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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2001

  • question_answer
    For the equilibrium\[{{N}_{2}}(g)+3{{H}_{2}}(g)\rightleftharpoons 2N{{H}_{3}}(g),\] The equilibrium constant,\[{{K}_{p}}\]is equal to:                      

    A)  \[{{K}_{c}}={{(RT)}^{2}}\]                           

    B) \[{{K}_{c}}(RT)\]

    C)   \[\frac{{{K}_{c}}}{{{(RT)}^{2}}}\]                             

    D) \[{{K}_{c}}\]

    Correct Answer: A

    Solution :

    \[{{N}_{2}}(g)+3{{H}_{2}}(g)\rightleftharpoons 2N{{H}_{3}}(g)\] We know that \[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta n}}\] Here \[\Delta n=2-4=-2\] Hence \[{{K}_{p}}={{K}_{c}}{{(RT)}^{-2}}=\frac{{{K}_{c}}}{{{(RT)}^{2}}}\]


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