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Punjab Medical Punjab - MET Solved Paper-2010

  • question_answer
    At\[700K\], the equilibrium constant\[{{K}_{p}}\]for the reaction\[2S{{O}_{3}}(g)2S{{O}_{2}}(g)+{{O}_{2}}(g)\]is\[1.80\times {{10}^{-3}}k\,\,Pa\]. What is the numerical value of\[{{K}_{c}}\]in moles per litre for this reaction at the same temperature?

    A) \[3.13\times {{10}^{-7}}\]                            

    B) \[3.13\times {{10}^{7}}\]

    C) \[3.13\times {{10}^{-10}}\]                         

    D)  None of these

    Correct Answer: A

    Solution :

    Here\[{{n}_{p}}=3mol,\,\,{{n}_{r}}=2mol\] \[\therefore \]  \[\Delta {{n}_{g}}={{n}_{p}}-{{n}_{r}}=3-2=1\,\,mol\] \[{{K}_{p}}=1.80\times {{10}^{-3}}kPa=1.80\,\,Pa=\frac{1.80}{{{10}^{5}}}bar\]                 \[=1.80\times {{10}^{-5}}bar\] \[\because \]\[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta {{n}_{g}}}}\] \[{{K}_{c}}=\frac{{{K}_{p}}}{RT}=\frac{1.80\times {{10}^{-5}}bar}{0.0821\,\,L\,\,bar\,\,{{K}^{-1}}\,\,mo{{l}^{-1}}\times 700K}\]       \[=3.13\times {{10}^{-7}}\,\,mol\,\,{{L}^{-1}}\]


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