Purpose of Refining, Product Specifications and Benzene–Toluene VLE: 23-Chem-B6 Petroleum Refining and Petrochemicals, December 2015, Q6

Question 6: Purpose of Refining, Product Specifications and Benzene–Toluene VLE (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) Why crude oil is refined, and how (4 marks)

Why. Crude oil as produced is an unusable mixture of thousands of hydrocarbons together with contaminants (sulphur, nitrogen, metals, salt, water) that has essentially no direct end use and does not meet any product specification. Refining converts it into a slate of marketable, on-specification products — LPG, gasoline, jet fuel, diesel, heating and fuel oils, lubricants, asphalt and petrochemical feedstocks — that satisfy performance and environmental limits.

How. In four stages: (1) Separation — desalting, then atmospheric and vacuum distillation into boiling-range cuts; (2) Conversion — cracking (FCC, hydrocracking, coking), catalytic reforming, alkylation and isomerization to upgrade heavy or low-octane streams; (3) Treating/finishing — hydrotreating/desulphurization and sweetening to strip sulphur, nitrogen and metals; and (4) Blending of the finished streams to meet each product’s specifications.

(b) Leading specification features (6 × 1 mark)

Product	Leading specification feature(s)
(i) Gasoline	Octane number (RON/MON/AKI); also volatility (RVP) and sulphur
(ii) Naphtha & kerosene	Naphtha: boiling range / PONA composition (reformer feed). Kerosene/jet: smoke point, flash point, freeze point
(iii) Gas oils (diesel)	Cetane number; also sulphur and cloud/pour point
(iv) Fuel oils	Viscosity; also sulphur, flash point and pour point
(v) Lubricating oils	Viscosity and viscosity index (VI); also pour point and flash point
(vi) Asphalts	Penetration and softening point (ring & ball); also ductility / performance grade

(c) Benzene–toluene vapour–liquid equilibrium at 40 °C

Given. An equimolar liquid ($x_B = x_T = 0.5$) is in equilibrium with its vapour at 40 °C. Antoine constants ($\log_{10} p^* = A - B/(T+C)$, $p^*$ in mmHg, $T$ in °C) are supplied on the paper.

Substance	A	B	C
Benzene (C₆H₆)	6.906	1211.033	220.790
Toluene (C₇H₈)	6.953	1750.286	235.0

Find. (i) the system (total) pressure $P$; (ii) the vapour composition $y_B, y_T$.

Approach. Both components are volatile and chemically similar, so Raoult’s law applies: get each pure-component vapour pressure from Antoine, sum the partial pressures for the total pressure, then take the partial-pressure ratio for the vapour mole fractions.

Pure-component vapour pressures (Antoine at 40 °C).$$\log_{10}p^*_B = 6.906 - \frac{1211.033}{40+220.790} = 2.262 \Rightarrow p^*_B = 182.9\ \text{mmHg},$$$$\log_{10}p^*_T = 6.953 - \frac{1750.286}{40+235.0} = 0.588 \Rightarrow p^*_T = 3.88\ \text{mmHg}.$$
Total pressure by Raoult’s law. $P = x_B p^*_B + x_T p^*_T$:$$P = 0.5(182.9) + 0.5(3.88) = \boxed{93.4\ \text{mmHg}}.$$
Vapour composition. $y_B = x_B p^*_B / P$:$$y_B = \frac{0.5(182.9)}{93.4} = \boxed{0.979}, \qquad y_T = 1 - y_B = 0.021.$$

Check — toluene Antoine constants look mistranscribed

Using the toluene constants exactly as printed gives $p^*_T = 3.9$ mmHg at 40 °C, which is non-physical — toluene’s true vapour pressure at 40 °C is ~59 mmHg (its normal boiling point is 110.6 °C). The benzene and water rows match the standard Felder Table B.4 values exactly, but the toluene $B,C$ (1750.286, 235.0) do not; the correct values are $A=6.9546,\ B=1344.8,\ C=219.48$. With those, $p^*_T = 59.2$ mmHg, giving a more realistic $P = 0.5(182.9)+0.5(59.2)= \mathbf{121\ mmHg}$ and $y_B = 91.4/121 = \mathbf{0.756}$ (so $y_T=0.244$). The boxed answers above are faithful to the numbers on the exam sheet; the values in this box are the physically correct result and are the ones a graded answer should flag.

Quantity	As-printed constants	Corrected toluene constants
$p^*_B$ (40 °C)	182.9 mmHg	182.9 mmHg
$p^*_T$ (40 °C)	3.9 mmHg	59.2 mmHg
(i) System pressure $P$	93.4 mmHg	121 mmHg
(ii) Vapour $y_B$ / $y_T$	0.979 / 0.021	0.756 / 0.244

Concept & Theory

Topic. Ideal (Raoult’s-law) vapour–liquid equilibrium of a nearly ideal binary; the Antoine correlation.

Key relations. Antoine $\log_{10}p^* = A - B/(T+C)$; Raoult’s law $y_i P = x_i p^*_i$; bubble-point pressure $P = \sum_i x_i p^*_i$.

Why this works. Benzene and toluene are chemically similar (both aromatics), so the liquid is close to ideal and each partial pressure is just its mole fraction times its pure vapour pressure. The vapour is always richer in the more volatile component (benzene), which is why $y_B > x_B$.

Common pitfall. Trusting a tabulated constant without a sanity check — here the toluene $p^*$ should be a good fraction of benzene’s, not 2% of it. Always confirm each $p^*$ against a known boiling point before combining them.

Source. Smith, Van Ness & Abbott, Introduction to Chemical Engineering Thermodynamics §10.1–10.4 (Raoult’s law, bubble/dew calculations); Antoine constants per Felder Table B.4.

Question 6 of 6: Purpose of Refining, Product Specifications and Benzene–Toluene VLE

Question 6: Purpose of Refining, Product Specifications and Benzene–Toluene VLE (20 marks — equal value)

(a) Why crude oil is refined, and how (4 marks)

(b) Leading specification features (6 × 1 mark)

(c) Benzene–toluene vapour–liquid equilibrium at 40 °C