JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Topic Test - Chemical Equilibrium (6-5-21)

  • question_answer
    The reaction between \[{{N}_{2}}\] and \[{{H}_{2}}\] to form ammonia has \[{{K}_{c}}=6\times {{10}^{-2}}\] at the temperature 500°C. The numerical value of \[{{K}_{p}}\] for this reaction is 

    A) \[1.5\times {{10}^{-5}}\]

    B) \[1.5\times {{10}^{5}}\]\[\]

    C) \[1.5\times {{10}^{-6}}\]

    D) \[1.5\times {{10}^{6}}\]

    Correct Answer: A

    Solution :

    [a] \[{{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta n}}\]; \[\Delta n=2-4=-2\]
                \[{{K}_{p}}=6\times {{10}^{-2}}\times {{(0.0812\times 773)}^{-2}}\]
                            \[{{K}_{p}}=\frac{6\times {{10}^{-2}}}{{{(0.0812\times 773)}^{2}}}=1.5\times {{10}^{-5}}\].

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