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JEE Main & Advanced Sample Paper JEE Main Sample Paper-31

  • question_answer
    If \[0.5\] mole \[{{H}_{2}}\], is reacted with \[0.5\] mole I, in a ten-litre container at \[{{444}^{o}}C\] and at same temperature value of equilibrium constant \[{{K}_{c}}\] is  49, the ratio of \[[HI]\] and \[[{{I}_{2}}]\] will be :

    A) 7

    B) \[\frac{1}{7}\]

    C) \[\sqrt{\frac{1}{7}}\]                                     

    D) 49

    Correct Answer: A

    Solution :

    \[{{H}_{2}}(g)+{{I}_{2}}(g)2HI(g)\] \[{{K}_{c}}=\frac{{{[HI]}^{2}}}{[{{H}_{2}}][{{l}_{2}}]}\]if \[[{{H}_{2}}]=[{{I}_{2}}]\] \[{{K}_{c}}=\frac{{{[Hl]}^{2}}}{{{[{{l}_{2}}]}^{2}}}\]or\[{{K}_{c}}\frac{[Hl]}{[{{l}_{2}}]}=\sqrt{{{K}_{c}}}=\sqrt{49}=7\]


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