Question

Calculate the standard free energy change for a reaction at 298 kelvin the equilibrium constant of the reaction is 50​

Answer 1

Answer:

The standard free energy change was -9.69 KJ

Explanation:

ΔG° =  $-RT\ln{(K_{eq}) }$

⇒ΔG° = -8.314×298×ln(50) =  -9.69 KJ