In the activity, click on the Keq and ΔG∘ quantities to observe how they are related. Calculate ΔG∘using this relationship and the equilibrium constant (Keq) obtained in Part A at T=298K:Keq=1.24×1020Express the Gibbs free energy (ΔG∘) in joules to three significant figures.

Respuesta :

Answer: The Gibbs free energy of the reaction is -114629.4 J

Explanation:

To calculate the Gibbs free energy of the reaction, we use the equation:

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

where,

[tex]\Delta G^o[/tex] = Gibbs free energy of the reaction = ?

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

T = temperature of the reaction = 298 K

[tex]K_{eq}[/tex] = equilibrium constant of the reaction = [tex]1.24\times 10^{20}[/tex]

Putting values in above equation, we get:

[tex]\Delta G^o=-(8.314J/mol.K\times 298K\times \ln (1.24\times 10^{20}))\\\\\Delta G^o=-114629.4J[/tex]

Hence, the Gibbs free energy of the reaction is -114629.4 J