Calculate ΔG° for the reduction of the oxides of iron and copper by carbon at 650K, represented by the equations:
a) 2Fe_2 O_3 (s) + 3C (graphite) ---> 4Fe (s) + 3CO_2 (g)
b) 2CuO (s) + C (graphite) ---> 2Cu (s) + CO_2 (g)
Values of ΔG°f at 700 K are -92 kJ/mol for CuO (s), -632 kJ/mol for Fe_2 O_3 (s), and -395 kJ/mol for CO_2 (g).
1 Answer | Add Yours
We know for any reaction, ΔG°reaction = ΔG°(f)products – ΔG°(f)reactants
a) For the first reaction, at 700K,
ΔG° = 3*ΔG°(f)CO2 + 4*ΔG°(f)Fe– 2*ΔG°(f)Fe2O3– 3*ΔG°(f)C
= 3*(-395) + 4*0 – 2*(-632) – 3*0 (since ΔG°(f) of all pure elements, in their standard state is zero).
= -1185+1264 = 79 kJ/mol.
By similar methods,
b) For the second reaction, at 700K,
ΔG° = ΔG°(f)CO2 + 2*ΔG°(f)Cu– 2*ΔG°(f)CuO– ΔG°(f)C
= (-395) + 2*0 – 2*(-92) – 1*0
= -395+184 = -211 kJ/mol.
Join to answer this question
Join a community of thousands of dedicated teachers and students.Join eNotes