Reactions carry 7–11 questions. The pattern is steady: get the rate law, put it in the design equation of the right reactor, and keep conversion, yield and selectivity apart.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Chemical
questions on the exam7–11
problems on this page16
of 17 knowledge areas12
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What the specification covers
The outline lists six subtopics. In our words:
Rates and reaction order.
The rate constant and its temperature dependence (Arrhenius).
Question 1 of 16 · Chemical Reaction Engineering · 7–11 items
An engineer is sizing a continuous reactor system for a liquid-phase autocatalytic reaction of the form A + B → 2B. The design objective is to achieve a final conversion of 90% while minimizing the total combined reactor volume. Based on Levenspiel plot evaluations for autocatalytic kinetics, which reactor configuration is optimal?
Solution
Autocatalytic rate −rA = kCACB is small at both ends, so 1/(−rA) versus X has a minimum. Minimum total volume: a CSTR exits at the conversion of the minimum, then a PFR integrates up to X = 0.90.
Starting in a PFR wastes volume, because the rate is lowest at low conversion when little B is present.
Two PFRs in series are the same as one PFR of the combined volume, with the same slow start.
A lone PFR has to integrate through the region where B is scarce and the rate is very slow, so it needs more volume.
On the Levenspiel plot 1/(−rA) has a minimum: a CSTR operating at that minimum covers the early part cheaply, and a PFR finishes the rising high-conversion part to 90 %.
A CSTR train approaches PFR behavior, which is inefficient in the slow early region; one CSTR at the rate maximum then a PFR does better.
Question 2 of 16 · Chemical Reaction Engineering · 7–11 items
A continuous reactor has a physical volume of 500 L and operates at a steady volumetric flow rate of 10 L/min. A tracer experiment is conducted to evaluate non-ideal flow patterns via Residence Time Distribution (RTD). The experimentally measured mean residence time (tm) is calculated from the exit age distribution to be 40 min. Based on this RTD metric, what is the primary flow anomaly occurring, and what is the effective active volume?
Given
Reactor volume V = 500 L
Volumetric flow rate v0 = 10 L/min
Measured mean residence time tm = 40 min
Find
Flow anomaly and effective active volume (L)
Solution
τ = V/v0 = 500 L/(10 L/min) = 50 min > tm = 40 min → dead (stagnant) volume
Vactive = v0·tm = 10 L/min × 40 min = 400 L (100 L dead)
If the whole 500 L were active the tracer would give tm = 50 min, not 40 min.
Bypassing alone does not lower the mean residence time, and an active volume larger than the vessel (500 L) is impossible.
τ = V/v0 = 50 min but tm = 40 min, so part of the vessel does not take part in flow: Vactive = v0·tm = 10 × 40 = 400 L.
The volume is right, but bypass with the whole volume active keeps tm = τ; a shorter mean residence time points to stagnant (dead) volume.
Dead zones is the right diagnosis, but 625 L inverts the ratio (500 × 50/40); the active volume must be smaller than the vessel.
Question 3 of 16 · Chemical Reaction Engineering · 7–11 items
For a reversible, strongly exothermic reaction, what is the optimal temperature profile along the length of a plug flow reactor to maximize both the reaction rate and final product conversion?
Solution
Exothermic and reversible: K falls with T, while k rises. The optimum progression follows the locus of maximum rate: hot at the inlet (far from equilibrium), then progressively cooler (interstage cooling in SO2, NH3 converters).
A constant low temperature allows high equilibrium conversion but is slow at the inlet, so it needs a much longer reactor.
A constant high temperature is fast at first but caps conversion at a low equilibrium value near the outlet.
Rising temperature is the profile for an endothermic reversible reaction; for an exothermic one it lowers equilibrium conversion exactly where conversion is high.
Alternating heating and cooling has no basis in the rate-equilibrium trade-off; the optimum falls steadily along the reactor.
Far from equilibrium near the inlet a high temperature gives a fast rate; as conversion rises, cooling raises the equilibrium limit so the reaction can keep going.
