Respuesta :
Answer : the correct answer for ksp = 1.59 * 10⁻⁹
Following are the steps to calculate the ksp of reaction
BaCO₃ →Ba ²⁺ + CO₃²⁻ :
Step 1 : To find ΔG° of reaction :
ΔG° of reaction can be calculates by taking difference between ΔG° of products and reactants as :
ΔG° reaction =Sum of ΔG° ( products ) - Sum of Δ G° ( reactants ) .
Given : ΔG° for Ba²⁺ ( product )= -560.7 [tex] \frac{KJ}{Mol} [/tex]
ΔG° for CO₃²⁻ (product ) =- 528.1 [tex] \frac{KJ}{Mol} [/tex]
ΔG° BaCO₃ ( reactant) = –1139 [tex] \frac{KJ}{Mol} [/tex]
Plugging value in formula :
ΔG° for reaction = ( ΔG° of Ba ²⁺ + ΔG° of CO₃²⁻ ) - (ΔG° of BaCO₃ )
⁻ = ( -560.7 [tex] \frac{KJ}{Mol} [/tex] + 528.1 [tex] \frac{KJ}{Mol} [/tex] ) - ( -1139 [tex] \frac{KJ}{Mol} [/tex] )
= ( -1088.8 [tex] \frac{KJ}{Mol} [/tex]) - (-1139 [tex] \frac{KJ}{Mol} [/tex] )
= - 1088.8 [tex] \frac{KJ}{Mol} [/tex] + 1139 [tex] \frac{KJ}{Mol} [/tex]
ΔG° of reaction = 50.2 [tex] \frac{KJ}{Mol} [/tex]
Step 2: To calculate ksp from ΔG° of reaction .
The relation between Ksp and ΔG° is given as :
ΔG° = -RT ln ksp
Where ΔG° = Gibb's Free energy R = gas constant T = Temperature
Ksp = Solubility constant product .
Given : ΔG° of reaction = 50.2 [tex] \frac{KJ}{Mol} [/tex]
T = 298 K R = 8.314 [tex] \frac{J}{Mol * K} [/tex]
Plugging values in formula
[tex] 50.2 \frac{KJ}{mol} = - 8.314 \frac{J}{mol * K} * 298 K * ln ksp
[/tex]
[tex] 50.2 \frac{KJ}{mol} = - 2477.572 \frac{J}{mol} * ln K [/tex]
((Converting 2477 [tex] \frac{J}{mol} to \frac{KJ}{mol} [/tex]
Since , 1 KJ = 1000 J So , [tex] 2477 \frac{J}{mol} * \frac{1 KJ}{1000J} = 2.477 \frac{KJ}{mol} [/tex] ))
Dividing both side by [tex] - 2.477 \frac{KJ}{mol} [/tex]
[tex] \frac{50.2\frac{KJ}{mol}}{-2.477 \frac{KJ}{mol}} = \frac{-2.477 \frac{KJ}{mol}}{-2.477 \frac{KJ}{mol}} * ln ksp [/tex]
ln ksp = [tex] ln ksp = -20.27 \frac{KJ}{mol} [/tex]
Removing ln :
ksp = 1. 59 * 10⁻⁹
