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.
|
|
Eqn |
C8H18 ----------> 8 CO2 + 9 H2O | Combustion (known), Hc(alkane) |
|
8 C + 9 H2 -----> 8 CO2 + 9 H2O | Combustion (calculate), Hc(elements) |
|
8 C + 9 H2 ----------> C8H18 | Formation (required) Hf(alkane) |
|
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.