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))/4p/K_(p)]^(-1//2)`
where p is equilibrium pressure. If `K_(p)` of the above reaction is `78.1 atm` at `400^(@)C`, calculate `K_(c )`.

Explanation

`NH_(3)(g)hArr(1)/(2)N_(2)+(3)/(2)H_(2)(g)` Total moles
`{:(t=0,1,0,0,1),(t_(eq),1-alpha,alpha//2,3alpha//2,1+alpha):}`
`p_(i) {(1-alpha)/(1+alpha)}p, {(alpha)/(2(1+alpha))}p, {(3alpha)/(2(1+alpha))}p`
`K_(p)=((P_(N_(2)))^(1//2)(P_(H_(2)))^(3//2))/((P_(NH_(3))))`
`=([alpha/(2(1+lpha))p]^(1//2)[(3alpha)/(2(1+alpha))p]^(3//2))/([(1-alpha)/(1+alpha)p])=(palpha^(2)sqrt(27))/(4(1-alpha^(2)))`
Solving for `alpha`, we get `alpha=[1+(3sqrt(3))/4p/K_(p)]^(1//2)`
`K_(c )` can be caiculate by using `K_(p)=K_(c )(RT)^(Deltan)`
`K_(p)=78.1, T=673, Deltan=1
K_(c )=K_(p)(RT)^(-1)=78.1/((0.082xx673))`
`K_(c )=78.1/55.18=1.415`