A) 8 min
B) 16 min
C) 20 min
D) 24 min
Correct Answer: B
Solution :
[b] Given: 75% reaction gets completed in 32 min Thus, \[k=\frac{2.303}{t}\log \frac{a}{(a-x)}\] \[=\frac{2.303}{32}\log \frac{100}{(100-75)}\] \[=\frac{2.303}{0.0433}\log 4=0.0433\,mi{{n}^{-1}}\] Now we can use this value of k to get the value of time required for 50% completion of reaction \[t=\frac{2.303}{k}\log \frac{a}{(a-x)}=\frac{2.303}{0.0433}\log \frac{100}{50}\] \[=\frac{2.303}{0.0433}\log 2=16\,\min \]
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