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The equilibrium constant for a reaction ...

The equilibrium constant for a reaction is 100 what will be the value of `DeltaG^(@)` ? `R=8.314JK^(-1)mol^(-1),T=300`K :-

A

`-11488KJ`

B

`-11.488KJ`

C

`-12KJ`

D

`-12000KJ`

Text Solution

Verified by Experts

The correct Answer is:
B
`DeltaG^(@)=-2.303RTlogK_(eq)`
`=-2.303xx8.314xx300log100J`
`DeltaG^(@)=-11.488KJ`
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