2

I'm reading Clayden's Organic Chemistry textbook and in Chapter 12 on equilibrium he writes: enter image description here

I am confused by the usage of $\Delta G$ as opposed to $\Delta G^\circ$. Isn't the equilibrium constant defined in terms of the standard-state Gibbs energy? Is there an unstated approximation in his definition of the equilibrium constant in terms of the bare $\Delta G$?

pQ12branch
  • 311
  • 1
  • 7
  • Yes, you are right. That equation should read $\Delta G^\circ = -RT \ln K_\text{eq}$. $\Delta G$ refers generally to the Gibbs free energy change for any process. – Buck Thorn Jan 16 '23 at 05:14
  • Suppose the reaction is $A\rightleftharpoons B$ then the equation is $\Delta G'=\Delta G^0 +RT \ln(P_B/P_A)$ and at equilibrium the pressures are the equilibrium values and $\Delta G'=0$ so that $\Delta G^0 =-RT\ln(K_p)$ where $K_p$ is the equilibrium constant. – porphyrin Jan 16 '23 at 17:29

0 Answers0