JEE Main & Advanced Sample Paper JEE Main Sample Paper-19

  • question_answer \[1\,mol\] of \[{{N}_{2}}\] and \[3\text{ }mol\] of \[{{H}_{2}}\] are placed in a closed container at a pressure of 4 atm. The pressure falls to 3 atm at the same temperature when the following equilibrium is attained \[{{N}_{2}}(g)+3{{H}_{2}}(g)2N{{H}_{3}}(g).\] The \[{{K}_{P}}\] for the dissociation of \[N{{H}_{3}}\] is

    A)  \[\frac{3\times 3}{0.5\times {{(1.5)}^{3}}}at{{m}^{-2}}\]         

    B)  \[0.5\times {{(1.5)}^{3}}at{{m}^{2}}\]

    C)  \[\frac{0.5\times {{(1.5)}^{3}}}{3\times 3}at{{m}^{2}}\]

    D)  \[\frac{{{(1.5)}^{3}}}{0.5}at{{m}^{-2}}\]

    Correct Answer: B

    Solution :

     \[{{N}_{2}}(g)+3{{H}_{2}}(g)2N{{H}_{3}}(g)\] \[1-x\]   \[3-3x\] \[2x\] at equilibrium Total moles, \[1-x+3-3x+2x=4-2x=3\] (given) (Since, 4 moles = 4 atm given) \[\therefore \]   \[x=0.5\] \[{{K}_{p}}\] for dissociation of \[N{{H}_{3}}=\frac{P{{N}_{2}}\times {{p}^{2}}{{H}_{2}}}{{{p}^{2}}N{{H}_{3}}}\] \[=\frac{{{\left. \left( \frac{1-0.5}{3}\times 3 \right)\times \left( \frac{3-3\times 0.5}{3} \right)\times 3 \right]}^{3}}}{{{\left[ \frac{2\times 0.5\times 3}{3} \right]}^{2}}}\] \[=0.5\times {{(1.5)}^{2}}\,at{{m}^{2}}\]


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