Optimum Temperature for a Reversible Exothermic Isomerization: 23-Chem-A4 Chemical Reactor Engineering, May 2014, Q2

Question 2: Optimum Temperature for a Reversible Exothermic Isomerization (20 marks: a 7, b 7, c 6)

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. Reversible first-order/first-order $A\rightleftharpoons B$, forward $E_1=6500$ cal/mol, $-\Delta H=4500$ cal/mol (exothermic). $R=1.98$ cal/(mol·K). $\Delta G^\circ_{298}=-710$ cal fixes the equilibrium constant at 298 K. Feed $A{:}B=2{:}1$, required product $A{:}B=1{:}2$.

Quantity	Value
Forward activation energy $E_1$	6500 cal/mol
Reverse activation energy $E_2=E_1-\Delta H$	11000 cal/mol
$K_{298}=\exp(-\Delta G^\circ/RT_0)$	3.33
Pre-exponential ratio $A_1/A_2$	$1.62\times10^{-3}$
Feed / product molar ratio $A{:}B$	2:1 → 1:2

Find. (a) $T_{opt}(X)$ and the single optimal CSTR temperature; (b) the bracketing limits for the best isothermal PFR temperature; (c) the initial and final temperatures of the optimal batch programme.

Figure 2 — The rate-maximising temperature falls monotonically as A is converted. A CSTR sits at the single exit point (297 K); an isothermal PFR’s best temperature lies between the inlet and exit optima; the batch programme tracks the whole curve from 363 K down to 297 K.

Approach. Write the net rate for the reversible reaction, set $\partial(-r_A)/\partial T=0$ at fixed conversion to obtain the locus of maximum rate, then evaluate it at the compositions that each reactor type “sees.”

Reverse-reaction parameters from thermodynamics. Since $K=k_1/k_2$ and $K=\exp(-\Delta G^\circ/RT)$, the reverse activation energy is $E_2=E_1-\Delta H=6500-(-4500)=11000$ cal/mol. From $K_{298}=\exp(710/(1.98\cdot298))=3.33$ and $K=(A_1/A_2)\exp(-\Delta H/RT)$, the pre-exponential ratio is $A_1/A_2=K_{298}/\exp(4500/(1.98\cdot298))=1.62\times10^{-3}$.
Rate and the maximum-rate locus (part a derivation). The net forward rate is $-r_A=k_1C_A-k_2C_B$. Setting $\partial(-r_A)/\partial T=0$ at fixed composition gives $E_1k_1C_A=E_2k_2C_B$, i.e. the optimum occurs where$$K(T_{opt})=\frac{k_1}{k_2}=\frac{C_B}{C_A}\,\frac{E_2}{E_1}.$$Because $K(T)=(A_1/A_2)\exp(-\Delta H/RT)$, this inverts to $T_{opt}=\dfrac{-\Delta H}{R\,\ln\!\big[K_{opt}/(A_1/A_2)\big]}$, a function of the local composition $C_B/C_A$ only. In terms of the degree of conversion $X$ of A, the 2:1 feed gives $C_A=C_{A0}(1-X)$ and $C_B=C_{A0}(0.5+X)$, so$$T_{opt}(X)=\frac{-\Delta H}{R\,\ln\!\Big[\dfrac{E_2}{E_1}\,\dfrac{0.5+X}{(1-X)\,(A_1/A_2)}\Big]},$$which falls from 363 K at $X=0$ to 297 K at $X=0.5$, where the required 1:2 product ratio is reached.
Optimal CSTR temperature (part a result). A CSTR reacts entirely at its exit composition, the required product ratio $A{:}B=1{:}2$ ($C_B/C_A=2$): $K_{opt}=2\,(11000/6500)=3.39$, so$$T_{opt,\text{CSTR}}=\frac{4500}{1.98\,\ln(3.39/1.62\times10^{-3})}=\boxed{297\ \text{K}}\;(\approx24\,{}^{\circ}\text{C}).$$
Isothermal PFR limits (part b). A PFR runs through every composition from the inlet ($A{:}B=2{:}1$, $C_B/C_A=0.5$) to the exit ($1{:}2$, $C_B/C_A=2$). Evaluating the same locus at the two ends: inlet $K_{opt}=0.5(11000/6500)=0.846\Rightarrow T=363$ K; exit $\Rightarrow T=297$ K. A single isothermal temperature cannot be optimal everywhere, so the best fixed PFR temperature must lie between 297 K and 363 K.
Batch temperature programme (part c). A temperature-programmed batch reactor can follow the maximum-rate locus exactly: start hot where little B is present and cool as B accumulates. Hence begin at the inlet optimum $\boxed{T_i\approx363\ \text{K}}$ and finish at the exit optimum $\boxed{T_f\approx297\ \text{K}}$ — a monotonically declining temperature schedule, as Figure 2 shows.

Reactor	Optimal temperature
(a) CSTR (at exit comp 1:2)	297 K
(b) Isothermal PFR — limits	between 297 K and 363 K
(c) Batch programme (initial → final)	363 K → 297 K (declining)

Concept & Theory

Topic. Optimum temperature progression for a reversible exothermic reaction.

Key relations. Net rate $-r_A=k_1C_A-k_2C_B$; maximum-rate locus $K_{opt}=(C_B/C_A)(E_2/E_1)$; van’t Hoff / Arrhenius link $K=(A_1/A_2)e^{-\Delta H/RT}$; endothermic-reverse constraint $E_2=E_1-\Delta H$.

Why this works. Raising $T$ speeds both directions, but for an exothermic reaction it speeds the reverse more ($E_2>E_1$), so at each conversion there is a temperature that maximises the net forward rate. A CSTR operates at one composition (its exit), giving one optimal $T$; a PFR spans a range, so the best single temperature is bracketed by the end-point optima; a batch reactor can be programmed to ride the locus down.

Common pitfall. Forgetting the $E_2/E_1$ factor (using the plain equilibrium $K=C_B/C_A$), or using the inlet composition for the CSTR — a CSTR sees only its exit stream.

Source. Fogler, Elements of Chemical Reaction Engineering §12 (steady-state nonisothermal design; optimum temperature); Levenspiel, Chemical Reaction Engineering §9 (optimum temperature progression).

Question 2 of 5: Optimum Temperature for a Reversible Exothermic Isomerization

Question 2: Optimum Temperature for a Reversible Exothermic Isomerization (20 marks: a 7, b 7, c 6)