Question

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 1

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

Step-by-step explanation:

Relation between standard Gibbs free energy and equilibrium constant follows:

`\Delta G^o=-RT\ln K_(eq)`

where,

`\Delta G^o` = Standard Gibbs free energy = ?

R = Gas constant =
`8.314J/K mol`

T = temperature = 298 K

`K_(eq)` = equilibrium constant = 581

Putting values in above equation, we get:

`\Delta G^o=-(8.314J/Kmol)* 298K* \ln (581)\\\\\Delta G^o=-15769.13J/mol=-15.8kJ/mol`

Conversion factor used: 1 kJ = 1000 J

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