# Integrated Rate Laws

Last Updated : 25 Aug, 2023

Integrated Rate Law is one of the fundamental concepts in the field of chemical kinetics, which is the branch of chemistry that deals with the speed or rate of reactions and various other factors affecting them. Integrated Rate Law tells us about the rate of the reaction for various different reactions such as zeroth order, first order, and second order, etc. Rate Law helps us from measuring rates to predicting the concentration of the reactants, which further helps scientists and scholars to unfold the mysteries of chemical transformations.

This article helps us learn about Integrated Rate Law and its derivation for various reactions. We will also learn how to solve numerical problems based on the Rate Law.

## What is Rate Law?

Rate Law is defined as the molar concentration of the reactants with each term raised to some powers which may or may not be the same as the stoichiometric coefficient of the reactant in a balanced chemical equation.

For a Chemical Balanced reaction i.e., A + B

Rate = k[A]α[B]β

Where,

• α and β are the concentration of the A and B respectively
• k is Rate Constant

### Rate Constant

The rate constant is defined as the rate of reaction when the molar concentration of all reactants is unity. For a balanced chemical equation of second order, i.e., A + B ⇢ C + D, if [A]=[B]= 1 mol/liter.

Therefore, Rate = K.

Note: Rate Constant depends on the molar concentration of the reactant, so it means the greater the value of the rate constant, the more quickly the reaction takes place.

Unit of Rate Constant: The unit of the rate constant depends on the order of the reaction (time, concentration).

## Order of Reaction

The sum of power to which the molar concentrations in the rate law equation are raised to express the observed rate of the reaction is known as the order of the reaction.

Example: What is the Order of Reaction given as follows:

2N2O5 ⇢ 4NO2  + O2

Solution:

For the given Reaction,

Rate of Reaction, R = [N2O5]1

As, rate of reaction only depends on the concentration of N2O5

Thus, Order of Reaction is 1

Note: Order of the reaction is basically integers numbers like 0,1,2 but for some complex species, it can be in fractions as well.

## What is Integrated Rate Law?

Integrated Rate Law provides a mathematical representation of the rate of the reaction using an initial and actual concentration of one or more reactants at any given time (t). Using this law we can calculate the rate constant as well as the mechanics of the reactions as well. For each order of reaction, the rate raw is given differently.  Let’s discuss the Integrated Rate Law for zeroth, first and second order reaction in detail.

### Integrated Rate Law Zero Order

Integrated Rate Law for zero order reaction states that the rate of reaction does not depend on the concentration of the reactants. For a reaction, A ⇢ Product zero order Integrated Rate Law is represented as:

Rate ∝ [A]0

Where,

[A] is the concentration of the Reactant

After integrating and simplifying the above condition, we get the integrated rate law for a zero-order reaction, i.e.,

[A] = [A0] – kt

Where,

• [A] is the concentration of the reactant A at time t
• [A0] is the initial concentration of reactant A
• k is the rate constant of the reaction
• t is the reaction time

### Integrated Rate Law First Order

This law states that the rate of the first-order reaction depends upon the first power of the reactant. For a reaction, A ⇢ Product first order Integrated Rate Law is represented as:

Rate ∝ [A]1

After integrating and simplifying the above condition, we get the integrated rate law for a first-order reaction, i.e.,

[A] = [A₀]e-kt

Where,

• [A] is the concentration of the reactant A at time t
• [A0] is the initial concentration of reactant A
• k is the rate constant of the reaction
• t is the reaction time
• e is the mathematical constant i.e., Euler number (e = 2.71828)

### Integrated Rate Law Second Order

Integrated Rate Law Second Order states that the rate of the second order reaction depends upon either the second power of the concentration. For a reaction, 2A ⇢ Product first order Integrated Rate Law is represented as:

Rate ∝ [A]2

After integrating and simplifying the above condition, we get the integrated rate law for a second-order reaction, i.e.,

Where,

• [A] is the concentration of the reactant A at time t
• [A0] is the initial concentration of reactant A
• k is the rate constant of the reaction
• t is the reaction time

## Integrated Rate Equations

For the general reaction aA + bB ⇢ cC + dD, the Rate of this reaction is given as:

Rate = -dR/dt = k[A]a[B]b

Where negative sign signify the decrease in concentration.

This form of the equation is called the differential rate equation. This form is not convenient to determine the rate law and hence the order of the reaction. This is because the instantaneous rate has to be determined to form the slope of the tangent at time t in the plot of concentration versus time.

To overcome the above difficulty, we integrate the differential equation for the reaction of any order. This gives us an equation related directly to the experimental data, i.e., time, concentrations at different times, and rate constant.

### Integrated Rate Equation for Zero-Order Reaction

Any reaction for which the rate doesn’t depend on the concentration of the reactant is called Zero Order Reaction. In other words, form A ⇢ Product; is Zero Order Reaction if its rate is independent of the concentration of A.

Consider the general reaction: A ⇢ products

If it is a reaction of zero-order,

Rate = – d[A]/dt = k[A]0 = k

⇒ d[A] = – k dt

Integrating both sides, we get:

[A] = – kt + C                . . . (i)

Where C is the constant of Integration.

at t = 0, the Concentration of the reaction is [A0], thus

[A0] = 0 + C ⇒ C = [A0]

Substituting this value of C in Eqn. (i), we get

[A] = – kt + [A0               . . . (ii)

⇒ kt = [A0] – [A]

⇒ k = {[A0] – [A]}/t                 . . . (iii)

### Characteristics of Reactions of Zero Order

The characteristics of the zero-order reaction are,

• Any reaction of zero-order must obey equation, (ii) As it is an equation of a straight line (y = mx + c), the plot of [A] versus t will be a straight line with slope = – k and intercept on the concentration axis = [A0],
• Half-Life Period: The half-life period (t1/2) is the time in which half of the substance has reacted.

This implies that when [A] = [A0]/2, t = t1/2.

Substituting these values in Eqn. (iii), we get

t1/2 = 1/k  { [A0] – [A0]/2 } = [A0]/2k

⇒ t1/2 =[A0]/2k                . . . (iv)

Thus, the half-life period of a zero-order reaction is directly proportional to the initial concentration, i.e., t1/2 ∝ [A]0.

• Units of Rate Constant (k) for Zeroth Order Reaction = molar conc./time = mol L-1 /time = mol L-1  time-1 .

### Integrated Rate Equation for First Order Reactions

A reaction is said to be of the first order if the rate of the reaction depends upon one concentration term only. Thus, we may have

• For the reaction: A ⇢ products, rate the reaction ∝ [A].
• For the reaction: 2A ⇢  products, rate the reaction ∝ [A] only.
• For the reaction: A + B ⇢  products, rate the reaction ∝ [A] or [B] only.

Let us consider the simplest case,

A  ⇢  Products

Suppose we start with moles per liter of the reactant A. After time t, suppose x moles per liter of it have decomposed. Therefore, the concentration of A after time t = (a – x) moles per liter. Then according to the law of mass action,

Rate of reaction ∝ (a – x),

i.e.,

dx/dt  ∝ (a – x)

⇒ dx/dt = k (a – x)                . . . (i)

where k is called the rate constant or the specific reaction rate for the reaction of the first order.

The expression for the rate constant k may be derived as follows:

Equation (i) may be rewritten in the form

dx/a-x =k dt                . . . (ii)

Integrating equation (ii), we get

-In (a – x) = kt + I                 . . . (iii)

where I is a constant of integration.

In the beginning, when t=0, x=0

Putting these values in equation (iii), we get

– In (a – 0) = k x 0 + I

⇒ – In a = I                . . . (iv)

Substituting this value of I in equation (iii), we get

– In (a – x) = kt + (- In a)

⇒ kt = a – In (a -x)

⇒ kt = In a/a-x                . . . (v)

⇒ k = 1/ t In a/a-x

⇒ k=2.303/t log a/a-x                . . . (vi)

If the initial concentration is [A]0 and the concentration after time t is [A], then putting a = [A]0 and (a – x ) = [A],

Equation (vi) becomes:

k = 2.303/t log [A0]/[A]                . . . (vii)

Putting a = [A0] and (a – x) = [A] in eqn. (v),

We get,

kt = In [A0]/[A]                . . . (viii)

which can be written in the exponential form as:

[A0]/[A] = ekt

⇒ [A]/[A0] =e-kt

[A] = [A0] e-kt                    . . . (ix)

### Characteristics of First-Order Reactions

The characteristics of the zero-order reaction are,

• Any reaction of the first order must obey equations (vi), (vii), and (ix).
• Half-life Period: The time in which half of the substance is reacted with each other and is converted into the product is called the half-life of the reaction.

