Learning CHEMISTRY for IIT JEE: Chemical equilibrium

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Sunday, December 29, 2019

JEE Main - Core Points for Revision - Ch. 8. Equilibrium II – Ionic Equilibrium in Solutions

8.3 Ionization of Water – Ionic Product of Water

8.4 Expressing Hydrogen Ion Concentration – pH Scale

pH

pH is related to hydrogen ion concentration. Since the H+ ion concentration in solution is often small, the concentration is generally expressed as the logarithm of its reciprocal, which is called a pH value. Therefore, pH is defined as

pH = -log[H+]

For a ten times increase in H+ ion concentration there is a decrease in the pH value of one unit.

Given the pH of a solution, its H+ concentration can be found:

[H+(aq)] = antilog -pH

or

[H+(aq)] = 10^-pH

8.5 Polyprotic Acids and Bases

8.6 Hydrolysis of Salts
Hydrolysis is the name for a substance chemically reacting with water.

Hydrolysis should be distinguished from solvation, which is the process of water molecules associating themselves with individual solute molecules or ions.

When a chemical reaction comes to equilibrium, there is a mixture of all involved substances in the reaction vessel. This mixture is characterized by a constant composition. (constant composition DOES NOT imply equal composition.)

The key point that makes a reaction come to equilibrium is that it is reversible. This means that both the forward reaction and the reverse reaction can happen,
The reaction comes to equilibrium when the rates of the two reactions (forward and reverse) become equal.
More details https://iit-jee-chemistry.blogspot.com/2008/02/jee-revision-hydrolysis-of-salts.html

8.7 Acid base Titrations and Indicators

8.8 Solubility Product

Saturday, December 28, 2019

JEE Main - Core Points for Revision - 7. Equilibrium I – Equilibrium Process and Phase Equilibria

Importance of Core Revision Points: Core Revision Points are important because if you remember them strongly, many more points related to them will come out of your memory and help you to answer question and problems. Read them many times and make sure you remember them very strongly.

7.1 Equilibrium and Its dynamic nature

In most of the reaction carried out in closed vessels, reaction does not go to completion under given set of conditions of temperature and pressure. Initially, in the vessel, only reactants are present, and as the reaction proceeds, the concetration of reactants will decrease and that of products will increase.

After some time a stage is reached when no further change in concetrations of reactants and products is observed. This state is called equilibrium state and some of the important questions regarding this phenomenon are:

1. why do reactions seem to stop before they reach completion?
2. What is the extent to which a reaction proceed?
3. Can we modify the conditions to improve the yield of products?

Equilibrium - The phenomenon

Equilibrium is the state at which the concentrations of reactants and products do not change with time.

It is important to remember that equilibrium is achieved in closed vessel reactions only.

The important aspect of reaction equilibrium is the reversibility. The products combine and form reactants. At equilibrium, both the forward and backward reactions are taking place. The rates of forward and backward reactions are same or equal at the equilibrium. As a result, the concentration of each species becomes constant.

The equilibrium is termed as dynamic reaction equilibrium. Dynamic means at a microscopic level, the system is in motion. But at macroscopic level, concentrations are not changing.

Chemical reactions may be classified as reversible reactions and irreversible reactions.

Example of irrereversible reaction

Decomposition of potassium chlorate into potassium chloride and oxygen. Even in a closed vessel this reaction is not reversible.

Example of reversible reaction

1. Decompositon of calcium carbonate. When solid calcium carbonate is heated in a closed vessel at 1073 K, it decomposes into solid calcium oxide and gaseous carbon dioxide. Due to gaseous CO2 there is pressure of gas in the vessel which can be measured. At a constant temperature it can be observed that pressure becomes constant after some time, which means no further CO2 is being produced even though calcium carbonate is still there in the vessel. The constant pressure indicates to us that reaction equilibrium is reached.

Characteristics of chemical equilibrium

1. Chemical equilibrium is dynamic in nature (already explained).

2.The properties of the system become constant at equilibrium and remain unchanged thereafter unless external or internal conditions are changed.

3. The equilibrium is attained only if the system is closed one.

4. As the reactions are reversible and happen under the same conditions, equilibrium can be attained from either direction.

