An organic compound undergoes first-order decomposition. The time taken for its decomposition to 1/8 and 1/10 of its initial concentration at $ {{t}_{\dfrac{1}{8}}} $ and $ {{t}_{\dfrac{1}{10}}} $ respectively. What is the value of $ \dfrac{{{t}_{\dfrac{1}{8}}}}{{{t}_{\dfrac{1}{10}}}}\times 10 $ ? (take $ {{\log }_{10}}2=0.3 $ )A. 9B. 8C. 10D. 15

Question

An organic compound undergoes first-order decomposition. The time taken for its decomposition to 1/8 and 1/10 of its initial concentration at  $ {{t}_{\dfrac{1}{8}}} $  and  $ {{t}_{\dfrac{1}{10}}} $  respectively. What is the value of  $ \dfrac{{{t}_{\dfrac{1}{8}}}}{{{t}_{\dfrac{1}{10}}}}\times 10 $ ? (take  $ {{\log }_{10}}2=0.3 $ )A. 9B. 8C. 10D. 15

Vedantu Content Team · Accepted Answer

Hint: Order of reaction depends upon the number of concentrations of the reactants i.e. if only one reactant is present then it is known as first order reaction. The half-life of a chemical reaction is the time taken for the initial concentration of the reactant to reach half of its original value.Complete Step by step solution: Half-life reaction is denoted by  $ {{t}_{\dfrac{1}{2}}} $ which actually tells the time taken for the initial concentration of the reactant to reach half of its original value. Therefore at t =  $ {{t}_{\dfrac{1}{2}}} $ , [A] =  $ {{A}_{o}}/2 $ ; Where [A] denotes the concentration of the reactant and [A]0 denotes the initial concentration of the reactant.So according to this concept,  $ {{t}_{\dfrac{1}{8}}} $ and  $ {{t}_{\dfrac{1}{10}}} $  tells the time taken for the initial concentration of the reactant to reach 1/8 and 1/10 of its original value.According to the concept of first order reaction of kinetics $ Kt=\ln \dfrac{{{A}_{o}}}{{{A}_{t}}} $  $ K{{t}_{\dfrac{1}{8}}}=\ln \dfrac{{{A}_{o}}}{{{A}_{o}}_{\dfrac{1}{8}}} $  $ K{{t}_{\dfrac{1}{8}}}=2.303({{\log }_{10}}2)\times 3 $ Similarly for  $ {{t}_{\dfrac{1}{10}}} $ ,  $ K{{t}_{\dfrac{1}{10}}}=2.303({{\log }_{10}}10) $ As we know log 10 = 1 $ \therefore K{{t}_{\dfrac{1}{10}}}=2.303 $ Now put the values in the equation; $ \dfrac{{{t}_{\dfrac{1}{8}}}}{{{t}_{\dfrac{1}{10}}}}\times 10 $  $ \dfrac{2.303({{\log }_{10}}2)\times 3}{2.303}\times 10 $ log 2 = 0.3 $ 0.3\times 3\times 10 $ = 9Thus the option (A) is correct.Note: The half-life term is commonly used in nuclear physics to describe how quickly unstable atoms undergo, or how long stable atoms survive radioactive decay. The term is also used more generally to characterize any type of exponential or non-exponential decay. For example, the medical sciences refer to the biological half-life of drugs and other chemicals in the human body. The converse of half-life is doubling time.

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