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

Step-by-step 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* 10^(20)`

Putting values in above equation, we get:

`\Delta G^o=-(8.314J/mol.K* 298K* \ln (1.24* 10^(20)))\\\\\Delta G^o=-114629.4J`

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