The value of DeltaG^(@) for a reaction i...

The value of `DeltaG^(@)` for a reaction is 7.97 `KJ//"mole"`. Its equilibrium constant `K_(c)` at 298K is `4 xx 10^(-x)` What is x?

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The correct Answer is:

2

`Delta G^(@)= -RTlnK`
`log K = -1.4= -2+0.6=log(4 xx 10^(-2)) implies K=2`

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