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Question: The rate of change of concentration of A for reaction\(A\to \Pr oduct\) is given by \(-\dfrac{d[A]}{...

The rate of change of concentration of A for reactionA→Pr⁡oductA\to \Pr oduct is given by −d[A]dt=k[A]13-\dfrac{d[A]}{dt}=k{{[A]}^{\dfrac{1}{3}}}half-life period of the reaction will be
A. 3[Ao]23[(2)23−1]2(2)53k\dfrac{3{{[{{A}_{o}}]}^{\dfrac{2}{3}}}{{[{{(2)}^{\dfrac{2}{3}}}-1]}^{2}}}{{{(2)}^{\dfrac{5}{3}}}k}
B. 3/2[Ao]23[(2)23−1]k\dfrac{3/2{{[{{A}_{o}}]}^{\dfrac{2}{3}}}[{{(2)}^{\dfrac{2}{3}}}-1]}{k}
C. 3/2[Ao]23[(2)23−1](2)23k\dfrac{3/2{{[{{A}_{o}}]}^{\dfrac{2}{3}}}[{{(2)}^{\dfrac{2}{3}}}-1]}{{{(2)}^{\dfrac{2}{3}}}k}
D. 2/3[Ao]32[(2)23−1]k\dfrac{2/3{{[{{A}_{o}}]}^{\dfrac{3}{2}}}[{{(2)}^{\dfrac{2}{3}}}-1]}{k}

Explanation

Solution

There is a special branch of chemistry which deals with the rate of reaction and the order of reaction and order of any reaction can be defined as the power which depends on the rate of concentration of all reactants.

Complete answer:
Rate constant can be defined as the proportionality constant which explains the relationship between the molar concentration of the reactants and the rate of a chemical reaction.
Half-life is defined as the amount of time taken for half of a particular sample for react and it can also be explained on the basis on the time which it requires to reduce its initial value to half.
According to the given equation
32[A0]2/3[(2)2/3−1](2)2/3k=t1/2\dfrac{\dfrac{3}{2}{{[{{A}_{0}}]}^{2/3}}[{{(2)}^{2/3}}-1]}{{{(2)}^{2/3}}k}={{t}_{1/2}}
This equation can also be written as
−d[A][A]13=kdt-\dfrac{d[A]}{{{[A]}^{\dfrac{1}{3}}}}=kdt
To calculate the half-life integrate both side with limits A0{{A}_{0}}to A0/2{{A}_{0}}/2and 0 to t1/2{{t}_{1/2}}, this can be written as:
-∫d[A][A]13\int{\dfrac{d[A]}{{{[A]}^{\dfrac{1}{3}}}}}= ∫kdt\int{kdt}
−32[A]2/3-\dfrac{3}{2}{{[A]}^{2/3}}having limits in between A0{{A}_{0}}to A0/2{{A}_{0}}/2now put the values of limits and the equation will be:
−32[(A0)2/3−(A02)2/3]-\dfrac{3}{2}[{{({{A}_{0}})}^{2/3}}-{{(\dfrac{{{A}_{0}}}{2})}^{2/3}}], on the right hand side it becomes kt and have the limits 0 to t1/2{{t}_{1/2}}now put the value of limit the equation will be equal to kt1/2k{{t}_{1/2}}
Equation becomes
−32[(A0)2/3−(A02)2/3]=kt1/2-\dfrac{3}{2}[{{({{A}_{0}})}^{2/3}}-{{(\dfrac{{{A}_{0}}}{2})}^{2/3}}]=k{{t}_{1/2}}
−32[(A0)2/3−(A02)2/3]k=t1/2-\dfrac{\dfrac{3}{2}[{{({{A}_{0}})}^{2/3}}-{{(\dfrac{{{A}_{0}}}{2})}^{2/3}}]}{k}={{t}_{1/2}}
Take the value of (A0)2/3{{({{A}_{0}})}^{2/3}}common from L.H.S and the equation becomes
32[A0]2/3[(2)2/3−1](2)2/3k=t1/2\dfrac{\dfrac{3}{2}{{[{{A}_{0}}]}^{2/3}}[{{(2)}^{2/3}}-1]}{{{(2)}^{2/3}}k}={{t}_{1/2}}

This suggests that option C is the correct answer.

Note:
Half-life time is generally used in nuclear physics to describe what time period an unstable atom undergoes through the process of radioactive decay and we can also calculate how long an atom survives to be stable.