Solid NH4HS decomposes in an evacuated rigid vessel: NH4HS(s) NH3(g) + H2S(g). The total pressure at equilibrium is 0.60 atm. Extra NH3 is then injected at constant temperature, and at the new equilibrium (solid still present) p(NH3) = 0.50 atm. The new total pressure (in atm) is

JEE Advanced levelSingle correct optionDifficulty 3 of 3ce037

Solid $NH X_{4} HS$ decomposes in an evacuated rigid vessel: $NH X_{4} HS (s) NH X_{3} (g) + H X_{2} S (g)$ . The total pressure at equilibrium is 0.60 atm. Extra $NH X_{3}$ is then injected at constant temperature, and at the new equilibrium (solid still present) $p (NH X_{3}) = 0.50$ atm. The new total pressure (in atm) is

Initially

p (NH X_{3}) = p (H X_{2} S) = 0.30

atm, so

K_{p} = (0.30)^{2} = 0.09 at m^{2}

. With

p (NH X_{3}) = 0.50

atm,

p (H X_{2} S) = 0.09/0.50 = 0.18

atm. Total pressure

= 0.50 + 0.18 = 0.68

atm.