Chemical Kinetics Question 175
Question: The rate for a first order reaction is $ 0.6932\times {10^{-2}}mol,{l^{-1}}mi{n^{-1}} $ and the initial concentration of the reactants is 1M, $ {T_{1/2}} $ is equal to [JIPMER (Med.) 2001]
Options:
A) $ 6.932 $ min
B) 100 min
C) $ 0.6932\times {10^{-3}} $ min
D) $ 0.6932\times {10^{-2}} $ min
Show Answer
Answer:
Correct Answer: B
Solution:
$ r=k{{[reactant]}^{-1}} $
$ \therefore k=\frac{0.693\times {10^{-2}}}{1} $ also $ {t_{1/2}}=\frac{0.693}{k}=\frac{0.693}{0.693\times {10^{-2}}}=100,\min $ .
