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Natasha2012 [34]

1 year ago

8

A chemist fills a reaction vessel with 0.750 M lead (II) (Pb2+) aqueous solution, 0.232 M bromide (Br) aqueous solution, and 0.9

56 g lead (II) bromide (PbBr2 solid at a temperature of 25.0°C. Under these conditions, calculate the reaction free energy AG for the following chemical reaction: Pb2+ (aq) + 2Br (aq) = PbBr2 (s) Use the thermodynamic information in the ALEKS Data tab. Round your answer to the nearest kilojoule.

1 answer:

Ronch [10]1 year ago

8 0

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

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