Given that:
`DeltaG_(F)^(@)(CuO) =-30.4 "Kcal"//"mole"`
`DeltaG_(f)^(@)(Cu_(2)O)=-34.98 Kcal//"mole" " "T=298K`
Now on the basis of above data which of the following predications will be most appropriate under the standard conditons and reversible reaction.

Options

A. Finely divided form fof CuO Kept in excess `O_(2)` would be completely converted to `Cu_(2)O`
B. Finely divided form of `Cu_(2)O` kept in excess `O_(2)` would be Completely converted to CuO
C. Finely divided form of CuO kept in excess `O_(2)` would be converted to a mixture of CuO and `Cu_(2)O` (having more of CuO)
D. Finely divided form of CuO kept in excess `O_(2)` would be converted to a mixture of CuO and `Cu_(2)O` (having more of `Cu_(2)O`)

Answer

B

Explanation

`(2) Cu_(2)O(s) +(1)/(2)O_(2)(g) hArr 2CuO(s)`
`DeltaG_("reaction")^(@) =[2xx(-30.4)]-[-34.98]=-25.82 `Kcal
and `-25.82 xx 10^(3) =2.303 xx 2xx 298 `logK
`therefore K~~10^(19), ` a very high value, hence reaction will be almost complete with a trace of `Cu_(2)O`.