# If the total equilibrium pressure of ${\mathrm{NH}}_{4}\mathrm{HS}\left(\mathrm{s}\right)⇌{\mathrm{NH}}_{3}\left(\mathrm{g}\right)+{\mathrm{H}}_{2}\mathrm{S}\left(\mathrm{g}\right)$  is 4 atm, the equilibrium constant is :

1. A

4 atm

2. B

2 atm

3. C

4 atm2

4. D

2 atm2

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### Solution:

$N{H}_{4}SH\left(s\right)⇌N{H}_{3}\left(g\right)+{H}_{2}S\left(g\right)$

p                       p (at equilibrium)

Total pressure at equilibrium = p+ p

4 = 2p       or            p  = 2

${K}_{p}={p}_{N{H}_{3}}×{p}_{{H}_{2}S}$ $=2×2=4\text{\hspace{0.17em}\hspace{0.17em}}at{m}^{2}$

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