Part 8 : THERMODYNAMICS

The question can be answered by considering the following three equations that can be use to create a "triangle" that relates the alkanes, their component elements and their combustion products.

Reaction
Heat of
Eqn
C8H18 ----------> 8 CO2 + 9 H2O Combustion (known), Hc(alkane)
1
8 C + 9 H2 -----> 8 CO2 + 9 H2O Combustion (calculate), Hc(elements)
2
8 C + 9 H2 ----------> C8H18 Formation (required) Hf(alkane)
3

From the "triangle" we get that Hf (alkane) = Hc(elements) - Hc(alkane)

Hc(elements)= 8 x Hc(C) + 9 x Hc(H2)

Hc(elements)= 8 x -94.05 + 9 x -57.8 = -1272.6 kcal/mol.

So for 2,2-dimethylhexane, Hf = -1272.6 - - 1304.6 = 32 kcal/mol

and for 2,2,3,3-tetramethylbutane, Hf = -1272.6 - -1303.0 = 30.4 kcal/mol

2,2,3,3-tetramethylbutane is the more stable. This is evident since it has the less exothermic Hc, the most exothermic Hf and is the most branched of the two hydrocarbons.