One reaction involved in the conversion of iron ore to themetal is FeO(s)+CO(g)-

Question

One reaction involved in the conversion of iron ore to themetal is FeO(s)+CO(g)------>Fe(s)+CO2(g) Calculate the standard enthalpy change for this reaction fromthese reactions of iron oxides with CO: 3 Fe2O3(s)+CO(g)-----> 2Fe3O4(s)+CO2(g)  H= -47 kj Fe2O3(s)+ 3CO(g)------>2 Fe(s) + 3CO2(g)    H= -25kj Fe3O4(s)+CO(g)----->3FeO(s)+CO2(g) H= 19kj Show how you did each step and explain to recieve kharmapoints thanks for the help One reaction involved in the conversion of iron ore to themetal is FeO(s)+CO(g)------>Fe(s)+CO2(g) Calculate the standard enthalpy change for this reaction fromthese reactions of iron oxides with CO: 3 Fe2O3(s)+CO(g)-----> 2Fe3O4(s)+CO2(g)  H= -47 kj Fe2O3(s)+ 3CO(g)------>2 Fe(s) + 3CO2(g)    H= -25kj Fe3O4(s)+CO(g)----->3FeO(s)+CO2(g) H= 19kj Show how you did each step and explain to recieve kharmapoints thanks for the help

Explanation / Answer

3 Fe2O3(s)+CO(g)-----> 2Fe3O4(s)+CO2(g)  H= -47 kj     ----(1) Fe2O3(s)+ 3CO(g)------>2 Fe(s) + 3CO2(g)    H=-25kj         ----(2) Fe3O4(s)+CO(g)----->3FeO(s)+CO2(g) H=19kj                ----(3) now, we needFeO(s)+CO(g)------>Fe(s)+CO2(g) i.e FeO and CO in reaction side and Fe and CO2 inproduct side so we need to cancel rest all other products. so, eq1 - 3*eq2 gives 3 Fe2O3(s)+CO(g)- 3Fe2O3(s)-9CO(g) -----> 2Fe3O4(s)+CO2(g) -6Fe(s) - 9 CO2(g) 6Fe + 9CO2 -->  2Fe3O4(s)+CO2(g) +8CO   -----(4) now adding eq 4 and 2*eq 3 we get 2Fe3O4(s)+2CO(g) + 6Fe + 9CO2----->6FeO(s)+2CO2 + 2Fe3O4(s)+CO2(g) +8CO   this gives 6FeO + 6CO --> 6Fe + 6CO2 dividing this by 6 we get our desired reaction i.e FeO(s)+CO(g)------>Fe(s)+CO2(g) so now H for this reaction is = (H1 - 3H2 + 2H3) /6 = (-47 + 25*3 +19*2)/6 = 19.33 J standard enthalpy change for this reaction =19.33 J dividing this by 6 we get our desired reaction i.e FeO(s)+CO(g)------>Fe(s)+CO2(g) so now H for this reaction is = (H1 - 3H2 + 2H3) /6 = (-47 + 25*3 +19*2)/6 = 19.33 J standard enthalpy change for this reaction =19.33 J

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