I have the following problem:

The following reversible system is at equilibrium: $$\ce{P_{4(s)} + 6Cl_{2(g)} <=> 4PCl_{3(l)}} $$ What happens with the equilibrium if $\ce{P_4}$ is added.

Normally, by applying Le Chatelier's principle the equilibrium should be shifted to the right, however, for some reason this does not apply to the above equation.

Why does increasing or decreasing the quantities does not change the equilibrium in the above system?

  • 6
    $\begingroup$ It would help yourself, if you wrote down the equation for the equilibrium constant. $\endgroup$
    – Poutnik
    Jan 27, 2021 at 20:41
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    $\begingroup$ The solution twists on the fact that $\ce{P4}$ is a solid and in a different phase than the $\ce{Cl2}$. By definition in this situation the activity of the $\ce{P4}$ is 1 in the equilibrium equation no matter if a microgram or a metric ton is left. $\endgroup$
    – MaxW
    Jan 27, 2021 at 21:08
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    $\begingroup$ @Ziezi - We typically write equilibrium expression with concentrations, but really activities should be used. Think of it this way. You dissolve sugar in water until the solution is saturated. Once saturated it doesn't matter if there is a single grain of sugar or a kilogram of sugar undissolved. No more sugar will dissolve. $\endgroup$
    – MaxW
    Jan 27, 2021 at 22:43
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    $\begingroup$ This question might help. The answer explains why pure solids and liquids have constant concentration. $\endgroup$
    – user102687
    Jan 28, 2021 at 7:06
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    $\begingroup$ The maximum concentration (or the pressure) of $\ce{P4}$ in the vapor phase does not depend on the amount of $\ce{P4}$ in the solid phase. It depends only on the temperature. It is like water in the atmosphere. The vapor pressure of water vapor does not depend on the amount of liquid vapor in contact with the humid air. $\endgroup$
    – Maurice
    Feb 2, 2021 at 21:30

1 Answer 1


Firstly, as an approximation, we can assume that only gases and species in aqueous solution are input into the equilibrium constant expression.

Since $\ce{Cl_2}$ is the only species in the reaction that qualifies (as it's in gaseous state), $K_p$ will be:


Because no change in temperature is taking place, $K_p$ is a constant, which means $P_{\ce{Cl_2}}$ is also a constant since:


Now, since $\ce{Cl_2}$ is the only gas, the total equilibrium pressure $P$ will only depend on $P_{\ce{Cl_2}}$:


This means that no matter how much phosphorus we add, the total equilibrium pressure will not change, since $P$ does not depend on phosphorus.

Let $P_1=$ total equilibrium pressure before adding more phosphorus.

Let $P_2=$ total equilibrium pressure after adding more phosphorus.

$$\Delta P=P_2-P_1=0$$

When we say "equilibrium shifts", (at constant temperature) what we mean is that $K_p$ remains constant, but $K_x$ does not. $K_x$ is the equilibrium expression in terms of molar fractions.

$K_x$ is a "false" or "fake" equilibrium constant in the sense that, while it has an analogous expression with respect to $K_p$, it is not actually always a constant (even when temperature is not changed).

This is due to the fact that $K_x$ changes at the expense of $K_p$ in order for the latter to remain constant, and the relationship between them is:

$$K_p=K_x\times P^{\Delta n}$$

In this case, if we can show that $K_x$ does not change after adding phosphorus, then there is no shift in equilibrium taking place.


$K_{x1}$ represent $K_x$ before phosphorus is added.

$K_{x2}$ represent $K_x$ after phosphorus is added.

$$K_p=K_{x1}\times P_1^{\Delta n}=K_{x2}\times P_2^{\Delta n}$$


$$\frac{K_{x2}}{K_{x1}}=\left(\frac{P_1}{P_2}\right)^{\Delta n}$$

$$K_{x2}=K_{x1}\left(\frac{P_1}{P_2}\right)^{\Delta n}$$

Here we can observe that $K_{x2}$ = $K_{x1}$ when at least one of two things happens:

(1) $\Delta n=0$

(2) $P_2 = P_1$

In our case:

(1) is false since $\Delta n=-6$


(2) is true because earlier we said $\Delta P=0$, so $P_2=P_1$

In conclusion, $\Delta K_x=0$, which means there is no shift in equilibrium for this reaction after adding more phosphorus.


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