The value of `log_(10)K` for a reaction `A hArr B` is (Given: `Delta_(f)H_(298K)^(Theta) =- 54.07 kJ mol^(-1)`,
`Delta_(r)S_(298K)^(Theta) =10 JK^(=1) mol^(-1)`, and `R = 8.314 J K^(-1) mol^(-1)`

Options

A. 5
B. 10
C. 95
D. 100

Answer

B

Explanation

For the equilibrium, `A hArr B`
`Delga G^(@) = Delta H^(@) - T Delta S^(@)`
`Delta G^(@) =-2.303 RT log_(10) K` (K is equilibrium constant) `-2.303 RT log_(10) K= Delta H^(@) - T Delta S^(@)`
`2.303 RT log_(10) K = T Delta S^(@)- Delta H^(@)`
`log _(10) K=(T Delta S^(@)- Delta H^(@))/(2.303 RT)`
`=(298xx10+ 54.07xx1000)/(2.303xx8.314xx298)=10`