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The equilibrium constant of a reaction i...

The equilibrium constant of a reaction is 0.008 at 298 K. The standard free energy change of the reaction at the same temperature is :

A

`+11.96 kJ`

B

`-11.96 kJ`

C

`-5.43 kJ`

D

`-8.46 kJ`

Text Solution

Verified by Experts

The correct Answer is:
A
`Delta G^(@)=-2.303 RT log K`
`=-2.303xx8.314xx298 log 0.008`
`=-2.303xx8.314xx298xx(-2.1)`
`=+11.960 J =+11.96 kJ`.
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