If I increase P, to compensate that change the equilibrium will move to the right if the forward reaction produces fewer moles.Then the change in pressure will be undone. Now my question is how is Kp staying constant while at the last pressure is just the same as before but number of moles are changed?Can it be explained with the equation of Kp involving pressures and mole fractions?

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    $\begingroup$ Have a look at this answer, its not exactly the same but closely related.chemistry.stackexchange.com/questions/73644/… $\endgroup$
    – porphyrin
    May 14, 2017 at 19:58
  • $\begingroup$ If mole fraction changes, partial pressure will change and total pressure will change too.That means the increase of pressure will be undone.Then why are we using the increased pressure in the last mentioned equation instead of using the unchanged pressure? $\endgroup$ May 14, 2017 at 20:27

1 Answer 1


Mathematically speaking, $$\Delta G_\mathrm{r} = \Delta {G_\mathrm{r}}^{\circ} + RT \ln Q$$

Here, $\Delta G_\mathrm{r}$ is the molar Gibb's free energy of the reaction and $\Delta {G_\mathrm{r}}^{\circ}$ is the standard molar Gibb's free energy of the reaction.

At equilibrium, $Q = K$ and $\Delta G_\mathrm{r} = 0$, thus $\Delta {G_\mathrm{r}}^{\circ} = -RT \ln K$. From this equation, you can clearly see that $K$ is dependent only on temperature and the molar Gibb's free energy of the system. So, adding or removing reactants, products, or changing pressure or volume won't affect $K$ or $K_p$.

For $K_p$ you can replace concentrations of reactants/products in the expression for $K$ by their partial pressures divided by the standard pressure $P^\circ = 1 ~\mathrm{bar}$. And as you know, partial pressure of a gas is equal to the total pressure multiplied by the mole fraction of the particular gas.

  • $\begingroup$ Can you please see the edited question? $\endgroup$ May 14, 2017 at 19:55
  • $\begingroup$ I've edited the answer accordingly. $\endgroup$ May 14, 2017 at 20:46

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