5. A catalyst does not alter the equilibrium point. The catalyst increases the rate of reaction, and at equilibrium it increases both forward and backward reaction rates. But it does not alter equilibrium point, the concentrations of products and reactants at a given set of conditions. But the equilibrium is reached earlier in the presence of a catalyst.

7.2 Equilibrium in Physical Processes

7.3 Equilibria involving Chemical Systems

7.4 Law of Chemical equilibrium and equilibrium Constant

There are reactions where one can see the reverse action also to be active and see the equilibrium point. In this case

A + B → C + D and

C+D → A+B both reactions keep taking place.

At the point of equilibrium the rate of both reactions is same. Formation of A+B is equal to consumption of A+B.

Rate of forward reaction = k-f[A][B]

Rate of reverse reaction or backward reaction = k-r[C][D]

Therefore k-f[A][B] = k-r[C]{D]

This gives k-f/k-r = [C]{D]/[A][B]

The equilibrium constant is always written as products by reactants.

For the a general reaction

aA + bB ↔ cC+dD (Normal two arrows are used for reversible reaction. Only one arrow with heads on both sides is used here to tide over the inability to show two arrows.)

k-eq = [C]^c[D]^d/[A]^a[B]^b

The equilibrium constant may or may not have units.

In the case of 2A ↔ 2B +C

The units of equilibrium constant are going to be: (mol/l)^2(mol/l)/(mol/l)^2

= mol/l

7.5 Types of Chemical Equilibria
7.6 Applications of Equilibrium Constant
7.7 Factors Which change the State of Equilibrium – Le Chatelier’s Principle
7.8 Applications of Le Chatelier’s Principle of Physical Equilibrium

Contents

7.1 Equilibrium and Its dynamic nature
7.2 Equilibrium in Physical Processes
7.3 Equilibria involving Chemical Systems
7.4 Law of Chemical equilibrium and equilibrium Constant
7.5 Types of Chemical Equilibria
7.6 Applications of Equilibrium Constant
7.7 Factors Which change the State of Equilibrium – Le Chatelier’s Principle
7.8 Applications of Le Chatelier’s Principle of Physical Equilibrium

21 May 2015

Wednesday, March 11, 2009

JEE - Study Guide - 7. Equilibrium I – Equilibrium Process and Phase Equilibria

Contents

7.1 Equilibrium and Its dynamic nature
7.2 Equilibrium in Physical Processes
7.3 Equilibria involving Chemical Systems
7.4 Law of Chemical equilibrium and equilibrium Constant
7.5 Types of Chemical Equilibria
7.6 Applications of Equilibrium Constant
7.7 Factors Which change the State of Equilibrium – Le Chatelier’s Principle
7.8 Applications of Le Chatelier’s Principle of Physical Equilibrium

Conceptual Questions with Answers: 20
Additional Numerical Problems for Practice: 10
Revision Exercises
Very Short Answer questions: 18
Short Answer Questions: 26
Long Answer Questions: 6

Competition File
Numerical Problems:10
Objective Questions: 31
Fill in the blanks: 5
True or False: 6

Study Plan

Regular Study 16 days, Revision: 9 days

Day 1

7.1 Equilibrium and Its dynamic nature
7.2 Equilibrium in Physical Processes

Day 2
7.2 contd.
Ex. 7.1

Day 3
7.3 Equilibria involving Chemical Systems

Day 4
7.4 Law of Chemical Equilibrium and Equilibrium Constant

Day 5
7.5 Types of Chemical Equilibria
Ex. 7.3

Day 6
7.5 contd.
Ex. 7.4
Day 7
7.6 Applications of Equilibrium Constant
Ex. 7.4 to 7.7

Day 8
Practice Problems 7.1 to 7.8

Day 9
7.6 contd.Ex. 7.8 to 7.19

Day 10
P.P 7.8 to 7.15

Day 11
P.P. 7.16 to 7.22

Day 12
7.7 Factors Which change the State of Equilibrium – Le Chatelier’s Principle

Day 13
7.7 contd.
7.8 Applications of Le Chatelier’s Principle of Physical Equilibrium
Ex. 7.20

