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JEE Main & Advanced JEE Main Paper (Held On 9 April 2014)

  • question_answer
    At a certain temperature, only 50% HI is dissociated into \[{{H}_{2}}\]and \[{{I}_{2}}\] at equilibrium. The equilibrium constant is:   [JEE Main Online Paper ( Held On 09 Apirl  2014  )

    A) 1.0                                         

    B) 3.0

    C) 0.5                                         

    D) 0.25

    Correct Answer: A

    Solution :

                    \[HI\frac{1}{2}{{H}_{2}}+\frac{1}{2}{{I}_{2}}\] at t = 0 at eq. \[(10.5)(0.5)(0.5)\]\[\left( \because \alpha =\frac{50}{100}=0.5 \right)\] Now\[{{K}_{C}}=\frac{{{\left( 0.5 \right)}^{1/2}}{{\left( 0.5 \right)}^{1/2}}}{0.5}=1\] \[=\frac{-13.6\times 9}{4}=-30.6eV\]


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