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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2000

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

    A) \[\text{20}\text{.0}\]                                    

    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\[HI=?\] \[{{H}_{2}}+{{I}_{2}}=2HI\] \[{{K}_{a}}=50,\,{{K}_{b}}=\frac{1}{50}=0.02\]


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