Question

If 0.40 moles of PCl5 is heated in a 10.0 L container, equilibrium is established in which 0.25 moles of Cl2 are present. The reaction is: PCl5(g)\Longleftrightarrow ⟺ PCl3(g) + Cl2(g) What is the value of the equilibrium constant?

Answer 1

Answer: 0.0036

Step-by-step explanation:

Initial moles of
`PCl_5` = 2 mole

Moles of
`Cl_2` at equilibrium= 0.25 mole

Volume of container = 10 L

Initial concentration of
`PCL_5=(moles)/(volume)=(2moles)/(10L)=0.2M`

equilibrium concentration of
`Cl_2=(moles)/(volume)=(0.25moles)/(10L)=0.025M`

The given balanced equilibrium reaction is,

`PCL_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)`

Initial conc. 0.2 M 0 0

At eqm. conc. (0.2-x) M xM xM

The expression for equilibrium constant for this reaction will be,

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

`K_c=(x* x)/(0.2-x)`

We are given : x = 0.025 M

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

`K_c=(0.025* 0.025)/(0.2-0.025)`

`K_c=0.0036`

Thus the value of the equilibrium constant is 0.0036.