Question 4 of 16 · Chemical Reaction Engineering · 7–11 items
An engineer proposes modifying an existing packed-bed reactor by significantly decreasing the solid catalyst particle size. The goal is to move the system further into the kinetically-controlled regime. What is the primary operational trade-off that will result from this design change?
Solution
φ = R√(k/De) falls with R, so η rises (toward kinetic control). Ergun: ΔP/L ∝ 1/Dp2 (viscous) and ∝ 1/Dp (inertial), so the bed pressure drop rises.
A smaller pellet lowers the Thiele modulus (φ ∝ R), raising η toward 1, while Ergun's ΔP grows roughly as 1/Dp to 1/Dp2.
Smaller particles raise the pressure drop through a packed bed; they do not lower it.
Particle size does not change the intrinsic rate constant; it changes how much of the pellet interior is used.
The pressure drop part is right, but internal mass transfer improves, not worsens, with shorter diffusion paths.
Smaller pellets do ease pore diffusion, but at the same flow and bed volume the residence time hardly changes; the cost is pressure drop.
Question 5 of 16 · Chemical Reaction Engineering · 7–11 items
A catalytic reaction A → B follows a Langmuir-Hinshelwood mechanism with a single-site surface reaction acting as the rate-limiting step. The observed rate equation is given by −rA = (k KA PA)/(1 + KA PA + KB PB). At very high partial pressures of both A and B, where catalyst surface coverage is nearly complete, what is the apparent overall reaction order with respect to the total pressure if the molar ratio of A to B remains constant?
Solution
−rA = kKAPA/(1 + KAPA + KBPB) → kKAPA/(KAPA + KBPB) at high P
With PA = yAP and PB = yBP at a fixed ratio, P cancels: apparent order 0 in total pressure (saturated surface).
Inverse order would need the denominator to grow faster than the numerator, which needs a squared adsorption term (dual-site); here both scale with P.
At high coverage the rate → kKAPA/(KAPA + KBPB); with PA/PB fixed, both scale with total pressure, so the rate is independent of P.
Fractional orders show up at intermediate coverage; at saturation the 1 in the denominator is negligible and the order goes to zero.
Second order would need two adsorbed reactant sites in the numerator; the single-site mechanism gives at most first order.
First order applies at low pressure, where the denominator is about 1 and the rate is kKAPA.
Question 6 of 16 · Chemical Reaction Engineering · 7–11 items
Consider the parallel reactions A → B and A → C with rate constants k1 and k2. If raising the reactor temperature lowers the selectivity toward product B, which inference about the activation energies is most likely correct?
Solution
SB/C = k1/k2 = (A1/A2)·exp[−(E1 − E2)/RT]
S falls with T when E2 > E1: high temperature favors the path with the higher activation energy.
Selectivity to B falls on heating only if k2 grows faster than k1, i.e. E2 > E1.
Equal activation energies make k1/k2 independent of temperature, so selectivity would not change.
SB/C = k1/k2 = (A1/A2)e−(E1−E2)/RT; activation energies are exactly what make it temperature dependent.
If E1 were larger, heating would speed up the B path more and selectivity to B would rise, not fall.
Question 7 of 16 · Chemical Reaction Engineering · 7–11 items
In a CSTR (Continuous Stirred Tank Reactor) operating at steady state with a first-order reaction, if the space time is doubled while maintaining the same inlet concentration, what happens to the conversion?
Solution
X = kτ/(1 + kτ) (first-order CSTR)
example kτ = 1: X = 0.50 → 2kτ = 2: X = 0.67 (up, but not doubled)
More space time means more time for reaction; conversion cannot fall.
X = kτ/(1 + kτ): doubling τ gives 2kτ/(1 + 2kτ), larger than before but less than twice the original value.
Conversion depends on the Damköhler number kτ, so doubling τ must raise it.
X = kτ/(1 + kτ) is not linear in τ; doubling τ always gives less than double (X can never exceed 1).
