Chemical Reaction Engineering practice questions

Answer: A) 2

• ACorrect. Units of \(\text{concentration}^{1-n}\,\text{time}^{-1}\) with n = 2.
• BIncorrect. First order has units of s⁻¹.
• CIncorrect. Zero order has units of mol/(L·s).
• DIncorrect. Third order has units of L²/(mol²·s).

Answer: C) 1 mol/L

• AIncorrect. Only 1 mol/L is consumed.
• BIncorrect. That's first-order decay.
• CCorrect. C = C₀ − kt = 2 − 1.
• DIncorrect. Multiply k by 10 min.

Answer: C) 2 min

A zero-order reaction loses concentration linearly, so t½ = C_A0/(2k) = 2 min. Unlike first-order kinetics, the half-life depends on the starting concentration.

• AIncorrect. C_A0/k = 2/0.5 = 4 min is the time for a zero-order reaction to consume all of the reactant, not half of it.
• BIncorrect. ln 2/k = 0.693/0.5 = 1.39 is the first-order half-life formula. It doesn't apply to a zero-order reaction, and the units don't work with k in mol/(L·min).
• CCorrect. For zero order, C_A = C_A0 − kt. Setting C_A = C_A0/2 gives t½ = C_A0/(2k) = 2/(2×0.5) = 2 min.
• DIncorrect. 1/(kC_A0) = 1/(0.5×2) = 1 min comes from the second-order half-life formula, which doesn't apply to a zero-order reaction.

Answer: B) A single spike

• AIncorrect. That's a CSTR.
• BCorrect. There's no axial mixing.
• CIncorrect. That isn't plug flow.
• DIncorrect. It's defined.

Answer: C) 0.5 mol/L

• AIncorrect. Use 1/C = 1/C₀ + kt.
• BIncorrect. That's first-order decay.
• CCorrect. 1/C = 1/C₀ + kt = 1 + 1 = 2.
• DIncorrect. Recheck: 1/C = 2.

Answer: B) 2

• AIncorrect. That's the rate factor.
• BCorrect. \(2^n = 4\).
• CIncorrect. First order would double the rate.
• DIncorrect. That would raise the rate by √2.

Answer: A) 80%

• ACorrect. X = kτ/(1 + kτ) = 4/5.
• BIncorrect. That's the plug-flow result, 1 − e⁻⁴.
• CIncorrect. That's 1/kτ.
• DIncorrect. A CSTR can't reach full conversion.

Answer: D) 0.2 min⁻¹

Space velocity is the reciprocal of space time: SV = 1/τ = 1/5 min = 0.2 min⁻¹ (which equals 12 h⁻¹).

• AIncorrect. 60/5 = 12 is the space velocity in h⁻¹ (0.2 min⁻¹ × 60 min/h). Labeling it min⁻¹ is a unit error.
• BIncorrect. This treats space velocity as equal to space time. Space velocity is the reciprocal of space time, so its units are inverse time.
• CIncorrect. This squares the space time (5²) instead of taking its reciprocal.
• DCorrect. Space velocity SV = 1/τ = 1/(5 min) = 0.2 min⁻¹. This is the number of reactor volumes of feed processed per minute.

Answer: A) rate ≈ k1[A], set by the first step

In consecutive elementary steps, the step with the much smaller rate constant is the rate-determining step. Here that is A → I, so the overall rate ≈ k1[A].

• ACorrect. The first step is far slower (k1 << k2), so I is consumed almost as soon as it forms. The first step is the bottleneck and rate-determining step, so the rate of forming P ≈ k1[A].
• BIncorrect. Releasing the product does not make a step rate-controlling. Here [I] stays very small, and the second step can only go as fast as the first step supplies I.
• CIncorrect. The step with the largest rate constant is the fastest, and it waits on the slower step. Also, k2 multiplies [I], not [A].
• DIncorrect. Consecutive steps act in series, not in parallel, so their rate constants do not add. The slow step limits the overall rate.

Answer: A) Stop the batch at the time when the concentration of B reaches its maximum

In series reactions the intermediate B rises to a maximum and then declines as it converts to C. The batch time should therefore be set to the time when C_B peaks; for first-order steps, t_max = ln(k2/k1)/(k2 − k1).

• ACorrect. B builds up and is then used up as it converts to C, so its concentration passes through a maximum (dC_B/dt = 0). For first-order steps this happens at t_max = ln(k2/k1)/(k2 − k1). Stopping at that time maximizes B.
• BIncorrect. B keeps converting to C the whole time. By the time A is nearly gone, much of the B has already been lost to C, so the yield of B is past its peak.
• CIncorrect. The net rate of B formation is highest early in the batch, when A is most concentrated and little B exists. B is still building up at that point, so stopping then gives less B than the optimum.
• DIncorrect. C levels off only after nearly all of the B has reacted to C. That outcome would suit a process where C is the product, not B.