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

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 $PCL_5=\frac{moles}{volume}=\frac{2moles}{10L}=0.2M$

equilibrium concentration of $Cl_2=\frac{moles}{volume}=\frac{0.25moles}{10L}=0.025M$

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,

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

$K_c=\frac{x\times x}{0.2-x}$

We are given : x = 0.025 M

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

$K_c=\frac{0.025\times 0.025}{0.2-0.025}$

$K_c=0.0036$

Thus the value of the equilibrium constant is 0.0036.