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JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Law of equilibrium and Equilibrium constant

  • question_answer
    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 [AIEEE 2005]

    A)                 0.30       

    B)                 0.18

    C)                 0.17       

    D)                 0.11

    Correct Answer: D

    Solution :

           \[\underset{\text{a}-\text{x}}{\mathop{\underset{\text{a}}{\mathop{N{{H}_{4}}HS}}\,}}\,\] ⇌ \[\underset{0.5+\text{x}\,\,\,\,\,\,\,\,\,\,\,\text{x}\,\,\,\,\,\,}{\mathop{\underset{0.5\text{atm}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}{\mathop{N{{H}_{3\,(g)}}+{{H}_{2}}{{S}_{(g)}}}}\,}}\,\]                     Total pressure \[=0.5+2x=0.84\]                     i.e., \[x=0.17\]                                 \[{{K}_{p}}={{P}_{N{{H}_{3}}}}.{{P}_{{{H}_{2}}S}}\] \[=(0.67).\,(0.17)\]=0.1139


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