For the reaction
`NH_(3)(g) hArr 1/2 N_(2)(g)+3/2 H_(2)(g)`
Show that the degree of dissociation of `NH_(3)` is given as `alpha=[1+(3sqrt(3))/4 P/K_(p)]^(-1//2)`, where P is the equilibrium pressure and `alpha` is the degree of dissociation. If `K_(p)` of the above reaction is `82.1` atm at `727^(@)C`, determine the value of `K_(c)`.

Explanation

Let `alpha` be the degree of dissociation of `NH_(3)(g)`,
`{:(,NH_(3)(g),hArr,1/2N_(2)(g),+,3/2H_(2)(g)),("Initial moles",1,,0,,0),("At equilibrium",1-alpha,,alpha//2,,3alpha//2):}`
Total number of moles at equilibrium `=1+alpha`
Now,
`p_(NH_(3))=(1-alpha)/(1+alpha)P, p_(N_(2))=(alpha//2)/(1+alpha)P` and `p_(H_(2))=(3alpha//2)/(1+alpha)P`
`K_(p)=((p_(N_(2)))^(1//2)(p_(H_(2)))^(3//2))/p_(NH_(3))`
`rArr ([alpha/(2(1+alpha))P]^(1//2)[(3alpha)/(2(1+alpha))P]^(3//2))/((1-alpha)/(1+alpha)xxP)`
`rArr (1-alpha^(2))/(alpha^(2))=(3sqrt(3))/4P/K_(p)`
`rArr 1/alpha^(2)=[1+(3sqrt(3))/4P/K_(p)]` or `alpha=[1+(3sqrt(3))/4P/K_(p)]^(-1//2)`
`Deltan_(g)`, change in the number of the given reaction `=+1`
`K_(p)=K_(c )(RT)^(Deltan_(g)) rArr K_(c )=K_(p)(RT)^(-Deltan_(g))`
`rArr K_(c )=82.1xx[0.0821xx1000]^(-1)=1.0 "mol"//L`