Discussion Forum

1)

$PCl_{5}\rightleftharpoons PCl_{3}+PCl_{2}$

 if the equilibrium constant $(K_{C})$  for the above reaction at 500 K is 1.79 and the equilibrium concentration of $PCl_{5}$  and $PCl_{3}$  are 1.41 M  and 1.59 M . respectively , then the concentration of $Cl_{2}$  is approximately 

Answer:

Option D

Explanation:

 Equilibrium constant $(K_{C})=  \frac{ [PCl_{3}][Cl_{2}]}{[PCl_{5}]}$

 $1.79= \frac{1.59 \times [Cl_{2}]}{1.41}$

 $[Cl_{2}]= \frac{1.79 \times 1.41}{1.59}= 1.5877$ ≈ 1.59 m