Question 1 of 6: Hydrogen Production, Reactions and a Benzene-Hydrogenation Recycle

Question 1: Hydrogen Production, Reactions and a Benzene-Hydrogenation Recycle (20 marks — equal value)

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.

(a) Hydrogen production in refineries — processes, reactions and conditions

(i)–(iii) Steam methane reforming (SMR). The dominant on-purpose route. Natural gas (or refinery light ends), after hydrodesulphurisation, is reacted with steam over a nickel catalyst in externally-fired tubes. The strongly endothermic reforming reaction is followed by the mildly exothermic water–gas shift, which converts the CO to additional hydrogen:

$$CH_4 + H_2O \rightleftharpoons CO + 3H_2 \qquad (\Delta H^\circ_{298} = +206\ \text{kJ/mol})$$ $$CO + H_2O \rightleftharpoons CO_2 + H_2 \qquad (\Delta H^\circ_{298} = -41\ \text{kJ/mol})$$

Typical conditions: reformer 700–900 °C and 15–30 bar over Ni/Al₂O₃; high-temperature shift ~350 °C (Fe–Cr) followed by low-temperature shift ~200 °C (Cu–Zn); the hydrogen is finally purified in a pressure-swing adsorption (PSA) unit.

Catalytic reforming (as a hydrogen by-product source). Naphtha reforming, run primarily to raise gasoline octane, also produces large volumes of hydrogen because its principal reactions are dehydrogenations. Naphthenes are dehydrogenated to aromatics, releasing three moles of hydrogen per ring:

$$C_6H_{12}\ (\text{cyclohexane}) \rightarrow C_6H_6\ (\text{benzene}) + 3H_2$$

Typical conditions: 480–525 °C and 5–25 bar over a bifunctional Pt (or Pt–Re) on chlorided-alumina catalyst. Partial oxidation / gasification of heavy residues ($C_nH_m + \tfrac{n}{2}O_2 \rightarrow nCO + \tfrac{m}{2}H_2$, followed by shift) is a third route used where cheap heavy feed is available. Any two of these, with their equations and conditions, answer part (a).

(b) Recycle-to-fresh-feed ratio for benzene hydrogenation

Given. Reaction $C_6H_6 + 3H_2 \rightarrow C_6H_{12}$; overall benzene conversion 95%; single-pass (per-reactor) conversion 25%; fresh feed carries 20% excess hydrogen; the recycle is 30 mol% benzene and 70 mol% hydrogen. The block flow is fresh feed → mixer → reactor → separator, with unreacted material recycled.

Find. The ratio of the recycle stream to the total fresh feed stream, $R/F_{\text{fresh}}$.

Approach. Take a basis of 100 mol fresh benzene, fix the fresh hydrogen from the 20% excess, use the overall conversion to fix the moles of benzene reacted (all in the reactor at steady state), then apply the single-pass conversion on the combined reactor-inlet benzene to solve for the recycle.

1. Fresh feed on a 100-mol benzene basis. Stoichiometric hydrogen for the fresh benzene is $3\times100=300$ mol; 20% excess gives $$\dot n_{\text{H}_2,\text{fresh}} = 1.20\times300 = 360\ \text{mol},\qquad F_{\text{fresh}} = 100 + 360 = \boxed{460\ \text{mol}}.$$
2. Benzene reacted overall. At 95% overall conversion, and since reaction occurs only in the reactor, the moles of benzene converted per pass through the reactor equal the overall amount reacted: $$n_{\text{reacted}} = 0.95\times100 = 95\ \text{mol}.$$ Only 5 mol benzene leaves with the net product, consistent with 95% overall conversion.
3. Reactor-inlet benzene from the single-pass conversion. The reactor inlet benzene is the fresh benzene plus the recycled benzene, $100 + 0.30R$. A single-pass conversion of 25% acting on that stream must supply the 95 mol reacted: $$0.25\,(100 + 0.30R) = 95.$$
4. Solve for the recycle. $$100 + 0.30R = \frac{95}{0.25} = 380 \;\Rightarrow\; 0.30R = 280 \;\Rightarrow\; \boxed{R = 933.3\ \text{mol}}.$$ Check: recycled benzene $=0.30\times933.3=280$ mol, so reactor-inlet benzene $=380$ mol, of which 25% (95 mol) reacts and 285 mol leaves — the separator returns 280 mol to the loop and passes 5 mol to product, exactly closing the benzene balance.
5. Recycle-to-fresh-feed ratio. $$\frac{R}{F_{\text{fresh}}} = \frac{933.3}{460} = \boxed{2.03}.$$