1.

On the basis of ellingham diagram which of the following is not correct ?

Answer»

Entropy change for all OXIDES is roughly same.
Below the BOILING point, `'T DeltaS'` factor DECOMPOSES into metal & oxygen.
Above `DeltaG=0` line, oxide decomposes into metal & oxygen.
If randomness increases the slope increases.

Solution :`(1) (DeltaG)/(T)=-DeltaS,(DeltaG)/(T)` is slope in Ellingham DIAGRAM. Which is same below the boiling point.
`(2)` Below the boiling point slope is same as factor `T DeltaS` is same.
`(3)` Above `DeltaG=0` line free energy BECOMES positive so oxide decomposes.
`(4)` Random increases `i.e. DeltaS` increases, so slope also increases.


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