# Will the total equilibrium pressure increase in gas phase reaction PCl5(g) ⇌ PCl3(g) + Cl2(g) after chlorine is removed?

Will the total pressure be greater than the initial equilibrium pressure if all $$\ce{Cl2}$$ is removed and the following reaction is allowed to re-equilibrate?

$$\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}$$

Assume constant temperature and volume.

I know that when $$\ce{Cl2}$$ is removed, the reaction will go forward and I also know expressions of $$K_p$$ and $$K_c$$. I also know Le Chatelier's principle. I have thought about it for a long time but have not made any progress.

• The final pressure will simply not change, if the temperature is maintained constant, whatever the value of the equilibrium constant.. If the system contains $n$ moles $\ce{PCl5}$ at the beginning, and if $x$ moles $\ce{PCl5}$ are transformed into $\ce{PCl3}$ and if $\ce{Cl2}$ is removed by reaction with a metal for example, the final amount of $\ce{PCl5}$ is $n-x$ and the amount of $\ce{PCl3}$ is $x$. As a consequence, the total amount of gas is $n-x + x = n$. So it does not change. Commented May 17, 2021 at 10:00
• @Maurice note the question has been edited to make it more clear. Commented May 17, 2021 at 18:00
• This answer may help chemistry.stackexchange.com/questions/150785/… Commented May 17, 2021 at 19:39
• @Buck Thorn. Whatever the changes in the text, the number of gaseous moles does not change as the reaction proceeds, if the $\ce{Cl2}$ is eliminated from the gaseous phase (for example by reaction with iron producing solid $\ce{FeCl3}$). Commented May 18, 2021 at 7:54

First it is necessary to understand that due to the stoichiometry of the reaction, if you move the equilibrium to the products side, pressure will be increased. This is based on the fact that 1 mole of gas produces 2 moles of gas. In other words the number of molecules in the system is increased. By removing all of the $\ce{Cl2}$ you force the reaction to move to the products side (Le Chatelier's Principle). Thus more $\ce{PCl3}$ will be produced along with $\ce{Cl2}$. Eventually you will end up with more atoms inside the system and that will lead to increase of the pressure.

• @Kelesedes Adonios I agree that the total number of molecules generated increases, but we are removing the chlorine fraction entirely, right? So, a product of partial pressures equal to the earlier one will develop gradually, right? But, will this increase the number of molecules present in the reaction mixture? Would you please support your answer with maths? Commented Dec 25, 2016 at 6:04

After giving this problem more thought, I have come up with a better demonstration for why equilibrium pressure will decrease after removing the chlorine gas.

Let:

A = $$\ce{PCl5}$$

C = $$\ce{PCl3}$$

D = $$\ce{Cl2}$$

The change in moles for this reaction is:

$$\Delta n=1+1-1=1$$

The subscript (0) will represent the initial non-equilibrium state (i.e. immediately after removing all the chlorine gas, but before reaching the second equilibrium state).

The subscript (1) will represent the initial equilibrium state (i.e. before removing all the chlorine gas).

The subscript (2) will represent the final equilibrium state (i.e. after removing all the chlorine gas).

Ok, so we place pure A in a sealed flask at constant temperature and let it dissociate into C and D until equilibrium in state (1) is obtained. Since the products are in a 1:1 ratio with no initial amounts, their mole fractions in such state will be equal:

$$\boxed{X_{C_1}=X_{D_1}}$$

Since only D is being removed in (0), the partial pressures of A and C between (0) and (1) remain constant:

$$P_{A_1}=P_{A_0}$$

$$P_{C_1}=P_{C_0}$$

The equilibrium partial pressures in state (2) are:

$$P_{A_2}\;\;=\;\;P_{A_0}-x\;\;=\;\;P_{A_0}-\alpha\;P_{A_0}\;\;=\;\;P_{A_0}\left(1-\alpha\right)\;\;=\;\;\pmb{P_{A_1}\left(1-\alpha\right)}$$

$$P_{C_2}\;\;=\;\;P_{C_0}-x\;\;=\;\;P_{C_0}-\alpha\;P_{A_0}\;\;=\;\;P_{A_0}\left(\theta_{C_0}-\alpha\right)\;\;=\;\;\pmb{P_{A_1}\left(\theta_{C_1}-\alpha\right)}$$

$$P_{D_2}\;\;=\;\;x\;\;=\;\;\alpha\;P_{A_0}\;\;=\;\;\pmb{\alpha\;P_{A_1}}$$

So the total equilibrium pressure in state (2) is:

$$P_2=P_{A_1}\left(1+\theta_{C_1}+\alpha\right)$$

Dividing both sides by the total equilibrium pressure in state (1):

$$\frac{P_2}{P_1}=X_{A_1}\left(1+\theta_{C_1}+\alpha\right)$$

The pressure ratio of C in (0) and (1) is equal to the mole fraction ratio of C in (0) and (1):

$$\theta_{C_0}=\frac{X_{C_0}}{X_{A_0}}=\frac{P_{C_0}}{P_{A_0}}=\frac{P_{C_1}}{P_{A_1}}=\frac{X_{C_1}}{X_{A_1}}=\theta_{C_1}$$

