In the reaction
`PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`, the equilibrium concentrations of `PCl_(5)` and `PCl_(3)` are 0.4 and 0.2 mole `//` litre respectively. If the value of `K_(c)` is 0.5 , what is the concentration of `Cl_(2)` in moles `//` litre ?

Options

A. 2
B. 1.5
C. 1
D. 0.5

Answer

C

Explanation

`PCl_(5)(g) hArr PCl_(3)(g) +Cl_(2)(g)`
`K_(c )=([PCl_(3)][Cl_(2)])/([PCl_(5)])`
`0.5=((0.2)[Cl_(2)])/((0.4))`
` [Cl_(2)]=(0.5xx0.4)/(0.2)=(0.20)/(0.2)=1.0 "mol " L^(-1)`