PCL5=PCl3+Cl2
a. Initially 0.72 mole PCl5 are placed in 1.50 L flask. At equilibrium, there is 0.16 mole PCl3 in the flask. What is the equilibrium concentration of the PCl5?
b. What is the equilibrium concentration of the Cl2?
c.What is the equilibrium constant for the reaction?
0.72/1.50L = 0.48 M = (PCl5)
0.16/1.50L = 0.107 M = (PCl3)
.........PCl5 ==> PCl3 + Cl2
I........0.48......0......0
C........-x........x.......x
E.......0.48-x.....x......x
x = 0.107M
(PCl3) = 0.107M
(Cl2) = 0.107M
(PCl5) = 0.48-0.107 = ?
Plug these values into the Kc expression and solve for Kc.
To solve this problem, we'll use the equation for the equilibrium constant, Kc:
Kc = [PCl3] * [Cl2] / [PCl5]
a. We are given that initially 0.72 moles of PCl5 are placed in a 1.50 L flask. At equilibrium, there is 0.16 moles of PCl3. We need to find the equilibrium concentration of PCl5.
[PCl5] = (0.72 - 0.16) / 1.50 = 0.40 M
b. We need to find the equilibrium concentration of Cl2. Since Cl2 is not mentioned in the initial or final moles given, we'll assume that it is produced in the same amount as PCl3:
[Cl2] = 0.16 M
c. To find the equilibrium constant Kc, we can use the given values from part a and b:
Kc = [PCl3] * [Cl2] / [PCl5] = 0.16 * 0.16 / 0.40
Kc = 0.064 / 0.40
Kc = 0.16
To find the equilibrium concentrations of PCl5, PCl3, and Cl2, we can use the given data as well as the stoichiometry of the balanced equation.
a. To find the equilibrium concentration of PCl5, we need to subtract the moles of PCl3 formed from the initial moles of PCl5. Given that initially there are 0.72 moles of PCl5 and at equilibrium there is 0.16 moles of PCl3, we can calculate the equilibrium moles of PCl5 as follows:
Equilibrium moles of PCl5 = Initial moles of PCl5 - Moles of PCl3 formed
= 0.72 moles - 0.16 moles
= 0.56 moles
To find the equilibrium concentration of PCl5, we divide the moles of PCl5 by the volume of the flask:
Equilibrium concentration of PCl5 = Moles of PCl5 / Volume of flask
= 0.56 moles / 1.50 L
= 0.373 M
Therefore, the equilibrium concentration of PCl5 is 0.373 M.
b. To find the equilibrium concentration of Cl2, we use the stoichiometry of the balanced equation. The equation tells us that 1 mole of PCl5 reacts to form 1 mole of Cl2. So, the moles of Cl2 formed would be the same as the moles of PCl3 formed:
Moles of Cl2 formed = Moles of PCl3 formed
= 0.16 moles
To find the equilibrium concentration of Cl2, we divide the moles of Cl2 by the volume of the flask:
Equilibrium concentration of Cl2 = Moles of Cl2 / Volume of flask
= 0.16 moles / 1.50 L
= 0.107 M
Therefore, the equilibrium concentration of Cl2 is 0.107 M.
c. The equilibrium constant (Kc) can be calculated using the equilibrium concentrations of the reactants and products. The balanced equation is PCl5 = PCl3 + Cl2.
Kc = [PCl3] * [Cl2] / [PCl5]
Substituting the given equilibrium concentrations, we get:
Kc = (0.16 M) * (0.107 M) / (0.373 M)
= 0.046
Therefore, the equilibrium constant for the reaction is 0.046.