A) \[1\times {{10}^{-4}}mol\,\,{{l}^{-1}}{{s}^{-1}}\]
B) \[3\times {{10}^{-4}}mol\,\,{{l}^{-1}}{{s}^{-1}}\]
C) \[4\times {{10}^{-4}}mol\,\,{{l}^{-1}}{{s}^{-1}}\]
D) \[6\times {{10}^{-4}}mol\,{{l}^{-1}}{{s}^{-1}}\]
Correct Answer: B
Solution :
\[{{N}_{2}}+3{{H}_{2}}\]⇌\[2N{{H}_{3}}\] \[\frac{-\Delta [{{N}_{2}}]}{\Delta t}=-\frac{1}{3}\frac{\Delta [{{H}_{2}}]}{\Delta t}=\frac{1}{2}\frac{\Delta [N{{H}_{3}}]}{\Delta t}\] \[\therefore \frac{\Delta [{{H}_{2}}]}{\Delta t}=\frac{3}{2}\times \frac{\Delta [N{{H}_{3}}]}{\Delta t}=\frac{3}{2}\times 2\times {{10}^{-4}}\] \[=3\times {{10}^{-4}}mol\,\,litr{{e}^{-1}}se{{c}^{-1}}\]You need to login to perform this action.
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