I just studied the formula for equilibrium constant in thermodynamics. But when an equilibrium is established $\Delta G$ is always $0$ as told by my teacher. Then $k$ has to be always $1$ whenever an equilibrium is established. But that does not happen always as equilibrium constant is not always $1$. I want to know where I am going wrong here. And this $k$ is $k_\mathrm p$ or $k_\mathrm c$ ? Any help will be really appreciated. Thanks in advance.


You are confused between $\Delta G$ and $\Delta G^{\varnothing}$, the standard free energy change.

The equation you are describing is actually:

$$\Delta G^{\varnothing} = -RT \ln K$$

These numbers are fixed for some reference temperature and pressure.

The specific $\Delta G$ you are referring to is a function of $Q$, the reaction quotient, which describes the position of your equilibrium. Its value, relative to $K$, will determine whether $\Delta G$ is positive, negative, or zero.


$$\Delta G = \Delta G^{\varnothing} + RT \ln Q$$

$Q$ is computed the same as $K$ except that you plug in the actual concentrations you have, not the values at equilibrium.

There are three cases to consider:

  1. $Q = K$. This corresponds to $\Delta G = 0$. We are equilibrium.
  2. $Q > K$. $\Delta G > 0$. We are favoring products to much. We need to shift the reaction back to decrease the value of $Q$.
  3. $Q < K$. $\Delta G < 0$. We are favoring reactants to much. We need to shift the reaction forward to increase the value of $Q$.
  • $\begingroup$ Thank you Sir. But I am still confused about this k . Is it kp or kc? $\endgroup$ – Avinash Sharma Oct 9 '18 at 2:30
  • $\begingroup$ Technically, it is neither, but it's closer to $K_{\mathrm{c}}$ than $K_{\mathrm{p}}$. $\endgroup$ – Zhe Oct 9 '18 at 13:20

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