams — a 110-question, computer-based test, taken near graduation, that is the first step toward becoming a licensed Professional Engineer (PE). Every question is meant to be answered from the FE Reference Handbook and engineering fundamentals, so practice that drills the right Chemical topic mix, in the right units, with worked solutions is exactly how you build exam-day speed.

Crameleon builds your practice exam on demand — pick the pathway and module you're studying, and we generate a fresh paper with step-by-step solutions (the first 2 free; all of them with Premium). Every question comes with a follow-up study chat, so you can dig into any concept that trips you up without leaving the page. The generation takes under 2 minutes and your first exam is free.

Sample FE Chemical exam questions

Three original questions on FE Chemical exam topics, each with a worked solution.

Question 1Multiple choiceMaterial/Energy Balances

Two aqueous sodium hydroxide solutions are blended continuously in a steady-state mixer with no reaction. Stream 1 contains 20 wt% NaOH and stream 2 contains 50 wt% NaOH. The product stream is 1,000 kg/h of 30 wt% NaOH. What is the required mass flow rate of stream 2 (the 50 wt% solution)?

  1. A.

    300 kg/h

  2. B.

    333 kg/h

  3. C.

    600 kg/h

  4. D.

    667 kg/h

Worked solution
Answer:B

A steady-state mixer with no reaction has two independent balances for a two-component system: a total mass balance and a balance on one component. Let m1m_1 and m2m_2 be the mass flow rates of streams 1 and 2 in kg/h.

Total mass balance (in = out):

m1+m2=1,000m_1 + m_2 = 1{,}000

NaOH balance, using the mass fraction of NaOH in each stream:

0.20 m1+0.50 m2=0.30(1,000)=3000.20\,m_1 + 0.50\,m_2 = 0.30(1{,}000) = 300

Eliminate m1m_1 with the total balance, m1=1,000−m2m_1 = 1{,}000 - m_2:

0.20(1,000−m2)+0.50 m2=300⇒200+0.30 m2=3000.20(1{,}000 - m_2) + 0.50\,m_2 = 300 \quad \Rightarrow \quad 200 + 0.30\,m_2 = 300 m2=1000.30=333 kg/h,m1=667 kg/hm_2 = \frac{100}{0.30} = 333\,\text{kg/h}, \qquad m_1 = 667\,\text{kg/h}

The lever rule gives the same split directly: the fraction of the product that must come from stream 2 is (0.30−0.20)/(0.50−0.20)=1/3(0.30 - 0.20)/(0.50 - 0.20) = 1/3.

Stream 2 flow rate ≈ 333 kg/h (option B)

The most tempting wrong answer is D (667 kg/h), which is the flow rate of stream 1; the two streams have been swapped. Because the product composition (30%) is much closer to stream 1 (20%) than to stream 2 (50%), the mixer must use more of stream 1, so the 50% stream must be the smaller flow. Option C (600 kg/h) divides the 300 kg/h of NaOH in the product by 0.50, forgetting that stream 1 also supplies NaOH. Option A (300 kg/h) is the mass of NaOH in the product, not a stream flow rate.

Question 2Short answerChemical Reaction Engineering

The liquid-phase reaction A → B is first order in A with a rate constant of 0.25 min⁻¹ at the operating temperature. A feed with CA0=2.0 mol/LC_{A0} = 2.0\,\text{mol/L} enters an ideal continuous stirred-tank reactor (CSTR) at a volumetric flow rate of 10 L/min. Assume constant density, isothermal operation and steady state. What reactor volume is required to achieve 80% conversion of A?

Worked solution
Answer:

V = 160 L (τ = 16 min)

The design equation for an ideal CSTR comes from a steady-state mole balance on A over the whole reactor, where the contents are perfectly mixed at the outlet conditions:

V=FA0X−rAV = \frac{F_{A0} X}{-r_A}

For a first-order reaction the rate is evaluated at the outlet concentration, −rA=kCA=kCA0(1−X)-r_A = kC_A = kC_{A0}(1 - X), and the molar feed rate is FA0=v0CA0F_{A0} = v_0 C_{A0}. Substituting,

V=v0CA0XkCA0(1−X)=v0Xk(1−X)V = \frac{v_0 C_{A0} X}{k C_{A0}(1 - X)} = \frac{v_0 X}{k(1 - X)}

The feed concentration cancels, which is characteristic of first-order kinetics: the required volume does not depend on CA0C_{A0}.

