Comparing Solubility of SrSO4 and BaSO4: Calculating ΔS Total for Justification

[RCB]

Homework Statement

"Which of SrSO4 and BaSO4 is more soluble. Justify your answer using the following data to calculate the difference between ΔS total:

ΔHsolution SrSO₄ (s) = -9 kJ/mol
ΔHsolution BarSO₄ (s) = +19 kJ/mol

S^∅ Sr2+ = -33 kJ/mol
S^∅ Ba2+ = +10 kJ/mol

Homework Equations

The Attempt at a Solution

So the difference in ΔH solution is 28 kJ/mol
this equates to a difference of 93.6 J/K/mol (finding ΔS surrounding)

ΔS total = ΔS surr + ΔS system

ΔS system = S(products) - S(reactants)

I can understand why I don't need the entropy for the SO_42- ion BUT how can I find ΔS total if I don't have S(products)

Once I have this it is obvious how to continue

thanks

 

[nate808]

I would approach this problem by first considering the thermodynamic principles involved in solubility. Solubility is a measure of how much of a substance can dissolve in a given solvent at a given temperature. It is influenced by several factors, including the enthalpy and entropy changes that occur during the dissolution process.

In this case, we are comparing the solubility of SrSO4 and BaSO4, two ionic compounds with similar structures and molecular weights. The first step would be to calculate the difference in enthalpy changes (ΔH) for their dissolution, which is given as -9 kJ/mol for SrSO4 and +19 kJ/mol for BaSO4. This means that the dissolution of BaSO4 is more energetically favorable than that of SrSO4, as indicated by the positive value of ΔH.

Next, we need to consider the entropy changes (ΔS) that occur during the dissolution process. Entropy is a measure of the disorder or randomness in a system, and its change during a reaction can provide insight into the spontaneity of the reaction. In this case, we are given the standard entropy values (S∅) for the cations, which are -33 kJ/mol for Sr2+ and +10 kJ/mol for Ba2+. These values represent the entropy of the ions in their standard state, which is a solid for both compounds.

To find the difference in ΔS surrounding, we can use the equation ΔS surr = -ΔH/T, where T is the temperature in Kelvin. Using the given values for ΔH and assuming a temperature of 298 K, we can calculate a difference in ΔS surrounding of 93.6 J/K/mol. This indicates that the dissolution of BaSO4 results in a greater increase in entropy in the surrounding environment compared to SrSO4.

Finally, to calculate the total entropy change (ΔS total) for each compound, we can use the equation ΔS total = ΔS surr + ΔS system. Since we do not have the entropy values for the products, we can assume that they are equal to the entropy of the ions in their standard state. This assumption is reasonable since both compounds will dissociate into their respective ions upon dissolution. Using this approach, we can calculate a ΔS total of -22.4 J/K/mol for SrSO4 and +110.4 J/K