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

  • question_answer
    For the reaction : \[{{H}_{2(g)}}+C{{O}_{2(g)}}\]  ⇌\[C{{O}_{(g)}}+{{H}_{2}}{{O}_{(g)}}\], if the initial concentration of \[[{{H}_{2}}]=[C{{O}_{2}}]\] and \[x\] moles/litre of hydrogen is consumed at equilibrium, the correct expression of \[{{K}_{p}}\] is   [Orissa JEE  2005]

    A)                 \[\frac{{{x}^{2}}}{{{(1-x)}^{2}}}\]             

    B)                 \[\frac{{{(1+x)}^{2}}}{{{(1-x)}^{2}}}\]

    C)                 \[\frac{{{x}^{2}}}{{{(2+x)}^{2}}}\]            

    D)                 \[\frac{{{x}^{2}}}{1-{{x}^{2}}}\]

    Correct Answer: A

    Solution :

                                  \[{{H}_{2(g)}}+C{{O}_{2(g)}}\]   ⇌ \[C{{O}_{(g)}}+{{H}_{2}}{{O}_{(l)}}\]
    Initial conc. 1 1 0 0
    At equili (1? x) (1 ? x) x x
                \[{{K}_{p}}=\frac{{{p}_{CO}}.{{p}_{{{H}_{2}}O}}}{{{p}_{{{H}_{2}}}}.{{p}_{C{{O}_{2}}}}}=\frac{x.x}{(1-x)(1-x)}=\frac{{{x}^{2}}}{{{(1-x)}^{2}}}\]


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