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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2000

  • question_answer
     At \[{{490}^{o}}C\], the equilibrium constant for the synthesis of \[HI\] is 50, the value of K for the dissociation of \[HI\] will be:

    A)  \[20.0\]          

    B)  \[2.0\]                    

    C)  \[0.2\]             

    D)  \[0.02\]                   

    Correct Answer: D

    Solution :

    The equilibrium constant for the synthesis of \[HI=50\] K for dissociation of \[HI=?\] \[{{H}_{2}}+{{I}_{2}}\rightleftharpoons 2HI\] \[{{K}_{a}}=50,\,{{K}_{b}}=\frac{1}{50}=0.02\]


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