As we know, the time taken for substance from concentration a to a-x is given by:

t = 2.303/k log [a/(a – x)]

When half of the reaction is completed, x = a/2. Representing the time taken for half of the reaction to be completed by t1/2

t1/2 = 2.303/k log a/(a-a/2)

⇒ t1/2 = 2.303/k log 2

t1/2 = 0.693/k

Where,

• t1/2 is the half life of reaction, and
• k is the rate constant.

### Integrated Rate Equation for First-Order Gas-Phase Reaction

Consider the general first-order gas-phase reaction:

A (g) ⇢ B (g) + C (g)

Suppose the initial pressure of A = P0 atm. After time t, suppose the pressure of A decreases by p atm.

Now, as 1 mole of A decomposes to give 1 mole of B and 1 mole of C, the pressure of B and C will increase by p each. Hence, we have

A(g)  ⇢  B (g) + C (g)
Initial Pressure

P0 atm

00
Pressure After Time, t

P0 – p

p atmp atm

The total pressure of the reaction mixture after time t,

Pt = (P0 – p) + p + p = P0 + p atm

⇒ p = Pt – P0

So, pressure of A after time t (PA) = P0 – p = P0 –  (Pt – P0) = 2 P0 – Pt

But initial pressure of A (P0) ∝ initial conc. of A, i.e., [A]0

and pressure of A after time t(PA) ∝ conc. of A at time t, i.e., [A]

Substituting these values in the first-order rate equation,

k = 2.303/t log [A]0/[A], We get

k = 2.303/t log [P0/(2P0 – Pt)]

Where,

• P0 is the Intial Pressure of Reactant,
• Pt is the total pressure after time t,
• t is the time in which pressure changes from P0 to Pt, and
• k is the rate constant.

### Integrated Rate Equation for Second Order Reactions

Let’s consider a second-order reaction,

2A ⇢ Product

Thus, the rate of the reaction is given by

Let’s consider that the initial concentration of the reactant is [A] and in time t, the concentration of the reactant becomes [A0], thus integrating the above equation under the suitable limits, we get

### Characteristics of Second-Order Reactions

The characteristics of the Second-Order reaction are,

• Half-life Period: Let’s consider the initial concentration of the reactant to be [A0] and after t1/2 time, the concentration becomes [A0]/2, thus half-life is given as follows:

## Summary of Integrated Rate Law

This is the most common method for studying the kinetics of a chemical reaction. For example, consider the reaction:

nA ⇢ products

If we start with a moles/liter of A and in time t, x moles/liter have reacted so that the concentration after time t is (a – x) moles/liter, then

if the reaction is of the first order, dx/dt = k(a – x), and if the reaction is of the second order, dx/dt = k (a – x)2, and so on.

These differential equations can be integrated to get expressions for the rate constants. These are given below for zero, first, and second-order reactions:

• For zero-order reaction, k = 1/t {[A0] – [A]}
• For first-order reaction, k = 2.303/t log  [A0]/[A]
• For second-order reaction, k = 1/t {1/[A] – 1/[A0]}

The advantage of the integrated method is that these integrated forms of equations contain the concentration of a reactant at different times and hence can be solved to find the value of k from the data of the run of one experiment only and need not start with different initial concentrations. Moreover, they can be used to find the time for any fraction of the reaction to complete.

## Differential vs Integrated Rate Laws

There are some key differences between both differential and integrated rate laws, which are listed in the following table:

Differential Rate Law

Integrated Rate Law

Describes the rate of change of concentration with respect to time.Describes the relationship between concentration and time.
Represents the rate equation in terms of initial concentrations and rate constants.Represents the concentration of reactants or products as a function of time.
Provides information about the instantaneous rate of a reaction at a specific moment in time.Provides information about the overall change in concentration over a given time interval.
Can vary with time and is dependent on the concentration of reactants.Remains constant for a specific reaction under constant conditions.
Typically expressed as a differential equation, involving derivatives.Usually expressed as a mathematical equation, relating concentrations and time.
Helpful in determining the order of a reaction and the reaction rate constant.Useful for determining reaction orders and obtaining information about reaction mechanisms.
Represents the rate of change at a specific point on the reaction progress curve.Represents the cumulative effect of the reaction at different time points.

## Sample Problems on Integrated Rate Law

Problem 1: At 373 K, the half-life period for the thermal decomposition of N2O5 is 4.6 sec and is independent of the initial pressure of N2O5. Calculate the specific rate constant at this temperature.

Solution:

Since the half-life period is independent of the initial pressure, this shows that the reaction is of the first order.

For a reaction of the first order, we know that t1/2 = 0.693/k

or

k = 0.693/t1/2 = 0.693/4.6s

= 0.1507 s-1

Problem 2: A first-order reaction is found to have a rate constant, k = 5.5 x 10-14s-1. Find the half-life of the reaction.

Solution:

For a first order reaction, t1/2 = 0.693/k

t1/2 = 0.693/5.5×10-14s-1

= 1.26 x 1013 s-1.

Problem 3: Show that in the case of a first-order reaction, the time required for 99.9% of the reaction to take place is about ten times that required for half the reaction.

Solution:

For reaction of first order,

t1/2 = 2.303/k log a/a – a/2

= 2.303/k log 2

= 2.303/k (0.3010)

t99.9% = 2,303/k log

= a/a-0.999a

t99.9% = 2.303/k log 10-3

= 2.303/k x 3

Therefore, t99.9%/t1/2 = 3/0.3010 ≅ 10

Problem 4: The initial concentration of N2O5 in the first-order reaction, N2O5(g) ⇢ 2 NO2(g)+ 1/2 O2(g), was 1.24 x 10-2 mol L-1 at 318 K. The concentration of N2O5 after 60 minutes was 0.20 x 10-2 mol L-1. Calculate the rate constant of the reaction at 318 K.

Solution:

k = 2.303/t log [A]0/[A] = 2.303/t log [N2O5]0/[N2O5]t

2.303/60min log 1.24 x 10-2 mol L-1/0.2 x 10-2 mol L-1

= 2.303/60 log 6.2 min-1 = 2.303/60 x 0.7924min-1

= 0.0304 min-1.

Problem 5: A first-order reaction is found to have a rate constant k = 7.39 x 10-5 sec-1. Find the half-life of the reaction (log 2 = 0.3010).

Solution:

For a first order reaction, k = 2.303/t log a/a-x

For t = t1/2, x = a/2

t1/2 = 2.303/k log a/ a-a/2

= 2.303/k log 2

= 2.303/7.39×10-5s-1 x 0.3010

= 9.38 x 103 s-1.

## FAQs on Integrated Rate Law

### 1. What is Integrated Rate Law?

An integrated rate law is an equation that describes the relationship between the concentration of a reactant and time during a chemical reaction. It integrates the rate equation, which is typically derived from the rate of change of reactant concentrations with respect to time.

### 2. What are Integrated Rate Laws for Different Order Reactions?

For different order reactions, rate law is given as follows:

• For Zero-order Reaction: Rate ∝ [A]0
• For First-order Reaction: Rate ∝ [A]
• For Second-order Reaction: Rate ∝ [A]2

Where [A] is the concentration of the reactant.

### 3. What are Common Forms of Integrated Rate Equations?

The common form for  integrated rate Equation for various differnt order reactions as follows:

• For Zero-order reaction: [A] = [A₀] – kt
• For First-order reaction: ln[A] = ln[A₀] – kt
• For Second-order reaction: 1/[A] = 1/[A₀] + kt

Where,

• k is the rate constant,
• [A₀] is the inital concentration of the reactant,
• [A] is the required concentration after time t, and
• ln represents the natural logarithm.

### 4.  Can Integrated Rate Laws be Used for All Types of Reactions?

Integrated rate laws are typically derived for elementary reactions (reactions that occur in a single step) and reactions with simple rate expressions. They may not be directly applicable to complex reaction mechanisms involving multiple steps or reactions with non-elementary rate laws. In such cases, alternative mathematical approaches or computational methods may be required to analyze the reaction kinetics.

### 5. What are Units for Rate Constant in Integrated Rate Laws?

The units of the rate constant (k) in integrated rate laws depend on the order of the reaction, and given as follows:

• For zero-order reactions, the units of k are concentration/time (e.g., M/s).
• For first-order reactions, the units of k are 1/time (e.g., 1/s).
• For second-order reactions, the units of k are 1/(concentration·time) (e.g., 1/(M·s)).

### 6. Can Integrated Rate Laws be Applied to Reversible Reactions?

Integrated rate laws can be used for reversible reactions if the reaction is in a steady state or if the forward and backward reactions have reached equilibrium. In such cases, the concentrations of reactants or products can be treated as constant, allowing for the application of integrated rate laws. However, for reversible reactions that are not at equilibrium, integrated rate laws may not accurately describe the kinetics, and more complex mathematical models are necessary.

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