Question

Derive the integrated rate equation for expressing the rate constant of a first order:
$R \to P$

Answer 1

Hint: The first order reaction is defined as the chemical reaction where the reaction rate is linearly dependent on the concentration of only one reactant. The integrated rate expression is used to calculate the rate constant of the reaction.

Complete step by step answer:
In first order reaction, In rate law the sum of the powers of concentration of reactant is equal to 1. The rate of reaction is directly proportional to the power of the concentration of reactant.
The reaction is given below.
$R \to P$
Here, the rate of reaction is directly proportional to [R].
The rate of reaction is given as shown below.
Rate of reaction = \[\dfrac{{ - d[R]}}{{dt}} = {k_1}{[R]^{1}}\]
Where, t is the time, k is the rate constant, R is the reactant.
At the initial state the t = 0, the concentration of R is ‘r’ mol $li{t^{ - 1}}$. After the reaction has been proceeded for some time ‘t’, let the concentration of R which has been reacted be x mol $li{t^{ - 1}}$. The concentration of the remaining reactant R at time t will be (R-x) mol $li{t^{ - 1}}$. The rate of reaction is given by dx/dt.
For first order reaction, the rate is given as shown below.
Rate = $\dfrac{{dx}}{{dt}} = {k_1}(R - x)$ ……..(i)
Integrate equation (i) from both the sides.
$\int {\dfrac{{dx}}{{(R - x)}}} = {k_1}\int {dt}$
$- \ln (R - x) = {k_1}t + c$ …….(ii)
Where, c is integration constant
At t = 0, x = 0.
Substitute the values in equation (ii).
$- \ln (R - 0) = {k_1} \times 0 + c$
$c = \operatorname{lnR}$
Substitute the value of c in equation (ii).
$- \ln (R - x) = {k_1}t - \ln R$
Rearranging the above equation, we get
${k_1} = \dfrac{1}{t}\ln \dfrac{R}{{R - x}}$
${k_1} = \dfrac{{2.303}}{t}\ln \dfrac{R}{{R - x}}$

Note: The unit of rate constant in first order reaction is ${\sec ^{ - 1}}$. The half life of the first order reaction is the time taken for the reactant’s initial concentration to reach half of the original value.