Question

Phosphorus pentachloride decomposes at higher temperatures. PCl5(g) ⇄ PCl3(g) + Cl2(g) An equilibrium mixture at some temperature consists of 3.74 g PCl5, 208.23 g/mol 4.86 g PCl3, 137.33 g/mol 3.59 g Cl2, 70.91 g/mol in a 1.00-L flask. If you add 1.31 g of Cl2, how will the equilibrium be affected and what will the concentration of PCl5 be when equilibrium is reestablished? shift left shift right no shift will occur [PCl5] = mol/L

Answer 1

Answer:

The reaction shifts from right to left. The new concentration of PCl₅ in equilibrium is 1.133 M.

Explanation:

The stress applied to the system is the addition of Cl₂. To offset this  stress, some Cl₂ reacts with PCl₃ to produce PCl₅ until a new equilibrium is established.

The net reaction therefore shifts from right to left, that is,

PCl₅ (g) ← PCl₃ (g) + Cl₂ (g)

To work with the Kc expression, we need to convert the grams to moles. Since the volume of the flask is 1 liter, the moles can be expressed as molarity.

For PCl₅:

208.23 g ---------- 1 mol

3.74 g -------------- x= 1.80 x10⁻² mol ⇒ 1.80 x10⁻² M

For PCl₃:

137.33 g ---------- 1 mol

4.86 g ------------- x= 3.54 x10⁻² mol ⇒ 3.54 x10⁻² M

For Cl₂:

70.91 g ---------- 1 mol

3.59 g ----------- x= 5.06 x10⁻² mol ⇒ 5.06 x10⁻² M

Now, taking into account the shift of the equation, we make an ICE chart:

                   PCl₅ (g)       ←            PCl₃ (g) +                           Cl₂ (g)

i)             1.80 x10⁻²                   3.54 x10⁻²                           5.06 x10⁻²

c)                   +x                                 -x                             +1.31 x10⁻² -x

e)           1.80 x10⁻² +x           3.54 x10⁻² -x             5.06 x10⁻²+1.31 x10⁻² -x

Now, we write the Kc formula:

Kc = [PCl₃] [Cl₂] / [PCl₅]

Kc = [ 3.54 x10⁻² -x] [5.06 x10⁻²+1.31 x10⁻² -x] / [1.80 x10⁻² +x]

After working with the expression, we find that x = 1.115.

Therefore, the new concentration of  PCl₅ will be 1.133 M.

Answer 2

Answer:

The new concentration of PCl5 will be 0.0212637 M

Explanation:

Step 1: Data given

Mass of Cl2 added = 1.31 grams

Molar mass Cl2 = 70.9 g/mol

Original Equilibrium Mixture:

3.74 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 = 3.74 grams / 208.23 g/mol

Moles PCl5 = 0.0180 moles

[PCl5] = 0.0180 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] = 0.0180 + 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)/0.0180

Kc = 0.0995

Step 5: Calculate [PCl5]

Kc = 0.0995 = ((0.0691 - X)(0.0354 - X)) / (0.0180 + X)

X = 0.0032637

[PCl5] = 0.0180 + 0.0032637 M = 0.0212637 M

[PCl3 ] =  0.0354 - 0.0032637 M = 0.0321363 M

[Cl2] = 0.0691 - 0.00313 M = 0.0658363

The new concentration of PCl5 will be 0.0212637 M