At `25^(@)C`, the thermal decomposition of `PCl_5` occuring in a closed vessel of volume 1L gives rise to the equilibrium: `PCl_5(g)hArr PCl_3(g)+Cl_2(g)`. How many moles of `PCl_5` is to be added so that the concentration of `Cl_2(g)` at equilibrium will be `0.1mol*L^(-1)` ? [Given : `K_p=1.78` (at `250^(@)C`]

Explanation

Let initial number of moles of `PCl_5` =x mol. So, initial `[PCl_5]=xmol*L^(-1)`
Equilibrium concentration : `PCl_5(g) `(mol*L^(-1))x-0.1`
For this reaction, `/_\n=(1+1)-1=1`
`K_p=K_c(RT)^(/_\n)=K_c(RT)`
`1.78=K_cxx0.0821xx523:.K_c=0.0414`
`:. (0.1xx0.1)/(x-0.1)=0.0414" or "x=0.341`