Day 14
P.P. 7.23 to 7.28

Day 15

Conceptual Questions with Answers: 20

Day 16
Additional Numerical Problems for Practice: 10

Revision Period Start

Day 17
Revision Exercises: Very Short Answer questions: 1 to 9

Day 18
Revision Exercises: Very Short Answer questions: 9 to 18

Day 19
Revision Exercises: Short Answer Questions: 1 to 13

Day 20
Revision Exercises: Short Answer Questions: 14 to 26

Day 21
Competition File: Numerical Problems: 1 to 10

Day 22
Objective Questions: 1 to 15
Objective Questions: 16 to 31

Day 23
Fill in the blanks: 5
True or False: 6

Day 24
Concept Revision

Day 25
Formula Revision

IIT JEE - Study Guide - 8. Equilibrium II – Ionic Equilibrium in Solutions

Sections in the Chapter

8.1 Acid-base Concepts
8.2 Acid- base Equilibria and Ionization of Acids and Bases
8.3 Ionization of Water – Ionic Product of Water
8.4 Expressing Hydrogen Ion Concentration – pH Scale
8.5 Polyprotic Acids and Bases
8.6 Hydrolysis of Salts
8.7 Acid base Titrations and Indicators
8.8 Solubility Product
8.9 Buffer Solution

Conceptual Questions with Answers: 18
Additional Numerical Problems for Practice: 25
Revision Exercises
Very Short Answer questions: 20
Short Answer Questions 25
Long Answer Questions 9

Competition File
Numerical Problems: 19
Objective Questions: 41
Fill in the blanks: 10
True or False: 10

Study Plan

Regular Study 16 days, Revision 11 days

Day 1

8.1 Acid-base Concepts

Day 2
8.1 contd.
Ex. 8.1 and 8.2
Practice Problems 8.1 to 8.8

Day 3
8.2 Acid- base Equilibria and Ionization of Acids and Bases
Ex. 8.3 to 8.6
P.P. 8.9 t0 8.10

Day 4
8.3 Ionization of Water – Ionic Product of Water
Ex. 8.7
P.P. 8.11 to 8.13

Day 5

8.4 Expressing Hydrogen Ion Concentration – pH Scale
Ex. 8.8 to 8.12

Day 6
Ex. 8.13 to 8.20

Day 7
P.P. 8.14 to 8.34

Day 8
8.5 Polyprotic Acids and Bases
Ex. 8.21 to 8.24
P.P. 8.35 to 8.38

Day 9
8.6 Hydrolysis of Salts

Day 10
Ex. 8.25 to 8.29
P.P. 8.39 to 8.46

Day 11

8.7 Acid base Titrations and Indicators
Ex. 8.30 and 8.31
P.P. 8.47 and 8.48

Day 12
8.8 Solubility Product
Ex. 8.32, 8.33
P.P. 8.49 to 8.54
Day 13
8.8 contd.
Ex. 8.35 to 8.40, 8.41 to 44
P.P. 8.55 to 8.58, 8.59 to 8.63

Day 14
8.9 Buffer Solution
Ex. 8.45 to 8.48
P.P. 8.64 to 8.66

Day 15

Conceptual Questions with Answers: 1 to 18

Day 16
Additional Numerical Problems for Practice: 25

Revision Period

Day 17
Revision Exercises: Very Short Answer questions: 20

Day 18
Revision Exercises :Short Answer Questions 1 to 15

Day 19
Revision Exercises :Short Answer Questions 16 to 25

Day 20
Competition File: Numerical Problems: 1 to 10
Day 21
Competition File: Numerical Problems: 11 to 19
Day 22
Competition File: Objective Questions: 1 to 20
Day 23
Competition File: Objective Questions: 21 to 41

Day 24
Fill in the blanks: 10
Day 25
True or False: 10
Day 26
Concept Review
Day 27
Formula Review

Sunday, December 28, 2008

Chemical equilibrium- Study Guide - IIT JEE

Law of mass action; Equilibrium constant, Le Chatelier's principle (effect of concentration, temperature and pressure); Significance of DG and DGo in chemical equilibrium;

Solubility product, common ion effect, pH and buffer solutions; Acids and bases (Bronsted and Lewis concepts); Hydrolysis of salts.

Sunday, February 3, 2008

IIT JEE Revision - Ch. 7. CHEMICAL EQUILIBRIA - Core Points

Chemical equilibrium

JEE Syllabus

Chemical equilibrium:
Law of mass action;
Equilibrium constant,
Le Chatelier's principle (effect of concentration, temperature and pressure); Significance of ΔG and ΔGo in chemical equilibrium;

Solubility product, common ion effect,
pH and buffer solutions;
Acids and bases (Bronsted and Lewis concepts);
Hydrolysis of salts.

