Answer:
Prob 9.1 Every substance is endowed with a certain amount
of energy and a certain amount of entropy. While the latter is
well defined, the former—the energy—has no absolute value; only
changes in energy can be measured. For this reason (entirely by
convention), the enthalpy of formation of elements in their natural
state is taken as zero.
Consider aluminum and oxygen. In their natural states, their
standard enthalpy of formation (i.e., the energy of formation at
STP) is, as we said, zero. Every kilogram of aluminum has (at
STP) an entropy of 1.05 kJ/K, whereas every kilogram of oxygen
has an entropy of 6.41 kJ/K.
Aluminum burns fiercely forming an oxide (Al2O3) and releasing energy. The standard enthalpy of formation of the oxide
is -1.67 GJ/kmole. The entropy of the oxide is 51.0 kJ/K per
kilomole.
According to the second law of thermodynamics, the entropy
of a closed system suffering any transformation can not diminish.
It can, at best, remain unchanged as the result of the transformation or else, it must increase. If you add up the entropies
of the aluminum and of the oxygen, you will discover that the
sum exceeds the entropy of the oxide formed. This means that
when aluminum combines with oxygen, only part of the chemical
energy can be transformed into electricity, while the rest must
appear as the heat associated with the missing entropy. That
part that can (ideally) be converted into electricity is called the
free energy.
Calculate the free energy of the aluminum/oxygen reaction.
Here are the data required:
MATERIAL ENTROPY STANDARD MOLECULAR
ENTHALPY MASS
OF FORM.
kJ/K GJ/K daltons
per kmole
Aluminum 1.05 per kg 0 26.98
Oxygen 6.41 per kg 0 32.00
AL2O3 51.0 per kmole -1.67
Remember that we defined RTP (reference temperature and pressure)
as a pressure of 1 atmosphere and a temperature of 298.15 K. This differs
from the usual STP (standard temperature and pressure) which is defined
as a pressure of 1 atmosphere and a temperature of 273.15 K.
Solution of Problem 9.1
Explanation:
