FE Chemical exam practice: practice problems and exam questions

Practice problems for the Fundamentals of Engineering (FE) Chemical exam: material and energy balances, thermodynamics and equilibrium, kinetics and reactor design, heat transfer and fluid flow.

Cram pack (PDF)Formula sheets, method steps and practice problems with full solutions, ready to print.

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Topics in this course

  1. Mass balancemixing two streams · blending to a target · evaporatorcalculations
  2. Energy balanceheating a stream · heat exchanger · exchanger with heat losscalculations
  3. Ideal gas lawP from n, V, T · from mass · state change or densitycalculations
  4. Stoichiometrymass to mass · limiting reactant · percent yieldcalculations
  5. Chemical equilibriumKc from concentrations · ICE table · ICE table with a quadraticcalculations
  6. Calorimetry & enthalpyheating with q = m·c·ΔT · calorimeter · enthalpy of formationcalculations
  7. Reaction kineticsfirst order · conversion and second order · Arrheniuscalculations
  8. Reactor sizing (CSTR & PFR)CSTR first order · PFR first order · conversion from volumecalculations
  9. Heat exchangers (LMTD)LMTD · required area · full designcalculations
  10. Heat conductionsingle wall · composite wall · wall with convectioncalculations
  11. Pipe friction & head lossReynolds number · Darcy–Weisbach head loss · minor losses and total headcalculations
  12. Pumps & pump powerhydraulic power · pump efficiency · system head and powercalculations

Exam tips for FE Chemical exam practice

  • Choose a basis and draw the flowsheet before writing balances.
  • For first-order reactors, remember that a PFR always needs less volume than a CSTR for the same conversion.
  • Keep track of whether ΔT is a difference (K or °C) or an absolute temperature (K).

Worked examples

Mass balanceEasy
MixerStream 1300 kg/h · 30% saltStream 2250 kg/h · 40% saltStream 3x₃ = ? %

Two salt solutions are mixed at steady state. Find the mass fraction of salt in the outlet x3.

ṁ1 = 300 kg/hx1 = 30 %ṁ2 = 250 kg/hx2 = 40 %

Find x3 (%)

Show hints
  1. At steady state, what goes in comes out: write a total balance and a balance for the dissolved component.
  2. Total: ṁ₁ + ṁ₂ = ṁ₃. Component: ṁ₁x₁ + ṁ₂x₂ = ṁ₃x₃.
  3. ṁ3 = 550 kg/h.
Show the answer and the step-by-step solution

Steady-state mass balance: in = out, for the total mass and separately for each component (no reaction).

Total balance

ṁ3 = ṁ1 + ṁ2 = 300 + 250 = 550 kg/h

Salt balance

ṁ1x1 + ṁ2x2 = ṁ3x3
x3 = (300·30 + 250·40)/550 = 34.55 %

Answer

x3 = 34.55 %
Energy balanceMedium
hot sidecold sideglycol mix · 150 °C120 °Cwater · 15 °Ct = ? °C

Hot glycol mix cools water in a counter-flow heat exchanger with no heat loss. Find the water outlet temperature tout.

ṁh = 1 kg/scp,h = 3.6 kJ/(kg·K)Th: 150 → 120 °Cṁc = 1 kg/scp,c = 4.18 kJ/(kg·K)tin = 15 °C

Find tout (°C)

Show hints
  1. Heat lost by the hot stream equals heat gained by the cold stream.
  2. Find Q from the hot side first, since you know both of its temperatures.
  3. Q = 108 kW.
Show the answer and the step-by-step solution

Exchanger energy balance: with no losses, ṁhcp,h(Th,in − Th,out) = ṁccp,c(tout − tin).

Heat released by the hot stream

Q = 1·3.6·(150 − 120) = 108 kW

Heat picked up by the water

tout = tin + Q/(ṁccp,c) = 15 + 108/(1·4.18) = 40.84 °C

Answer

tout = 40.84 °C

How practising works

  • “Practise this course” mixes problems from all its topics, like a real exam.
  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Mock exams are timed and have no hints, and you get your score and a full review at the end.

Calculators for this course

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