Integrated Rate Equations and Halflife

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

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:

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

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

Integrated Rate Equation:

$[R]_{t} = [R]_{0} - k t$

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.

$t_{1/2} = \frac{[ R ] _{0}}{2 k}$

Substituting this into the integrated rate equation:

$[R]_{0} /2 = - k t_{1/2} + [R]_{0}$

$k t_{1/2} = [R]_{0} - [R]_{0} /2 = [R]_{0} /2$

$t_{1/2} = \frac{[ R ] _{0}}{2 k}$

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:

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

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

Integrated Rate Equation:

$ln [R]_{t} = - k t + ln [R]_{0}$

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:

$k = \frac{2.303}{t} lo g \frac{[ R ] _{0}}{[ R ] _{t}}$

$H a l f - L i f e f or F i rs t - O r d er R e a c t i o n s (t_{1/2}) :$

$t = t_{1/2}, [R]_{t} = [R]_{0} /2.$

Substituting this into the integrated rate equation:

$k = \frac{2.303}{t _{1/2}} lo g \frac{[ R ] _{0}}{[ R ] _{0} /2} = \frac{2.303}{t _{1/2}} lo g 2$

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

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

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:

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

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

Integrated Rate Equation:

$\frac{1}{[ A ] _{t}} = k t + \frac{1}{[ A ] _{0}}$

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 $(t_{1/2})$ :

$A tt = t_{1/2}, [A]_{t} = [A]_{0} /2.$

$\frac{1}{[ A ] _{0} /2} = k t_{1/2} + \frac{1}{[ A ] _{0}}$

$\frac{2}{[ A ] _{0}} - \frac{1}{[ A ] _{0}} = k t_{1/2}$

$t_{1/2} = \frac{1}{k [ A ] _{0}}$

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:

$t = \frac{2.303}{k} lo g \frac{[ R ] _{0}}{[ R ] _{t}}$

Given:

$k = 5.5 \times 1 0^{- 14} s^{- 1}$

$[R]_{t} = 0.25 [R]_{0}$

Substitute the values into the equation:

$t = \frac{2.303}{5.5 \times 1 0 ^{- 14}} lo g \frac{[ R ] _{0}}{0.25 [ R ] _{0}}$

$t = \frac{2.303}{5.5 \times 1 0 ^{- 14}} lo g 4$

$t = \frac{2.303}{5.5 \times 1 0 ^{- 14}} \times 0.6020$

$t \approx 2.52 \times 1 0^{13} s$

7.0 Frequently Asked Questions (FAQs)

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

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:

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

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

Integrated Rate Equation:

$[R]_{t} = [R]_{0} - k t$

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.

$t_{1/2} = \frac{[ R ] _{0}}{2 k}$

Substituting this into the integrated rate equation:

$[R]_{0} /2 = - k t_{1/2} + [R]_{0}$

$k t_{1/2} = [R]_{0} - [R]_{0} /2 = [R]_{0} /2$

$t_{1/2} = \frac{[ R ] _{0}}{2 k}$

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:

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

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

Integrated Rate Equation:

$ln [R]_{t} = - k t + ln [R]_{0}$

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:

$k = \frac{2.303}{t} lo g \frac{[ R ] _{0}}{[ R ] _{t}}$

$H a l f - L i f e f or F i rs t - O r d er R e a c t i o n s (t_{1/2}) :$

$t = t_{1/2}, [R]_{t} = [R]_{0} /2.$

Substituting this into the integrated rate equation:

$k = \frac{2.303}{t _{1/2}} lo g \frac{[ R ] _{0}}{[ R ] _{0} /2} = \frac{2.303}{t _{1/2}} lo g 2$

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

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

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:

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

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

Integrated Rate Equation:

$\frac{1}{[ A ] _{t}} = k t + \frac{1}{[ A ] _{0}}$

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 $(t_{1/2})$ :

$A tt = t_{1/2}, [A]_{t} = [A]_{0} /2.$

$\frac{1}{[ A ] _{0} /2} = k t_{1/2} + \frac{1}{[ A ] _{0}}$

$\frac{2}{[ A ] _{0}} - \frac{1}{[ A ] _{0}} = k t_{1/2}$

$t_{1/2} = \frac{1}{k [ A ] _{0}}$

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:

$t = \frac{2.303}{k} lo g \frac{[ R ] _{0}}{[ R ] _{t}}$

Given:

$k = 5.5 \times 1 0^{- 14} s^{- 1}$

$[R]_{t} = 0.25 [R]_{0}$

Substitute the values into the equation:

$t = \frac{2.303}{5.5 \times 1 0 ^{- 14}} lo g \frac{[ R ] _{0}}{0.25 [ R ] _{0}}$

$t = \frac{2.303}{5.5 \times 1 0 ^{- 14}} lo g 4$

$t = \frac{2.303}{5.5 \times 1 0 ^{- 14}} \times 0.6020$

$t \approx 2.52 \times 1 0^{13} s$

Frequently Asked Questions (FAQs)

How do integrated rate equations differ from differential rate equations?

Can I use half-life to determine the order of a reaction?

What is the significance of the half-life of a first-order reaction being constant?

What is the difference between the Rate Law and the Specific Rate Constant?

Join ALLEN!

Integrated Rate Equations and Half-Life

1.0What are Integrated Rate Equations?

2.0Half-Life(t_1/2):

3.0Integrated Rate Equation for Zero-Order Reactions

4.0Integrated Rate Equation for First-Order Reactions

5.0Integrated Rate Equation for Second-Order Reactions

6.0Practice Problem

7.0

Table of Content

Frequently Asked Questions (FAQs)

How do integrated rate equations differ from differential rate equations?

Can I use half-life to determine the order of a reaction?

What is the significance of the half-life of a first-order reaction being constant?

What is the difference between the Rate Law and the Specific Rate Constant?

Join ALLEN!

Integrated Rate Equations and Half-Life

1.0What are Integrated Rate Equations?

2.0Half-Life(t_1/2):

3.0Integrated Rate Equation for Zero-Order Reactions

4.0Integrated Rate Equation for First-Order Reactions

5.0Integrated Rate Equation for Second-Order Reactions

6.0Practice Problem

7.0

Table of Content

Frequently Asked Questions (FAQs)

How do integrated rate equations differ from differential rate equations?

Can I use half-life to determine the order of a reaction?

What is the significance of the half-life of a first-order reaction being constant?

What is the difference between the Rate Law and the Specific Rate Constant?

Join ALLEN!

Integrated Rate Equations and Half-Life

1.0What are Integrated Rate Equations?

2.0Half-Life(t1/2):

3.0Integrated Rate Equation for Zero-Order Reactions

4.0Integrated Rate Equation for First-Order Reactions

5.0Integrated Rate Equation for Second-Order Reactions

6.0Practice Problem

7.0

Table of Content

Frequently Asked Questions (FAQs)

How do integrated rate equations differ from differential rate equations?

Can I use half-life to determine the order of a reaction?

What is the significance of the half-life of a first-order reaction being constant?

What is the difference between the Rate Law and the Specific Rate Constant?

Join ALLEN!

Integrated Rate Equations and Half-Life

1.0What are Integrated Rate Equations?

2.0Half-Life(t1/2):

3.0Integrated Rate Equation for Zero-Order Reactions

4.0Integrated Rate Equation for First-Order Reactions

5.0Integrated Rate Equation for Second-Order Reactions

6.0Practice Problem

7.0

Table of Content

2.0Half-Life(t_1/2):

2.0Half-Life(t_1/2):