`DeltaG^(Ө)` vs T plot in Ellingham diagram slopes downward for the reaction .

Options

A. `2CO(g) O_(2)(g) rarr 2CO_(2)g`
B. `C(s)+ O_(2)(g) rarr CO_(2)(g)`
C. `2C(s) + O_(2)(g) rarr 2CO(g)`
D. Both (2) and (3)

Answer

c

Explanation

Carbon in the form of coke, charkoal orcarbon monoxide is used as a reducing agent in pyrommetallurgical operations. Such a reduction process in the extraction of a metal in termed smelting
When carbon is to act as a reducing agent the following there reactions are possible
`C(s) + O_(2) (g) rarr CO_(2)(g) `.....(i)
`2C(s) + O_(2) (g) rarr 2CO(g)`.....(ii)
`2CO(g) + O_(2)(g) rarr 2CO_(2)(g)`......(iii)
In the first raetion (formation of `CO_(2))` there is hardly any entropy change i.e. `Delta_(T)S = 0` and therefore `Delta G` remins nearly the same with rise in temperature i.e. `Delta _(r) G^(Ө)` is independent of temperature
In the second (formation of `CO_(2))` there is increases in entropy `(Delta_(r)S^(Ө)` is positive ) and therefore, `Delta G`^(Ө)` becomes more megnative with increase of temperature. However ,in third reaction , there is decrease is energies `(DeltaS^(Ө)` is negative) and therefore, `Delta G^(Ө)` becomes less negative with increase in temperature

The three curve have been found to intersect at `673 K` it 6 imples that above this temperature the reaction (ii) is most stable .it mean ns that carbons can reduce any metal oxide at very high temperature and is then itself oxident is `CO`. However , the reduction with carbon at high temprature is not prefent in all cases due in the following reasons:
(a) It involves high cost
(b) Some metals reach with carbon at high temperature and form carbides.
(c ) There are many practical difference in the matelences of high temperature
From the plot for the reaction of carbon monoxide with oxygen , it is evident that carbon monooxide acts as a beter reducing agent carbon at temprature below `673K`.