2. Solved Papers
  3. NEET

NEET AIPMT SOLVED PAPER 1998

  • question_answer
                    If \[{{K}_{1}}\] and \[{{K}_{2}}\] are the respective equilibrium constants for the two reactions:                                 \[Xe{{F}_{6}}\left( g \right)+{{H}_{2}}O\left( g \right)~\rightleftharpoons XeO{{F}_{4}}\left( g \right)+2HF\left( g \right)\]                         \[Xe{{O}_{4}}\left( g \right)+Xe{{F}_{6}}\left( g \right)~\rightleftharpoons XeO{{F}_{4}}\left( g \right)+Xe{{O}_{3}}{{F}_{2}}\left( g \right)\]                 the equilibrium constant of the reaction                 \[Xe{{O}_{4}}\left( g \right)+2HF\left( g \right)\rightleftharpoons ~Xe{{O}_{3}}{{F}_{2}}\left( g \right)+{{H}_{2}}O\left( g \right)\] will be:

    A)                  \[{{K}_{2}}/{{({{K}_{2}})}^{2}}\]                                               

    B)                          \[{{K}_{1}}\,.\,{{K}_{2}}\]                           

    C)                 \[{{K}_{1}}/{{K}_{2}}\]                  

    D)                 \[{{K}_{2}}/{{K}_{1}}\]

    Correct Answer: D

    Solution :

                    \[Xe{{F}_{6}}(g)+{{H}_{2}}O(g)\rightleftharpoons XeO{{F}_{4}}(g)+2HF(g)\]                 \[{{K}_{1}}=\frac{[XeO{{F}_{4}}]\,{{[HF]}^{2}}}{[Xe{{F}_{6}}]\,[{{H}_{2}}O]}\]                      ....(i)                 \[Xe{{O}_{4}}(g)+Xe{{F}_{6}}(g)\rightleftharpoons XeO{{F}_{4}}(g)+Xe{{O}_{3}}{{F}_{2}}(g)\]                 \[{{K}_{2}}=\frac{[XeO{{F}_{4}}]\,[Xe{{O}_{3}}{{F}_{2}}]}{[Xe{{O}_{4}}]\,[Xe{{F}_{6}}]}\]                               ...(ii)                 For reaction,                 \[Xe{{O}_{4}}(g)+2HF(g)\rightleftharpoons Xe{{O}_{3}}{{F}_{2}}(g)+{{H}_{2}}O\,(g)\]                 \[K=\frac{[Xe{{O}_{3}}{{F}_{2}}]\,[{{H}_{2}}O]}{[Xe{{O}_{4}}]\,{{[HF]}^{2}}}\]                 \[\therefore \,\]              From Eqs. (i) and (ii)                 \[K=\frac{{{K}_{2}}}{{{K}_{1}}}\]


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