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Two moles of PCl(5) were heated to 327^(...

Two moles of `PCl_(5)` were heated to `327^(@)C` in a closed two-litre vessel, and when equilibrium was achieved, `PCl_(5)` was found to be `40%` dissociated into `PCl_(3)` and `Cl_(2)`. Calculate the equilibrium constant `K_(p)` and `K_(c)` for this reaction.

Text Solution

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`PCl_5` dissociates as :
`PCl_5 hArr PCl_3+ Cl_2`
Initial concentration of `PCl_5`=2mol
% age dissociation at equilibrium =40%
`therefore` Number of moles of `PCl_5` dissociated
`=(2xx40)/100`=0.8 mol
Now , 1 mol of `PCl_5` on dissociation gives 1 mole of `PCl_3` and 1 mole of `Cl_2`, therefore , concentration at equilibrium.
`PCl_5` = 2.0-0.8=1.2mol , `PCl_3` =0.8 mol , `Cl_2` =0.8 mol.
Since Volume of flask is 2L, molar concentrations are :
`[PCl_5]=12/2=0.6 "mol L"^(-1), [PCl_3]=0.8/2 =0.4 "mol L"^(-1), [Cl_2]=0.8/2="0.4 mol L"^(-1)`
Now, `K_c=([PCl_3][Cl_2])/([PCl_5])=((0.4)xx(0.4))/(0.6)=0.267 "mol L"^(-1)`
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