A) \[54mins\]
B) \[68mins\]
C) \[40mins\]
D) \[76mins\]
Correct Answer: C
Solution :
: It \[{{[A]}_{0}}=100\] then \[\text{ }\!\![\!\!\text{ A }\!\!]\!\!\text{ }=100-60=40\]in 20 minutes Rate constant \[(k)=\frac{2.303}{t}\log \frac{{{[A]}_{0}}}{[A]}\] \[k=\frac{2.303}{20}\log \frac{100}{40}\] \[k=\frac{2.303}{20}(1-0.6020)=\frac{2.303}{20}\times 0.397mi{{n}^{-1}}\] ?(i) The time for 84% completion of the reaction is \[{{t}_{84%}}=\frac{2.303}{k}\log \frac{100}{1000-84}\] substituting the value of k from eq. (1) we get \[{{t}_{84%}}=\frac{2.303\times 20}{2.303\times 0.397}\log \frac{100}{16}\] \[{{t}_{84%}}=\frac{20}{0.397}(2-1.20)\] \[{{t}_{84%}}=50.377\times 0.795=40.09min\]
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