A) \[1.8\times {{10}^{-2}}\]
B) \[1.2\times {{10}^{-2}}\]
C) \[4.0\times {{10}^{-2}}\]
D) \[3.6\times {{10}^{-2}}\]
Correct Answer: B
Solution :
\[\because \]In 10 min, the number of\[{{H}_{2}}\]disappears \[=6\times {{10}^{-2}}\] \[\therefore \]In 3 min, the number of moles of\[{{H}_{2}}\] \[disappears=\frac{6\times {{10}^{-2}}\times 3}{10}\] \[=1.8\times {{10}^{-2}}\] \[{{N}_{2}}+3{{H}_{2}}2N{{H}_{3}}\] \[\Rightarrow \] \[\frac{1}{3}[Rate\text{ }of\text{ }disappearance\text{ }of\text{ }{{H}_{2}}]\] \[=\frac{1}{2}[Rate\text{ }of\text{ }formation\text{ }of\text{ }N{{H}_{3}}]\] \[\therefore \]Rate of formation of\[N{{H}_{3}}=\frac{2}{3}\times 1.8\times {{10}^{-2}}\] \[=1.2\times {{10}^{-2}}mol\,N{{H}_{3}}\] \[\therefore \]\[1.2\times {{10}^{-2}}\]moles of NH3 are formed.
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