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JEE Main & Advanced AIEEE Solved Paper-2005

  • 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  Solved  Paper-2005

    A) 0.11

    B)                                        0.17                       

    C) 0.18                       

    D)        0.30

    Correct Answer: A

    Solution :

    \[N{{H}_{4}}HS(s)N{{H}_{3}}(g)+{{H}_{2}}S(g)\]
    Initially 1 0.5 0
    at equilibrium \[(1-x)\] \[(0.5+x)\] \[x\]
    Total pressure at equilibrium\[={{\rho }_{N{{H}_{3}}}}+{{\rho }_{{{H}_{2}}S}}\] \[=0.5+x+x=0.84\] \[\therefore \] \[x=0.17\text{ }aim\] \[\therefore \] \[{{p}_{N{{H}_{3}}}}=0.50+0.17=0.67\,atm\] \[{{p}_{{{H}_{2}}S}}=0.17\,atm\] \[\therefore \] \[{{k}_{p}}={{p}_{N{{H}_{3}}}}.{{p}_{{{H}_{2}}S}}\] \[=0.67\times 0.17=0.114\text{ }atm\]


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