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Given that: DeltaG(F)^(@)(CuO) =-30.4 ...

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.

A

Finely divided from of CuO kept in excess `O_(2)` would be completely converted to `Cu_(2)O`

B

Finely divided form `Cu_(2)O` of 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` )

Text Solution

Verified by Experts

The correct Answer is:
B
`Cu_(2)O(s) +1/2O_(2) (g) , 2Cuo (g)`
`DeltaG_("reaction")^(@) = [2 xx (-30.4)] -(-34.98) = -25.82` kcal
and `-25.82 xx 10^(3) = -2.303 xx 2 xx 298` log R
`therefore R=10^(19)`, a very high value, hence reaction will be almost complete with a trace of `Cu_(2)O`
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