If `DeltaG = -177`K 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)`?A. `+229.6` Kcal/molB. `-242.3` Kcal/molC. `-727` Kcal/molD. `-229.6` Kcal/mol

If `DeltaG = -177`K cal for `" "(1)2Fe(s)+(3)/(2)O_(2)(g) rarr Fe_(2)O

1.	If `DeltaG = -177`K 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)`?A. `+229.6` Kcal/molB. `-242.3` Kcal/molC. `-727` Kcal/molD. `-229.6` Kcal/mol
Answer» Correct Answer - B `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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