Question

Initially, 0.59 mol

of PCl5
is placed in a 1.0 L
flask. At equilibrium, there is 0.20 mol
of PCl3
in the flask. What is the equilibrium concentration of PCl5
?

Answer 1

The equilibrium concentration of PCl5 in a 1.0 L flask, given the initial amount of 0.59 mol and the formation of 0.20 mol of PCl3 at equilibrium, is 0.39 M.

Step-by-step explanation:

To determine the equilibrium concentration of PCl5, we must consider the equilibrium reaction and the initial amount of PCl5 present in the system.

The decomposition of PCl5 can be represented by the following equilibrium reaction:

PCl5(g) ⇌ PCl3(g) + Cl2(g)

Given that initially 0.59 mol of PCl5 is placed in a 1.0 L flask and at equilibrium, there is 0.20 mol of PCl3, we can use the stoichiometry of the reaction to determine the amount of PCl5 that has reacted. Since the formation of 1 mol of PCl3 comes from the reaction of 1 mol of PCl5, we can conclude that 0.20 mol of PCl5 was used to form PCl3.

Initial moles of PCl5: 0.59 mol
PCl3 formed: 0.20 mol (which means 0.20 mol of PCl5 reacted)

Unreacted PCl5:
Initial moles - reacted moles = 0.59 mol - 0.20 mol = 0.39 mol

Since the reaction takes place in a 1.0 L flask, the equilibrium concentration of PCl5 (unreacted moles/volume) is:

Equilibrium concentration of PCl5 = 0.39 M