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:

$A⟶B$

We have,

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

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

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

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:
  
  ${\log }_{\mathrm{e}}\mathrm{A}={\log }_{\mathrm{e}}{\mathrm{A}}_{\mathrm{o}}-\mathrm{kt}$

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

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

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

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