Adiabatic Batch Hydrolysis of Acetic Anhydride — Time for 80% Conversion: 23-Chem-A4 Chemical Reactor Engineering, May 2017, Q4

Question 4: Adiabatic Batch Hydrolysis of Acetic Anhydride — Time for 80% Conversion (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Pseudo first-order irreversible exothermic hydrolysis; adiabatic well-mixed batch.

Quantity	Value
$k$ vs $T$ (°C, s$^{-1}\times10^3$)	(15, 1.34), (20, 1.88), (25, 2.63), (30, 3.51)
Initial charge $C_{A0}$, $T_0$	300 mol/m³, 15 °C
Specific heat $c_p$, density $\rho$	3.8 kJ/kg·K, 1070 kg/m³
Heat of reaction (interpreted)	$-\Delta H_r=210$ MJ/kmol $=210$ kJ/mol anhydride (printed “210 MJ/kg”; see callout)
Target conversion	$X=0.80$

Find. The batch time to reach 80% conversion under adiabatic operation (temperature free to rise).

Figure 4.1 — Arrhenius plot of the pseudo first-order constant; the four data lie on a straight $\ln k$ vs $1/T$ line giving $E_a=46.9$ kJ/mol.

Approach. Fit the supplied $k(T)$ to Arrhenius form, close an adiabatic energy balance to express temperature as a linear function of conversion, then integrate the batch design equation numerically along that rising-temperature path.

Check (heat-of-reaction units): the source prints “210 MJ/kg”, which cannot be meant literally — with $M_A=0.102$ kg/mol it would drive an adiabatic rise of $\sim$1580 K. Assumption: the unit is read as 210 MJ/kmol $=$ 210 kJ/mol (a kg↔kmol slip). This is the reading the rest of the data support: it gives $\Delta T_\text{ad}\approx15.5$ K, so the mixture runs from 15 °C to about 27 °C at 80% conversion, which is exactly the 15–30 °C span over which $k$ is tabulated. (The alternative reading, 210 kJ/kg, would give a rise of only 1.6 K and $t\approx1133$ s, making the temperature data nearly redundant.) Per exam instruction #1, this assumption is stated explicitly.

Fit the Arrhenius line. Regressing $\ln k$ on $1/T$ (K) gives slope $-E_a/R=-5635$ K, hence $$E_a=(5635)(8.314)=\boxed{46.9\ \text{kJ/mol}},\qquad A=4.2\times10^{5}\ \text{s}^{-1},$$ so $k(T)=4.2\times10^{5}\exp(-5635/T)$ reproduces all four tabulated values within 3%.
Adiabatic operating line. With constant properties the batch energy balance $\rho c_p\,dT=(-\Delta H_r)(-dC_A)$ integrates to a straight $T$–$X$ relation $$T=T_0+\Delta T_\text{ad}\,X,\qquad \Delta T_\text{ad}=\frac{(-\Delta H_r)\,C_{A0}}{\rho c_p}.$$ Numerically (J/mol, mol/m³, kg/m³, J/kg·K) $$\Delta T_\text{ad}=\frac{(210{,}000)(300)}{(1070)(3800)}=\boxed{15.5\ \text{K}.}$$ At $X=0.8$ the mixture has warmed by $0.8\times15.5=12.4$ K, from 15 °C to about 27.4 °C — inside the tabulated range, so $k(T)$ is an interpolation throughout.
Batch design equation. For first order, $\dfrac{dX}{dt}=k(T)\,(1-X)$, so $$t=\int_0^{0.80}\frac{dX}{k\big(T(X)\big)\,(1-X)},\qquad T(X)=288.15+15.5\,X.$$
Integrate. Carrying $k(T)$ along the rising adiabatic line ($k$ climbs from $1.35\times10^{-3}$ at $X=0$ to $3.02\times10^{-3}$ s$^{-1}$ at $X=0.8$) and integrating numerically (trapezoidal rule, fine grid; interpolating the table directly instead of using the Arrhenius fit gives 733 s), $$t_\text{adiabatic}=\boxed{736\ \text{s}\approx12.3\ \text{min}.}$$
Isothermal benchmark. Had the reactor been held at the 15 °C charge temperature, $t=\dfrac{\ln[1/(1-0.8)]}{k(15\,{}^\circ\text{C})}=\dfrac{\ln5}{1.34\times10^{-3}}=1201\ \text{s}=20.0\ \text{min}.$ The 12 K self-heating more than doubles $k$ by the end of the run and cuts the batch time by about 39% relative to isothermal operation.

Quantity	Value
Activation energy $E_a$	46.9 kJ/mol
Pre-exponential $A$	$4.2\times10^{5}$ s$^{-1}$
Adiabatic temperature rise (full conversion)	15.5 K
Time for 80% conversion (adiabatic)	736 s ≈ 12.3 min
Isothermal benchmark at 15 °C	1201 s ≈ 20.0 min

Concept & Theory

Topic. Non-isothermal batch reactor; coupled mole and energy balances with Arrhenius kinetics.

Key relations. Arrhenius $k=A\,e^{-E_a/RT}$; adiabatic operating line $T=T_0+\Delta T_\text{ad}X$ with $\Delta T_\text{ad}=(-\Delta H_r)C_{A0}/(\rho c_p)$; batch design $t=\int_0^X dX/[k(T)(1-X)]$.

Why this works. In an adiabatic batch every unit of conversion releases a fixed quantity of heat that has nowhere to go but the mixture, so temperature is an exact linear function of conversion — the two balances decouple into “$T$ from $X$” and “time from $X$.” Here the 15.5 K adiabatic rise lifts the temperature through the whole tabulated range, so the rate constant more than doubles and the batch finishes well ahead of the isothermal benchmark.

Common pitfall. Accepting the printed “MJ/kg” heat of reaction uncritically (a 1580 K rise is unphysical); the molar reading is the one consistent with the 15–30 °C rate data, or evaluating $k$ only at the initial temperature when the balance actually calls for $k$ along the operating line.

Source. Fogler, Elements of Chemical Reaction Engineering, Ch. 11–13 (energy balances, non-isothermal reactors); Levenspiel, Ch. 9.

Question 4 of 5: Adiabatic Batch Hydrolysis of Acetic Anhydride — Time for 80% Conversion

Question 4: Adiabatic Batch Hydrolysis of Acetic Anhydride — Time for 80% Conversion (25 marks)