Respuesta :
Answer:
1. Solubility will increase, because as T increases the − Δ H ∘ R T −ΔH∘RT term becomes smaller therefore K will get larger.
2. To ensure the dissolution process was at equilibrium.
Explanation:
Given that;
ΔG°= -RTlnK
and
ΔG° = ΔH° - TΔS°
So;
-RTlnK = ΔH° - TΔS°
lnK = ΔH°/-RT - TΔS°/-RT
lnK = -(ΔH°/RT) + ΔS°/R
K = e^-(ΔH°/RT) + ΔS°/R
Hence, Solubility will increase, because as T increases the − Δ H ∘ R T −ΔH∘RT term becomes smaller therefore K will get larger.
2.
Since solubility is an equilibrium process, it means that some undissolved solute must be present in order to determine the solubility product correctly.
Solubility is the ability of the solute to dissolve in the solution. As temperature increases, the solubility increases, and the presence of solid ensure dissolution.
What is Gibbs energy?
The Gibbs free energy is the change in the energy of a system from the energy present in the reactant towards the product formation.
It is given as,
[tex]\rm \Delta G^{\circ}= \rm -RTlnK \\\\\rm \Delta G ^{\circ} = \Delta H ^{\circ} - T\Delta S ^{\circ}[/tex]
Equating both equations we get:
[tex]\begin{aligned} \rm -RTlnK &= \rm \Delta H ^{\circ} - T\Delta S ^{\circ}\\\\\rm lnK &=\rm -(\dfrac{\Delta H ^{\circ}}{RT}) + \dfrac{\Delta S ^{\circ}}{R}\\\\\rm K &=\rm e^{-(\frac{\Delta H ^{\circ}}{RT}) + \frac{\Delta S ^{\circ}}{R}}\end{aligned}[/tex]
Thus, the solubility of the substance increases with an increase in temperature.
The solubility product is estimated correctly by the equilibrium process that ensures the presence of the undissolved solute particles.
Therefore, 1. option A. solubility increases and 2. option B. to ensure the equilibrium process are the correct options.
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