1.

For a reversible reactionn at 298 K the equilibrium constant K is 200. What is value of `DeltaG^(@)` at 298 K ?A. `-13.13kcal`B. `-0.13kcal`C. `-3.158kcal`D. `-0.413kcal`

Answer» Correct Answer - C
Applying `DeltaG^(@)=-2.303RTxxlogK`
`" "=-2.303xx2xx298xxlog200`
`" "=-3158.4" cal "=-3.158" kcal"`


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