23-Chem-A4 Chemical Reactor Engineering · December 2016
Question 3 of 5: Half-Order Batch Kinetics with an Arrhenius Temperature Shift
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams / EGBC — December 2016 — 04-Chem-A4 Chemical Reactor Engineering. Three-hour open-book exam; one textbook of the candidate’s choice (Fogler or Levenspiel), personal unit-conversion / mathematical tables (CRC Handbook) and a non-communicating programmable calculator are permitted. Five questions are printed and any four constitute a complete paper (each worth 25 marks; Q1 and Q4 are split 5 / 12 / 8, Q5 is 9 / 5 / 11); all five are solved below for completeness. No credit is given for re-deriving standard rate expressions, so the batch / CSTR / PFR design equations are quoted and applied, and significant formulae are cited by origin as the rubric requests. Property look-ups not printed on the paper (the gas constant $R$) are stated explicitly in each Given block as permitted open-book references.
Reference texts: H. S. Fogler, Elements of Chemical Reaction Engineering (5th ed., Prentice Hall) — Levenspiel plots and the mixed-flow / plug-flow design equations (Ch. 2), the stoichiometric table (Ch. 3), adiabatic energy balances (Ch. 11–12), internal-diffusion effectiveness factors and the generalized Thiele modulus (Ch. 15), and residence-time distributions (Ch. 16–17). O. Levenspiel, Chemical Reaction Engineering (3rd ed., Wiley, 1999) — the "which reactor is better" rate-curve reasoning (Ch. 5–6) and pulse-tracer RTD analysis (Ch. 11–13).
Question 3: Half-Order Batch Kinetics with an Arrhenius Temperature Shift (25 marks)
Given. Half-order batch kinetics with two temperature–time data points used to extract the activation energy, then extrapolated to 40 °C.
Condition
Value
Rate law
$-r_A = k C_A^{0.5}$ (valid $2.0 \ge C_A \ge 0.25$)
Datum 1 (25 °C)
$2.0 \to 1.0$ mol/L in 15 min
Datum 2 (50 °C)
$2.0 \to 1.0$ mol/L in 20 s
Target (40 °C)
$C_A$ after 10 min, $C_{A0}=2.0$ mol/L
Gas constant $R$ (open-book)
1.987 cal/(mol·K)
Find. $C_A(t=10\ \text{min})$ at 40 °C.
Approach. Integrate the half-order batch equation to get $k$ at 25 and 50 °C from the two half-drop times, take the Arrhenius ratio for $E_a$, interpolate $k$ to 40 °C, then integrate forward — watching for the finite completion time that fractional orders produce.
Integrate the half-order batch law. $-\dfrac{dC_A}{dt} = kC_A^{0.5}$ separates to $\int C_A^{-0.5}dC_A = -k\int dt$, giving $$C_{A0}^{0.5} - C_A^{0.5} = \tfrac{k}{2}\,t.$$ This is the working equation for every part below.
Rate constants at 25 °C and 50 °C. For the $2.0\to1.0$ drop, $C_{A0}^{0.5}-C_A^{0.5}=\sqrt2-1=0.4142$, so $k = 2(0.4142)/t$. At 25 °C ($t=15$ min): $k_{25}=0.0552\ (\text{mol/L})^{0.5}\text{min}^{-1}$; at 50 °C ($t=1/3$ min): $k_{50}=2.485\ (\text{mol/L})^{0.5}\text{min}^{-1}$. Note the ratio $k_{50}/k_{25}=t_{25}/t_{50}=45$ is independent of reaction order — the same concentration change is compared.
Activation energy from the Arrhenius ratio. $\ln\dfrac{k_{50}}{k_{25}} = \dfrac{E_a}{R}\Big(\dfrac{1}{298.15}-\dfrac{1}{323.15}\Big)$. With $\ln 45 = 3.807$ and the temperature factor $2.594\times10^{-4}\ \text{K}^{-1}$, $E_a/R = 1.468\times10^{4}$ K and $$\boxed{E_a \approx 29.2\ \text{kcal/mol}}.$$
Rate constant at 40 °C. $$k_{40} = k_{25}\exp\!\big[\tfrac{E_a}{R}(\tfrac{1}{298.15}-\tfrac{1}{313.15})\big] = 0.0552\,e^{2.358} = \boxed{0.583\ (\text{mol/L})^{0.5}\text{min}^{-1}}.$$
Integrate forward — and catch the completion time. A half-order reaction reaches $C_A=0$ in a finite time $t_{\text{end}} = 2C_{A0}^{0.5}/k = 2\sqrt2/0.583 = 4.85$ min. Formally solving for $C_A$ at $t=10$ min gives $C_A^{0.5} = \sqrt2 - \tfrac{0.583}{2}(10) = -1.51 < 0$, which is unphysical — the reactant is already exhausted. The rate law is validated only down to $C_A=0.25$ mol/L (reached at $t=3.13$ min); beyond that the model predicts full depletion by 4.85 min. Either way, at 10 min $$\boxed{C_A \approx 0\ \text{(reaction complete)}}.$$
