Answer to Question #346240 in General Chemistry for Ashter

Question

Calculate the ΔG⁰_f at 298 K for the PbCI₂(s) from the following information. ΔG⁰ for the reactions below is -58.4 kJ at 298 K.

PbS(s) + 2HCI(g) > PbCI₂(s) + H₂S(g)

ΔG⁰_f (kJ/mol) 97.82 149.0 0 -228.6

Answer 1

If the reaction is carried out under standard conditions (unit concentrations and pressures) and at a temperature that corresponds to a table of thermodynamic values (usually 298 K or 25° C ), then you can subtract the standard Gibbs Free Energy of Formation ( ΔGf ) of the reactants from those of the products:

∆G ⁰ = ∆G(H2S) + ∆G(PbCl ₂ ) - 2*∆G(HCl) - ∆G(PbS);

∆G(PbCl ₂ ) = ∆G ⁰ - ∆G(H2S) + 2*∆G(HCl) + ∆G(PbS);

∆G(PbCl ₂ ) = (-58,4) - (-228,6) + 2*149 + 97,82 = 108,82 kJ

ANSWER: 108,82 kJ