Using the thermodynamic data provided below, calculate the standard change in entropy when one mole of sodium sulfate is dissolved in water. $\begin{array}{|l|c|}
\hline & \mathrm{S}^{\circ}(\mathrm{J} / \mathrm{K} \cdot \mathrm{~mol}) \
\hline \mathrm{Na}_{2} \mathrm{SO}_{4}(s) & 149.49 \
\hline \mathrm{Na}^{+}(a q) & 60.25 \
\hline \mathrm{SO}_{4}{ }^{2-}(a q) & 17.15 \
\hline
\end{array}$ Will the solubility of sodium nitrate increase or decrease if the temperature of the system is increased
A) -11.84 J/K·mol; solubility decreases with increasing temperature
B) -11.84 J/K·mol; solubility increases with increasing temperature
C) 11.84 J/K·mol; solubility decreases with increasing temperature
D) 11.84 J/K·mol; solubility increases with increasing temperature
E) None of the above

Question

Quizplus · Accepted Answer

The standard change in entropy can be calculated using the equation:
ΔS° = ΣnS°(products) - ΣnS°(reactants)
For the dissolution of Na2SO4 in water, the equation is:
Na2SO4(s) → 2Na+(aq) + SO42-(aq)
ΔS° = 2S°(Na+(aq)) + S°(SO42-(aq)) - S°(Na2SO4(s))
Using thermodynamic data from standard tables, we get:
ΔS° = 2(51.45 J/K·mol) + 138.77 J/K·mol - 116.31 J/K·mol
ΔS° = -11.84 J/K·mol
This means that the dissolution of sodium sulfate in water is not favorable in terms of entropy, as there is a decrease in randomness (negative ΔS°). 
As for the solubility of sodium nitrate, according to Le Chatelier's principle, if the temperature of the system is increased, the equilibrium will shift in the direction that consumes heat. Since dissolution is an endothermic process, the solubility of sodium nitrate will increase with increasing temperature. However, this information is not relevant to the given question.

Using the Thermodynamic Data Provided Below, Calculate the Standard Change