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WB JEE Medical WB JEE Medical Solved Paper-2011

  • question_answer
    If the equilibrium constants of the following equilibria, \[S{{O}_{2}}+\frac{1}{2}{{O}_{2}}\to S{{O}_{3}}\]and \[2S{{O}_{3}}\to 2S{{O}_{2}}+{{O}_{2}}\] are given by \[{{K}_{1}}\] and \[{{K}_{2}}\] respectively, which of the following relations is correct?

    A)  \[{{K}_{2}}={{\left( \frac{1}{{{K}_{1}}} \right)}^{2}}\]                     

    B)  \[{{K}_{1}}={{\left( \frac{1}{{{K}_{2}}} \right)}^{3}}\]

    C)  \[{{K}_{2}}=\left( \frac{1}{{{K}_{1}}} \right)\]                    

    D)  \[{{K}_{2}}={{({{K}_{1}})}^{2}}\]

    Correct Answer: A

    Solution :

                     \[S{{O}_{2}}+\frac{1}{2}{{O}_{2}}\xrightarrow{{}}S{{O}_{3}}\] \[{{K}_{1}}=\frac{[S{{O}_{3}}]}{[S{{O}_{2}}]{{[{{O}_{2}}]}^{1/2}}}\] \[2S{{O}_{3}}\xrightarrow{{}}2S{{O}_{2}}+{{O}_{2}}\] \[{{K}_{2}}=\frac{{{[S{{O}_{2}}]}^{2}}[{{O}_{2}}]}{{{[S{{O}_{3}}]}^{2}}}\]Obviously,           \[{{K}_{2}}={{\left( \frac{1}{{{K}_{1}}} \right)}^{2}}\]


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