1.

If DeltaG = -177K cal for ""(1)2Fe(s)+(3)/(2)O_(2)(g) rarr Fe_(2)O_(3)(s) and DeltaG =- 19 K cal for ""(2)4Fe_(2)O_(3)(s)+ Fe(s) rarr 3Fe_(3)O_(4)(s) What is the Gibbs free energy of formation fo Fe_(3)O_(4)(s)?

Answer»

`+229.6` Kcal/mol
`-242.3` Kcal/mol
`-727` Kcal/mol
`-229.6` Kcal/mol

Solution :`DeltaG^(@)` for `3Fe(s) + 2O_(2)(g) rarr Fe_(3)O_(4)(s)` can be obtained by taking
`[(2) + 4 xx (1)] xx (1)/(3)`
Hence , we GET `DeltaG_(f) =[-19 + 4 xx (-177)] xx (1)/(3)=-242.3 K " cal for " 1 mol Fe_(3)O_(4)`


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