23-Chem-A4 Chemical Reactor Engineering · May 2015
Question 5 of 5: Catalytic Mixed Reactor — Determining the Rate Equation
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams / EGBC — May 2015 — 04-Chem-A4 Chemical Reactor Engineering. Three-hour open-book exam; the designated Fogler textbook (any edition), 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 20 marks); all five are solved below for completeness. No credit is given for re-deriving standard rate expressions, so the batch / MFR / PFR design equations are quoted and applied. Property look-ups not printed on the paper (molar masses, the gas constant) are stated explicitly in each Given block as permitted open-book references.
Reference texts: H. S. Fogler, Elements of Chemical Reaction Engineering (4th/5th ed., Prentice Hall) — the designated open-book text: batch/CSTR/PFR design equations, the stoichiometric table with expansion factor $\varepsilon$ for gas-phase reactions with a change in moles, and reversible-reaction kinetics; O. Levenspiel, Chemical Reaction Engineering (3rd ed., Wiley) — non-ideal flow (dead-zone / bypass models), the dispersion and tanks-in-series RTD models, and rate-equation determination from a differential (mixed) catalytic reactor; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.).
Given. Catalytic $A\rightarrow3R$ in a gradientless (well-mixed) reactor, $W=1$ g catalyst, $P=8$ atm, $T=700\,{}^{\circ}\text{C}=973.15$ K, pure-A feed ($y_{A0}=1$). Because 1 mol A makes 3 mol R, the mole change is $\delta=2$ and $\varepsilon_A=y_{A0}\delta=2$. Five $(v_0,\,p_A/p_{A0})$ data pairs are supplied.
$v_0$ (L/hr)
100
22
4
1
0.6
$p_A/p_{A0}=r$
0.8
0.5
0.2
0.1
0.05
Find. A rate equation $-r_A'$ (per gram of catalyst) — its reaction order and rate constant — that fits the five runs.
Figure 5 — Log–log plot of the mixed-reactor rate $-r_A'$ against exit concentration $C_A$; the near-unit slope (1.04) identifies a first-order rate law, and the intercept fixes the constant.
Approach. A gradientless reactor is a differential-CSTR: each run gives one rate directly from the mass balance. Convert the exit pressure ratio to conversion (accounting for the mole change), compute each rate and exit concentration, then take a log–log fit to read the order and constant.
Exit conversion from the pressure ratio. For $A\rightarrow3R$ with $\varepsilon_A=2$, the mole fraction of A is $y_A=\dfrac{1-X}{1+2X}=r.$ Solving, $X=\dfrac{1-r}{1+2r}.$ For example $r=0.5\Rightarrow X=0.25$ and $r=0.2\Rightarrow X=0.571.$
Rate from the mixed-reactor (differential-CSTR) balance. The catalytic CSTR balance is $\dfrac{W}{F_{A0}}=\dfrac{X}{-r_A'}$, so $-r_A'=\dfrac{F_{A0}X}{W}$ with $F_{A0}=C_{A0}v_0$ and $C_{A0}=P/RT=8/(0.082057\times973.15)=0.1002$ mol/L. Thus $-r_A'=C_{A0}v_0X$ [mol/(g·hr)] — e.g. at $v_0=100$, $-r_A'=0.1002(100)(0.0769)=0.771.$
Exit concentration for each run. $C_A=p_A/RT=rP/RT=r\,C_{A0}$, so $C_A$ ranges from $0.080$ (run 1) down to $0.0050$ mol/L (run 5). Tabulating $(C_A,\,-r_A')$ gives a rate that falls smoothly with $C_A$.
Order and constant by log–log regression. Plotting $\ln(-r_A')$ vs $\ln C_A$ (Figure 5) is a straight line of slope $n=1.04\approx1$ and intercept giving $k=11.4$ L/(g·hr). The reaction is therefore first order:$$\boxed{-r_A'=11.4\,C_A\quad[\text{mol}/(\text{g}\cdot\text{hr}),\ C_A\text{ in mol/L}]}.$$
Pressure form of the rate law. Since $C_A=p_A/RT$, the same law in partial-pressure units is $-r_A'=k'p_A$ with $k'=k/RT=11.4/(0.082057\times973.15)=\boxed{0.142\ \text{mol}/(\text{g}\cdot\text{hr}\cdot\text{atm})}.$ Either form represents the data on 3-mm catalyst.
