If equilibrium constant of a process is `3.8 xx 10^(-3) " at" 25^(@)C`, standard free energy change of the process is
`R = 8.314 J mol^(-1) K^(-1), Log 0.0038 =-2.42)`

Options

A. `5.7 kJ mol^(-1)`
B. `9.9 kJ mol^(-1)`
C. `13.8 kJ mol^(-1)`
D. `15.6 kJ mol^(-1)`

Answer

C

Explanation

`Delta G^(@) = - RT ln K`
`Delta G^(@) = -2.303 RT log K`
`Delta G^(@) = -2.303 xx 8.314 mol^(-1) xx 298 log 3.8 xx 10^(-3)`
`Delta G^(@) = 13.8 xx 10^(3) J mol^(-1)`
`Delta G^(@) = 13.8 kJ mol^(-1)`