Chemical Kinetics Question 79
Question: In a first order reaction the concentration of reactant decreases from $ 800,mol/dm^{3} $ to $ 50,mol/dm^{3} $ is $ 2\times 10^{2}\sec $ . The rate constant of reaction in $ {{\sec }^{-1}} $ is [IIT-JEE (Screening) 2003]
Options:
A) $ 2\times 10^{4} $
B) $ 3.45\times {10^{-5}} $
C) $ 1.386\times {10^{-2}} $
D) $ 2\times {10^{-4}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ k=\frac{2.303}{t}{\log_{10}}\frac{a}{a-x} $ ; $ t=2\times 10^{2},a=800,a-x=50 $
$ k=\frac{2.303}{2\times 10^{2}}{\log_{10}}\frac{800}{50}=\frac{2.303}{2\times 10^{2}}{\log_{10}}16 $
$ =\frac{2.303}{2\times 10^{2}}{\log_{10}}2^{4}=\frac{2.303}{2\times 10^{4}}\times 4\times 0.301 $
$ =1.38\times {10^{-2}}{s^{-1}} $
