At what temperature, Teq, do the forward and reverse corrosion reactions occur in equilibrium? The chemical reaction that causes aluminum to corrode in air is given by
4Al+3O2?2Al2O3
in which at 298 K
?Hrxn? = ?3352 kJ
?Srxn = ?625.1 J/K
What is the standard Gibbs free energy for this reaction? Assume the commonly used standard reference temperature of 298 K?
What is the Gibbs free energy for this reaction at 5975K ? Assume that ?H and ?S do not change with temperature.
At what temperature, Teq, do the forward and reverse corrosion reactions occur in equilibrium?

See explanation below.

Explanation:

We can obtain the Gibb's free energy from the formula;

∆G= ∆H - T∆S

Where;

∆G = change in free energy= the unknown

∆H= change in enthalpy = 3352 kJ

∆S= change in entropy of the solution= 625.1 J/K

T= absolute temperature = 298 K

Substituting values;

∆G= 3352 ×10^3 J - (298 K × 625.1 J/K)

∆G= 3352 ×10^3 J - 186279.8

∆G= 3.16 × 10^6 J

At 5975K,

∆G= ∆H - T∆S

∆G= 3352 ×10^3 J - (5975K × 625.1 J/K)

∆G= 3.352 ×10^6 J - 3.735 × 10^6

∆G= -3.83×10^5 J

At equilibrium, ∆G=0, Teq is given by;

0= 3352 ×10^3 J - (Teq × 625.1 J/K)

0= 3352 ×10^3 - 625.1Teq

625.1Teq = 3352 ×10^3

Teq= 3352 ×10^3/625.1

Teq= 5362.3 K

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explanation: