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First Order Reaction, Half Life of First Order Reaction, Graphs of First Order Kinetics are considered the most difficult concepts.
153 Questions around this concept.
Which of the following is the unit of rate constant for first order reaction?
can be taken as the time taken for the concentration of a reactant to drop to 3/4 of its initial value. If the rate constant for a first order reaction is
the
can be written as
The rate equation for the reaction is found to be : rate
The correct statement in relation to this reaction is that the
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The half-life period of a first-order reaction is 15 minutes. The amount of substance left after one hour will be :
Units of the rate constant of first and zero-order reactions in terms of molarity M unit are respectively.
In a reaction, , the X disappears at
Identify the correct order of reaction for which initial concentration vs half-time plot is -
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A reaction that is of the first order with respect to reactant has a rate constant
. If we start with
, when would
reach the value
?
Which of the following plot does not represent first order reaction? ( $r=$ rate at time ${ }^{\prime} t{ }^{\prime}$ and $r_0$ is the initial rate)
A bacterial infection in an internal wound grows as $N^{\prime}(t)=N_o \exp (t)$, where the time $t$ is in hours. A dose of antibiotic, taken orally, needs 1 hour to reach the wound.
Once it reaches there, the bacterial population goes down as $\frac{d N}{d t}=-5 N^2$.
What will be the plot of $\frac{N_o}{N}$ vs. t after 1 hour?
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:
$
\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, $\mathrm{A}=\mathrm{A}_{\mathrm{o}} \mathrm{e}^{-\mathrm{kt}}$
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 t1/2.
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 t1/2 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 t1/2 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 $\quad t_{1 / 2}=\frac{2.303}{k} \log 2$
$t_{1 / 2}=\frac{0.693}{k}$
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