Question

An amount of solid $N{H}_{4}HS$ is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm. pressure. Ammonium hydrogen sulphide decomposes to yield $N{H}_3$ and $H_{2}S$ gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm. The equilibrium constant for $N{H}_{4}HS$ decomposition at this temperature is

• A. 0.30
• B. 0.18
• C. 0.17
• D. 0.11

Answer 1

Answer: (D)

$N{H}_{4}HS(s)\rightleftharpoons N{H}_3(g)+H_{2}S(g)$
If $x$ is the partial pressure of $H_{2}S$ formed at equilibrium.
$(p_{N{H}_3}{)}_{eqm}=(0.50+x)atm$
$(p_{H_2}{)}_{eqm}=xatm$
$(p_{N{H}_3}{)}_{eqm}+(p_{H_{2}S}{)}_{eqm}=0.50+x+x=0.84$
$x0.17$ atm
$K_p=(p_{N{H}_3}{)}_{eqm}\times (p_{H_{2}s}{)}_{eqm}$
$=(0.50+0.17)\times 0.17=0.1139=0.11$