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A chemist fills a reaction vessel with a 0.105g mercurous chloride Hg2Cl2 solid, 0.926 M mercury (I) (Hg2^2+) aqueous solution, and 0.573 M chloride (Cl^-) aqueous solution at a temperature of 25.0°C.

Under these conditions, calculate the reaction free energy ΔG for the following chemical reaction:
Hg2Cl2(s) ---->Hg2^2+ (aq) + 2Cl^- (aq)

User Horkavlna
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1 Answer

3 votes

Answer:

ΔG = 98.67 kJ/mol

Step-by-step explanation:

Let' s consider the following reaction.

Hg₂Cl₂(s) → Hg₂²⁺(aq) + 2Cl⁻(aq)

The standard Gibbs free energy (ΔG°) for the reaction is:

ΔG° = 1 mol × ΔG°f(Hg₂²⁺(aq)) + 2 mol × ΔG°f(Cl⁻(aq)) - 1 mol × ΔG°f(Hg₂Cl₂(s))

where,

ΔG°f: standard Gibbs free energy of formation

ΔG° = 1 mol × (154.72 kJ/mol) + 2 mol × (-134.08 kJ/mol) - 1 mol × (-215.06 kJ/mol)

ΔG° = 101.62 kJ

This is standard Gibbs free energy change per mole of reaction.

The Gibbs free energy of the reaction (ΔG) can be calculated using the following expression.

ΔG = ΔG° - R.T.lnQ

where,

R: ideal gas constant

T: absolute temperature

Q: reaction quotient

ΔG = ΔG° - R.T.ln([Hg₂²⁺].[Cl⁻]²)

ΔG = 101.62 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) . (298.2 K) . ln [(0.926).(0.573)²]

ΔG = 98.67 kJ/mol

User Bigfanjs
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