3
$\begingroup$

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 $K_\text{p}$ and $K_\text{c}$. I also know Le Chateliers Principle. I have thought about it for a long time but have not made any progress.

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$
    – Maurice
    May 17 at 10:00
  • $\begingroup$ @Maurice note the question has been edited to make it more clear. $\endgroup$
    – Buck Thorn
    May 17 at 18:00
  • $\begingroup$ This answer may help chemistry.stackexchange.com/questions/150785/… $\endgroup$
    – porphyrin
    May 17 at 19:39
  • 1
    $\begingroup$ @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 $\ce{FeCl3}$). $\endgroup$
    – Maurice
    May 18 at 7:54
1
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ @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? $\endgroup$ Dec 25 '16 at 6:04
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.