Integrated Rate Equations and Half-Life

The rate law provides the instantaneous rate of a reaction, but to determine the concentration of a reactant at a specific time or the time required to complete a certain fraction of the reaction, we need integrated rate equations. These equations are derived by integrating the differential rate laws. The concept of half-life is directly related to these equations and is a crucial tool for understanding reaction kinetics, especially for JEE aspirants.

1.0What are Integrated Rate Equations?

Integrated rate equations are mathematical expressions that relate the concentration of a reactant to time. They are derived by integrating the differential rate law. These equations are invaluable because they allow us to predict:

• The concentration of a reactant at any given time.
• The time required for a reaction to reach a specific concentration.
• The half-life of a reaction.

2.0Half-Life(t_1/2):

Half-life t_1/2 is the time required for a reactant's concentration to be reduced to one-half of its initial value. It is a critical parameter for characterizing the rate of a reaction.

• Zero-Order:

$t_{1/2}\propto [R{]}_0$

Half-life decreases as the reaction proceeds.

• First-Order: ​$t_{1/2}=\frac{\ln 2}{k}=\frac{0.693}{k}$

Half-life is constant, regardless of the initial concentration. This is a key characteristic of first-order reactions, such as radioactive decay.

• Second-Order:  $t_{1/2}\propto \frac{1}{[R{]}_0}$

Half-life increases as the reaction proceeds.

3.0Integrated Rate Equation for Zero-Order Reactions

A zero-order reaction is a reaction whose rate is independent of the concentration of the reactants. The rate law is given by:

$\text{Rate}=k[R{]}^0=k$

Integrating this equation gives the integrated rate equation for a zero-order reaction:

Integrated Rate Equation:

$[R{]}_t=[R{]}_0-kt$

Where:

• [R]_t is the concentration of the reactant at time t.
• [R]_0​ is the initial concentration of the reactant.
• k is the rate constant.

The graph of [R]_t versus time (t) for a zero-order reaction is a straight line with a negative slope equal to −k.

Half-Life for Zero-Order Reactions t_1/2 

Half-life is the time required for the concentration of a reactant to decrease to half of its initial value.

$t_{1/2}=\frac{[R{]}_0}{2k}$

Substituting this into the integrated rate equation:

$[R{]}_0/2=-k{t}_{1/2}+[R{]}_0$

$k{t}_{1/2}=[R{]}_0-[R{]}_0/2=[R{]}_0/2$

$t_{1/2}=\frac{[R{]}_0}{2k}$

For a zero-order reaction, the half-life is directly proportional to the initial concentration of the reactant.

4.0Integrated Rate Equation for First-Order Reactions

A first-order reaction is a reaction whose rate is directly proportional to the concentration of a single reactant. The rate law is given by:

$\text{Rate}=-\frac{d[R]}{dt}=k[R]$

Integrating this equation gives the integrated rate equation for a first-order reaction:

Integrated Rate Equation:

$\ln [R{]}_t=-kt+\ln [R{]}_0$

This equation is in the form y=mx+c, where a plot of ln[R]t​ versus time (t) is a straight line with a negative slope equal to −k.

The equation can also be expressed as:

$k=\frac{2.303}{t}\log \frac{[R{]}_0}{[R{]}_t}$

$Half-LifeforFirst-OrderReactions(t_{1/2}):$

$t=t_{1/2},[R{]}_t=[R{]}_0/2.$

Substituting this into the integrated rate equation:

$k=\frac{2.303}{t_{1/2}}\log \frac{[R{]}_0}{[R{]}_0/2}=\frac{2.303}{t_{1/2}}\log 2$

$k=\frac{2.303\times 0.3010}{t_{1/2}}=\frac{0.693}{t_{1/2}}$

$t_{1/2}=\frac{0.693}{k}$

For a first-order reaction, the half-life is independent of the initial concentration of the reactant.

5.0Integrated Rate Equation for Second-Order Reactions

A second-order reaction is a reaction whose rate depends on the square of a single reactant concentration or the product of two reactant concentrations. For the reaction 2A→ Products, the rate law is:

$\text{Rate}=-\frac{d[A]}{dt}=k[A{]}^2$

Integrating this gives the integrated rate equation for a second-order reaction:

Integrated Rate Equation:

$\frac{1}{[A{]}_t}=kt+\frac{1}{[A{]}_0}$

A plot of $\frac{1}{[A{]}_t}$ versus time (t) gives a straight line with a positive slope equal to k.

Half-Life for Second-Order Reactions $(t_{1/2})$:

$Att=t_{1/2},[A{]}_t=[A{]}_0/2.$

$\frac{1}{[A{]}_0/2}=k{t}_{1/2}+\frac{1}{[A{]}_0}$

$\frac{2}{[A{]}_0}-\frac{1}{[A{]}_0}=k{t}_{1/2}$

$t_{1/2}=\frac{1}{k[A{]}_0}$

For a second-order reaction, the half-life is inversely proportional to the initial concentration of the reactant.

6.0Practice Problem

Problem: A first-order reaction has a rate constant (k) 5.5 \times 10^{-14}\text{ s}^{-1}  at a certain temperature. How long will it take for the reactant to be reduced to 25% of its initial concentration?

Solution:

For a first-order reaction, the integrated rate equation is:

$t=\frac{2.303}{k}\log \frac{[R{]}_0}{[R{]}_t}$

Given:

$k=5.5\times 10^{-14}{\text{ s}}^{-1}$

$[R{]}_t=0.25[R{]}_0$

Substitute the values into the equation:

$t=\frac{2.303}{5.5\times 10^{-14}}\log \frac{[R{]}_0}{0.25[R{]}_0}$

$t=\frac{2.303}{5.5\times 10^{-14}}\log 4$

$t=\frac{2.303}{5.5\times 10^{-14}}\times 0.6020$

$t\approx 2.52\times 10^{13}\text{ s}$

7.0