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Question: An organic compound undergoes first-order decomposition. The time taken for its decomposition to 1/8...

An organic compound undergoes first-order decomposition. The time taken for its decomposition to 1/8 and 1/10 of its initial concentration at t18{{t}_{\dfrac{1}{8}}} and t110{{t}_{\dfrac{1}{10}}} respectively. What is the value of t18t110×10\dfrac{{{t}_{\dfrac{1}{8}}}}{{{t}_{\dfrac{1}{10}}}}\times 10 ? (take log102=0.3{{\log }_{10}}2=0.3 )

A. 9

B. 8

C. 10

D. 15

Explanation

Solution

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 t12{{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 = t12{{t}_{\dfrac{1}{2}}} , [A] = Ao/2{{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, t18{{t}_{\dfrac{1}{8}}} and t110{{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=lnAoAtKt=\ln \dfrac{{{A}_{o}}}{{{A}_{t}}}

Kt18=lnAoAo18K{{t}_{\dfrac{1}{8}}}=\ln \dfrac{{{A}_{o}}}{{{A}_{o}}_{\dfrac{1}{8}}}

Kt18=2.303(log102)×3K{{t}_{\dfrac{1}{8}}}=2.303({{\log }_{10}}2)\times 3

Similarly for t110{{t}_{\dfrac{1}{10}}} , Kt110=2.303(log1010)K{{t}_{\dfrac{1}{10}}}=2.303({{\log }_{10}}10)

As we know log 10 = 1

Kt110=2.303\therefore K{{t}_{\dfrac{1}{10}}}=2.303

Now put the values in the equation;

t18t110×10\dfrac{{{t}_{\dfrac{1}{8}}}}{{{t}_{\dfrac{1}{10}}}}\times 10

2.303(log102)×32.303×10\dfrac{2.303({{\log }_{10}}2)\times 3}{2.303}\times 10

log 2 = 0.3

0.3×3×100.3\times 3\times 10

= 9

Thus 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.