Question

Derive the equation of rate constant for the first order reaction.Answer it in detail.

Answer 1

please see the attached photograph

Answer 2

Rate constant for the first order reaction:

When the rate of the reaction is proportional to the first power of concentration of the reactants, then it is called first order reaction.

$A\overset{k_{1}}{\rightarrow}Product$

Now, the rate of the above reaction is:

$R=-\frac{d\left [ A \right ]}{dt}=k_{1}\left [ A^{10} \right ]$

Where, $k_1$ is the rate constant of the first order reaction

At the beginning of the reaction:

Time = t = 0

Let the concentration of A be 'a' $mole \ lit^{-1}$

After some time:

Time = t = t

Let the concentration of A that has reacted be ‘x’ $mole \ lit^{-1}$

The unreacted reactant:

The remaining at time is 't'

The concentration will be (a-x)(a−x) $mole \ lit^{-1}$

The rate of the reaction will be dx/dt.

For a first order reaction,

$Rate=\frac{dx}{dt}=k_{1}\left ( a-x \right )\rightarrow \left ( 1 \right )$

Integrating (1) on both sides,

$\int \frac{dx}{dt}=k_{1}\int dt$

$-ln\left ( a-x \right )=k_{1}t+c\rightarrow \left ( 2 \right )$

c = integration constant.

Now,

At time, t = 0, x = 0

In equation (2)

$-ln\left ( a-0 \right )=\left ( k_{1}\times 0 \right )+ c$

Or

$c=-ln \ a$

Substituting ‘c’ value in equation (2)

$-ln\left ( a-x \right )=k_{1}t-ln \ a$

Rearranging,

$k_{1}=\frac{1}{t}ln\frac{a}{a-x}$

Now,

$k_{1}=\frac{2.303}{t}log\frac{a}{a-x}$

Unit of $k_{1}$ is $sec ^{-1}$

This equation is known as the first order rate constant equation.