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. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are 77.1 k/mole and −39.5 kJ/mole respectively, and the solubility product for Ag2S(s) is 8 10−51.

Respuesta :

Answer: The standard free energy change of formation of [tex]S^{2-}(aq.)[/tex] is 92.094 kJ/mol

Explanation:

We are given:

[tex]K_{sp}\text{ of }Ag_2S=8\times 10^{-51}[/tex]

Relation between standard Gibbs free energy and equilibrium constant follows:

[tex]\Delta G^o=-RT\ln K[/tex]

where,

[tex]\Delta G^o[/tex] = standard Gibbs free energy = ?

R = Gas constant = [tex]8.314J/K mol[/tex]

T = temperature = [tex]25^oC=[273+25]K=298K[/tex]

K = equilibrium constant or solubility product = [tex]8\times 10^{-51}[/tex]

Putting values in above equation, we get:

[tex]\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ[/tex]

For the given chemical equation:

[tex]Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)[/tex]

The equation used to calculate Gibbs free change is of a reaction is:  

[tex]\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}][/tex]

The equation for the Gibbs free energy change of the above reaction is:

[tex]\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})][/tex]

We are given:

[tex]\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ[/tex]

Putting values in above equation, we get:

[tex]285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol[/tex]

Hence, the standard free energy change of formation of [tex]S^{2-}(aq.)[/tex] is 92.094 kJ/mol

pstnonsonjoku

The enthalpy of formation of the S^2- is 92.1 kJ/mol.

The standard free energy change can be computed from the information provided as follows;

ΔGrxn = -RTlnKsp

ΔGrxn = -(8.314 × 298 × ln8 × 10−51) = 285.79 kJ/mol

Now;

Ag2S(s) -----> 2Ag^+(aq) + S^2-(aq)

285.79 kJ/mol =  [2(77.1kJ/mole) + S^2-] - [-39.5kJ/mole] -

285.79 kJ/mol = 193.7 kJ/mol + S^2-

S^2- = -193.7 kJ/mol +   285.79 kJ/mol

S^2- = 92.1 kJ/mol

Learn more about enthalpy of formation: https://brainly.com/question/13164491

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