Question 1 of 5: Reversible Hydrodealkylation — Equilibrium Conversion, Adiabatic Rise, Reactor Design

Question 1: Reversible Hydrodealkylation — Equilibrium Conversion, Adiabatic Rise, Reactor Design (25 marks: a 10, b 10, c 5)

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. Gas-phase equilibrium $C_7H_8 + H_2 \leftrightarrow C_6H_6 + CH_4$ at $T = 900$ K, feed ratio $H_2:\text{toluene} = 2:1$, $K_p = 227$, exothermic with $-\Delta H_{rxn} = 50{,}000$ kJ/kmol. Molar heat capacities:

Find. (a) equilibrium conversion $X$ of toluene; (b) adiabatic temperature rise $\Delta T$; (c) a reactor configuration that keeps $\Delta T \le 100$ K.

Approach. The reaction is mole-conserving ($\Delta n = 0$), so $K_p$ equals the mole-fraction quotient and is pressure-independent — solve a single quadratic for $X$; then close an adiabatic energy balance for $\Delta T$, and compare it against the 100 K ceiling to select a staged design.

1. Set the basis and mole table. Take 1 mol toluene and 2 mol $H_2$ (total 3 mol). Because each side of the equation has two molecules, the total moles stay at 3 for any conversion $X$: toluene $=1-X$, hydrogen $=2-X$, benzene $=X$, methane $=X$.
2. Write the equilibrium relation. With $\Delta n = 0$ the pressure and the total-mole factor cancel, so $K_p = K_y = \dfrac{y_{C_6H_6}\,y_{CH_4}}{y_{C_7H_8}\,y_{H_2}} = \dfrac{X^2}{(1-X)(2-X)}.$ Setting this equal to 227 gives $$\frac{X^{2}}{(1-X)(2-X)} = 227.$$
3. Solve the quadratic. Expanding, $(227-1)X^{2} - 3(227)X + 2(227) = 0$, i.e. $226X^{2} - 681X + 454 = 0$. The physically admissible root ($X<1$) is $$X = \frac{681-\sqrt{681^{2}-4(226)(454)}}{2(226)} = \boxed{0.996.}$$ The large excess of hydrogen and the large $K_p$ drive toluene almost to extinction.
4. Release of reaction heat (part b). Per kmol of toluene fed, the heat liberated at equilibrium is $Q = (-\Delta H_{rxn})\,X = (50{,}000)(0.996) = 49{,}800$ kJ. This heat has nowhere to go in an adiabatic reactor, so it raises the temperature of the exit stream.
5. Choose the enthalpy path consistently. Because $-\Delta H$ is quoted at 900 K, which is the outlet temperature, take the path: heat the feed from $T_0$ to 900 K, then react at 900 K. The sensible heat is therefore carried by the feed mixture (per kmol toluene fed): $$\sum \theta_i C_{p,i} = 240 + 2(30) = 300\ \tfrac{\text{kJ}}{\text{K}}.$$ (The exit-stream sum, $(1-X)240 + (2-X)30 + X(198+67) = 295$ kJ/K $= 300 + X\Delta C_p$ with $\Delta C_p = -5$, belongs with $\Delta H$ quoted at the inlet temperature.)
6. Adiabatic temperature rise. $$\Delta T = \frac{(-\Delta H_{rxn})_{900}X}{\sum \theta_i C_{p,i}} = \frac{49{,}800}{300} = \boxed{166\ \text{K}.}$$ The inlet is thus about $900-166 = 734$ K. Using the exit-stream heat capacity instead gives $49{,}800/295 = 169$ K, a common shortcut within 2% — the conclusion for part (c) is the same either way.

(c) Reactor design for a 100 K ceiling. A single adiabatic bed would rise $\approx166$ K — well above the 100 K limit — so the heat must be removed in stages. Since $\Delta T = (-\Delta H)X/\sum\theta_i C_{p,i}$, the conversion allowed in one adiabatic bed before hitting 100 K is $X_{bed} \le \dfrac{(100)(300)}{50{,}000} = 0.60$. A practical answer is therefore a multi-bed adiabatic reactor with interstage cooling (two beds suffice: bed 1 taken to $X\approx0.6$ with a $\approx100$ K rise, the stream cooled back in an intercooler, then bed 2 completing to $X\approx0.996$). Equivalent industrial options are cold-shot (quench) reactors that inject cold feed between beds, or a multitubular cooled reactor with molten-salt coolant. Alternatively the thermal mass can be raised so a single pass stays under 100 K: $\sum nC_p$ must reach $50{,}000/100 = 500$ kJ/K, i.e. about 6.7 kmol extra $H_2$ (or inert) per kmol toluene — a feed ratio near 8.7:1 — at the cost of larger downstream separation.