Question 8 of 16 · Chemical Reaction Engineering · 7–11 items
An elementary exothermic gas-phase reaction A → B occurs in an adiabatic plug flow reactor. The feed consists of pure A at an initial concentration of 50 mol m−3, a superficial velocity of 2.5 m/s, and an inlet temperature of 400 K. The rate constant at this temperature is 0.5 s−1. The heat of reaction is constant at ΔHrxn = −80,000 J mol−1. The molar heat capacity of both A and B depends on temperature according to Cp = 40 + 0.1T in J mol−1 K−1, where T is in Kelvin. What is the initial temperature gradient dT/dz at the reactor inlet?
1250 K/m multiplies by the superficial velocity instead of dividing by it.
500 K/m leaves out the superficial velocity; dT/dz needs the molar flux u·CA0 in the denominator.
400 K/m keeps only the constant 40 J/(mol·K) of the heat capacity and drops the 0.1T term.
304 K/m evaluates Cp with the temperature in degrees Celsius (127 °C); the correlation is written for kelvin.
At the inlet −rA = kCA0 = 25 mol/(m3·s) and Cp = 40 + 0.1(400) = 80 J/(mol·K), so dT/dz = (80,000 × 25)/(2.5 × 50 × 80) = 200 K/m.
Question 9 of 16 · Chemical Reaction Engineering · 7–11 items
The intrinsic activation energy of a first-order catalytic reaction is 100 kJ mol−1. The enthalpy of internal diffusion is considered negligible. If the system is operating in a regime that is strongly limited by internal pore diffusion (where the Thiele modulus φ≫ 1 ), what is the apparent activation energy observed experimentally?
25 kJ/mol halves the activation energy twice; the strong-diffusion falloff halves it once.
With φ ≫ 1, η ≈ 3/φ ∝ k−1/2, so the observed rate ∝ ηk ∝ k1/2 and Eapp = Ea/2 = 50 kJ/mol.
100 kJ/mol is the intrinsic value, which is what you measure only in the kinetic regime (φ ≪ 1).
150 kJ/mol adds half the intrinsic value instead of taking half of it.
200 kJ/mol doubles the activation energy; diffusion limitation lowers the observed temperature sensitivity rather than raising it.
Question 10 of 16 · Chemical Reaction Engineering · 7–11 items
For a reaction A → products, the rate doubles when the concentration of A is doubled. What is the most likely reaction order with respect to A?
Solution
r2/r1 = (C2/C1)n → 2 = 2n → n = 1
r ∝ CAn: 2n = 2 gives n = 1.
Half order would raise the rate by √2 ≈ 1.41 when the concentration doubles.
A zero-order rate does not change with concentration at all.
Second order would make the rate four times larger (22) when the concentration doubles.
Question 11 of 16 · Chemical Reaction Engineering · 7–11 items
For an elementary reaction 2A + B → products, what is the rate law?
Solution
Elementary 2A + B → products: r = k[A]2[B] (overall third order). Only for elementary steps do the orders equal the stoichiometric coefficients.
This drops the coefficient 2 on A; in an elementary step two A molecules collide, giving [A]2.
The coefficient becomes an exponent, not a multiplier inside the bracket: k[2A][B] = 2k[A][B] is still first order in A.
B appears once in the elementary step, so its exponent is 1, not 2.
For an elementary step the orders equal the molecularities: second order in A, first order in B.
Question 12 of 16 · Chemical Reaction Engineering · 7–11 items
For a batch reactor with constant volume operating at constant temperature, what is the design equation for a first-order reaction in terms of conversion?
Solution
CA0·dX/dt = kCA0(1 − X) → t = (1/k)·ln[1/(1 − X)]
FE Reference Handbook, Chemical Reaction Engineering: batch reactor, first-order irreversible.
1/X − 1 has no basis in the first-order mole balance; it even makes the time fall as conversion rises.
t = X/k is the zero-order batch time with CA0 = 1; first order needs the logarithm.
dX/dt = k(1 − X) integrates to kt = −ln(1 − X), so t = (1/k)ln[1/(1 − X)].
