(i) 2Fe2O3(s)→4Fe(s)+3O2(g);∆rG0=+1487.0kJmol−1 (ii) 2CO(g)+O2(g)→2CO2(g);∆rG0=−514.4kJmol−1 Multiply above reaction with 3 (iii) 6CO(g)+3O2(g)→6CO2(g); ∆rG0=3×−514.4=−1543.2kJmol−1 When we add reaction (i) and reaction (iii), we get reaction (iv) (iv) 2Fe2O3(s)+6CO(g)→4Fe(s)+6CO2(g) Free energy change, ∆rG0 for the reaction will be, 1487.0−1543.2=−56.2kJmol−1