Question

In the given reaction, ${\text{A}} \to $Product, \[{\left[ {\mathbf{A}} \right]_{\mathbf{0}}} = {\mathbf{2M}}\]. After 10 minutes the reaction is $10\% $ completed. If $\dfrac{{{\mathbf{d}}\left[ {\mathbf{A}} \right]}}{{{\mathbf{dt}}}} = {\mathbf{k}}\left[ {\mathbf{A}} \right]\;$ then \[{{\text{t}}_{1/2}}\] ​ is approximately:
A.\[0.693{\text{ min}}\]
B.\[69.3{\text{ min}}\]
C.\[{\text{66}}{\text{.0 min}}\]
D.\[0.0693{\text{ min}}\]

Answer 1

Hint: To answer this question, you should recall the concept of the specific activity of a labeled sample. The specific activity can be defined as the activity per quantity of a radionuclide and it is a physical property of that radionuclide. Use the formula of decay to calculate the decay constant and then use it to calculate half-life.

Formula used:
\[{\text{N = }}{{\text{N}}_{\text{0}}}{{\text{e}}^{{ - \lambda t}}}\] where ${\text{N}}$ = No. of particles left after time t, ${{\text{N}}_{\text{o}}}$ = No. of particles initially, $\lambda $ = decay constant and ${\text{t}}$= time. ---(i)
${{\text{t}}_{1/2}} = \dfrac{{\ln 2}}{\lambda } $where \[{{\text{t}}_{1/2}}\] is the half-life.

Complete Step by step solution:
We will use differential rate law:
\[\dfrac{{d\left[ {\text{A}} \right]}}{{dt}} = {\text{K}}\left[ {\text{A}} \right]\]
which clearly depicts that the rate of the reaction $A \to $Product, is dependent on the concentration of the reactant.
So, it is a first order reaction.
 Now, Given that \[{\left[ {\text{A}} \right]_o} = 2{\text{M}}\]and after \[{\text{t}} = 10{\text{min}}\],
\[\left[ {\text{A}} \right] = \dfrac{{90}}{{100}} \times 2{\text{M}} = 1.8{\text{M}}\]
Now, for first order reaction we have:
\[{\text{K = }}\dfrac{{\text{1}}}{{\text{t}}}{ \times ln}\left( {\dfrac{{{{\left[ {\text{A}} \right]}_{\text{o}}}}}{{\left[ {\text{A}} \right]}}} \right)\]
The values of the above variables: \[{\text{ t}}\]=10 min, \[{\left[ {\text{A}} \right]_o}\]==2, \[\left[ {\text{A}} \right] = 1.8\] , ${\text{K}}$ =?
using the above given data, and solving:
\[{\text{K}} = \dfrac{1}{{10}} \times {\text{ln}}(\dfrac{2}{{1.8}})\]
$\Rightarrow$ ${\text{K = }}\dfrac{1}{{10}} \times \log \left( {\dfrac{{10}}{9}} \right)$
Solving this, we get:
$\Rightarrow$ ${\text{K = }}0.0105{\text{ mi}}{{\text{n}}^{ - 1}}$
Now, for first order reaction we have,
\[{{\text{t}}_{{\text{1/2}}}} = \dfrac{{0.693}}{{\text{K}}}\]
Solving and rearranging we get:
$\Rightarrow$ \[{{\text{t}}_{{\text{1/2}}}} = \dfrac{{0.693}}{{0.0105}} = 69.3\;{\text{min}}\]

Therefore, we can conclude that the correct answer to this question is option B.

Note: The emissions in most of the spontaneous radioactive decay involves alpha $(\alpha )$particle, the beta $(\beta )$ particle, the gamma-ray, and the neutrino. The alpha particle is the nucleus of doubly charged ${\text{He}}_2^4$. Beta particles can be beta minus beta plus. Beta minus is an electron created in the nucleus during beta decay. Beta plus particle is also known as a positron, is the antiparticle of the electron; when brought together, two such particles will mutually annihilate each other. Radioactivity is also considered as a first order reaction as its rate of disintegration only depends on the concentration of the reactant. The rate equations used are also similar to the first order rate equation.