Question

Calculate `K_(p)` for the equilibrium,
`NH_(4)HS_((s))hArrNH_(3(g))+H_(2)S_((g))`
if the total pressure inside reaction vessel s 1.12 atm at `105.^(@)C`.
A. 0.56
B. 1.25
C. 0.31
D. 0.63

Answer 1

Correct Answer - C
`{:(,NH_(4)HS_((s)),hArr,NH_(3(g)),+,H_(2)S_((g))),("Initial moles",1,,0,,0),("At equilibrium",(1-x),,x,,x):}`
Total gaseous moles at equilibrium `=x+x=2x`
We know, `K_(p)=p_(NH_(3))xxp_(H_(2)S)`
But, partial pressure (p) = mole fraction `xx` total pressure (P)
`K_(p)=((x)/(2x)xxP)((x)/(2x)xxP)=((P)/(2))^(2)=((1.12)/(2))^(2)=0.3136`