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}} $



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