Chemical Kinetics Question 52
Question: The rate for a first order reaction is $ 0.6932\times {10^{-2}}mol,{L^{-1}} $ and the initial concentration of the reactants is 1 M, $ {T_{1/2}} $ is equal to
Options:
A) 6.932 mm
B) 100 min
C) $ 0.6932\times {10^{-3}}\min $
D) $ 0.6932\times {10^{-2}}\min $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ 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 $
