Question

In the reaction
`PCI_(5)(g)hArr PCI_(3)(g)+CI_(2)(g)`
the equilibrium concentrations of `PCI_(3)` and `PCI_(3)` are `0.4` and `0.2 mol^(-1)`, respectively. If the value of `K_(c)` is `0.5`, what id the concentration of `CI_(2)` in moles per litre ?
A. `2.0`
B. `1.5`
C. `1.0`
D. `0.5`

Answer 1

Correct Answer - C
`PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g)`
According to the law of chemical equilibrium,
`K_(C)=(C_(PCl_(3))C_(Cl_(2)))/(C_(PCl_(5)))`
or `C_(Cl_(2))=K_(C)(C_(PCl_(5)))/(C_(PCl_(5)))`
`=(0.5)((0.4))/((0.2))`
`=0.1`