Answer :
To determine the solubility product constant [tex]\( K_{sp} \)[/tex] for [tex]\(\text{PbCl}_2(s)\)[/tex], we need to use the given standard free energies of formation ([tex]\(\Delta G_f^\circ\)[/tex]) for [tex]\( \text{Pb}^{2+}(aq) \)[/tex] and [tex]\( \text{Cl}^{-}(aq) \)[/tex].
### Steps to Find [tex]\( \Delta G \)[/tex] for the Dissolution Reaction
1. Write-down the dissolution reaction of [tex]\(\text{PbCl}_2(s)\)[/tex]:
[tex]\[ \text{PbCl}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{Cl}^{-}(aq) \][/tex]
2. Given Data:
- [tex]\(\Delta G_f^\circ\)[/tex] for [tex]\(\text{Pb}^{2+}(aq)\)[/tex] is [tex]\(-24.3 \, \text{kJ/mol}\)[/tex]
- [tex]\(\Delta G_f^\circ\)[/tex] for [tex]\(\text{Cl}^{-}(aq)\)[/tex] is [tex]\(-131.2 \, \text{kJ/mol}\)[/tex]
3. Calculate [tex]\(\Delta G_{reaction}\)[/tex]:
The standard free energy change for the reaction ([tex]\(\Delta G_{reaction}\)[/tex]) is given by:
[tex]\[ \Delta G_{reaction} = \Delta G_f^\circ(\text{Pb}^{2+}) + 2 \cdot \Delta G_f^\circ(\text{Cl}^{-}) \][/tex]
Substituting the given values:
[tex]\[ \Delta G_{reaction} = (-24.3 \, \text{kJ/mol}) + 2 \cdot (-131.2 \, \text{kJ/mol}) \][/tex]
[tex]\[ \Delta G_{reaction} = -24.3 \, \text{kJ/mol} + (-262.4 \, \text{kJ/mol}) \][/tex]
[tex]\[ \Delta G_{reaction} = -286.7 \, \text{kJ/mol} \][/tex]
4. Converting [tex]\(\Delta G_{reaction}\)[/tex] from kJ to J:
[tex]\[ -286.7 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = -286700 \, \text{J/mol} \][/tex]
### Linking [tex]\(\Delta G_{reaction}\)[/tex] to [tex]\(K_{sp}\)[/tex]
5. The relationship between [tex]\(\Delta G_{reaction}\)[/tex] and [tex]\(K_{sp}\)[/tex] is given by the equation:
[tex]\[ \Delta G_{reaction} = RT \ln K_{sp} \][/tex]
where:
- [tex]\(R\)[/tex] is the universal gas constant ([tex]\(8.314 \, \text{J/(mol·K)}\)[/tex])
- [tex]\(T\)[/tex] is the temperature in Kelvin (Assuming standard temperature [tex]\(T = 298 \, \text{K}\)[/tex])
6. Rearranging to solve for [tex]\(K_{sp}\)[/tex]:
[tex]\[ \ln K_{sp} = \frac{\Delta G_{reaction}}{-RT} \][/tex]
[tex]\[ \ln K_{sp} = \frac{-286700 \, \text{J/mol}}{-8.314 \, \text{J/(mol·K)} \times 298 \, \text{K}} \][/tex]
[tex]\[ \ln K_{sp} = \frac{-286700}{-2476.172} \approx 115.76 \][/tex]
7. Exponentiating both sides to find [tex]\(K_{sp}\)[/tex]:
[tex]\[ K_{sp} = e^{\ln K_{sp}} = e^{115.76} \][/tex]
[tex]\[ K_{sp} \approx 1.80 \times 10^{50} \][/tex]
### Final Answer
The solubility product constant [tex]\(K_{sp}\)[/tex] for [tex]\( \text{PbCl}_2(s) \)[/tex] is approximately:
[tex]\[ K_{sp} \approx 1.80 \times 10^{50} \][/tex]
### Steps to Find [tex]\( \Delta G \)[/tex] for the Dissolution Reaction
1. Write-down the dissolution reaction of [tex]\(\text{PbCl}_2(s)\)[/tex]:
[tex]\[ \text{PbCl}_2(s) \rightleftharpoons \text{Pb}^{2+}(aq) + 2\text{Cl}^{-}(aq) \][/tex]
2. Given Data:
- [tex]\(\Delta G_f^\circ\)[/tex] for [tex]\(\text{Pb}^{2+}(aq)\)[/tex] is [tex]\(-24.3 \, \text{kJ/mol}\)[/tex]
- [tex]\(\Delta G_f^\circ\)[/tex] for [tex]\(\text{Cl}^{-}(aq)\)[/tex] is [tex]\(-131.2 \, \text{kJ/mol}\)[/tex]
3. Calculate [tex]\(\Delta G_{reaction}\)[/tex]:
The standard free energy change for the reaction ([tex]\(\Delta G_{reaction}\)[/tex]) is given by:
[tex]\[ \Delta G_{reaction} = \Delta G_f^\circ(\text{Pb}^{2+}) + 2 \cdot \Delta G_f^\circ(\text{Cl}^{-}) \][/tex]
Substituting the given values:
[tex]\[ \Delta G_{reaction} = (-24.3 \, \text{kJ/mol}) + 2 \cdot (-131.2 \, \text{kJ/mol}) \][/tex]
[tex]\[ \Delta G_{reaction} = -24.3 \, \text{kJ/mol} + (-262.4 \, \text{kJ/mol}) \][/tex]
[tex]\[ \Delta G_{reaction} = -286.7 \, \text{kJ/mol} \][/tex]
4. Converting [tex]\(\Delta G_{reaction}\)[/tex] from kJ to J:
[tex]\[ -286.7 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = -286700 \, \text{J/mol} \][/tex]
### Linking [tex]\(\Delta G_{reaction}\)[/tex] to [tex]\(K_{sp}\)[/tex]
5. The relationship between [tex]\(\Delta G_{reaction}\)[/tex] and [tex]\(K_{sp}\)[/tex] is given by the equation:
[tex]\[ \Delta G_{reaction} = RT \ln K_{sp} \][/tex]
where:
- [tex]\(R\)[/tex] is the universal gas constant ([tex]\(8.314 \, \text{J/(mol·K)}\)[/tex])
- [tex]\(T\)[/tex] is the temperature in Kelvin (Assuming standard temperature [tex]\(T = 298 \, \text{K}\)[/tex])
6. Rearranging to solve for [tex]\(K_{sp}\)[/tex]:
[tex]\[ \ln K_{sp} = \frac{\Delta G_{reaction}}{-RT} \][/tex]
[tex]\[ \ln K_{sp} = \frac{-286700 \, \text{J/mol}}{-8.314 \, \text{J/(mol·K)} \times 298 \, \text{K}} \][/tex]
[tex]\[ \ln K_{sp} = \frac{-286700}{-2476.172} \approx 115.76 \][/tex]
7. Exponentiating both sides to find [tex]\(K_{sp}\)[/tex]:
[tex]\[ K_{sp} = e^{\ln K_{sp}} = e^{115.76} \][/tex]
[tex]\[ K_{sp} \approx 1.80 \times 10^{50} \][/tex]
### Final Answer
The solubility product constant [tex]\(K_{sp}\)[/tex] for [tex]\( \text{PbCl}_2(s) \)[/tex] is approximately:
[tex]\[ K_{sp} \approx 1.80 \times 10^{50} \][/tex]
