• # question_answer In a catalytic conversion of ${{N}_{2}}$ to $N{{H}_{3}}$ by Haber's process, the rate of reaction was expressed as change in the concentration of ammonia per time is $40\times {{10}^{-3}}\,mol\,litr{{e}^{-1}}{{s}^{-1}}$. If there are no side reaction, the rate of the reaction as expressed in terms of hydrogen is     (in mol $litr{{e}^{-1}}{{s}^{-1}}$) A)                 $60\times {{10}^{-3}}$                B)                 $20\times {{10}^{-3}}$ C)                 1.200     D)                 $10.3\times {{10}^{-3}}$

$\frac{-\,d({{N}_{2}})}{dt}=-\frac{1}{3}\frac{d({{H}_{2}})}{dt}=\frac{1}{2}\frac{d(N{{H}_{3}})}{dt}$=$\frac{3}{2}\times 40\times {{10}^{-3}}$                    $=60\times {{10}^{-3}}.$