Write the relation between `Delta G^(@)` and equilibrium constant.

Options

A. `Delta G^(@) = -2.303 RT log K`
B. ` Delta G^(@) = - RT log K`
C. ` Delta G^(@) = 2.303 RT log K`
D. ` Delta G^(@) = -2.303 log K`

Answer

A

Explanation

Consider a reversible reaction, `A + B hArr C + D`. The relationship between Gibb's free energy change `(Delta G)` at any time and standard Gibb's free energy change `(Delta G^(@))` of the reaction is `Delta G = Delta G^(@) + RT ln Q` where `Q = (p_(C ).p_(D))/(p_(A).p_(B))` at any given time.
This equation is called "Van't Hoff's reaction isotherm.". When the reaction attains equilibrium, `Delta G = 0` and `(p_(C ).p_(D))/(p_(A).p_(B)) = K`. The above equation becomes,
`0 = Delta G^(@) + RT ln K`
or `Delta G^(@) = -RT ln K`
`:. Delta G^(@) = -2.303 RT log K`
`Delta G^(@)` is the standard free energy change of the reaction, R is the gas constant, T is the temperature in absolute scale and `K_(p)` is the equilibrium constant
In general, for `aA + bB hArr c C + d D`
`Delta G^(@) = -2.303 RT log K_(p)` where `K_(p) = (p_(C ).p_(D))/(p_(A).p_(B))`