Solving Equilibrium Problem: PCl5 <--> PCl3 + Cl2

[tornzaer]

Hi, I have no clue how to solve this problem:

Given that the initial n of PCl5 is 0.40 mol and the equilibrium n of Cl2 = 0.25 mol for the following equilibrium in a 10.0 L flask, find the Ke for these conditions.

PCl5(g) <-----> PCl3(g) + Cl2(g)


Thanks a lot of your help. :)

 

[tornzaer]

Bump. Can someone please help me?

 

[Blueshift5]

To solve this equilibrium problem, we can use the equilibrium constant (Ke) expression, which is Ke = [PCl3][Cl2]/[PCl5]. We can also use the given information to set up an ICE table (initial, change, and equilibrium).

Initial: PCl5 = 0.40 mol, PCl3 = 0 mol, Cl2 = 0 mol
Change: -x, +x, +x
Equilibrium: PCl5 = 0.40-x, PCl3 = x, Cl2 = x

Substituting these values into the Ke expression, we get:
Ke = (x)(x)/(0.40-x)
Since the equilibrium concentrations of PCl3 and Cl2 are both 0.25 mol, we can set up an equation:
0.25 = x and solve for x, giving us x = 0.25 mol.

Substituting this value back into the Ke expression, we get:
Ke = (0.25)(0.25)/(0.40-0.25) = 0.625/0.15 = 4.17

Therefore, the equilibrium constant (Ke) for this reaction under these conditions is 4.17. This means that at equilibrium, the concentrations of PCl3 and Cl2 are 4.17 times greater than the concentration of PCl5. This information can be useful in predicting how the equilibrium will shift if any changes are made to the system. I hope this helps!