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Some solid NH(4)HS is placed in flask co...

Some solid `NH_(4)HS` is placed in flask containing `0.5` atm of `NH_(3)`. What would be the pressure of `NH_(3)` and `H_(2)S` when equilibrium is reached.
`NH_(4)HS(g) hArr NH_(3)(g)+H_(2)S(g), K_(p)=0.11`

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0.6653 atm, 0.1653 atm
`NH_4HS(s) hArr NH_3(g) + H_2S(g) , K_P` = 0.11
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Some acid NH_(4)HS is placed in flask containing 0.5 atm of NH_(3) . What would be pressures of NH_(3) and H_(2)S when equilibrium is reached? NH_(4)HS_((s))hArrNH_(3(g))+H_(2)S_((g)) , K_(p)=0.11

On adding NH_4HS in following equilibrium NH_4HS(s) leftrightarrow NH_3(g)+H_2S(g)

" The equilibrium constant "Kp" for the reaction NH_(4)HS_((s))to NH_(3(g))+H_(2)S_((g)) is

The equilibrium constant Kp for the reaction NH_(4)HS_((s)) Leftrightarrow NH_(3(g))+H_(2)S_((g)) is

Which of the following statement(s) is/are correct for the following equilibrium? NH_(4)HS(s)hArrNH_(3)(g)+H_(2)(g)

A definite amount of solid NH_(4)HS is placed in a flask already containing ammonia gas at a certain temperature and 0.1 atm pressure. NH_(4)HS decompses to give NH_(3) and H_(2)S and at equilibrium total pressure in flask is 1.1 atm. If the equilibrium constant K_(P) for the reaction NH_(4)HS(s) iff NH_(3)(g)+H_(2)S(g) is represented as zxx10^(-1) then find the value of z.

One "mole" of NH_(4)HS(s) was allowed to decompose in a 1-L container at 200^(@)C . It decomposes reversibly to NH_(3)(g) and H_(2)S(g). NH_(3)(g) further undergoes decomposition to form N_(2)(g) and H_(2)(g) . Finally, when equilibrium was set up, the ratio between the number of moles of NH_(3)(g) and H_(2)(g) was found to be 3 . NH_(4)HS(s) hArr NH_(3)(g)+H_(2)S(g), K_(c)=8.91xx10^(-2) M^(2) 2NH_(3)(g) hArr N_(2)(g)+3H_(2)(g), K_(c)=3xx10^(-4) M^(2) Answer the following: Assuming the volume due to solid NH_(4)HS is negligible what will be the density of the gaseous mixture in the above equilibrium system?

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