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In the reaction PCl(5)(g) hArr PCl(3)(...

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 ?

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The correct Answer is:
1
` PCl _(s) (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)] = (0.5 xx 0.4)/(0.2) = (0.20)/(0.2) = 1.0 mol L^(-1)`
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