A) \[3.44\times {{10}^{-3}}mol\,{{L}^{-1}}{{\min }^{-1}}\]
B) \[3.44\times {{10}^{3}}mol\,{{L}^{-1}}{{\min }^{-1}}\]
C) \[1.86\times {{10}^{-3}}mol\,{{L}^{-1}}{{\min }^{-1}}\]
D) \[1.86\times {{10}^{3}}mol\,{{L}^{-1}}{{\min }^{-1}}\]
Correct Answer: A
Solution :
Rate constant in\[mi{{n}^{-1}}\]indicates that, the reaction is of 1st order. Hence, \[k=\frac{2.303}{t}\log \frac{{{[R]}_{0}}}{[R]}\] \[4.5\times {{10}^{-3}}{{\min }^{-1}}=\frac{2.303}{60\min }\log \frac{1}{[R]}\] \[\log [R]=-0.1172=\overline{1}.8821\] or \[[R]=anti\log \,\overline{1}.8828\] \[[A]=7635\text{ }mol\text{ }{{L}^{-1}};\]this is the concentration after 60 min or one hour,
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