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Which is correct about DeltaG?...

Which is correct about `DeltaG`?

A

`DeltaG=DeltaH-TDeltaS`
`=49.4-(300xx336xx10^(-3))`
`=-51.4kJ`
Since, the free energy change is negative, the given reaction is spontaneous.

B

At equilibrium `DeltaG^(@)=0`

C

At eq. `DeltaG=-RTlogK`

D

`DeltaG=DeltaG^(@)+RT" log "K`

Text Solution

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The correct Answer is:
A, D
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