General Chemistry Principles, Patterns, and Applications, 2011

glucose and oxygen to the elements (+1273.3 kJ) and for the conversion of the elements to carbon dioxide and water
(−4075.8 kJ):
Equation 5.31
DHocomb = +1273.3 kJ + (-4075.8 kJ) = -2802.5 kJ
This is the same result we obtained using the “products minus reactants” rule () and values. The two results must be
the same because is just a more compact way of describing the thermochemical cycle shown in.
E X A M P L E 5
Long-chain fatty acids such as palmitic acid [CH 3(CH 2) 14CO 2H] are one of the two major sources of energy in our diet.
Use the data in to calculate for the combustion of palmitic acid. Based on the energy released in combustion per
gram, which is the better fuel—glucose or palmitic acid?
Given: compound and values DHo f
Asked for: DHocomb per mole and per gram
Strategy:
A After writing the balanced chemical equation for the reaction, use and the values from to calculate the energy
released by the combustion of 1 mol of palmitic acid.
B Divide this value by the molar mass of palmitic acid to find the energy released from the combustion of 1 g of
palmitic acid. Compare this value with the value calculated infor the combustion of glucose to determine which is the
better fuel.
Solution:
A To determine the energy released by the combustion of palmitic acid, we need to calculate it. As always, the first
requirement is a balanced chemical equation:
C 16 H 32 O 2 (s) + 23O 2 (g) → 16CO 2 (g) + 16H 2 O(l)
Using (“products minus reactants”) with values from (and omitting the physical states of the reactants and products to
save space) gives
DHocomb = SmDHo f products ( ) - SnDHo f reactants ( )
= [16(-393.5 kJ / mol CO2) + 16(-285.8 kJ / mol H 2O]
-[(-891.5 kJ / mol C16H 32O2) + 23(0 kJ / mol O2)]
= -9977.3 kJ / mol
This is the energy released by the combustion of 1 mol of palmitic acid.
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