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BHU PMT BHU PMT (Screening) Solved Paper-2011

  • question_answer
    The equilibrium constant for the reaction\[S{{O}_{3}}(g)S{{O}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\]is\[{{K}_{c}}=4.9\times {{10}^{-2}}\]. The value of\[{{K}_{c}}\]for the reaction \[2S{{O}_{2}}(g)+{{O}_{2}}(g)2S{{O}_{3}}(g)\]will be

    A)  416                                       

    B)  \[2.40\times {{10}^{-3}}\]

    C)   \[9.8\times {{10}^{-2}}\]            

    D) \[4.9\times {{10}^{-2}}\]

    Correct Answer: A

    Solution :

                     \[S{{O}_{2}}(g)+\frac{1}{2}{{O}_{2}}S{{O}_{3}}(g)\] \[{{K}_{c}}=\frac{1}{4.9\times {{10}^{-2}}}\] \[2S{{O}_{2}}(g)+{{O}_{2}}(g)2S{{O}_{3}}(g)\]                 \[{{K}_{c}}={{\left( \frac{1}{4.9\times {{10}^{-2}}} \right)}^{2}}\]                 \[=\frac{{{10}^{4}}}{{{(4.9)}^{2}}}=416.49\]


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