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

Answer:

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

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:

$\text{Moles of }PCl_5=\frac{0.105g}{208.24g/mol}=5.04\times 10^{-4}mol$

• 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:

$\text{Moles of }PCl_3=\frac{0.220g}{137.33g/mol}=1.60\times 10^{-3}mol$

• 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:

$\text{Moles of }Cl_2=\frac{2.12g}{71.0g/mol}=0.029mol$

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:

$K_c=\frac{PCl_5}{PCl_3\times Cl_2}$

We are given:

$[PCl_5]=\frac{5.04\times 10^{-4}mol}{25L}$

$[PCl_3]=\frac{1.60\times 10^{-3}mol}{25L}$

$[Cl_2]=\frac{0.029mol}{25L}$

Putting values in above equation, we get:

$K_c=\frac{(\frac{5.04\times 10^{-4}}{25})}{(\frac{1.60\times 10^{-3}}{25})\times (\frac{0.029}{25})}\\\\K_c=271.6$

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:

$K_p=271.6\times (0.0821\times 523)^{-1}\\\\K_p=6.32$

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