Free FE Chemical practice problems

Ten real questions from our FE Chemical bank — one from each knowledge area, with the same exam-realistic figures and fully worked solutions our members get. No signup, no email. Try each one first, then reveal the solution.

8 with figures10 knowledge areasFull worked solutions

1
medium
A. Molecular & convective diffusion
Species A diffuses at steady state through a stagnant film of thickness δ = 0.005 m. The molecular diffusivity is D_AB = 0.00002 m²/s, and the molar concentration of A is C_A1 = 50 mol/m³ at one face (z = 0) and C_A2 = 20 mol/m³ at the other (z = δ), as shown. The molar flux N_A is most nearly:
1. 0.2 mol/(m²·s)
2. 0.12 mol/(m²·s)
3. 0.28 mol/(m²·s)
4. 0.14 mol/(m²·s)
Show the answer & worked solution
Answer: B.0.12 mol/(m²·s)
N_A = D_AB·(C_A1 − C_A2)/δ = 0.00002·(50 − 20)/0.005 = 0.12 mol/(m²·s). Distractors: the two concentrations ADDED (C_A1 + C_A2 = 70) used as the driving force, 0.28; only the UPSTREAM C_A1 = 50 (forgot to subtract the downstream value), 0.2; and the MEAN concentration (C_A1 + C_A2)/2 = 35 used in place of the difference, 0.14.
2
medium
A. Process dynamics & transfer functions
A pressure loop responds as an under-damped second-order system with damping ratio $\zeta = 0.2$ . The percent overshoot of its unit-step response is most nearly:
1. 53.3 %
2. 80 %
3. 20 %
4. 52.7 %
Show the answer & worked solution
Answer: D.52.7 %
$\sqrt{1-\zeta^2} = 0.9798$ , so $OS = e^{-\zeta\pi/\sqrt{1-\zeta^2}}\times100 = e^{-0.641}\times100 = 52.7 %$ . Distractors: the $\sqrt{1-\zeta^2}$ DROPPED, $e^{-\pi\zeta}\times100 = 53.3 %$ (always an overstatement); the damping ratio read as the overshoot, $100\zeta = 20 %$ ; and $100(1-\zeta) = 80 %$ .
3
medium
A. Statics, pipe flow & flow measurement
Water flows steadily through a pipe that contracts from a diameter $D_1 = 200$ mm to $D_2 = 100$ mm. The upstream velocity is $V_1 = 2$ m/s. The velocity in the smaller section is most nearly:
1. 8 m/s
2. 4 m/s
3. 32 m/s
4. 1 m/s
Show the answer & worked solution
Answer: A.8 m/s
Continuity $A_1V_1=A_2V_2$ with $A\propto D^2$ gives $V_2=V_1(D_1/D_2)^2=2\,(200/100)^2=8$ m/s. Distractors: the diameter ratio raised to the FOURTH power ( $V_1(D_1/D_2)^4$ , the area-ratio squaring applied a second time); the DIAMETER ratio used un-squared ( $V_1 D_1/D_2$ , forgetting $A\propto D^2$ ); and the ratio both inverted and un-squared ( $V_1 D_2/D_1$ ).
4
medium
A. Conduction, convection & radiation
A plane wall of thermal conductivity $k=0.8$ W/m·K has a face area $A=10$ m² and thickness $L=0.2$ m, with its two faces held at a temperature difference $\Delta T=20$ K. By Fourier's law, the steady rate of heat conduction through the wall is most nearly:
1. 160 W
2. 800 W
3. 1600 W
4. 80 W
Show the answer & worked solution
Answer: B.800 W
Fourier's law: $q=\dfrac{kA\,\Delta T}{L}=\dfrac{0.8\cdot10\cdot20}{0.2}=800$ W. Distractors: used half the thickness — the distance to the mid-plane — in the denominator ( $kA\Delta T/(L/2)$ , which doubles the rate); dropped the area $A$ (reported the heat flux $k\Delta T/L$ as a rate); and omitted the thickness entirely ( $kA\Delta T$ , as if $L=1$ m).
5
medium
A. Steady-state material balances
A feed of 1000 kg/h containing 12 wt% solute is concentrated in an evaporator. Pure water leaves overhead as vapor V and a concentrated liquid leaves at 40 wt% solute. The water evaporation rate V is most nearly:
1. 700 kg/h
2. 300 kg/h
3. 880 kg/h
4. 120 kg/h
Show the answer & worked solution
Answer: A.700 kg/h
Solute balance: L = F·w_F/w_L = 1000·0.12/0.4 = 300 kg/h; overall balance: V = F − L = 700 kg/h. Distractors: the CONCENTRATE L reported instead of the evaporated water, 300; ALL the feed except the entering solute assumed to evaporate (product's retained water ignored), F·(1 − w_F) = 880; and the SOLUTE THROUGHPUT F·w_F reported, not the water removed, 120.
