Question

A chemist fills a reaction vessel with 0.486g calcium phosphate (Ca3(PO4)2) solid, 0.528 M calcium (Ca2+) aqueous solution, and 0.189M phosphate (PO43-) aqueous solution at a temperature of 25.0 degrees celsius.

under these conditions, calculate the reaction free energy delta G for the following chemical reaction

Ca3(PO4)2 (s)--> 3Ca2+ (aq)+2PO43-(aq)

thermodynamic info:

Ca3(PO4)2: -3899.0 Ca2+: -553.6 PO43-: -1013.

Answer 1

ΔG = 199.2 kJ

Step-by-step explanation:

Let's consider the following reaction.

Ca₃(PO₄)₂(s) → 3 Ca²⁺(aq) + 2 PO₄³⁻(aq)

We can calculate the standard Gibbs free energy of reaction (ΔG°) using the following expression.

ΔG° = 3 mol × ΔG°f(Ca²⁺(aq)) + 2 mol × ΔG°f(PO₄³⁻(aq) ) - 1 mol × ΔG°f(Ca₃(PO₄)₂(s) )

ΔG° = 3 mol × (-553.6 kJ/mol) + 2 mol × (-1013 kJ/mol) - 1 mol × (-3899.0 kJ/mol)

ΔG° = 212.2 kJ

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

We can calculate the Gibbs free energy of reaction (ΔG) using the following expression.

ΔG = ΔG° + R.T.lnQ

where,

R: ideal gas constant

T: absolute temperature (25.0 + 273.15 = 298.2 K)

Q: reaction quotient

ΔG = ΔG° + R.T.ln [Ca²⁺]³.[PO₄³⁻]²

ΔG = 212.2 kJ/mol + (8.314 × 10⁻³ kJ/K.mol) × 298.2 K × ln (0.528³ × 0.189²)

ΔG = 199.2 kJ