A) 1.732
B) 3
C) \[1.02\times {{10}^{-4}}\]
D) \[3.4\times {{10}^{5}}\]
Correct Answer: B
Solution :
\[2{{N}_{2}}{{O}_{5}}\xrightarrow{{}}4N{{O}_{2}}+{{O}_{2}}\] \[\frac{-d[{{N}_{2}}{{O}_{5}}]}{dt}=k\,[{{N}_{2}}{{O}_{5}}]\] \[1.02\times {{10}^{-4}}=3.4\times {{10}^{-5}}\,{{s}^{-1}}\times \,[{{N}_{2}}{{O}_{5}}]\] \[\therefore [{{N}_{2}}{{O}_{5}}]=\frac{1.02\times {{10}^{-4}}}{3.4\times {{10}^{-5}}}=3\]
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