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The standard free energy change of a rea...

The standard free energy change of a reaction is `DeltaG^(@)=-115` at 298K. Calculate the equilibrium constant `K_(P)` in log `K_(P).(R=8.314JK^(-1)mol^(-1))`

A

`20.16`

B

`2.303`

C

`2.016`

D

`13.83`

Text Solution

Verified by Experts

The correct Answer is:
A
The Standard free energy change of a reaction `DeltaG^(@)=-2.303RTlogK_(p)`
`-115xx10^(3)=-2.303RT log K_(p)`
`logK_(p)=(115xx10^(3))/(2.303xx8.314xx298)logK_(p)=20.16.`
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