X/(1 − X) is the first-order CSTR result (τ = X/[k(1 − X)]), not the batch result.
Question 13 of 16 · Chemical Reaction Engineering · 7–11 items
For the series reactions A → B → C with rate constants k1 and k2, when is the maximum concentration of intermediate B achieved?
1/k2 is the time of the maximum only when k1 = k2; in general tmax = ln(k1/k2)/(k1 − k2).
1/k1 would be the time of the maximum only if k1 = k2; in general tmax = ln(k1/k2)/(k1 − k2).
Question 14 of 16 · Chemical Reaction Engineering · 7–11 items
A rate constant for the consumption of reactant A is reported in units of L0.5·mol−0.5·s−1. What is the overall order of the reaction?
Given
Units of k: L0.5·mol−0.5·s−1
Find
Reaction order n
Solution
−rA = kCAn → [k] = (mol/L)1−n·s−1
(mol/L)−0.5 → 1 − n = −0.5 → n = 1.5
−0.5 is the exponent on mol in the units of k, not the reaction order.
0.5 takes the exponent on L as the order, or writes the units as (mol/L)n−1 with the sign reversed.
1.0 assumes that a rate constant in s−1 terms means first order; the concentration factor in the units says otherwise.
For order n, k has units (mol/L)1−n·s−1; L0.5·mol−0.5·s−1 = (mol/L)−0.5·s−1, so 1 − n = −0.5 and n = 1.5.
Question 15 of 16 · Chemical Reaction Engineering · 7–11 items
For the first-order parallel reactions A → B (rate constant k1) and A → C (rate constant k2), where B is the desired product, what is the instantaneous selectivity of B with respect to C (moles of B formed per mole of C formed)?
Solution
SB/C = rB/rC = k1CA/(k2CA) = k1/k2
The fractional yield of B, k1/(k1 + k2), is a different ratio. FE Reference Handbook, Chemical Reaction Engineering: selectivity and yield.
k2/(k1 + k2) is the fraction of reacted A that goes to the undesired product C.
k2/k1 is the selectivity of C over B, the inverse of the ratio asked for.
k1/(k1 + k2) is the fractional (instantaneous) yield of B, moles of B per mole of A reacted; the definition in the stem divides by moles of C instead.
SB/C = rB/rC = k1CA/(k2CA) = k1/k2 for two first-order parallel reactions.
Question 16 of 16 · Chemical Reaction Engineering · 7–11 items
In the reversible first-order reaction A ⇌ B with forward rate constant kf = 0.8 min−1 and reverse rate constant kr = 0.2 min−1, what is the equilibrium conversion of A, starting from pure A?
At equilibrium kfCA0(1 − X) = krCA0X, so Xeq = kf/(kf + kr) = 0.8/1.0 = 0.80.
0.60 subtracts the rate constants (kf − kr); at equilibrium the forward and reverse rates are equal, which gives a ratio.
0.25 is kr/kf, the inverse of the equilibrium constant, not a conversion.
0.20 is the fraction of A left unconverted at equilibrium, kr/(kf + kr).
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Common traps
Units in Arrhenius. Temperature in kelvin, Ea in J/mol with R = 8.314 J/(mol·K), or cal/mol with R = 1.987 cal/(mol·K). A rate constant read at one temperature and reused at another without the exponential correction is a slip no unit check will catch.
Which reactor is bigger. For a single reaction of positive order, a CSTR needs more volume than a PFR for the same conversion, because it runs at the outlet concentration:
VCSTR = FA0·X/(−rA), VPFR = FA0·∫0X dX/(−rA)
Space time taken for residence time. τ = V/v0 uses the inlet flow; in a gas-phase reaction that changes the number of moles, the actual residence time differs.
Rate units. −rA per unit volume in homogeneous reactors, per unit mass of catalyst in packed beds; mixing the two changes the answer by the bed density.
Sources: Fogler, Elements of Chemical Reaction Engineering; Levenspiel, Chemical Reaction Engineering.