C102B
Problem of the Week (ANSWERS)
27–31 January, 2003
Chemistry of Petroleum Cracking
Crude oil is a mixture of thousands of individual chemical compounds, ranging from very low molecular
weight hydrocarbons such as propane (C3H8) to hydrocarbons with 30 or
more carbon atoms. At the turn of the last century, chemists developed the
refinery practice of “cracking”; i.e., heating the crude oil to very high temperatures in the absence of air to produce additional amounts of the lower
n-octane; mw = 114.2 g mol–1
boiling fractions, the part we now call gasoline.
1. Last week you calculated H° for the combustion of octane, an important component of gasoline (i.e.,
C8H18(g) + 121⁄2O2(g) Æ 8CO2(g) + 9H2O(g); ∆H° = –5470 kJ mol–1). Another important constituent of
gasoline is n-hexane, C6H14 (86.2 g mol–1), which on combustion releases –4170 kJ mol–1 of heat. Octane
might seem to be more energy producing than hexane, but in practice, fuels
are sold and used by weight or volume, not by mole. Which hydrocarbon produces the more energy per gram? How do they compare on a per carbon
atom basis (i.e., H/n, where n = 6 for hexane, n = 8 for octane?) What do your
n-hexane; mw = 86.2 g mol–1
results say about the amount of heat available from the combustion of hydrocarbons?
Octane: –5470 kJ mol–1/114.2 g mol–1 = 47.9 kJ g–1
Hexane: –4170 kJ mol–1/86.2 g mol–1 = 48.4 kJ g–1 (they are very similar on an energy/g basis)
Octane: –5470 kJ mol–1/8 carbon unit = 684 kJ/carbon unit
Hexane: –4170 kJ mol–1/6 carbon unit = 695 kJ/carbon unit
The energy from burning long chain hydrocarbons is roughly constant when measured on a
per gram basis. This stems from the fact that about ~690 kJ are evolved per CHx unit (the
presence of both CH2 and CH3 groups makes the numerical value inexact).
2. n-Hexane is only one member of a family of compounds with the formula C6H14.
Such six-carbon molecules can be found in a number of isomeric forms, and the
early fuel engineers learned that “branched chain” versions displayed improved fuel
performance. Is it possible to convert n-hexane to one of its branched isomers, 2,2dimethylbutane, by simple cracking? Calculate ∆G° for the reaction: hexane (∆Hf° =
–167 kJ/mol; S° = +389 J/mol-K) Æ 2,2-dimethylbutane (∆Hf° = –186 kJ/mol; S° =
+359 J/mol-K) at 25 °C.
2,2-dimethylbutane
∆H° = –186 kJ/mol – (–167 kJ/mol) = –19 kJ/mol; S° = 359 J/mol-K – 389 J/mol-K = –30 J/mol-K
∆G° = ∆H° – T∆S° = –19 kJ/mol – (298K)(–0.030 kJ/mol-K) = –10 kJ/mol (slightly spontaneous)
3. Reactions at room temperature can sometimes be too slow to be of commercial use. Petroleum cracking is
therefore carried out at much higher temperatures. What is ∆G° for the hexane Æ 2,2-dimethylbutane reaction at 400 °C?
assuming ∆H and ∆S don’t change appreciably with temperature,
∆G° = ∆H° – T∆S° = –19 kJ/mol – (673K)(–0.030 kJ/mol-K) = +1 kJ/mol (almost at equilibrium;
in any case, high temperatures do not improve the thermodynamics of cracking)
4. As your results from #3 suggest, high-temperature cracking may not be a feasible way to produce
branched hydrocarbons. In fact, hexane will break down to form smaller molecules such as propane (H3CCH2-CH3; ∆Hf° = –105 kJ/mol; S° = +270 J/mol-K) and propene (H3C-CH=CH2); ∆Hf° = +20.4 kJ/mol; S° = +267
J/mol-K). Calculate ∆G° for the reaction hexane Æ (propane + propene) at 25 °C.
∆H° = (–105 + 20.4)kJ/mol – (–167 kJ/mol) = 82.4 kJ/mol
∆S° = (270 + 267) J/mol-K – 389 J/mol-K = 148 J/mol-K
∆G° = ∆H° – T∆S° = 82.4 kJ/mol – (298K)(0.148 kJ/mol-K) = +38 kJ/mol (nonspontaneous)
5. Repeat the G° calculation in #4 at the more realistic cracking temperature of 400 °C. Is the reaction feasible at such temperatures?
assuming H and S don’t change appreciably with temperature,
∆G° = ∆H° – T∆S° = 82.4 kJ/mol – (673K)(0.148 kJ/mol-K) = –17 kJ/mol (spontaneous; high
temperatures help)