The value of *K**c *for a reaction does not depend on the rate of the reaction. However, as you have studied in Unit 6, it is directly related to the thermodynamics of the reaction and in particular, to the change in Gibbs energy, ∆*G*. If,

∆*G *is negative, then the reaction is spontaneous and proceeds in the forward direction.

∆*G *is positive, then reaction is considered non-spontaneous. Instead, as reverse reaction would have a negative _*G*, the products of the forward reaction shall be converted to the reactants

∆*G *is 0, reaction has achieved equilibrium; at this point, there is no longer any free energy left to drive the reaction.

A mathematical expression of this thermodynamic view of equilibrium can be described by the following equation.

∆*G *= + RT ln*Q *where, is standard Gibbs energy

At equilibrium, when ∆*G *= 0 and *Q *= *K**c*, the above equation becomes, ∆*G *= + RT In K=0

InK = -∆*G */ RT

Taking antilog of both sides, we get, K = e^{-}^{∆}^{G }^{/ RT}

Hence, using the equation K = e^{-}^{∆}^{G }^{/ RT},

the reaction spontaneity can be interpreted in terms of the value of

- If < 0, then R
*T*is positive, and e^{-}^{∆}^{G }^{/ RT}>1, making*K*>1, which implies a spontaneous reaction or the reaction which proceeds in the forward direction to such an extent that the products are present predominantly. - If > 0, then R
*T*is negative, and e^{-}^{∆}^{G }^{/ RT}< 1, that is ,*K*< 1, which implies a non-spontaneous reaction or a reaction which proceeds in the forward direction to such a small degree that only a very minute quantity of product is formed.