Aluminum forms a layer of aluminum oxide when exposed to air which protects the bulk metal from further corrosion. 4 Al(s) + 3 O2(g) LaTeX: \rightarrow → 2 Al2O3(s) Using the thermodynamic data provided below, calculate LaTeX: \Delta ΔS° for this reaction.

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Tringa0 Tringa0 · Answer 1 · 2019-11-26T06:52:48+01:00

Answer:

The ΔS° for this reaction is -626.22 J/K.

Explanation:

[tex]4Al(s)+3O_2\rightarrow 2Al_2O_3(s)[/tex]

The equation used to calculate ΔS° is of a reaction is:

[tex]\Delta S^o_{rxn}=\sum [n\times \Delta S^o_f(product)]-\sum [n\times \Delta S^o_f(reactant)][/tex]

The equation for the enthalpy change of the above reaction is:

[tex]\Delta S^o_{rxn}=(2 mol\times \Delta S^o_f_{(Al_2O_3(s))})-(4 mol\times \Delta S^o_f_{(Al(s))}+3 mol\times \Delta S^o_f_{(O_2(g))})[/tex]

We are given:

[tex]\Delta S^o_f_{(Al(s))}=28.3J/K mol\\\Delta S^o_f_{(O_2(g))}=205.0 J/K mol[/tex]

[tex]\Delta S^o_f_{(Al_2O_3(s))}=50.99 J/K mol[/tex]

Putting values in above equation, we get:

[tex]\Delta S^o_{rxn}=(2 mol\times 50.99 J/K mol)-(4 mol\times 28.3J/K mol-3 mol\times 205.0 J/K mol)=22.5kJ/mol[/tex]

[tex]\Delta S^o_{rxn}=-626.22 J/K[/tex]

The ΔS° for this reaction is -626.22 J/K.