Free energies of formation (Delta_(f)G) of MgO(s) and Co(g) at 1273 K and 2273 K are given below : Delta_(f)G^(@) (MgO(s))=-941 kJ//"mol at" 1273 K Delta_(f)G^(@) (MgO(s))=-344 kJ//"mol at" 2273 K Delta_(f)G^(@) (CO (g)) =- 439 kJ//"mol at" 1273 K Delta_(f)G^(@) (CO(g))=-628 kJ//"mol at" 2273 K On the basis of the above data, predict the temperature at which carbon can be used as reducing agent for MgO(s).

Free energies of formation (Delta_(f)G) of MgO(s) and Co(g) at 1273 K

1.	Free energies of formation (Delta_(f)G) of MgO(s) and Co(g) at 1273 K and 2273 K are given below : Delta_(f)G^(@) (MgO(s))=-941 kJ//"mol at" 1273 K Delta_(f)G^(@) (MgO(s))=-344 kJ//"mol at" 2273 K Delta_(f)G^(@) (CO (g)) =- 439 kJ//"mol at" 1273 K Delta_(f)G^(@) (CO(g))=-628 kJ//"mol at" 2273 K On the basis of the above data, predict the temperature at which carbon can be used as reducing agent for MgO(s).
Answer» Solution :The redox reaction for MgO (s) is : `MgO(s) +C(s) to Mg(s) +CO(g)` In order that it MAY proceed, `DeltaG^(@)` must be negative. According to GIBBS Helmholtz EQUATION. `DeltaG^(@)=Delta_(f)G^(@)("products")-Delta_(f)G^(@) ("reactants")` At 1273 K, `DeltaG^(@)=Delta_(f)G^(@)CO(g)-Delta_(f)G^(@)MgO (s)` `=(-439)-(-941)=502 " kJ mol"^(-1)` The reaction will not be feasible as `DeltaG^(@)` is positive. At 2273 K , `DeltaG^(@)=Delta_(f)G^(@)CO(g)-Delta_(f)G^(@)MgO(s)` `=(-628)-(-344)=-284 " kJ mol"^(-1)` The reaction is feasible which means that MgO can be reduced by coke at 2273 K.