
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
568.8k+ 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.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Basicity of sulphurous acid and sulphuric acid are

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

How much time does it take to bleed after eating p class 12 biology CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

