Question

The two common chlorides of phosphorus, PCl3, and PCl5, both important for the production of the other phosphorus compounds, coexist in equilibrium through the reaction

PCl3(g) + Cl2(g) = PCl5(g)
At 250 ᵒC , an equilibrium mixture in a 25.0 L flask contains 0.105 g PCl5, 0.220 g PCl3 and 2.12 g of Cl2. What are the values of
a) Kc
b) Kp for this reaction at 250 ᵒC ?

Answer 1

For a: The value of
`K_c` for the given reaction is 271.6

For b: The value of
`K_p` for the reaction is 6.32

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

• For
  `PCl_5` :

Given mass of
`PCl_5` = 0.105 g

Molar mass of
`PCl_5` = 208.24 g/mol

Putting values in equation 1, we get:

`\text{Moles of }PCl_5=(0.105g)/(208.24g/mol)=5.04* 10^(-4)mol`

• For
  `PCl_3` :

Given mass of
`PCl_3` = 0.220 g

Molar mass of
`PCl_5` = 137.33 g/mol

Putting values in equation 1, we get:

`\text{Moles of }PCl_3=(0.220g)/(137.33g/mol)=1.60* 10^(-3)mol`

• For
  `Cl_2` :

Given mass of
`Cl_2` = 2.12 g

Molar mass of
`Cl_2` = 71.0 g/mol

Putting values in equation 1, we get:

`\text{Moles of }Cl_2=(2.12g)/(71.0g/mol)=0.029mol`

Volume of the flask = 25.0 L

For the given chemical equation:

`PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)`

• For a:

The equation used to calculate concentration of a solution is:

`\text{Molarity}=\frac{\text{Moles}}{\text{Volume (in L)}}`

The expression of
`K_c` for above reaction follows:

`K_c=(PCl_5)/(PCl_3* Cl_2)`

We are given:

`[PCl_5]=(5.04* 10^(-4)mol)/(25L)`

`[PCl_3]=(1.60* 10^(-3)mol)/(25L)`

`[Cl_2]=(0.029mol)/(25L)`

Putting values in above equation, we get:

`K_c=(((5.04* 10^(-4))/(25)))/(((1.60* 10^(-3))/(25))* ((0.029)/(25)))\\\\K_c=271.6`

Hence, the value of
`K_c` for the given reaction is 271.6

• For b:

Relation of
`K_p` with
`K_c` is given by the formula:

`K_p=K_c(RT)^(\Delta ng)`

where,

`K_p` = equilibrium constant in terms of partial pressure = ?

`K_c` = equilibrium constant in terms of concentration = 271.6

R = Gas constant =
`0.0821\text{ L atm }mol^(-1)K^(-1)`

T = temperature =
`250^oC=250+273=523K`

`\Delta n_g` = change in number of moles of gas particles =
`n_(products)-n_(reactants)=1-2=-1`

Putting values in above equation, we get:

`K_p=271.6* (0.0821* 523)^(-1)\\\\K_p=6.32`

Hence, the value of
`K_p` for the reaction is 6.32