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

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.