Respuesta :
Explanation:
It is known that relation between Gibb's free energy and equilibrium or reaction quotient is as follows.
[tex]\Delta G = \Delta G^{o} + 2.303 RT logQ[/tex] ..... (1)
At equilibrium, [tex]\Delta G[/tex] = 0. Also, Q = K this means that reaction is at equilibrium.
Therefore, formula in equation (1) will become as follows.
[tex]\Delta G^{o} = -2.303 RT log K[/tex] ....... (2)
It is known that for a sparingly soluble salt, K = [tex]K_{sp}[/tex]
Thus, equation (2) will become as follows.
[tex]\Delta G^{o} = -2.303 RT log K_{sp}[/tex] ....... (3)
So, when value of [tex]K_{sp}[/tex] increases then there will be decrease in the value of [tex]\Delta G^{o}[/tex] according to equation (3).
Hence, we can conclude that the calculated value of [tex]K_sp[/tex] would be too big.
