Question

Derive an expression for the rate constant for a first order reaction.

Answer 1

Rate constant of a first order reaction: The reaction in which, the overall rate of the reaction is proportional to the first power of concentration of one of the reactants only are called first order reaction.
$A\stackrel{k_1}{\to }$ products
Rate of reaction $=\frac{-d\left[A\right]}{dt}=k_1{\left[A\right]}^{10}$
where $k_1$ is the rate constant of the first order reaction.
At the beginning of the reaction, time ${}^{\prime }{t}^{\prime }=0$, let the concentration of $A$ be '$a$' mole lit${}^{-1}$. After the reaction has proceeded for some time '$t$', let the concentration of $A$ that has reacted be $x$ mole lit${}^{-1}$. The concentration of unreacted $A$ remaining at time '$t$' will be $(a-x)$ mole lit${}^{-1}$. The rate of the reaction will be $dx/dt$. For a first order reaction,
rate $=dx/dt=k_1(a-x)$ ...... (1)
Integrating $\left(1\right)$, both sides
$\int \frac{dx}{(a-x)}=k_1\int dt$
Which is $-ln(a-x)=k_{1}t+c$ ...... (2)
$C=$ integration constant.
at time, $t=0,x=0$
In equation $\left(2\right)$
$-ln(a-0)=k_1\times 0+c$
or
$c=ln$ $a$
Substituting $c$ value in equation $\left(2\right)$
$-ln(a-x)=k_{1}t-ln$ $a$
rearranging,
$k_1=\frac{1}{t}ln\frac{a}{a-x}$
$k_1=\frac{2.303}{t}\log \frac{a}{a-x}$
Unit of $k_1$ is ${\sec }^{-1}$
This equation is known as the first order rate constant equation.