First Order Reaction - Practice Questions & MCQ

First Order Reactions

The rate of the reaction is proportional to the first power of the concentration of the reactant

Let us consider a chemical reaction which occurs as follows:

$A \longrightarrow B$

We have,

$\operatorname{rate}(\mathrm{r})=\mathrm{K}[\mathrm{A}]^1$

$\frac{-\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\mathrm{K}[\mathrm{A}]$

$\Rightarrow \frac{\mathrm{d}[\mathrm{A}]}{[\mathrm{A}]}=-\mathrm{kdt}$

Integrating both sides and putting the limits

$\Rightarrow \int_{[\mathrm{A}]_0}^{[\mathrm{A}]_{\mathrm{t}} \mathrm{d}[\mathrm{A}]}[\mathrm{A}]=-\mathrm{k} \int_0^{\mathrm{t}} \mathrm{dt}$

$\Rightarrow \ln \left(\frac{[\mathrm{A}]_{\mathrm{t}}}{[\mathrm{A}]_0}\right)=-\mathrm{kt}$

Simplifying the above expression we have, 

$\Rightarrow \ln \left(\frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}}\right)=\mathrm{kt}$

In case we are dealing in terms of a and x (where a is the initial concentration of A and x is the amount of A dissociated at any time t)

$\Rightarrow \mathrm{k}=\frac{1}{\mathrm{t}} \ln \left(\frac{[\mathrm{A}]_0}{[\mathrm{~A}]_{\mathrm{t}}}\right)=\frac{1}{\mathrm{t}} \ln \left(\frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}}\right)$

Graphical Representation of first order reaction

Units of $\mathrm{k}=\mathrm{time}^{-1}$

We know that the first-order equation is given as follows:

$\log _{10} \mathrm{~A}=\log _{10} \mathrm{~A}_{\mathrm{o}}-\frac{\mathrm{kt}}{2.303}$

But there are other forms of rate law also available that we use for different purposes. These forms are mentioned below:

• Use to solve numerical:
  
  $\log _{10} \mathrm{~A}=\log _{10} \mathrm{~A}_{\mathrm{o}}-\frac{\mathrm{kt}}{2.303}$
  
  $\Rightarrow \log _{10}\left[\frac{\mathrm{~A}_0}{\mathrm{~A}}\right]=\frac{\mathrm{kt}}{2.303}$
  
  Thus, $\mathrm{t}=\frac{2.303}{\mathrm{k}} \log _{10}\left[\frac{\mathrm{~A}_{\mathrm{o}}}{\mathrm{A}}\right]$
• Exponential form:
  
  $\log _{\mathrm{e}} \mathrm{A}=\log _{\mathrm{e}} \mathrm{A}_{\mathrm{o}}-\mathrm{kt}$

  $
  \begin{aligned}
  & \Rightarrow \log \frac{\mathrm{A}}{\mathrm{~A}_{\mathrm{o}}}=-\mathrm{kt} \\
  & \Rightarrow \frac{\mathrm{~A}}{\mathrm{~A}_{\mathrm{o}}}=\mathrm{e}^{-\mathrm{kt}}
  \end{aligned}
  $

  
  Thus, $\mathrm{A}=\mathrm{A}_{\mathrm{o}} \mathrm{e}^{-\mathrm{kt}}$

  This equation is also known as exponential form.

The half-life of a reaction is the time in which the concentration of a reactant is reduced to one half of its initial concentration. It is represented as t_1/2.
For a zero order reaction, rate constant is given as:
$k=\frac{[\mathrm{R}]_0-[\mathrm{R}]}{t}$
At $t=t_{1 / 2}, \quad[\mathrm{R}]=\frac{1}{2}[\mathrm{R}]_0$
The rate constant at t_1/2 becomes:
$k=\frac{[\mathrm{R}]_0-1 / 2[\mathrm{R}]_0}{t_{1 / 2}}$

$t_{1 / 2}=\frac{[\mathrm{R}]_0}{2 \mathrm{k}}$
It is clear that t_1/2 for a zero order reaction is directly proportional to the initial concentration of the reactants and inversely proportional to the rate constant.
For the first order reaction,
$k=\frac{2.303}{t} \log \frac{[\mathrm{R}]_0}{[\mathrm{R}]}$

at $t_{1 / 2} \quad[\mathrm{R}]=\frac{[\mathrm{R}]_0}{2}$
So, the above equation becomes
$k=\frac{2.303}{t_{1 / 2}} \log \frac{[\mathrm{R}]_0}{[\mathrm{R}]_0 / 2}$

or $\quad t_{1 / 2}=\frac{2.303}{k} \log 2$

$t_{1 / 2}=\frac{0.693}{k}$