2. Sample Papers
  3. JEE Main & Advanced
  4. Chemistry
  5. Equilibrium / साम्यावस्था

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

  • question_answer
    \[=\frac{2/3}{2}=\frac{1}{3}=0.33\] Ammonium carbonate dissociates as \[N{{H}_{2}}COON{{H}_{4}}(s)2N{{H}_{3}}(g)+C{{O}_{2}}(g)\] in a closed vessel containing ammonium carbamate in equilibrium. Ammonia is added such that partial pressure of \[N{{H}_{3}}\] now becomes two times to the equilibrium pressure. Calculate the ratio of partial pressure of \[C{{O}_{2}}\] now to the original partial pressure of \[C{{O}_{2}}\].

    A) \[\frac{2}{9}\]

    B) \[\frac{4}{9}\]

    C) \[\frac{3}{9}\]

    D) \[\frac{1}{9}\]

    Correct Answer: D

    Solution :

    [d] \[N{{H}_{2}}CON{{H}_{4}}(s)2N{{H}_{3}}(g)+C{{O}_{2}}(g)\]
    Let, partial pressure of \[C{{O}_{2}}=p\]
    Then,                \[{{p}_{N{{H}_{3}}}}=2p\]
    Total pressure at equilibrium \[=2p+p=3p\]
    \[{{K}_{p}}={{(2p)}^{2}}\times p=4{{p}^{3}}\]
    After addition of \[N{{H}_{3}}\]pressure of \[N{{H}_{3}}=2\times 3p=6p\]
    \[{{K}_{p}}={{({{p}_{N{{H}_{3}}}})}^{2}}\times p{{'}_{C{{O}_{2}}}}\]
    \[4{{p}^{3}}={{(6p)}^{2}}\times p{{'}_{C{{O}_{2}}}}\]
    \[p{{'}_{C{{O}_{2}}}}=\frac{1}{9}p\]              \[\frac{p{{'}_{C{{O}_{2}}}}}{{{p}_{C{{O}_{2}}}}}=\frac{1}{9}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner