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 $ .



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