Question

Why is copper matte put in a silica lined converter?

Answer 1

Hint: In the roasting process, ore is heated in a regular supply of oxygen, for the conversion to their oxide form. The sulphide ores of copper are heated in a reverberatory furnace. If the ore contains iron, it is mixed with silica before heating. Iron oxide slags of iron oxide and copper oxide are produced in the form of copper matte which contains $C{u_2}S,FeS$.

Complete answer:
In the Ellingham diagram, the graph of $\Delta {G^ \odot }vs\,T$ for the formation of oxide, the $C{u_2}O$ line is almost at the top. So it is quite easy to reduce copper oxide to copper by simply heating with coke. But most of the copper ores are sulphide and in some way contains iron. The sulphide ores are roasted to give oxides as
$C{u_2}S + 3{O_2} \to 2C{u_2}O + 2S{O_2}$
The oxide is then converted to copper metal using coke as
$C{u_2}O + C \to 2Cu + C{O_{}}$
But in the actual process, the ore is heated in a reverberatory furnace after mixing with silica. In the reverberatory furnace iron oxide slags off as iron silicate and copper is produced as copper matte which contains $C{u_2}S,FeS$.
$FeO + Si{O_2} \to FeSi{O_3}$ (Slag)
Copper matte is then charged into a silica lined converter. Some silica is also added and hot air blast is blown to convert the remaining $FeS,FeO$ and $C{u_2}S/C{u_2}O$ to metallic copper.
The reaction take places in the silica lined convertor are as-
$
  C{u_2}O + C \to 2Cu + C{O_{}} \\
  FeO + Si{O_2} \to FeSi{O_3} \\
  2C{u_2}O + 3{O_2} \to 2C{u_2}O + 2S{O_2} \\
  2FeS + 3{O_2} \to 2FeO + 2S{O_2}
 $

Note:
The Ellingham diagram simply indicates a reaction is possible or not, that means the tendency of reduction with a reducing agent is indicated. This is because it is based only on thermodynamic concepts. And it does not say anything about the kinetics of the reduction reactions. So we cannot tell by looking at the Ellingham diagram how fast a reduction reaction can be.