Consider aluminum and oxygen. In their natural states, their standard enthalpy of formation (i.e., the energy...

Consider aluminum and oxygen. In their natural states, their standard enthalpy of formation (i.e., the energy of formation at RTP) is zero. Every kilogram of aluminum has (at RTP) 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/kmol. 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 cannot 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 available energy.

Calculate the available energy of the aluminum/oxygen reaction (MJ/kmol).

Calculate the percentage of available to total energy for this reaction (%).

Note: The molecular weights of Al and O2 are 27 and 32 kg/kmol, respectively.

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genius_generous answered 1 year ago

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