Calculate `K_p` for the equilibrium,
`NH_(4)HS_(s) Leftrightarrow NH_(3(g))+H_(2)S_(g)` if the total pressure inside the reaction vessel is 1.12 atm at `105^@C`

Options

A. 0.56
B. 1.25
C. 0.31
D. 0.63

Answer

C

Explanation

`"Let initial moles "underset(1)(NH_(4)HS_(s)) Leftrightarrow underset(0)(NH_(3(g)))+underset(0)(H_(2)S_(g))`
`"At equilibrium (1-x) x x"`
Total gaseous moles at equilibrium =x+x=2x
We know `K_(p)=p_(NH_(3)) xx p_(H_(2)S)`
but partial pressure (p)=mole fration `xx` total pressure (P)
`K_(p)=((x)/(2x) xx P) ((x)/(2x) xx P) =(P/2)^2 =((1.12)/(2))^2=0.3136`
IInd method : Both `NH_(3) and H_(2)S` have same number of mols at equilibrium so, have mole fraction and thus equal partial pressure.
i.e, `p_(NH_(3))=p_(H_(2)S)=(1.12)/(2)`
`K_(p)=p_(NH_(3)) xx p_(H_(2)S)=(1.12)/(2) xx (1.12)/(2)=0.3136`