Question

A certain quantity of pcl5 was heated in a ten litre vessel at 250°C. At equilibrium the vessel contains 0.1 mole of pcl5 0.2 mole of pcl3 and 0.2 mole of cl2;The equilibrium constant of the reaction pcl5(g) gives pcl3(g) and cl2(g)

Answer 1

Answer : The value of $K_c$ of the reaction is 0.04

Explanation : Given,

Moles of $PCl_3$ at equilibrium = 0.2 mole

Moles of $Cl_2$ at equilibrium = 0.2 mole

Moles of $PCl_5$ at equilibrium = 0.1 mole

Volume of solution = 10 L

First we have to calculate the concentration of $PCl_5,PCl_3\text{ and }Cl_2$ at equilibrium.

$\text{Concentration of }PCl_5=\frac{\text{Moles of }PCl_5}{\text{Volume of solution}}=\frac{0.1mole}{10L}=0.01mole=0.01M$

$\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}=\frac{0.2mole}{10L}=0.02mole=0.02M$

$\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}=\frac{0.2mole}{10L}=0.02mole=0.02M$

Now we have to calculate the value of equilibrium constant.

The balanced equilibrium reaction is,

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

The expression of equilibrium constant $K_c$ for the reaction will be:

$K_c=\frac{[PCl_3][Cl_2]}{[PCl_5]}$

Now put all the values in this expression, we get :

$K_c=\frac{(0.02)\times (0.02)}{(0.01)}$

$K_c=0.04$

Therefore, the value of $K_c$ of the reaction is 0.04

Answer 2

PCl5 - > PCl3 + Cl2

[PCl5] = 0.1/10 = 0.01M

[PCl3] = [Cl2] = 0.2/10 = 0.02M

Kc = [PCl3]×[Cl2]/[PCl5]

= 0.02×0.02/0.01

= 0.04.