The syllabus has two main components: Equilibrium among ions and equilibrium among compounds
---------

There are reactions where one can see the reverse action also to be active and see the equilibrium point. In this case

A + B → C + D and

C+D → A+B both reactions keep taking place.

At the point of equilibrium the rate of both reactions is same. Formation of A+B is equal to consumption of A+B.

Rate of forward reaction = k-f[A][B]

Rate of reverse reaction or backward reaction = k-r[C][D]

Therefore k-f[A][B] = k-r[C]{D]

This gives k-f/k-r = [C]{D]/[A][B]

The equilibrium constant is always written as products by reactants.

For the a general reaction

aA + bB ↔ cC+dD (Normal two arrows are used for reversible reaction. Only one arrow with heads on both sides is used here to tide over the inability to show two arrows.)

k-eq = [C]^c[D]^d/[A]^a[B]^b

The equilibrium constant may or may not have units.

In the case of 2A ↔ 2B +C

The units of equilibrium constant are going to be: (mol/l)^2(mol/l)/(mol/l)^2

= mol/l

Ionic Equilibrium – Introduction

Acids, basess and salts when dissolved n water dissociate to some extent and form ions. In the ion formation, an equilibrium is established between ionized and unionized (whole) molecules as this ionization is a reversible reaction. Such an equilibrium that involves ions is called ionic equilibrium.

Acids and bases definitions (Arrhenius, Bronsted and Lewis)

Arrhenius defined acid as a hydrogen compound which in water solution give hydrogen ions.
He defined base as a hydroxide compound which in water solution gives hydroxide ions.

Lowry and Bronsted
An acid is defined as a substance having a tendency of lose or to donate one or more protons.
A base is defined as a substance having a tendency to accept or add a proton

Lewis Theory of Acids and Bases

Acid: An acid is any substance (molecule, ion or atom) that can accept a lone pair of electrons to form a coordinate bond (*Remember coordinate bond and lone pair topics in chapter on Bonding)

Base: Base is any species (molecule, ion or atom) that can donate a lone pair of electrons to form a co-ordinate bond.

Ostwald’s Dilution law
α = SQRT(K/C)

α = Total mole of acid or base dissociated/Total mole of acid or base present in the solution

Chemical equilibrium - Introduction - Revision points

In most of the reaction carried out in closed vessels, reaction does not go to completion under given set of conditions of temperature and pressure. Initially, in the vessel, only reactants are present, and as the reaction proceeds, the concetration of reactants will decrease and that of products will increase.

After some time a stage is reached when no further change in concetrations of reactants and products is observed. This state is called equilibrium state and some of the important questions regarding this phenomenon are:

1. why do reactions seem to stop before they reach completion?
2. What is the extent to which a reaction proceed?
3. Can we modify the conditions to improve the yield of products?

Equilibrium - The phenomenon

Equilibrium is the state at which the concentrations of reactants and products do not change with time.

It is important to remember that equilibrium is achieved in closed vessel reactions only.

The important aspect of reaction equilibrium is the reversibility. The products combine and form reactants. At equilibrium, both the forward and backward reactions are taking place. The rates of forward and backward reactions are same or equal at the equilibrium. As a result, the concentration of each species becomes constant.

The equilibrium is termed as dynamic reaction equilibrium. Dynamic means at a microscopic level, the system is in motion. But at macroscopic level, concentrations are not changing.

Chemical reactions may be classified as reversible reactions and irreversible reactions.

Example of irrereversible reaction

Decomposition of potassium chlorate into potassium chloride and oxygen. Even in a closed vessel this reaction is not reversible.

Example of reversible reaction

1. Decompositon of calcium carbonate. When solid calcium carbonate is heated in a closed vessel at 1073 K, it decomposes into solid calcium oxide and gaseous carbon dioxide. Due to gaseous CO2 there is pressure of gas in the vessel which can be measured. At a constant temperature it can be observed that pressure becomes constant after some time, which means no further CO2 is being produced even though calcium carbonate is still there in the vessel. The constant pressure indicates to us that reaction equilibrium is reached.

