
When initial concentration of a reactant is doubled in a reaction, its half-life period is not affected. The order of the reaction is:
A. second
B. more than zero but less than first
C. zero
D. first
Answer
569.4k+ views
Hint: We know that the order of reaction can generally be described by the power of dependence of the rate in the concentration of all the reactant present in the system. The order of reaction usually depends on the rate of concentration.
Complete step by step answer:
As we all know, for zero order reaction the value of ${{\rm{t}}_{\dfrac{{\rm{1}}}{{\rm{2}}}}}$ is directly proportional to the initial molarity or concentration of the given reaction $\left[ {{{\rm{R}}_{\rm{0}}}} \right]$.
So, in the mathematical form the half life time can be expressed as shown below.
${{\rm{t}}_{\dfrac{{\rm{1}}}{{\rm{2}}}}}{\rm{\alpha }}\left[ {{{\rm{R}}_{\rm{0}}}} \right]$.
The first order reaction can be written as shown below.
\[{\rm{K}} = \dfrac{{{\rm{2}}{\rm{.303}}}}{{{{\rm{t}}_{\dfrac{{\rm{1}}}{{\rm{2}}}}}}}{\rm{log}}\dfrac{{\left[ {{{\rm{R}}_{\rm{0}}}} \right]}}{{\left[ {\rm{R}} \right]}}\]
\[\begin{array}{l}
{\rm{At}}\;{{\rm{t}}_{\dfrac{{\rm{1}}}{{\rm{2}}}}}\\
\left[ {\rm{R}} \right] = \dfrac{{\left[ {{{\rm{R}}_0}} \right]}}{2}
\end{array}\]
Substitute the value in the above equation.
\[\begin{array}{c}
{\rm{K}} = \dfrac{{{\rm{2}}{\rm{.303}}}}{{{{\rm{t}}_{\dfrac{{\rm{1}}}{{\rm{2}}}}}}}{\rm{log}}\dfrac{{\left[ {{{\rm{R}}_{\rm{0}}}} \right]}}{{\dfrac{{\left[ {{{\rm{R}}_0}} \right]}}{2}}}\\
{{\rm{t}}_{\dfrac{{\rm{1}}}{{\rm{2}}}}} = \dfrac{{2.303}}{{{{\rm{t}}_{\dfrac{{\rm{1}}}{{\rm{2}}}}}}}\log 2\\
{{\rm{t}}_{\dfrac{{\rm{1}}}{{\rm{2}}}}} = 0.3010
\end{array}\]
\[{{\rm{t}}_{\dfrac{{\rm{1}}}{{\rm{2}}}}} = \dfrac{{0.693}}{{\rm{K}}}\]
This clearly shows that the half-life period of the equation is independent from the initial concentration of the reactant. Thus, when the initial concentration of the given reactant gets doubled in the chemical reaction, then its half-life period is generally not affected. The order of that type of reaction is first order reaction.
Hence, the correct answer for this question is D.
Note:
The first order reaction is generally being used for the modulation or estimation of time in any chemical reaction. The hydrolysis of aspirin is the example of first order reaction.
Complete step by step answer:
As we all know, for zero order reaction the value of ${{\rm{t}}_{\dfrac{{\rm{1}}}{{\rm{2}}}}}$ is directly proportional to the initial molarity or concentration of the given reaction $\left[ {{{\rm{R}}_{\rm{0}}}} \right]$.
So, in the mathematical form the half life time can be expressed as shown below.
${{\rm{t}}_{\dfrac{{\rm{1}}}{{\rm{2}}}}}{\rm{\alpha }}\left[ {{{\rm{R}}_{\rm{0}}}} \right]$.
The first order reaction can be written as shown below.
\[{\rm{K}} = \dfrac{{{\rm{2}}{\rm{.303}}}}{{{{\rm{t}}_{\dfrac{{\rm{1}}}{{\rm{2}}}}}}}{\rm{log}}\dfrac{{\left[ {{{\rm{R}}_{\rm{0}}}} \right]}}{{\left[ {\rm{R}} \right]}}\]
\[\begin{array}{l}
{\rm{At}}\;{{\rm{t}}_{\dfrac{{\rm{1}}}{{\rm{2}}}}}\\
\left[ {\rm{R}} \right] = \dfrac{{\left[ {{{\rm{R}}_0}} \right]}}{2}
\end{array}\]
Substitute the value in the above equation.
\[\begin{array}{c}
{\rm{K}} = \dfrac{{{\rm{2}}{\rm{.303}}}}{{{{\rm{t}}_{\dfrac{{\rm{1}}}{{\rm{2}}}}}}}{\rm{log}}\dfrac{{\left[ {{{\rm{R}}_{\rm{0}}}} \right]}}{{\dfrac{{\left[ {{{\rm{R}}_0}} \right]}}{2}}}\\
{{\rm{t}}_{\dfrac{{\rm{1}}}{{\rm{2}}}}} = \dfrac{{2.303}}{{{{\rm{t}}_{\dfrac{{\rm{1}}}{{\rm{2}}}}}}}\log 2\\
{{\rm{t}}_{\dfrac{{\rm{1}}}{{\rm{2}}}}} = 0.3010
\end{array}\]
\[{{\rm{t}}_{\dfrac{{\rm{1}}}{{\rm{2}}}}} = \dfrac{{0.693}}{{\rm{K}}}\]
This clearly shows that the half-life period of the equation is independent from the initial concentration of the reactant. Thus, when the initial concentration of the given reactant gets doubled in the chemical reaction, then its half-life period is generally not affected. The order of that type of reaction is first order reaction.
Hence, the correct answer for this question is D.
Note:
The first order reaction is generally being used for the modulation or estimation of time in any chemical reaction. The hydrolysis of aspirin is the example of first order reaction.
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