Substitute the data:

V=(10 L/min)(0.80)(0.25 min−1)(0.20)=8.00.050=160 LV = \frac{(10\,\text{L/min})(0.80)}{(0.25\,\text{min}^{-1})(0.20)} = \frac{8.0}{0.050} = 160\,\text{L}

The corresponding space time is τ=V/v0=16\tau = V/v_0 = 16 min. For comparison, a plug-flow reactor would need only τ=−ln⁡(1−X)/k=6.4\tau = -\ln(1 - X)/k = 6.4 min, or 64 L, because the CSTR runs entirely at the low outlet concentration where the rate is slowest.

V = 160 L

Question 3Worked problemMass Transfer and Separation

A liquid feed of 100 mol/h containing 50 mol% benzene (B) and 50 mol% toluene (T) is flashed in a single-stage equilibrium flash drum. At the drum temperature and pressure the vapour-liquid equilibrium ratios are KB=yB/xB=1.8K_B = y_B/x_B = 1.8 and KT=yT/xT=0.6K_T = y_T/x_T = 0.6. The vapour and liquid leaving the drum are in equilibrium.

(a) Determine the fraction of the feed that is vapourised, V/FV/F.

(b) Determine the mole fractions of benzene and toluene in the liquid and vapour products.

(c) Determine the molar flow rates of the vapour and liquid products.

Worked solution
Answer:

(a) V/F = 0.625; (b) xBx_B = 0.333, xTx_T = 0.667, yBy_B = 0.600, yTy_T = 0.400; (c) V = 62.5 mol/h, L = 37.5 mol/h

(a) Vapour fraction. Let ψ=V/F\psi = V/F, so L/F=1−ψL/F = 1 - \psi. A balance on each component, Fzi=Vyi+LxiF z_i = V y_i + L x_i, combined with the equilibrium relation yi=Kixiy_i = K_i x_i, gives the liquid composition in terms of ψ\psi:

xi=zi1+ψ(Ki−1)x_i = \frac{z_i}{1 + \psi(K_i - 1)}

The liquid mole fractions must sum to 1 and so must the vapour mole fractions; subtracting the two sums gives the Rachford–Rice equation, which fixes ψ\psi:

∑izi(Ki−1)1+ψ(Ki−1)=0\sum_i \frac{z_i(K_i - 1)}{1 + \psi(K_i - 1)} = 0

For the two components:

0.5(0.8)1+0.8ψ−0.5(0.4)1−0.4ψ=0⇒0.4(1−0.4ψ)=0.2(1+0.8ψ)\frac{0.5(0.8)}{1 + 0.8\psi} - \frac{0.5(0.4)}{1 - 0.4\psi} = 0 \quad \Rightarrow \quad 0.4(1 - 0.4\psi) = 0.2(1 + 0.8\psi) 0.4−0.16ψ=0.2+0.16ψ⇒ψ=0.20.32=0.6250.4 - 0.16\psi = 0.2 + 0.16\psi \quad \Rightarrow \quad \psi = \frac{0.2}{0.32} = 0.625

(b) Compositions. Substitute ψ=0.625\psi = 0.625 into the expression for xix_i, then apply yi=Kixiy_i = K_i x_i:

xB=0.51+0.625(0.8)=0.51.5=0.333,xT=0.51−0.625(0.4)=0.50.75=0.667x_B = \frac{0.5}{1 + 0.625(0.8)} = \frac{0.5}{1.5} = 0.333, \qquad x_T = \frac{0.5}{1 - 0.625(0.4)} = \frac{0.5}{0.75} = 0.667 yB=1.8(0.333)=0.600,yT=0.6(0.667)=0.400y_B = 1.8(0.333) = 0.600, \qquad y_T = 0.6(0.667) = 0.400

Each set sums to 1.000, and the vapour is richer in benzene, the more volatile component (KB>1K_B > 1), as it must be.

(c) Product flow rates. V=ψF=0.625(100)=62.5V = \psi F = 0.625(100) = 62.5 mol/h and L=F−V=37.5L = F - V = 37.5 mol/h. The benzene balance closes: 62.5(0.600)+37.5(0.333)=37.5+12.5=5062.5(0.600) + 37.5(0.333) = 37.5 + 12.5 = 50 mol/h, equal to the 50 mol/h of benzene in the feed.

V/F = 0.625; xBx_B = 0.333, xTx_T = 0.667, yBy_B = 0.600, yTy_T = 0.400; V = 62.5 mol/h, L = 37.5 mol/h

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