`1`mol of `Cl_(2)` and `3` mol of `PCl_(5)` are placed in a `100 L` vessel heated to `227^(@)C`. The equilibrium pressure is `2.05` atm. Assuming ideal behaviour, calculate the degree of dissociation for `PCl_(5)` and `K_(p)` for the reaction.
`PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`

Explanation

Dissociation of `PCI_5` and value of `K_p` for its dissociation.
`PCI_5(g) hArr PCI_3 +CI_2(g)`
Let ,x be the no of moles of `PCI_5` decomposed at equilibrium
`{:("",PCI_5,hArr,PCI_3,+CI_2),("Intial Moles",3,"",0,0),("Moles at eqm",3-x,"",x,x):}`
Now, total gaseous moles in the container =`n_T`
`n_T`= moles of `(PCI_5+PCI_3+CI_2)` + moles of `N_2`
The mixture behaves ideally hence PV= `n_TRT`
Let, us calculate no. of moles by using gas equation
`rArr n_T=(PV)/(RT)=(2.05xx100)/(0.821xx400) rArr n_T=5`
Now, equating the two values of `n_T` ,we have
`4+x=5 rArr x=1 rArr` degree of dissociation =1/3 =0.333
Now, `K_p=(P_(PCI_3)xxP_(CI_2))/(P_(PCI_5))`
`P_(PCI_5)=(3-x)/(4+x)P=2/5xx2.05 =0.82 atm`
`P_(PCI_3)1/5xx2.05 =0.41 atm`
`P_(CI_2)=1/5xx2.05 =0.41 atm`
`K_p=(P_(PCI_3)xxP_(CI_2))/(P_(PCI_5))atm rArr K_p=(0.41xx0.41)/(0.82)=0.205 atm`
`K_p=0.205 atm`
NOTE: The inert gases like `N_2` or noble gases (He,Ne tec,) though do not take part in the reaction, but still they affect the degree of dissociation and equilibrium concentration for the reactions in which `Deltan ne 0`. They add to the total pressure of the equlibrium mixture `(p prop n)` .