So by distributing, we get:

$$\frac{P_2}{P_1}=X_{A_1}\alpha+X_{C_1}+X_{A_1}$$

In state (1), the mole fraction of A and D can be set in terms of the mole fraction of C:

$$X_{A_1}+X_{C_1}+X_{D_1}=1\;\;\implies\;\; X_{A_1}+2X_{C_1}=1\implies \;\;\boxed{X_{A_1}=1-2X_{C_1}}$$

So the pressure ratio equation can be transformed into:

$$\boxed{\frac{P_2}{P_1}=\left(1-2X_{C_1}\right)\alpha+1-X_{C_1}}$$

Since $$K_p$$ remains constant as there is no change in temperature and $$\Delta n=1$$, the ratio of equilibrium constants in terms of mole fractions between (2) and (1) is:

$$\frac{K_{x_2}}{K_{x_1}}=\frac{\require {cancel} \cancel{K_p}}{\cancel{K_p}}\left(\frac{P_1}{P_2}\right)^{\cancel{\Delta n}}\;\implies\; K_{x_2}=\frac{K_{x_1}}{\left(1-2X_{C_1}\right)\alpha+1-X_{C_1}}$$

And $$K_{x_1}$$ is defined as:

$$\boxed{K_{x_1}=\frac{X_{C_1}\;X_{D_1}}{X_{A_1}}}$$

Now we need a formula that generates the value of $$\alpha$$ only in terms of molar composition in state (1):

$$\frac{\left(\theta_{C_1}+\alpha\right)\alpha}{\left(1-\alpha\right)\left(\theta_{C_1}+1+\alpha\right)}=K_{x_2}=\frac{K_{x_1}}{\left(1-2X_{C_1}\right)\alpha+1-X_{C_1}}$$

Solving for $$\alpha$$:

$$\boxed{\alpha=\frac{\sqrt{4X_{A_1}\;K_{x_1}+X_{C_1}^2+2K_{x_1}\;X_{C_1}+K_{x_1}^2}-X_{C_1}-K_{x_1}}{2X_{A_1}}}$$

We can now tabulate all relevant formulas and plot $$\frac{P_2}{P_1}$$ as a function of $$X_{C_1}$$, keeping in mind that the mole fraction of $$C$$ in (1) can only adopt the following values:

$$0

$$\begin{array} {|r|r|}\hline \pmb{X_{C_1}} & \pmb{X_{D_1}} & \pmb{X_{A_1}} & \pmb{K_{x_1}} & \pmb{\alpha} & \pmb{\frac{P_2}{P_1}} \\ \hline \to 0 & \to 0 & \to 1 & \to 0 & \to 0 & \to 1 \\ \hline 0.0500 & 0.0500 & 0.900 & 0.00278 & 0.0335 & 0.980 \\ \hline 0.100 & 0.100 & 0.800 & 0.0125 & 0.0731 & 0.9585 \\ \hline 0.150 & 0.150 & 0.700 & 0.0321 & 0.121 & 0.9585 \\ \hline 0.200 & 0.200 & 0.600 & 0.0667 & 0.178 & 0.907 \\ \hline 0.250 & 0.250 & 0.500 & 0.125 & 0.25 & 0.875 \\ \hline 0.300 & 0.300 & 0.400 & 0.225 & 0.340 & 0.836 \\ \hline 0.350 & 0.350 & 0.300 & 0.408 & 0.456 & 0.787 \\ \hline 0.400 & 0.400 & 0.200 & 0.800 & 0.606 & 0.721 \\ \hline 0.450 & 0.450 & 0.100 & 2.03 & 0.793 & 0.629 \\ \hline \to 0.500 & \to 0.500 & \to 0 & \to \infty & \to \infty & \to 0.5 \\ \hline \end{array}$$

As we can see, the pressure ratio is inversely proportional to the molar composition of C in (1) and $$\alpha$$, which means it is always lower than 1, but higher than 0.5, and decreases as the reaction progresses and the mole fraction of C in (1) increases:

$$0.5<\frac{P_2}{P_1}<1$$

In conclusion, the new equilibrium pressure will necessarily always be lower than the initial equilibrium pressure after removing all the chlorine gas, but forming an initial equilibrium mixture with low and equal molar compositions of $$\ce{PCl3}$$ and $$\ce{Cl2}$$, and high molar composition of $$\ce{PCl5}$$ will only result in a small decrease in total pressure.

