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WB JEE Medical WB JEE Medical Solved Paper-2009

  • question_answer
    For the reaction \[S{{O}_{2}}+\frac{1}{2}{{O}_{2}}\rightleftharpoons S{{O}_{3}},\]if we write\[{{K}_{p}}={{K}_{c}}{{(RT)}^{x}},\] then \[x\]becomes

    A)  \[-1\]                                   

    B)  \[-\frac{1}{2}\]

    C)  \[\frac{1}{2}\]                                  

    D)  1

    Correct Answer: B

    Solution :

                     In the relation, \[{{K}_{p}}={{K}_{c}}{{(RT)}^{x}},\] \[x={{n}_{p}}-{{n}_{r}}\] For the reaction, \[S{{O}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\rightleftharpoons S{{O}_{3}}(g)\] \[x=1-\left( 1+\frac{1}{2} \right)=-\frac{1}{2}\]


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