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JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Critical Thinking

  • question_answer
    Ammonia under a pressure of 15 atm at 27°C is heated to 347°C in a closed vessel in the presence of a catalyst. Under the conditions, \[N{{H}_{3}}\] is partially decomposed according to the equation, \[2N{{H}_{3}}\]⇌ \[{{N}_{2}}+3{{H}_{2}}\].The vessel is such that the volume remains effectively constant whereas pressure increases to 50 atm. Calculate the percentage of \[N{{H}_{3}}\] actually decomposed.      [IIT 1981; MNR 1991; UPSEAT 2001]

    A)                 65%       

    B)                         61.3%

    C)                 62.5%   

    D)                 64%

    Correct Answer: B

    Solution :

                                               \[2N{{H}_{3}}\] ⇌ \[{{N}_{2}}+\]  \[3{{H}_{2}}\]                    Initial mole                  \[a\]             0          0                    Mole at equilibrium  \[(a-2x)\]       \[x\]      \[3x\]                    Initial pressure of \[N{{H}_{3}}\] of a mole = 15 atm at \[{{27}^{o}}C\]                    The pressure of 'a' mole of \[N{{H}_{3}}=p\] atm at \[{{347}^{o}}C\]                    \[\therefore \]  \[\frac{15}{300}=\frac{p}{620}\]                    \[\therefore \]  \[p=31\]atm                    At constant volume and at \[{{347}^{o}}C\], mole \[\propto \] pressure                    \[a\propto 31\] (before equilibrium)                    \[\therefore \] \[a+2x\,\propto 50\] (after equilibrium)                    \[\therefore \] \[\frac{a+2x}{a}=\frac{50}{31}\]                    \[\therefore \] \[x=\frac{19}{62}a\]                    \[\therefore \] % of \[N{{H}_{3}}\] decomposed \[=\frac{2x}{a}\times 100\]                                 \[=\frac{2\times 19a}{62\times a}\times 100=61.33%\]


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