JEE MAIN - Chemistry (2005 - No. 1)
An amount of solid NH4HS is placed in a flask already containing ammonia gas at a
certain temperature and 0.50 atm. Pressure. Ammonium hydrogen sulphide
decomposes to yield NH3 and H2S 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 NH4HS decomposition at this temperature is :
0.30
0.11
0.17
0.18
Paaiškinimas
$$\mathop {N{H_4}HS\left( s \right)\,\rightleftharpoons\,}\limits_{\matrix{
{start} \cr
{At\,\,equib.} \cr
} } \,\,\mathop {N{H_3}\left( g \right)}\limits_{\matrix{
{0.5\,\,atm} \cr
{0.5 + x\,\,atm} \cr
} } \,\, + \mathop {{H_2}S\left( g \right)}\limits_{\matrix{
{0\,\,atm} \cr
{x\,\,atm} \cr
} } $$
Then $$0.5 + x + x = 2x + 0.5 = 0.84\,\,$$ (Given)
$$ \Rightarrow x = 0.17\,\,atm.$$
$${P_{N{H_3}}} = 0.5 + 0.17 = 0.67\,atm;$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $${P_H}_{_2S} = 0.17\,atm$$
$$K = {P_{N{H_3}}} \times {P_{{H_2}S}}$$
$$ = 0.67 \times 0.17\,at{m^2}$$
$$ = 0.1139 = 0.11$$
Then $$0.5 + x + x = 2x + 0.5 = 0.84\,\,$$ (Given)
$$ \Rightarrow x = 0.17\,\,atm.$$
$${P_{N{H_3}}} = 0.5 + 0.17 = 0.67\,atm;$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $${P_H}_{_2S} = 0.17\,atm$$
$$K = {P_{N{H_3}}} \times {P_{{H_2}S}}$$
$$ = 0.67 \times 0.17\,at{m^2}$$
$$ = 0.1139 = 0.11$$