For a particular reaction at 298 K, the equilibrium constant is equal to 581. Determine ΔG° in kJ/mol for the reaction. Do not include units. Report your answer to 3 significant figures.

Answer :

Answer: The standard Gibbs free energy of the reaction is -15.8 kJ/mol

Explanation:

Relation between standard Gibbs free energy and equilibrium constant follows:

[tex]\Delta G^o=-RT\ln K_{eq}[/tex]

where,

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

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

T = temperature = 298 K

[tex]K_{eq}[/tex]  = equilibrium constant = 581

Putting values in above equation, we get:

[tex]\Delta G^o=-(8.314J/Kmol)\times 298K\times \ln (581)\\\\\Delta G^o=-15769.13J/mol=-15.8kJ/mol[/tex]

Conversion factor used: 1 kJ = 1000 J

Hence, the standard Gibbs free energy of the reaction is -15.8 kJ/mol

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