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 \[{{\text{H}}_{\text{2}}}\]disappears\[\text{=}\,\text{6}\times {{10}^{-2}}\] \[\therefore \]In 3 min, the number of moles of \[{{\text{H}}_{\text{2}}}\] disappears \[=\frac{6\times {{10}^{-2}}\times 3}{10}\] \[=1.8\times {{10}^{-2}}\] \[{{N}_{2}}+3{{H}_{2}}\rightleftharpoons 2N{{H}_{3}}\] \[\Rightarrow \]\[\frac{1}{3}\] [Rate of disappearance of\[{{\text{H}}_{\text{2}}}\]] \[\text{=}\frac{1}{2}\] [Rate of formation of\[\text{N}{{\text{H}}_{\text{3}}}\]] \[\therefore \] Rate of formation of\[N{{H}_{3}}=\frac{2}{3}\times 1.8\times {{10}^{-2}}\] \[=1.2\times {{10}^{-2}}\,\text{mol}\,\text{N}{{\text{H}}_{3}}\] \[\therefore \]\[1.2\times {{10}^{-2}}\] moles of \[\text{N}{{\text{H}}_{\text{3}}}\]are formed.You need to login to perform this action.
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