Question

The equilibrium constant for a reaction is $10$. What will be the value of $∆G^Θ$?
$R= 8.314\ JK^{-1} mol^{-1}, T=300\ K$

Answer 1

From the expression,
$∆G^Θ = –2.303 RT\ log\ K_{eq}$
$∆G^Θ$ for the reaction,
= $(2.303) (8.314\ JK^{–1} mol^{–1}) (300\ K) log\ 10$
= $–5744.14\ Jmol^{–1}$
= $–5.744\ kJ mol^{–1}$