Answer:
The new concentration of PCl5 will be 0.01953 M
Explanation:
Step 1: Data given
Mass of Cl2 added = 1.31 grams
Molar mass Cl2 = 70.9 g/mol
Original Equilibrium Mixture:
3.42 g PCl5
4.86 g PCl3
3.59 g Cl2
Volume = 1.0 L
Step 2: The balanced equation
PCl5(g) ⇆ PCl3(g) + Cl2(g)
Step 3: Calculate the original moles and molarity
Moles = mass / molar mass
Moles PCL5 = 3.42 grams / 208.24 g/mol
Moles PCl5 = 0.0164 moles
[PCl5] = 0.0164 M
moles PCl3 = 4.86 grams / 137.33 g/mol
moles PCl3 = 0.0354 moles
[PCl3] = 0.0354 M
moles Cl2 = 3.59 grams / 70.9 g/mol
moles Cl2 = 0.0506 moles
[Cl2] = 0.0506 M
the new mass Cl2 = 3.59 + 1.31 = 4.9 grams
moles Cl2 = 0.0691 moles
[Cl2]= 0.0691 M
The new concentration at the equilibrium
[PCl5] = 0.0164 + X M
[PCl3 ] = 0.0354 - X M
[Cl2] = 0.0691 - X M
Step 4: Calculate Kc
Kc = [Cl2][PCl3] / [PCl5]
Kc = (0.0506*0.0354)/0.0164
Kc = 0.109
Step 5: Calculate [PCl5]
Kc = 0.109 = ((0.0691 - X)(0.0354 - X)) / (0.0164 + X)
X = 0.00313
[PCl5] = 0.0164 + 0.00313 M = 0.01953 M
[PCl3 ] = 0.0354 - 0.00313 M = 0.03227 M
[Cl2] = 0.0691 - 0.00313 M = 0.06597
The new concentration of PCl5 will be 0.01953 M
