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. 100
B. 95
C. 5
D. 10

Answer

D

Explanation

According the thermodynamics.
(i) `Delta G^(@) = Delta H^(@) = T Delta S^(@)`
(ii) `Delta G^(@) = - 2.303 RT log K`
Equating the two, we get
`-2.303 RT log K = Delta H^(@) = T Delta S^(@)`
`:. Log K = (Delta H^(@) - T Delta S^(@))/(-2.303 RT)`
`= ((-54.07 xx 10^(3)) - (298)(10))/(- 2.303 xx 8.314 xx 298)`
`= (57050)/(-5705) = 0`