Chemical Kinetics Question 166
Question: A reaction proceeds by first order, 75% of this reaction was completed in 32 min. The time required for 50% completion is
Options:
A) 8 min
B) 16 min
C) 20 min
D) 24 min
Show Answer
Answer:
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 $
BETA
