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

Options

A. `+ 229.6 ("kcal")/("mol")`
B. `-242.3 ("kcal")/("mol")`
C. `-727 ("kcal")/("mol")`
D. `-229.6 ("kcal")/("mol")`

Answer

B

Explanation

Reaction for formation of `Fe_(3)O_(4)` is
`3Fe_((s)) + 2O_(2(g)) rarr Fe_(3)O_(4(s))` …(1)
And we have reactions as follows: `2Fe_((s)) + (3)/(2) O_(2(g)) rarr Fe_(2)O_(3(s))` …(2)
`4Fe_(2)O_(3(s)) + Fe_((s)) rarr 3Fe_(3)O_(4)` ..(3)
Eq (1) `= (Eq(2) xx 4 + Eq (3) xx 1)/(3)`
So, `Delta G = (4 (-177) + (-19))/(3)`
`= -242.3` Kcal/mole