First Order Reaction MCQ - Practice Questions & Answers

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First Order Reaction, Half Life of First Order Reaction, Graphs of First Order Kinetics are considered the most difficult concepts.
89 Questions around this concept.

Concepts Covered - 4

First Order Reaction

The rate of the reaction is proportional to the first power.

The chemical reaction occurs as follows:

R $\rightarrow$ P

a 0

a-x x

We have,
$rate[r]=K[R]^{1}$

$\frac{-d(a-x)}{dt}=K(a-x)$

$\frac{-dx}{dt}=K(a-x)$ [differentiate rate law]

$ln \:[\frac{a}{a-x}]=kt \:(Integrated\: rate\: law)$

$\mathrm{k\: =\: \frac{1}{t}\, ln\left [ \frac{a}{a-x} \right ]}$

Unit of $k=sec^{-1}$

Other Forms of Rate Law

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

$\mathrm{log_{10}A\: =\: log_{10}A_{o}\: -\: \frac{kt}{2.303}}$

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

Use to solve numericals:

$\mathrm{log_{10}A\: =\: log_{10}A_{o}\: -\: \frac{kt}{2.303}}$

$\mathrm{\Rightarrow log_{10}\left [ \frac{A_{o}}{A} \right]\: =\: \frac{kt}{2.303}}$

$\mathrm{Thus, t\: =\: \frac{2.303}{k}\, log_{10}\left [ \frac{A_{o}}{A} \right]}$
Exponential form:

$\mathrm{log_{e}A\: -\: log_{e}A_{o}\: -kt}$
$\\\mathrm{\Rightarrow log\frac{A}{A_{o}}\: =\: -kt}\\\\\mathrm{\Rightarrow \frac{A}{A_{o}}\: =\: e^{-kt}}\\\\\mathrm{Thus,\: A\: =\: A_{o}e^{-kt}}$
This equation is also known as exponential form.

Half Life of First Order Reaction

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:
$k=\frac{[\mathrm{R}]_{0}-[\mathrm{R}]}{t}$
$\text { At } t=t_{1 / 2}, \quad[\mathrm{R}]=\frac{1}{2}[\mathrm{R}]_{0}$
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,

$k=\frac{2.303}{t} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}$

$\text { at } t_{1 / 2} \quad[\mathrm{R}]=\frac{[\mathrm{R}]_{0}}{2}$
So, the above equation becomes

$k=\frac{2.303}{t_{1 / 2}} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]_{0} / 2}$