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JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Kp and Kc Relationship and Characteristics of K

  • question_answer
    For the reversible reaction, \[{{N}_{2(g)}}+3{{H}_{2(g)}}\] ⇌\[2N{{H}_{3(g)}}\] at 500°C, the value of \[{{K}_{P}}\] is \[1.44\times {{10}^{-5}}\] when partial pressure is measured in atmospheres. The corresponding value of \[{{K}_{c}}\] with concentration in mole litre-1, is [IIT Screening 2000; Pb. CET  2004]

    A)                 \[1.44\times {{10}^{-5}}\]/\[{{\left( 0.082\times 500 \right)}^{-2}}\]        

    B)                 \[1.44\times {{10}^{-5}}\]/\[{{\left( 8.314\times 773 \right)}^{-2}}\]

    C)                 \[1.44\times {{10}^{-5}}\]/\[{{\left( 0.082\times 773 \right)}^{2}}\]

    D)                 \[1.44\times {{10}^{-5}}\]/\[{{\left( 0.082\times 773 \right)}^{-2}}\]

    Correct Answer: D

    Solution :

               \[\underset{4}{\mathop{{{N}_{2}}+3{{H}_{2}}}}\,\]⇌ \[\underset{2}{\mathop{2N{{H}_{3}}}}\,\]            \[\Delta n\]= 2 ? 4 = ? 2            \[{{K}_{p}}={{K}_{c}}{{[RT]}^{\Delta n}}\];  \[{{K}_{p}}={{K}_{c}}{{[RT]}^{-2}}\]                                 \[{{K}_{c}}=\frac{{{K}_{p}}}{{{[RT]}^{-2}}}=\frac{1.44\times {{10}^{-5}}}{{{[0.082\times 773]}^{-2}}}\]


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