• # 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)  $\frac{{{(1.5)}^{3}}}{0.5}at{{m}^{-2}}$

${{N}_{2}}(g)+3{{H}_{2}}(g)2N{{H}_{3}}(g)$ $1-x$   $3-3x$ $2x$ 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 $N{{H}_{3}}=\frac{P{{N}_{2}}\times {{p}^{2}}{{H}_{2}}}{{{p}^{2}}N{{H}_{3}}}$ $=\frac{{{\left. \left( \frac{1-0.5}{3}\times 3 \right)\times \left( \frac{3-3\times 0.5}{3} \right)\times 3 \right]}^{3}}}{{{\left[ \frac{2\times 0.5\times 3}{3} \right]}^{2}}}$ $=0.5\times {{(1.5)}^{2}}\,at{{m}^{2}}$