-1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

Specifically,

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

$Q$ 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 $Q$.
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.