Suppose I have a container of ideal gases $A,B$ and $C$ with a chemical reaction taking place, say the following,

$$ A+B \rightleftharpoons C $$

Now Le Chateliers principle states if the overall pressure is increased, the reaction moves towards the direction where there are less number of moles. Does this make an assumption that the partial pressures of the gases do not change?

Since, I can define the total pressure of the gas as,

$$ P_{tot}=n_{tot}\left(\frac{RT}{V}\right) $$ where, $n_{tot}=n_A+n_B+n_C$ So now if I were to look at the partial pressure of an individual gas say $C$, then $$ \frac{P_C}{P_{tot}}=\frac{n_{C}}{n_{tot}}\frac{RT/V}{RT/V}=\frac{n_{C}}{n_{tot}}=\chi_C $$ Hence, $$ P_C=\chi_i P_{tot} $$

Now if I increase the total pressure, say I double $P_{tot}$. A couple of things could happen looking at the equation above. I could double my partial pressure but keep the mole fraction the same, or I could halve my mole fraction and keep the partial pressure the same, or somewhere in the middle of these two.

Thus, by reducing my mole fraction of $C$ and having little to no change in the partial pressures of my gases, I could violates the Le Chatelier's principle. or I could keep the same mole fraction and thus but instead have the partial pressures of all gases double.

So what really happens if I double the total pressure? I would love some help understanding the partial pressure and Le Chatlier principle happening at play here.


1 Answer 1


If the equilibrium constant is $K_p$, we have $$\frac{p_C}{p_Ap_B}=\frac{1}{P}\frac{x_C}{x_Ax_B}=K_p$$Let the total number of moles be m, such that $m_C=mx_C$, $m_B=mx_B$, $m_A=mx_A$Then $$x_A=\frac{m_A}{m_A+m_B+m_C}$$$$x_B=\frac{m_B}{m_A+m_B+m_C}$$$$x_C=\frac{m_C}{m_A+m_B+m_C}$$Then $$\frac{m}{P}\frac{m_C}{m_Am_B}=K_p$$or, taking the natural log of both sides, and then taking the differential of each side, gives $$\frac{dm}{m}-\frac{dP}{P}+\frac{dm_C}{m_C}-\frac{dm_A}{m_A}-\frac{dm_B}{m_B}=0$$ where dP the change in the total pressure, and $dm_C$, $dm_B=-dm_C$, and $dm_A=-dm_C$ are the corresponding changes in the moles of C, B, and A, respectively. Then, in terms of the differentials of P and $m_C$, we have $$-\frac{dm_C}{m}-\frac{dP}{P}+\frac{dm_C}{m_C}+\frac{dm_C}{m_A}+\frac{dm_C}{m_B}=0$$or, $$\frac{dP}{P}=(m_{A}+m_{B})\left[\frac{1}{m m_{C}}+\frac{1}{m_{A}m_{B}}\right]dm_C$$ So if the total pressure increases, the number of moles of C increases.

  • $\begingroup$ Aah, I see, I think I better understand the Le Chatlier principle now. The equilibrium constant K becomes a function of total pressure in this scenario. So increasing the pressure, increases K and thus shifting the reaction to the right. But if I had A + B --> 2C, then K would have been independent of total pressure P. Thus doubling the pressure would have no effect on K and we would just have the partial pressures of each of the gases double with no changes in the number of moles on either side $\endgroup$
    – KugelBlitz
    Commented Nov 4, 2023 at 19:57
  • $\begingroup$ K is always a function of T only, not P. $\endgroup$ Commented Nov 5, 2023 at 11:48
  • $\begingroup$ aah yes. Above you had $$K_p = \frac{1}{P}\frac{x_c}{x_Ax_B}$$ so increasing P here would effect $K_p$ as we saw through your calculation. If I now had the scenario that $$A+B\rightleftharpoons 2C$$. Now If I double the total pressure, the partial pressure of each of the gases would double, since $dm/m=dm_c/m_c=0$, so no change in mole fraction? $\endgroup$
    – KugelBlitz
    Commented Nov 6, 2023 at 4:19
  • 1
    $\begingroup$ Yes, that is correct. $\endgroup$ Commented Nov 6, 2023 at 11:08

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