A) \[1.25 \times 1{{0}^{-}}^{4}\,and 3.75 \times 1{{0}^{-\,4}} respectively\]
B) \[3.75 \times 1{{0}^{-}}^{4}\,and 1.25 \times 1{{0}^{-\,4}} respectively\]
C) \[1.25 \times 1{{0}^{-}}^{3}\,and 3.75 \times 1{{0}^{-\,3}} respectively\]
D) \[3.75 \times 1{{0}^{-}}^{3}\,and 1.25 \times 1{{0}^{-\,3}} respectively\]
Correct Answer: A
Solution :
\[2N{{H}_{3}}\,\,\,{{N}_{2}}+3{{H}_{2}}\] By dividing the equation by 2 \[N{{H}_{3}}\,\xrightarrow{{}}\,\,\frac{1}{2}{{N}_{2}}+\frac{3}{2}{{H}_{2}}\] \[Rate=\frac{-d(N{{H}_{3}})}{dt}=+\frac{2d({{N}_{2}})}{dt}=+\frac{2}{3}\,\frac{d[{{H}_{2}}]}{dt}\] For zero order reaction, \[Rate=k\] So,\[\frac{-d[N{{H}_{3}}]}{dt}=\frac{2d[{{N}_{2}}]}{dt}=\frac{2}{3}\frac{d[{{H}_{2}}]}{dt}\] \[=\,\,2.5\,\,\times \,\,1{{0}^{-\,4}}\,mol\,\,{{L}^{-\,1}}\,{{s}^{-\,1}}\] \[\therefore \,\,Rate of production of {{N}_{2}}=\frac{d[{{N}_{2}}]}{dt}\] \[=\,\,\frac{2.5\times {{10}^{-\,4}}}{2}\,mol\,{{L}^{-\,1}}\,{{s}^{-\,1}}\] \[=\,\,\,1.25\times 1{{0}^{-\,4}}\,mol\,\,{{L}^{-\,1}}\,s{{\,}^{-\,1}}\] \[\therefore \] Rate of production of \[{{\operatorname{H}}_{2}}=\frac{3}{2}\times \,\,(2.5\times {{10}^{-\,4}}mol\,{{L}^{-\,1}}\,{{s}^{-\,1}})\] \[=\,\,\,3.75\times 1{{0}^{-\,4}}\,mol\,{{L}^{-\,1}}\,{{s}^{-\,1}}\]
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