Question

 

Derive the relation between `DeltaG^@`and equilibrium constant (K) for the reaction -

aA_bB ⇌ cC+dD.​

 

Answer 1

 

The free energy (G) of any substance at a temperature T is represented as

`G=G^@+RTln[`

A+B ⇌ C+D

G_A, G_B, G_C and G_D are standard free energies

`G_A=G_A^0+RTln[A]`

`G_B=G_B^0+RTln[B]`

`G_C=G_c^0+RTln[C]`

`G_D=G_D^0+RTln[D]`

`:.DeltaG=sumG_"product"-sumG_"reactant"`

`=[G_C+G_D]-[G_A+G_B]`

`={G_C^0+RTln[C]+G_D^0+RTLn[D]}-{G_A^0+RTln[A]+G_B^0+RTln[B]}`

 `{(G_C^0+G_D^0)-(G_A^0+G_B^0)}+{RTln[C]+RTln[D]-RTln[A]+RTln[B]}`

 if `DeltaG^0={(G_C^0+G_D^0)-(G_A^0+G_B^0)}`

 `:.DeltaG=DeltaG^0+(RTln[C]xx[D]-RTln[A]xx[B])`

`DeltaG=DeltaG^0+RTln`

 or `DeltaG=DeltaG^0+RTlnQ`

`Q=(["product"])/(["reactant"])`

`Q=K=([C]xx[D])/([A]xx[B])`

Hence from above equation

`DeltaG=DeltaG^0+RTlnk`

 since at equillibrium ΔG=0

`:.O=DeltaG^0+RTlnK`

`:.DeltaG^0=-RTlnK`

or

`:.DeltaG^0=-2.303RTlog_10K`