- Given the following data:
Fe 2 O 3 (s) + 3 CO(g) →2 Fe(s) + 3 CO 2 (g) ∆H°= −27 kJ/mole
3 Fe 2 O 3 (s) + CO(g) →2 Fe 3 O 4 (s) + CO 2 (g) ∆H°= −61 kJ/mole
Fe 3 O 4 (s) + CO(g) →3 FeO(s) + CO 2 (g) ∆H°= 38 kJ/moleSpecies ∆S°(J ⋅K–1⋅mole–1)
Fe 2 O 3 (s)87.0
CO(g) 190.0
Fe(s)27.0
CO 2 (g) 214.0
Fe 3 O 4 (s) 146.0
FeO(s)61.0Calculate the approximate ∆G°(at 25°C) for the reactionFeO(s) + CO(g) →Fe(s) + CO 2 (g)A. −26 kJ/mole
B. −13 kJ/mole
C. 13 kJ/mole
D. 26 kJ/mole
E. 39 kJ/moleAnswer: BTo solve this problem, use the Gibbs-Helmholtz equation:∆G°= ∆H°−T∆S°Step 1:Solve for ∆H°. Realize that you will have to use Hess’s law to determine ∆H°. Be sure
to multiply through the stepwise equations to achieve the lowest common denominator (6), and
reverse equations where necessary.
sg s g
sg sg
sg sgsgs gFe CO3969
36222
6666
Fe O CO Fe CO
2Fe O CO Fe O CO
FeO CO Fe O COFeO CO32
232422
34 23"""++
++
++++"^ _ ^ _
^ _ ^ _
^ _ ^ _^ _ ^ _h i h i
h i h i
h i h ih i h i8176kJ/molekJ/mole61 kJ/mole- 96 kJ/mole
∆H kJ mole/ kJ mole/
6%=-^96 =- 16
Energy and Spontaneity