Question

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.

Answer 1

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

Step-by-step explanation:

`4Al(s)+3O_2\rightarrow 2Al_2O_3(s)`

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

`\Delta S^o_(rxn)=\sum [n* \Delta S^o_f(product)]-\sum [n* \Delta S^o_f(reactant)]`

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

`\Delta S^o_(rxn)=(2 mol* \Delta S^o_f_((Al_2O_3(s))))-(4 mol* \Delta S^o_f_((Al(s)))+3 mol* \Delta S^o_f_((O_2(g))))`

We are given:

`\Delta S^o_f_((Al(s)))=28.3J/K mol\\\Delta S^o_f_((O_2(g)))=205.0 J/K mol`

`\Delta S^o_f_((Al_2O_3(s)))=50.99 J/K mol`

Putting values in above equation, we get:

`\Delta S^o_(rxn)=(2 mol* 50.99 J/K mol)-(4 mol* 28.3J/K mol-3 mol* 205.0 J/K mol)=22.5kJ/mol`

`\Delta S^o_(rxn)=-626.22 J/K`

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