(i) Name the method used for the refining of zirconium.
(ii) What is the role of CO in the extraction of Iron?
(iii) Reduction of metal oxide to metal becomes easier if the metal obtained is in liquid state. Why?
(i) Name the method used for the refining of zirconium.
(ii) What is the role of CO in the extraction of Iron?
(iii) Reduction of metal oxide to metal becomes easier if the metal obtained is in liquid state. Why?
(i) The Van Arkel method is used for the refining of zirconium.
(ii) CO acts as a reducing agent due to which iron in the ore reduces to iron metal.
(iii) Entropy of a metal in the liquid state is higher than that of the same metal in the solid state (Sliquid>Ssolid). So, when the metal formed is in the liquid state and the metal oxide being reduced is in the solid state, the value of entropy change (ΔS) for the reduction reaction is more on the positive side. When the value of TΔS increases and that of Δ H remains the same, the value of Δ G∘r for the reduction reaction becomes negative and thus reduction becomes easier.
(i) The Van Arkel method is used for the refining of zirconium.
(ii) CO acts as a reducing agent due to which iron in the ore reduces to iron metal.
(iii) Entropy of a metal in the liquid state is higher than that of the same metal in the solid state (Sliquid>Ssolid). So, when the metal formed is in the liquid state and the metal oxide being reduced is in the solid state, the value of entropy change (ΔS) for the reduction reaction is more on the positive side. When the value of TΔS increases and that of Δ H remains the same, the value of Δ G∘r for the reduction reaction becomes negative and thus reduction becomes easier.
