Answer:
The free energy = -20.46 KJ
Explanation:
given Data:
Pb²⁺ = 0.750 M
Br⁻ = 0.232 M
R = 8.314 Jk⁻¹mol⁻¹
T = 298K
The Gibb's free energy is calculated using the formula;
ΔG = ΔG° + RTlnQ -------------------------1
Where;
ΔG° = standard Gibb's freeenergy
R = Gas constant
Q = reaction quotient
T = temperature
The chemical reaction is given as;
Pb²⁺(aq) + 2Br⁻(aq) ⇄PbBr₂(s)
The ΔG°f are given as:
ΔG°f (PbBr₂) = -260.75 kj.mol⁻¹
ΔG°f (Pb²⁺) = -24.4 kj.mol⁻¹
ΔG°f (2Br⁻) = -103.97 kj.mol⁻¹
Calculating the standard gibb's free energy using the formula;
ΔG° = ξnpΔG°(product) - ξnrΔG°(reactant)
Substituting, we have;
ΔG° =[1mol*ΔG°f (PbBr₂)] - [1 mol *ΔG°f (Pb²⁺) +2mol *ΔG°f (2Br⁻)]
ΔG° =(1 *-260.75 kj.mol⁻¹) - (1* -24.4 kj.mol⁻¹) +(2*-103.97 kj.mol⁻¹)
= -260.75 + 232.34
= -28.41 kj
Calculating the reaction quotient Q using the formula;
Q = 1/[Pb²⁺ *(Br⁻)²]
= 1/(0.750 * 0.232²)
= 24.77
Substituting all the calculated values into equation 1, we have
ΔG = ΔG° + RTlnQ
ΔG = -28.41 + (8.414*10⁻³ * 298 * In 24.77)
= -28.41 +7.95
= -20. 46 kJ
Therefore, the free energy of reaction = -20.46 kJ
