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JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Sample Paper Topic Test - Chemical Equilibrium

  • question_answer
      When \[{{N}_{2}}{{O}_{5}}\] is heated at temperature T, it dissociates as \[{{N}_{2}}{{O}_{5}}(g)\rightleftharpoons {{N}_{2}}{{O}_{3}}(g)+{{O}_{2}}(g),\] \[{{K}_{c}}=2.5.\] At the same time \[{{N}_{2}}{{O}_{3}}\] decomposes as \[{{N}_{2}}{{O}_{3}}(g)\rightleftharpoons {{N}_{2}}O(g)+{{O}_{2}}(g).\] If initially 4.0 moles of \[{{N}_{2}}{{O}_{5}}\] are taken in 2.0 litre flask and allowed to attain equilibrium, concentration of \[{{O}_{2}}\] was formed to be 2.5 M. Equilibrium concentration of \[{{N}_{2}}O\] is -

    A) 2.0

    B) 1.0

    C) 0.334

    D) None of these

    Correct Answer: B

    Solution :

    [b] \[co{{n}^{n}}at\,e{{q}^{m}}\]
    \[\underset{2-x}{\mathop{{{N}_{2}}{{O}_{5}}}}\,(g)\underset{x-y}{\mathop{{{N}_{2}}{{O}_{3}}}}\,(g)+\underset{x+y}{\mathop{{{O}_{2}}(g)}}\,\]
    \[co{{n}^{n}}at\,e{{q}^{m}}\]
    \[\underset{x-y}{\mathop{{{N}_{2}}{{O}_{3}}}}\,(g)\underset{y}{\mathop{{{N}_{2}}O}}\,(g)+\underset{y+x}{\mathop{{{O}_{2}}(g)}}\,\]
    \[2.5=\frac{2.5\times (x-y)}{(2-x)}\]
    \[x-y=2-x\] or  \[2x-y=2\And \] as per given
    \[[{{O}_{2}}(g)]=x+y=2.5\] \[x=1.5\]
    and \[[{{N}_{2}}O(g)]=y=1.0\]


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