Question

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.

Answer 1

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:

$\Delta G^o=-RT\ln K_{eq}$

where,

$\Delta G^o$ = Gibbs free energy of the reaction = ?

R = Gas constant = $8.314 J/K.mol$

T = temperature of the reaction = 298 K

$K_{eq}$ = equilibrium constant of the reaction = $1.24\times 10^{20}$

Putting values in above equation, we get:

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

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