Ksp of PbCl2(s) = 1.7x10-5 Delta Gf of PbCl2 = -314KJ/mol Delta Gf of Pb2+ =...

Question

Question

Ksp of PbCl2(s) = 1.7x10-5

Delta Gf of PbCl2 = -314KJ/mol

Delta Gf of Pb2+ = -24.3KJ/mol

Calculate the standard free energy change for the formation of Cl-

Answer 1

Answer #1

The reaction that occurs is:

PbCl2 = Pb + 2 + 2 Cl-

The ΔG of the reaction is calculated:

ΔG = - R * T * ln K = - 0.008314 * 298 K * ln 1.7x10 ^ -5 = 27.21 kJ / mol

It has:

ΔG = ΔG products - ΔG reagents

27.21 = -24.3 + 2 * ΔG Cl- - (-314)

ΔG Cl- = - 131.2 kJ / mol is cleared

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Answer 2

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