Question

What is the relation between standard Gibbs energy change and equilibrium constant of a reaction

Answer 1

if the Gibbs free energy imhas more negative value then it value of equal barium constant will have larger value and Vic versa If the value of Gibbs energy is zero then the value of k is almost 1

Answer 2

$\huge\mathfrak{A{\underline{\underline{\textbf{\red{nswer :}}}}}}$


∆G = ∆G° + RT $lnQ_{c}$

∆G = Gibb's energy

∆G° = Gibb's energy at standard condition

R = gas constant = 8.314 JK-¹Mol-¹

$Q_{c}$ = Equilibrium quotient

Now...

∆G = 0

$Q_{c}\:=\:K_{c}$

0 = ∆G° + RT $lnK_{c}$

∆G° = - RT $lnK_{c}$

∆G° = - 2.303 RT log $K_{c}$

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