6
medium
A. Particle properties, comminution & crystallization
A cumulative sieve analysis gives percent undersize versus size: 0.1 mm → 8%, 0.25 mm → 30%, 0.5 mm → 58%, 1 mm → 82%, 2 mm → 100%. The median particle size d₅₀ (size at 50 % undersize) is most nearly:
1. 0.5 mm
2. 0.321 mm
3. 0.375 mm
4. 0.429 mm
Show the answer & worked solution
Answer: D.0.429 mm
50 % falls between 0.25 mm (30 %) and 0.5 mm (58 %): d₅₀ = 0.25 + (50 − 30)/(58 − 30)·(0.5 − 0.25) = 0.429 mm. Distractors: the NEAREST sieve size read without interpolating, 0.5 mm; interpolation from the WRONG end ((U_hi − 50) instead of (50 − U_lo)), 0.321 mm; and the MIDPOINT of the two sizes (percentages ignored), 0.375 mm.
7
medium
A. Time value of money & equivalence
An engineer borrows $200,000 to be repaid in 15 equal end-of-year payments at 6% per year. The annual payment is most nearly:
1. $20,593
2. $8,593
3. $12,000
4. $13,333
Show the answer & worked solution
Answer: A.$20,593
$(A/P,0.06,15)=\dfrac{i(1+i)^n}{(1+i)^n-1}=0.10296$ , so $A=$ $200,000 $\times\ 0.10296=$ $20,593. Distractors: $P/n$ ignoring interest, the interest-only payment $P\cdot i$ , and the sinking-fund factor (A/F) used in place of A/P.
8
medium
A. Distributions & descriptive statistics
The breakdown voltage of a batch of Zener diodes is normally distributed with mean $\mu=50$ V and standard deviation $\sigma=8$ V. Given $\Phi(0.50)=0.6915$ and $\Phi(1.50)=0.9332$ , the probability that a randomly selected diode breaks down between 54 V and 62 V is most nearly:
1. 0.9332
2. 0.2417
3. 0.7583
4. 0.6247
Show the answer & worked solution
Answer: B.0.2417
$z_1=(54-50)/8=0.50$ and $z_2=(62-50)/8=1.50$ . $P=\Phi(z_2)-\Phi(z_1)=0.9332-0.6915=0.2417$ . Distractors subtract the wrong lower tail ( $1-\Phi(z_1)$ instead of $\Phi(z_1)$ ), ignore the lower limit and report $\Phi(z_2)$ alone, and report the complement $1-P$ .
9
medium
A. Codes, contracts, liability & IP
An engineer preparing the switchgear specification for a hospital project owns a 20 percent interest in one of the three manufacturers likely to bid on the equipment. Under the NCEES Model Rules of Professional Conduct, the engineer should:
1. sell the ownership interest quietly after the contract is awarded to eliminate the conflict
2. write the specification around generic performance requirements so the ownership has no practical effect
3. continue normally, since the bids will be evaluated against objective criteria the client controls
4. disclose the ownership interest to the client in writing before continuing with the specification
Show the answer & worked solution
Answer: D.disclose the ownership interest to the client in writing before continuing with the specification
The Model Rules require prompt written disclosure of any business interest that could influence professional judgment; the informed client then decides how to proceed. Distractors: drafting around the interest conceals the conflict instead of disclosing it; objective bid criteria do not cure an undisclosed stake in a bidder who helped shape the spec; divesting after award removes the benefit only after the influence has already operated.
10
medium
A. Properties, material types & corrosion
A 2 kg block of aluminum is heated from 25 °C to 185 °C. Its specific heat is $c=900$ J/(kg·°C). The heat that must be added is most nearly:
1. 288 kJ
2. 824 kJ
3. 333 kJ
4. 144 kJ
Show the answer & worked solution
Answer: A.288 kJ
$Q=mc\,\Delta T=(2)(900)(160)=288000$ J $=288$ kJ. Distractors: the final temperature $T_2=185$ °C used in place of $\Delta T=160$ °C; the mass omitted ( $Q=c\,\Delta T$ ); and the absolute final temperature $T_2+273=458$ K used as the temperature change — a $\Delta T$ is identical in °C and K, so no 273 is added.

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