Which of the following statements are true :
` A to Mg(s) + 1/2 O_2 (g) overset(T_1)(to) MgO(s)`
`B to mg(?) + 1/2 O_2(g) overset(T_2)(to)MgO(s)`
`C to mg (g) + 1/2 O (g) overset(T_3)(to) MgO (s)`

I : Below `1350^@` Mg can reduce `Al_2O_3`
II : Above `1350^@ C` Mg Will reduce `Al_2O_3`
III : Below `1350^@` Al can reduce MgO
IV : Above `1350^@` Al can reduce MgO
V : At `1350^@ C` there is no change in free energy i.e., `Delta G = 0`

Explanation

`underset(Its Delta F " high " )(3MgO + 2Al) overset("then above " 1350^@C)(to) underset(Its Delta G "less" )( Al_2O_3 + 3Mg , DeltaG lt 0) `
`Al_2O_3 + 3Mg overset(Below 1350^@C)(to) 3MgO + 2Al`
At `1350^@C` both reactions have same G
` therefore Delta G = 0`
To carry out smelting below `800^@ C` , CO is used as reducing agent while above `800^@ C` , smelting is carried out by coke.
`{:(2C(s) + O_2(g) to 2CO(g), DeltaH = "- 221.0 kJ/,mole"),(" " , DeltaS = "+179.4 JkJ/mol"),(C(s) + O_2(g) to CO_2(g) , Delta H = "-395.5kJ/mol"),(" " , DeltaS = 2.89 JK^(-1) "mole"^(-1)):}`

• Aluminium can be extracted from Alumina by carbon reduction but the method is highly uneconomical because -
(i) As the smelting occurs above `200^@C` hence a part of the aluminium will go into vapour phase `(M.P. = 2520^@C) `
(ii) At this high temperature the liberated Al will combine with the carbon & aluminium carbide will be formed.
(iii) `Delta H_(for)` for of alumina is high – ve value
` therefore` It is thermodynamically more stable & reduction is more difficult
• To extract metal from sulphide ore is carried out by firstly roasting it into metal oxide & followed by its smelting. Metal sulphide or sulphide ore is not directly smelted to metal.
` 2PbS + C overset(Delta)(to) 2Pb + CS_2` (Thermodynamically Not feasible )
`PbS + 3/2 O_2 overset(Delta)(to) PbO + SO_2 uparrow`
` PbO + C overset(Delta )(to) Pb +CO uparrow}` Thermodynamically feasible
`DeltaG_f` of PbS = –21.9 kcal/mol
` DeltaG_f` of `CS_2` = +17.15 kcal/mol
` DeltaG_f` of PbO = –45.1 kcal/mol
` DeltaF_f` of `SO_2` = - 71.7 kcal/mol
` DeltaG_f` of = – 32.8 kcal/mol