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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
`log" "K_(P)=(-DeltaG^(@))/(2.303RT)`
`=(-(-115xx1000))/(2.303xx8.313xx298)`
`=20.16`
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