A) \[6.25\times {{10}^{-3}}mol\,{{L}^{-1}}{{s}^{-1}}\]and \[6.25\times {{10}^{-3}}mol\,{{L}^{-1}}{{s}^{-1}}\]
B) \[1.25\times {{10}^{-2}}mol\,{{L}^{-1}}{{s}^{-1}}\]and \[3.125\times {{10}^{-3}}mol\,{{L}^{-1}}{{s}^{-1}}\]
C) \[6.25\times {{10}^{-3}}mol\,{{L}^{-1}}{{s}^{-1}}\]and \[3.125\times {{10}^{-3}}mol\,{{L}^{-1}}{{s}^{-1}}\]
D) \[1.25\times {{10}^{-3}}mol\,{{L}^{-1}}{{s}^{-1}}\]and \[6.25\times {{10}^{-3}}mol\,{{L}^{-1}}{{s}^{-1}}\]
Correct Answer: B
Solution :
Key Idea Rate of disappearance of reactant = rate of appearance of product \[\text{or-}\frac{\text{1}}{\begin{align} & \text{stoichiometric}\,\text{coefficient} \\ & \text{of}\,\text{reactant} \\ \end{align}}\]\[\frac{d\text{ }\!\![\!\!\text{ reactant}]}{dt}\] \[\text{=}\,\text{+}\frac{\text{1}}{\begin{align} & \text{stoichiometric} \\ & \text{coefficient}\,\text{of}\,\text{product} \\ \end{align}}\]\[\frac{\text{d }\!\![\!\!\text{ product }\!\!]\!\!\text{ }}{\text{dt}}\] For the reaction, \[{{N}_{2}}{{O}_{5}}(g)\xrightarrow[{}]{{}}2N{{O}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\] \[\frac{-d[{{N}_{2}}{{O}_{5}}]}{dt}=+\frac{1}{2}\frac{d[N{{O}_{2}}]}{dt}\] \[=+\frac{2d[{{O}_{2}}]}{dt}\] \[\therefore \] \[\frac{d[N{{O}_{2}}]}{dt}=-2\frac{d[{{N}_{2}}{{O}_{5}}]}{dt}\] \[\text{=}\,\text{2 }\!\!\times\!\!\text{ }\,\text{6}\text{.25 }\!\!\times\!\!\text{ 1}{{\text{0}}^{-3}}\text{mol}\,{{\text{L}}^{-1}}{{\text{s}}^{-1}}\] \[\text{=}\,12.\text{5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{-3}}\text{mol}\,{{\text{L}}^{-1}}{{\text{s}}^{-1}}\] \[\text{=}\,1.\text{25 }\!\!\times\!\!\text{ 1}{{\text{0}}^{-2}}\text{mol}\,{{\text{L}}^{-1}}{{\text{s}}^{-1}}\] \[\frac{d[{{O}_{2}}]}{dt}=-\frac{d[{{N}_{2}}{{O}_{5}}]}{dt}\times \frac{1}{2}\] \[=\frac{6.25\times {{10}^{-3}}mol\,{{L}^{-1}}\,{{s}^{-1}}}{2}\] \[=3.125\times {{10}^{-3}}mol\,{{L}^{-1}}\,{{s}^{-1}}\]
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