Chemical Engineering practice questions (Jahziah exam)
One worked question for each of the 9 official subject areas of Chemical Engineering, from a bank built on ETEC’s academic standards. Work them here, then practise the full set — with spaced review and timed mocks — in the app, free.
In a reactor, the reaction A → 2 B takes place. The feed contains 60 mol/min of A and no B. The fractional conversion of A is 0.50. What is the total molar flow rate of the product stream?
A60 mol/min
90 mol/min
C120 mol/min
D30 mol/min
WHY
A reacted = 0.50 × 60 = 30 mol/min, so A remaining = 30 mol/min. Reaction A → 2 B gives B = 2 × 30 = 60 mol/min. Total product flow = 30 + 60 = 90 mol/min.
Sample — the full set, with spaced review and mocks, is in the app.
Area 02 · Fluid mechanics
A fluid flows steadily through a 90° reducing elbow. The pipe is horizontal and the flow enters at section 1 and leaves at section 2. In the control-volume momentum balance for the fluid, which term represents the rate at which momentum leaves the control volume?
The mass flow rate multiplied by the velocity vector at section 2
BThe pressure at section 2 multiplied by the area at section 2
CThe mass flow rate multiplied by the velocity vector at section 1
DThe mass flow rate multiplied by the pressure drop between section 1 and section 2
WHY
For steady flow through the control volume, the momentum balance is ΣF = ṁv₂ − ṁv₁. The term ṁv₂ is the rate at which momentum leaves the control volume at the outlet (section 2).
Sample — the full set, with spaced review and mocks, is in the app.
Area 03 · Heat transfer
A cylindrical pipe of inner radius 0.05 m and outer radius 0.10 m has thermal conductivity 0.20 W/(m·K). The inner surface is maintained at 500 K and the outer surface at 400 K. What is the conductive heat transfer rate per unit length of the pipe under steady-state conditions? (Use π ≈ 3.14.)
Sample — the full set, with spaced review and mocks, is in the app.
Area 04 · Thermodynamics
For the liquid-phase reaction A ⇌ B at 300 K, the standard Gibbs free energy change is ΔG° = −6.0 kJ/mol. A reactor initially contains only A at 1.0 mol/L. Assuming ideal behavior and that ΔG° is independent of temperature, evaluate whether the equilibrium mixture will contain more B than A. What is the equilibrium constant K at 300 K?
K = 11, and the equilibrium mixture will contain more B than A because ΔG° is negative.
BK = 0.091, and the equilibrium mixture will contain more A than B because the reaction is not spontaneous.
CK = 11, but the equilibrium mixture will contain more A than B because the initial concentration of A is high.
DK = 1.1, and the equilibrium mixture will contain more B than A because ΔG° is negative.
WHY
ln K = −ΔG°/(RT) = 6000/(8.314 × 300) ≈ 2.41, so K ≈ 11. Since K = [B]/[A] = 11 > 1, the equilibrium mixture contains more B than A, consistent with negative ΔG°. Answer A.
Sample — the full set, with spaced review and mocks, is in the app.
Area 05 · Separation processes
In convective mass transfer between a flowing fluid and a solid surface, the mass transfer coefficient k_c is best described as:
AThe proportionality constant between the diffusive flux and the concentration gradient inside the solid.
The proportionality constant between the mass flux at the fluid–solid interface and the concentration difference between the bulk fluid and the interface.
CThe ratio of the diffusivity to the boundary layer thickness, independent of fluid velocity.
DThe maximum possible mass flux that can be achieved when the interface concentration equals the bulk concentration.
WHY
By definition N_A = k_c (C_Ab − C_As), so k_c is the proportionality constant linking flux to the bulk-to-interface concentration difference.
Sample — the full set, with spaced review and mocks, is in the app.
Area 06 · Chemical reaction engineering
The catalytic decomposition of A over a solid catalyst is first order with respect to A and obeys the rate law −rA = k' C_A, where k' = 0.25 s⁻¹. A packed bed reactor is to treat a feed of pure A at a volumetric flow rate of 0.005 m³/s and inlet concentration C_A0 = 2.0 mol/m³. What catalyst weight W is required to achieve 80% conversion in an isothermal plug-flow packed bed?
A0.16 kg
0.032 kg
C0.08 kg
D0.014 kg
WHY
W = (v0/k′) × (−ln(1−X)) = (0.005/0.25) × 1.609 = 0.032 kg.
Sample — the full set, with spaced review and mocks, is in the app.
Area 07 · Process control
A feed flow disturbance enters a process through a disturbance transfer function G_d = 2 m³/h per m³/h. The controller transfer function is G_c = 4, the final control element is G_v = 0.5, and the process transfer function is G_p = 3. In a regulatory problem, what is the change in the controlled variable caused by a unit step disturbance when the loop is open?
2
B12
C6
D0.5
WHY
With the loop open, the disturbance acts only through Gd. Unit step gives ΔY = Gd × 1 = 2 × 1 = 2.
Sample — the full set, with spaced review and mocks, is in the app.
Area 08 · Process design & economics
A process flow diagram shows a liquid stream entering a fired heater at 10 kg/s and 25 °C. The heater adds 2000 kW to the stream. The heated stream then splits: one branch (4 kg/s) goes to a reactor, and the remaining stream goes to a distillation column. Assuming steady state with negligible heat loss and no phase change, what is the mass flow rate to the distillation column?
A4 kg/s
6 kg/s
C10 kg/s
D14 kg/s
WHY
At steady state, the total mass entering the split must equal the sum of the branches. Total inlet = 10 kg/s, branch to reactor = 4 kg/s, so the branch to the distillation column = 10 - 4 = 6 kg/s. The heat input (2000 kW) and temperatures do not affect the mass split because no phase change is assumed.
Sample — the full set, with spaced review and mocks, is in the app.
Area 09 · Safety, health & environment
A safety data sheet for a chlorinated solvent reports the following: flash point 4 °C, boiling point 61 °C, National Fire Protection Association health rating 3, and a permissible exposure limit of 25 ppm. A worker asks which SDS entry is the primary basis for requiring a supplied-air respirator and local exhaust ventilation, rather than ordinary safety glasses.
AThe flash point of 4 °C, because a low flash point indicates the greatest inhalation hazard.
The health rating of 3 (serious hazard) and the permissible exposure limit of 25 ppm, because these quantify the inhalation toxicity that respiratory protection controls.
CThe boiling point of 61 °C, because a low boiling point means the liquid evaporates and causes asphyxiation.
DNone of the listed entries govern respiratory protection; only the SDS section on first-aid measures is relevant.
WHY
Respiratory protection and ventilation are driven by inhalation toxicity, which the SDS expresses through the health hazard rating and the PEL. A low flash point and a low boiling point concern fire and evaporation, not the need for respiratory protection.
Sample — the full set, with spaced review and mocks, is in the app.
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