Since this reaction is endothermic, conducting it under low temperature will result in a negligible pressure drop $$(P_2\lesssim P_1)$$, and conducting it under high temperature will result in the pressure decreasing almost by half $$(P_2\gtrsim 0.5P_1).$$

## No.

For this problem I'm going to assume ideal gas conditions, mostly because looking into nonideal behavior of PCl3 and PCl5 at unspecified pressure and temperature sounds painful. There are some intermolecular forces with phosphorus compounds to consider, but I didn't see much about PCl3/PCl5 dimerization in a quick search and this table gives an entry for PF3 that doesn't seem extraordinary, though it doesn't have PCl3 or PCl5. My guess is PCl3 ought to be smaller than PCl5 if anything.

For ideal conditions, you can just say every particle smacks up against the walls of the container with a certain average energy that depends only on temperature, not what the particle is. 1 kPa = 1 J/L, representing the work needed to push back all the particles in one degree of freedom for a distance x area. (The energy in one degree of freedom is half that, but thermodynamics won't let you stop them in their tracks at absolute zero, so you have to pay double).

In other words, the amount of pressure is proportional to the number of particles ("Avogadro's law") If you remove the chlorine, the number of particles won't change when you turn PCl5 into PCl3. Assuming the question means first you remove the chlorine then you see the equilibrium reestablish itself, with more chlorine being removed, you ought to see no change in pressure. Otherwise you see a reduction in proportion to the original amount of chlorine.

• The statements in the last paragraph are incorrect. If you continuously remove chlorine, the pressure will decrease a lot. If you remove chlorine once, it will also decrease.
– Karsten
Commented Sep 30, 2022 at 21:50
• @Karsten - under an ideal gas approximation, a mole of PCl3, PCl5, Cl2, or burrito gas exerts the same pressure when at the same volume or temperature. Removing the Cl2 therefore cannot reduce the pressure below what it started at, unless the PCl3 would show non-ideal characteristics (such as dimerization) that reduce its pressure to a greater degree than PCl5. Commented Oct 3, 2022 at 21:39

I know that when $$\ce{Cl2}$$ is removed, the reaction will go forward and I also know expressions of $$K_p$$ and $$K_c$$. I also know Le Chatelier's principle. I have thought about it for a long time but have not made any progress.

Seven years have passed for thinking about the answer. Knowing the direction of the reaction is a good first step. We can also summarize what happens to the pressure when the reaction goes forward or reverse:

$$\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}$$

For every molecule of $$\ce{PCl5}$$ that reacts, we gain one molecule. Said differently, the increase in gas molecules total is equal to the increase in chlorine molecules.

So if you remove all the chlorine gas, the reaction would have to replace all of it by going forward to reach the same pressure (assuming ideal gas). Now we are back at the original pressure. Are we at equilibrium?

The chlorine concentration is the same, but the $$\ce{PCl5}$$ is lower than the starting point (we had to use up some to replenish the chlorine gas), and the $$\ce{PCl3}$$ concentration is higher. If you look at the equilibrium constant expression, the reaction is not at equilibrium but has to reverse to reach it, decreasing the pressure.

Here is a different way to think about this. You start with the initial state $$S_0$$. You take away the chlorine and let the system reach equilibrium. This is state $$S_1$$, which has a certain pressure and a certain concentration of chlorine $$c_1$$. Now you ask what happens when you add the chlorine again. To remain at the same pressure, all the added chlorine would have to react, using up $$\ce{PCl3}$$ and making $$\ce{PCl5}$$. The chlorine concentration would still be $$c_1$$, but there would be more $$\ce{PCl5}$$ and less $$\ce{PCl3}$$ now than in state $$S_1$$. To reach equilibrium, the reaction has to go forward, increasing the pressure when it goes back to the equilibrium state $$S_0$$.

Both ways of thinking about it give the same conclusion: Removing chlorine will reduce the equilibrium pressure. This is reflected in one way of explaining Le Chatelier. "The reaction will go in the direction that partially undoes the change imposed from the outside." The principle does not always give you the right answer, so it is good to check with a specific example using the equilibrium constant expression and the criterion $$Q$$ vs $$K$$.