Questions
1) List the reactants and product of the fuel cell.
Renctants →
and
uct

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: