The reaction in which the rate depends on the second power of concentration terms is called a second-order reaction.

Most biological processes can be described through second-order reactions.

Hydrolysis of the ester by sodium hydroxide is an example of a second-order reaction.

CH_{3}COOC_{2}H_{5}+NaOH→CH_{3}COONa+C_{2}H_{5}OH

In this reaction, the concentration of Na is decreasing with time. It is titrated against standard acid. The rate of reaction can be determined from the slope of the graph plotted between the concentration of NaOH on the y-axis and time on the x-axis.

**Examples of second-order reactions in the gas phase**

- Thermal decomposition of NO
_{2 }to give N_{2 }and O_{2}.

2NO_{2}→N_{2}+2O_{2}

- Thermal decomposition of HI to give H
_{2 }and I_{2}.

2HI→H_{2}+I_{2}

**Examples of second-order reactions in the Solution phase**

- The reaction between CH
_{3}I and C_{2}H_{5}ONa gives ether.

CH_{3}I+ C_{2}H_{5}ONa→C_{2}H_{5}OCH_{3}+NaI

- Formation of urea from NH
_{4}^{+ }and CNO^{–}.

NH_{4}^{+}+CNO^{–}→NH_{2}-CO-NH_{2}

**Prerequisite concept**

The original term, half-life period, dating to** Ernest Rutherford’s** discovery of principle in 1907, was shortened to half-life in the early 1950s. Rutherford applied the principle of a radioactive element’s half-life to studies of age determination of rocks by measuring the decay period of **radium** to **lead-206.**

Half-life is constant over the lifetime of an exponentially decaying quantity, and it is a characteristic unit for the exponential decay equation.

There are two cases for second-order reactions.

- 2nd order reaction with same initial concentrations
- 2nd order reaction with different initial concentrations

## Case -1

**When the concentration of reactants is the same**

Consider two reactants that combine to give products.

A+A→Product

At time t=0 concentration of reactants are _{oo}

A+A→Product

_{o o}→ 0

At time t=’t’sec

_{t}→_{o}–

Rate of reaction in term of reactants

=k_{2}^{2}

=-k_{2}^{2}→Eq(1)

This is a differential equation for second-order reactions.

Separating the variables from eq(1)

=-k_{2}^{2}

=-k_{2}dt

Integrating within limits

=-k_{2}dt

By solving and putting limits

-=-k_{2}t

=k_{2}t

+k_{2}t→Eq(2)

This is called an integrated form of second-order reaction.

## Case -2

**When the concentration of reactants are different**

A+B→Product

At time t=0sec concentration of reactants are

A+B→Product

a+b→0

At time t=tsec

a-x b-x→x

Rate of reaction in term of the product

=k_{2}(a-x)(b-x)→Eq(3)

This is a differential equation for second-order reactions.

Separating the variables from eq(3)

=k_{2}dt

Integrating

=k_{2}dt

=k_{2}t+c→Eq(4)

Resolve into partial fraction

=+→Eq(5)

Multiply both sides by L.C.M

1=A(b-x)+B(a-x)

Put b-x=0; x=b

1=B(a-b)

B=

B=

Put a-x=0; x=a

1=A(b-a)

A=

Put values of A and B in eq(5)

=-

Integrating

=-

=ln(a-x)+ln(b-x)

=

Put in eq(4)

=k_{2}t+c→Eq(6)

Apply boundary conditions; when t=0 x=0 in eq(6)

=c

Put value of c in eq(6)

=k_{2}t+→Eq(7)

By solving

-=k_{2}t

-=k_{2}t

=k_{2}t→Eq(8)

This is called an integrated form of a second-order reaction with different concentrations.

## Graphs

Graphs are useful in various ways.

1. They give information about the relation between the second-order order concentration of reactants or products with time.

2. From the slope of the graph we can calculate the value of the rate constant.

3. From the intercept we can calculate the value of c.

Eq(2), eq(7), and eq(8) can be plotted in various ways to show different relations.

## Half-life

Half-life is a time that is required for reactants to consume one-half of their initial concentration.

For example, if the concentration of the reactant is 12×10^{-3}moldm^{-3} at the start of the reaction, then during the first half-life concentration of reactant drops to 6×10^{-3}moldm^{-3}, during second half-life concentration of reactant drops to 3×10^{-3}moldm^{-3}, during third half-life concentration of reactant drop to 1.5×10^{-3}moldm^{-3}.

### The half-life for a second-order reaction

The integrated rate law for 2nd order reaction with the same concentration is

+k_{2}t

For half-, life second-order put t=t_{1/2} and _{t}=_{o}/2

=k_{2}t_{1/2} +

=k_{2}t_{1/2} +

-=k_{2}t_{1/2}

=k_{2}t_{1/2}

=k_{2}t_{1/2}

=t_{1/2}

**t _{1/2}=**→Eq(8)

k_{2}=rate constant for second-order reaction and _{o}=initial concentration of reactant.

### Explanation of Eq(8)

This equation shows that half-life depends on both

- Rate constant
- The initial concentration of reactant

From the equation, it is concluded that if the concentration of reactants is less, its half-life will be more, which means reactants exist for a longer period of time.

If the half-life is short means reactants will be consumed in a short period of time.

**The half-life for second-order reaction with different concentration of reactants **cannot be determined because , so the times takes to consume one half of will not be equal to time takes to consume one half of .

So we cannot define a general equation.

## Importance of half-life

- Half-life is important in nuclear physics to describe how long unstable atoms exist and how quickly they undergo radioactive decay.
- The concept of a half-life has also been applied for pesticides in plants describing dissipation from plants.
- Using the concept of half-life excretion rates and steady-state concentrations for any specific drug can be determined.
- Different drugs have different half-lives; but all follow the definition of the half-life: after one-half life has passed, 50% of the initial amount of drug is removed from the body.
- The order of reaction can also be determined by using half-life.

**Graphical method**

The half-life for a zero-order reaction is

t_{1/2}=

By plotting a graph between t_{1/2} on the y-axis and _{o} on the x-axis if a straight line is obtained the reaction is zero order.

**Graph **

The half-life for 1st order reaction is

t_{1/2}=

If the half-life is independent, the concentration of reactant reaction will be 1st order.

**Graph **

The half-life for 2nd order reaction is

t_{1/2}**=**

By plotting a graph between t_{1/2} on the y-axis and on the x-axis if a straight line is obtained the reaction is second order.

**Graph**

The half-life for 3rd order reaction is

t_{1/2}=

By plotting a graph between t_{1/2} on the y-axis and on the x-axis if a straight line is obtained the reaction is third order.

**Graph**

**Mathematical method**

The half-life for 1st order reaction,

t_{1/2}∝

The half-life for 2nd order reaction,

t_{1/2}∝

The half-life for 3rd order reaction,

t_{1/2}∝

The half-life for nth-order reaction,

t_{1/2}∝

Suppose we have a reaction whose order is to be determined. At initial concentration ‘a_{1}’ the half-life is ‘t_{1}’. At initial concentration ‘a_{2}’ the half-life is ‘t_{2}’. From the above equations, we can write

t_{1}∝→Eq(1)

t_{2}∝→Eq(2)

Divide both equations

=

ln=(n-1)ln

n-1=

n=1+

So, if we know two initial concentrations ‘a_{1}’ and ‘a_{2}’, and two half-lives ‘t_{1}’ and t_{2} of a certain reaction, the order can be calculated.

## Concepts Berg

- How to calculate the rate constant?

Rate constant ‘k’ for any order of reaction can be calculated by putting values in the equation or by plotting a graph.

For example, for zero-order reaction

=k

By putting values of ,and time the value of the rate constant can be calculated.

Also, by plotting the graph between on the y-axis and time on the x-axis, we can calculate the value of k from the slope of the graph.

- What is a Second Order Reaction?

The reaction in which the rate depends on the second power of concentration terms is called a second-order reaction.

- How do you find the half-life of a second-order reaction?

The half life for second-order reaction can be calculated from formula

**t _{1/2}=**

If the value of the initial concentration and rate constant are known we can calculate half-life.

- What does Second half-life mean?

As the first half life is a time required to decay from initial value to its half i.e 50%. Second half life is a time required to decay 50% to 25%.

- Is the half- life of a second- order reaction constant?

**t _{1/2}=**

From equation half -life depends on both initial concentration and rate constant.

- Why does half-life decrease when concentration increases?

As half-life of second-order reaction is

**t _{1/2}=**

From the equation, the half-life is inversely proportional to a concentration that’s why half-life decreases when concentration increases.

- What does 2nd order mean?

It is a reaction in which the sum of the exponent of the concentration term is 2.

Rate=

1+1=2

- What is 2nd order kinetics?

It is defined as the rate of reaction is directly proportional to the product of the concentration of reactants raised to some power.

- What’s the difference between first-order and second-order reactions?

In first order reaction, rate is proportional to first power of concentration term i.e sum of exponent of concentration term is 1. Where, in second order reaction the rate is proportional to second power of concentration terms i.e sum of exponent of concentration terms is 2.

- What is the first-order half-life equation?

The first order half life equation is

t_{1/2}=

Equation shows that half life is independent of concentration of reactants. It only depends on the rate constant.

- What is zero-order half-life?

The zero-order half life equation is

t_{1/2}=

Equation shows that half life depends on both concentration of reactants and rate constant.

- How would you calculate the half-life of a second-order reaction with the following data The initial concentration is 1M and k 0.14 M
^{-1}s^{-1}?

Second order half life is

t_{1/2}=

By putting these values in equation

=

=7.14second

- In a 2nd order reaction, why does the half-life increase with a decrease in substrate concentration?

Second order half life is

**t _{1/2}=**

It is clear from the equation that half life is inversely proportional to concentration of substrate. That’s why half-life increases with a decrease in substrate concentration.

- What is the order if half-life of a reaction is halved as the initial concentration of the reactant is doubled?

If half-life of reaction is halved as the initial concentration of reactant is doubled the reaction will be of second order.

- What percentage of the initial concentration of a second-order remains after the third half-life?

As the first half life of second-order reaction is

t_{1/2}=

The third half-life is

3t_{1/2}=

The integrated rate law for 2nd order reaction with the same concentration is

+k_{2}t

For third half-life put t=t_{1/2}

+k_{2}3t_{1/2}

Put value of 3t_{1/2}

+k_{2}

=+

=

_{t}=_{o}/4

So, the concentration of reactant remains 25% of its initial concentration after the third half-life.

- If the half period of a reaction is doubled as the initial concentration of reactant is also doubled how do you evaluate the order of reaction?

The half-lives expression for zero, first and second order reactions are

t_{1/2}=

t_{1/2}=

t_{1/2}=

From the following equations, if half life of reaction is doubled as the initial concentration of reactant is also doubled then the reaction will be zero-order as half life is directly proportional to concentration of reactants.

- What is the expression of a second-order reaction with the same and different initial concentrations?

The integrated rate law for 2nd order reaction with the same concentration is

+k_{2}t

The integrated form of a second-order reaction with different concentrations is

=k_{2}t+

- What is an example of a second-order reaction?

Hydrolysis of the ester by sodium hydroxide is an example of a second-order reaction.

CH_{3}COOC_{2}H_{5}+NaOH→CH_{3}COONa+C_{2}H_{5}OH

In this reaction, the concentration of Na is decreasing with time. It is titrated against standard acid. The rate of reaction can be determined from the slope of the graph plotted between the concentration of NaOH on the y-axis and time on the x-axis.

- How long will it take for a first-order reaction to be ended if the half-life time is 40 seconds?
- What is the order of saponification reaction?

Saponification reaction is the hydrolysis of ethyl acetate and sodium hydroxide.

CH_{3}COOC_{2}H_{5}+NaOH→CH_{3}COONa+C_{2}H_{5}OH

As two molecules are involved, reaction is 2nd order.

- The half-life for a first-order reaction is 32 s. What was the original concentration if after 2.0 minutes the reactant concentration is 0.062 M?

As

t_{1/2}=

Put the value of half-life

32=0.693/k

k=0.02165sec^{-1}

The integrated rate law for first order reaction is

ln _{t}=-kt+ln _{o}

Put the value of _{t}=0.062M and t=120 sec

ln (0.062)=-(0.021)(2×60)+ln _{o}

-2.7806+2.5987=ln _{o}

ln _{o}=-0.18185

Taking antilog on both sides

_{o}=0.8337M

- After 71.0 minutes 35.% of a compound has decomposed. What is the half-life of this reaction assuming first-order kinetics?

If initial concentration is 100% then concentration after 71 minutes is (100%-35%)=65%.

The integrated rate law for first order reaction is

ln _{t}=-kt+ln _{o}

Substitute the given values in this equation and find rate constant

ln(65%)=-k(71)+ln(100%)

ln(65%)-ln(100%)=-k(71)

ln=-k(71)

k=0.006067min^{-1}

The first order half life equation is

t_{1/2}=

t_{1/2}=0.693/0.006067

t_{1/2}=114.224minutes

- What is the rate constant of a first-order reaction that takes 408 seconds for the reactant concentration to drop to half of its initial value?

To find rate constant, use relation

t_{1/2}=

The time taken by reactants to drop to half of its initial value is called half life.

So, t_{1/2}=408 sec

t_{1/2}=

k=0.693/408

k=0.00169sec^{-1}

- The half-life of a first-order reaction is 6.0 hrs. How long will it take for the concentration of reactant to decrease from 0.8M to 0.25M?

First of all, calculate the rate constant.

Half life of first order reaction is

t_{1/2}=

k=0.693/6

k=0.1155hrs^{-1}

The rate law for first order reaction is

ln _{t}=-kt+ln _{o}

Rearrange equation for ‘t’

t=ln _{o}-ln _{t}/k

t=ln

_{t}=0.25M; _{o}=0.8M

t=

t=10.07hrs

So, it takes 10.07 hours to decrease the concentration of reactants from 0.8M to 0.25M.

- The half-life period of a decomposition of a liquid A by an irreversible first-order reaction is 12 minutes. What is the time required for 75% conversions of A?

First of all, calculate the rate constant.

half-life of first-order reaction is

t_{1/2}=

k=0.693/720

k=0.0009625sec^{-1}

The rate law for first order reaction is

ln _{t}=-kt+ln _{o}

ln _{t}-ln _{o}=-kt

ln=-kt

t=ln

If initial concentration is 1g, then concentration left after 12 min is 0.25g.

t=ln()/0.0009625

t=1440.3second

So the time required for 75% conversion is 1440.3seconds.

- The rate constant K for a first-order reaction was found to be 0.2 sec. What will be its half-life a) 10 seconds b) 5 seconds c) 25 seconds d) 15 seconds?

The half-life of the first-order reaction is

t_{1/2}=

=

t=3.46sec

None of these options is correct.

- Liquid A decomposes by first-order kinetics and in a batch reactor, 50 of A is converted in a 5-minute run. How much longer would it take to reach 75 conversions?
- How do you find the order of reaction from a chemical equation?

The order of a reaction is defined as the sum of exponents of concentration terms in the rate equation. for example

aA+bB→cC+dD

Rate=

The order of reaction will be a+b.

- The gas-phase decomposition of SO
_{2}Cl_{2}SO_{2}Cl_{2 (g)}→SO2_{(g) }Cl2_{(g)}is first order in SO_{2}Cl_{2}. At 600K the half-life for this process is 2.3×10^{5}s. What is the rate constant at this temperature?

The half-life of the first-order reaction is

t_{1/2}=

k=0.693/2.3×10^{5}

k=3.01×10^{-6}sec^{-1}

- The reaction 2A B is first order in A with a rate constant of 2.8 x10
^{-2}s^{-1}at 80 degrees C. How long in seconds will it take for A to decrease from 0.550M to 0.170M?

The integrated rate equation for first order reaction is

ln _{t}=-kt+ln _{o}

As from data, put the values of _{o}=0.170M, _{t}=0.550M and k=2.8×10^{-2}=0.28sec^{-1}

ln=-kt

1/0.28ln=-t

After solving we get

t=41.93seconds

So A will take 41.93sec to decrease from 0.55M to 0.170M.

- If k 8.52 x 10
^{-5}mol^{-1}Ls^{-1}and the initial concentration of A is 1.13 mol L^{-1}.What is the half life of this reaction in hours?

The unit of k indicates that the reaction is 2nd order.

The half life of 2nd order reaction is

t_{1/2}=

Put values of rate constant and initial concentration

=1/(8.52×10^{-5})(1.13)

=1/0.000096276

=10386.804sec

Convert seconds into hours

=10386.804/3600

=2.88hours

- How can a reaction have fractional or negative order?

Negative order of reaction means that rate of reaction is inversely proportional to concentration of reactants. If concentration of substance increases the rate of reaction decreases.

The order of reaction can be fractional that indicates that the reaction is a complex reaction.

- The half life of a first order reaction is 5 10 4s. What percentage of the initial reactant will react in 2 hours?
- The half life of a substance is 10 days. How much of the substance will be left after 40 days if the initial mass is 2gm?

Initial mass of substance is 2gm and its half life is 10 days means after 10 days the amount of substance will half of its initial value.

After 10 days amount of substance=1gm

After 20 days amount of substance=1/2gm

After 30 days amount of substance=1/4gm

After 40 days amount of substance=1/8gm

So, after 40 days the amount of substance left is 1/8gm.

- Why is radioactivity a first order reaction?

Radioactivity is a first order reaction because the rate of radioactive decay of nuclei depends on the first power of the radioactive atom.

- How to find K in a first order reaction?

The integrated rate equation for first order reaction is

ln _{t}=-kt+ln _{o}

By plotting a graph between ln _{t }on y-axis and time on x-axis, the slope of the graph gives the value of rate constant k.

- If the half-life of a reactant in a 1st order reaction is 25 minutes then what is the extent of reaction after 2 minutes?

The half-life of the first-order reaction is

t_{1/2}=

k=0.693/25

k=0.02772min^{-1}

The integrated rate equation for first order reaction is

ln _{t}=-kt+ln _{o}

ln=-kt

ln=-(0.02772)(2)

- What is the derivation of third order reaction and its half life?

Consider three reactants with the same concentration that are converted into products.

A+A+A→P

If initial concentration of A is a mole at time t=0 and concentration of product is 0mole.after time t concentration of product is x mole and concentration of reactant A left behind is a-x.

Rate law for third order reaction is

=k(a-x)^{3}

Separating the variables and integrating

=kdt

(a-x)^{-3}dx=kdt

After solving, we get

=kt+c→eq(1)

For the value of c apply boundary conditions. At t=0, x=0

c=put in (1)

=kt+

=2kt+→eq(2)

Rearrange the equation and solve

-=kt

=2kt

=2kt

**Half life**

For half life put t=t_{1/2} and x=a/2 in eq(2)

=2kt_{1/2}+

-=2kt_{1/2}

2kt_{1/2}=

t_{1/2}=

- Liquid a decomposes by first order kinetics and in a batch reactor 50% of a is converted in 5 minutes. What is the value of the rate constant?

In this reaction, 50% of reactants are converted into products in 5 minutes. It means 5min is a half-life of this reaction because half of reactants are converted into products in this time.

The half-life of the first-order reaction is

t_{1/2}=

k=0.693/5

k=0.1386min^{-1}

- Why do we use CSTR for negative order reactions?
- The reaction 3A 4B products in first order in A and first order in B. What is the overall order of the reaction?
- In a first order reaction the amount of reactant decayed in three half lives(let a be is initial amount) would be A) 7a/8 B) a/8 C) a/6 D) 5a /6?

In first order reaction, half life is a time in which half of reactants are converted into products.

If a is initial concentration, then after the first half life the amount of reactant decay is 1/2a.

After second half-life amount of reactant decay is (1/2a)(1/2a)=1/4a

After third half life amount of reactant decay is (1/2a)(1/2a)(1/2a)=1/8a

Total amount of reactant decay after three half lives would be

1/2a+1/4a+1/8a=7/8a

So option A is correct.

- A certain compound has an initial mass of 200 g. After 4 hours there is only 3.125 g remaining. What is the half life period of the compound?

For half life, first we calculate the rate constant k.

The integrated rate equation for first order reaction is

ln _{t}=-kt+ln _{o}

ln=-kt

As _{o}=200g and _{t}=3.125g

By Substituting the values

1/tln=-k

1/4ln=-k

k=1.04hours^{-1}

The half-life of the first-order reaction is

t_{1/2}=

t_{1/2}=0.693/1.04

t_{1/2}=0.666hours

- Explain this If a reaction is pseudo order the half life is always defined with respect to the species present in the smallest amount?

In pseudo order reaction, one of the reactants is taken in large amounts so that its concentration remains constant during the reaction and it does not take part in the rate of reaction. That’s why half life is always defined with respect to species present in the smallest amount in pseudo order reaction.

- What does the one third life of a first order reaction mean?

One third life means the time in which 33.3% of the reaction is to be completed.

Consider the reaction in which A is converted into B. Initially 3 moles of A are present and the amount of product B is 0mol. After one third life 2 moles of A are left and 1 mole of B is formed.

- What is the maximum order of reaction?

We deal upto 3rd order reaction but in actual practice there is no maximum order of reaction.

- What is the meaning of zero-order kinetics?

Zero-order kinetics means that the rate of reaction is independent of the concentration of reactants.

- How does the half-life period of 3rd order reactions vary with their initial concentration?

Half-life of 3rd order reaction is

t_{1/2}∝1/a^{2}

So, from the equation, it varies inversely as the square of the initial concentration of reactant.

- What is the cause of pseudo order reaction?

When one of the reactants is present in a larger amount than the other, the concentration of this reactant is kept constant so it does not take part in the rate of reaction that’s why it is called pseudo order reaction.

- Similar quid A decomposes by second order kinetics and in a batch reactor 50% of A is converted in a 5 min run. How much longer would it take to reach 75% conversion?

From the given information, the concentration of A after 5minutes is half of its initial concentration.

The integrated rate law for 2nd order reaction with the same concentration is

+k_{2}t

Put _{t}=_{o}/2

+k_{2}t

=k_{2}t

=k_{2}

k_{2}=1/5_{o}

Now solve the equation for _{t}=_{o}/4 because 75% of the reaction is completed.

Put values of k and _{t}

+k_{2}t

=k_{2}t

x5_{o}=t

t=15minutes

**References links**