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

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $K$.

Answer 4

Let:

$A$ represent $\ce{PCl5}$

$C$ represent $\ce{PCl3}$

$D$ represent $\ce{Cl2}$

Then the reaction becomes:

$$\ce{A <=> C +D}$$

Let:

(1) represent "before removing $\ce{Cl2}$"

(2) represent "after removing $\ce{Cl2}$"

Assuming that, initially, we have $P_{A0}$ amount of $A$ and no $C$ or $D$ present:

$$P{A_1= P_{A0}-x}$$

$$P{C_1=0+x}$$

$$P{D_1=0+x}$$

$$---------$$

$$P_1=P_{A0}+x$$

Then, we remove all the $\ce{Cl2}$, so more products are formed:

$$P{A_2=P_{A0}-x-y}$$

$$P{C_2=x+y}$$

$$P{D_2=y}$$

$$---------$$

$$P_2=P_{A0}+y$$

Then, we define the difference in total equilibrium pressure as:

$$\Delta P=P_2-P_1=y-x$$

Temperature is constant, so $K_P$ won't change after removing $\ce{Cl2}$:

$$K_P=\frac{P_{C2}\;P_{D2}}{P_{A2}}=\frac{P_{C1}\;P_{D1}}{P_{A1}}$$

Rearranging:

$$\left(\frac{P_{C2}}{P_{C1}}\right)\left(\frac{P_{D2}}{P_{D1}}\right)=\left(\frac{P_{A2}}{P_{A1}}\right)$$

Substituting in terms of $x$, $y$, and $P_{A0}$:

$$\left(\frac{x+y}{x}\right)\left(\frac{y}{x}\right)=\left(\frac{P_{A0}-x-y}{P_{A0}-x}\right)$$

Now, we define the dissociation fractions of $A$ for equilibrium system (1) and (2), respectively as:

$$\alpha=\frac{x}{P_{A0}}$$

$$\beta=\frac{y}{P_{A1}}=\frac{y}{P_{A0}-x}$$

Solving for $x$ and $y$:

$$x=P_{A0}\;\alpha$$

$$y=P_{A0}\;\beta-x\;\beta=P_{A0}\;\beta-P_{A0}\;\alpha\;\beta$$

Substituting $x$ and $y$ in the expression we derived earlier:

$$\left(\frac{P_{A0}\;\alpha+P_{A0}\;\beta-P_{A0}\;\alpha\;\beta}{P_{A0}\;\alpha}\right)\left(\frac{P_{A0}\;\beta-P_{A0}\;\alpha\;\beta}{P_{A0}\;\alpha}\right)=\left(\frac{P_{A0}-P_{A0}\;\alpha-P_{A0}\;\beta-P_{A0}\;\alpha\;\beta}{P_{A0}-P_{A0}\;\alpha}\right)$$

All the $P_{A0}$ terms cancel out:

$$\left(\frac{\alpha+\beta-\alpha\;\beta}{\alpha}\right)\left(\frac{\beta-\alpha\;\beta}{\alpha}\right)=\left(\frac{1-\alpha-\beta-\alpha\;\beta}{1-\alpha}\right)$$

After expanding, we get:

$$\alpha^3-\alpha^3\beta^2+3\alpha^2\beta^2-\alpha^2\beta-\alpha^2-3\alpha\beta^2+\alpha\beta+\beta^2=0$$

The positive solution for $\beta$ in terms of $\alpha$ is:

$$\beta=\frac{\alpha\left(\sqrt{4\alpha^2-8\alpha+5}-1\right)}{2(1-\alpha)^2}=f(\alpha)$$

Recalling our derived expression for $\Delta P$:

$$\Delta P=y-x=P_{A0}\;\beta-P_{A0}\;\alpha\;\beta-P_{A0}\;\alpha=P_{A0}(\beta-\alpha\beta-\alpha)$$

$$\Delta P=P_{A0}[(\beta(1-\alpha)-\alpha)]$$

Since $P_{A0}>0$, the term in between "[ ]" controls the sign of $\Delta P$, so we can evaluate 3 different outcomes.

Assuming just for a moment that $\Delta P=0$, we would get:

$$\beta(1-\alpha)-\alpha=0$$

Solving for $\beta$:

$$\beta=\frac{\alpha}{1-\alpha}=g(\alpha)$$

Then the three possible outcomes are:

(i) If $\beta>g(\alpha)$, then $\Delta P >0$

(ii) If $\beta=g(\alpha)$, then $\Delta P =0$

(iii) If $\beta<g(\alpha)$, then $\Delta P <0$

Now, we can compare $f(\alpha)$ vs $g(\alpha)$, noting that $f$ represents the actual value of $\beta$, and $g$ represents the exact boundary at which $\Delta P$ =0.

Considering $\alpha$ is a dissociation fraction, and the fact that for equilibrium to make sense neither 0% nor 100% of dissociation is possible, the domain of $\alpha$ is:

$$\alpha ∈ (0,1)$$

We can now tabulate $\alpha$ vs $f(\alpha)$ vs $g(\alpha)$:

\begin{array} {|r|r|}\hline \alpha & f(\alpha) & g(\alpha) \\ \hline \to 0 & \to 0 & \to 0 \\ \hline 0.1 & 0.06538 & 0.1111 \\ \hline 0.2 & 0.1386 & 0.25 \\ \hline 0.3 & 0.2206 & 0.4286 \\ \hline 0.4 & 0.3122 & 0.6667 \\ \hline 0.5 & 0.4142 & 1 \\ \hline 0.6 & 0.5262 & 1.5 \\ \hline 0.7 & 0.6463 & 2.333 \\ \hline 0.8 & 0.7703 & 4 \\ \hline 0.9 & 0.8912 & 9 \\ \hline \to 1 & \to 1 & \to \infty \\ \hline \end{array}

From the table, we can observe that for every value within the domain of $\alpha$, $f$ is smaller than $g$:

$$\alpha ∈ (0,1) \to f(\alpha)<g(\alpha)$$

In other words, condition (iii) is met, and $\Delta P$ is negative, so total equilibrium pressure decreases after removing all the $\ce{Cl2}$:

$$\beta<g(\alpha)\to \Delta P<0\to P_2<P_1$$

Although it's not very important, as a side note it also follows that:

$$\alpha ∈ (0,1) \to \beta<\alpha$$

I included a graph below to illustrate the solution of the problem, where I also added a $\beta=\alpha$ function as a reference line: