What will the concentration of PCl5 be when equilibrium is reestablished after addition of 1.31 g Cl2? PCl5(g) ⇆ PCl3(g) + Cl2(g) Original Equilibrium Mixture: 3.42 g PCl5 4.86 g PCl3 3.59 g Cl2 in a 1.00-L flask.

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