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

  • question_answer
    2 mol of \[{{N}_{2}}\] is mixed with 6 mol of \[{{H}_{2}}\]in a closed vessel of one litre capacity. If 50% of \[{{N}_{2}}\]is converted into \[N{{H}_{3}}\]at equilibrium, the value of \[{{K}_{c}}\]for the reaction \[{{N}_{2(g)}}+3{{H}_{2(g)}}\] ⇌\[2N{{H}_{3(g)}}\]is              [Kerala PMT 2004]

    A)                 \[4/27\]               

    B)                         \[27/4\]

    C)                 \[1/27\]               

    D)                 24

    E)                 9

    Correct Answer: A

    Solution :

               \[\underset{(a-x)}{\mathop{\overset{a}{\mathop{{{N}_{2}}}}\,}}\,+\underset{(b-3x)}{\mathop{\overset{b}{\mathop{3{{H}_{2}}}}\,}}\,\] ⇌  \[\underset{(2x)}{\mathop{\overset{0}{\mathop{2N{{H}_{3}}}}\,}}\,\]                    50% Dissociation of \[{{N}_{2}}\] take place so,                     At equilibrium \[\frac{2\times 50}{100}=1\]; value of \[x=1\]                 \[{{K}_{c}}=\frac{{{[2]}^{2}}}{[1]\,{{[3]}^{3}}}=\frac{4}{27}\] so, \[{{K}_{c}}=\frac{4}{27}\]


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