Question

`102 g` of solid `NH_(4)HS` is taken in the `2L` evacuated flask at `57^(@)`. Following two equilibrium exist simultaneously
`NH_(4)(s)iffNH_(3)(g)+H_(2)S(g)`
`NH_(3)(g)iff(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)`
one mole of the solid decomposes to maintain both the equilibrium and `0.75 mol e` of `H_(2)` was found at the equilibrium then find the equilibrium concentration of all the species and `K_(C)` for the both the reaction.

Answer 1

Moles of `NH_(4)HS=(102)/(51)=2`
`NH_(4)(s)iffNH_(3)(g)+H_(2)S(g)`
`{:(2,0,0),(1,1-x,1):}`
`NH_(3)(g)iff(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g) K_(C_(2))`
`1-x (x)/(2) (3x)/(2)`
Given that moles of `H_(2)=(3x)/(2)=0.75 iff x=(1)/(2)`
`K_(C_(2))=1/2 ((1-x))/(2)=1/8` [Since `V=2L`]
`K_(C_(2))=(((3x)/(4))^(3//2)((x)/(4))^(1//2))/(((1-x)/(2)))=(((3)/(8))^(3//2)((1)/(8))^(1//2))/(1/4)=(3)^(3//2)1/64xx4/1=(3)^(3//2)/(16)`