Characteristics of chemical equilibrium

1. Chemical equilibrium is dynamic in nature (already explained).

2. the properites of the system become constant at equilibrium and remain unchanged thereafter unless external or internal conditions are changed.

3. The equilibrium is attained only if the system is closed one.

4. As the reactions are reversible and happen under the same conditions, equilibrium can be attained from either direction.

5. A catalyst does not alter the equilibrium point. The catalyst increases the rate of reaction, and at equilibrium it increases both forward and backward reaction rates. But it does not alter equilibrium point, the concentrations of products and reactants at a given set of conditions. But the equilibrium is reached earlier in the presence of a catalyst.

Law of mass action - Revision Points

Law of Mass Action
For the reaction

2 NO2 = N2O4

in a sealed tube the ratio
[N2O4]
-------
[NO2]²
is a constant. This phenomenon is known as chemical equilibrium. The ratio is called equilibrium constant (K).
[N2O4] and [NO2] are molar concentrations of N2O4 and NO2.

Such a law of nature is called the law of mass action or mass action law.
Of course, when conditions, such as pressure and temperature, change, a period of time is required for the system to establish an equilibrium.
For systems that are not at equilibrium yet, the ratio calculated from the mass action law is called a reaction quotient Q. The Q values of a closed system have a tendency to reach a limiting value called equilibrium constant K over time. A system has a tendency to reach an equilibrium state.

The law of mass action may be written as:

The rate of a chemical reaction at any particular temperature is proportional to the product of the molar concentrations of reactants with each concentration term raised to the power equal to the number of molecules of the respective reactants taking part in the reaction.

In the chemical kinetics chapter we come to know that chemical reactions can be elementary reactions or complex reactions having number of elementary reactions.

Law of mass action is valid for elementary reactions.

Revision - Equilibrium constant

There are reactions where one can see the reverse action also to be active and see the equilibrium point. In this case

A + B → C + D and

C+D → A+B both reactions keep taking place.

At the point of equilibrium the rate of both reactions is same. Formation of A+B is equal to consumption of A+B.

Rate of forward reaction = k-f[A][B]

Rate of reverse reaction or backward reaction = k-r[C][D]

Therefore k-f[A][B] = k-r[C]{D]

This gives k-f/k-r = [C]{D]/[A][B]

The equilibrium constant is always written as products by reactants.

For the a general reaction

aA + bB ↔ cC+dD (Normal two arrows are used for reversible reaction. Only one arrow with heads on both sides is used here to tide over the inability to show two arrows.)

k-eq = [C]^c[D]^d/[A]^a[B]^b

The equilibrium constant may or may not have units.

In the case of 2A ↔ 2B +C

The units of equilibrium constant are going to be: (mol/l)^2(mol/l)/(mol/l)^2

= mol/l

JEE Revision - Le Chatelier's principle

Le Chatelier's principle (effect of concentration, temperature and pressure);

In 1884, the French Chemist Henri Le Chatelier suggested that equilibrium systems tend to compensate for the effects of perturbing influences.

When a system at equilibrium is disturbed, the equilibrium position will shift in the direction which tends to minimise, or counteract, the effect of the disturbance.

If the concentration of a reactant is increased, the equilibrium position shifts to use up the added reactants by producing more products.

If the pressure on an equilibrium system is increased, then the equilibrium position shifts to reduce the pressure.

If the volume of a gaseous equilibrium system is reduced (equivalent to an increase in pressure) then the equilibrium position shifts to increase the volume (equivalent to a decrease in pressure)

If the temperature of an endothermic equilibrium system is increased, the equilibrium position shifts to use up the heat by producing more products.

Solubility product

Solubility product of a salt at a given temperature is equal to the product of the concentrations of its ions in the saturated solution, with each concentration term raised to the power equal to the number of moles of ions produced on dissociation of one mole of the substance.

Relationship between solubility (S) and solubility product (K_sp)

Consider M_qA_r a sparingly soluble salt.
Where
q = Number of cations (M^r+) and
r = Number of anions (A^q-)

That is we have in dissolved state

M_qA_r ↔ qM^r+ + r A^q-

Then

K_sp = [M^r+]^q [A^q-]^r

If solubility is S, according to the definition of solubility product

We have
[M^r+]^q = q.S mol/dm³
[A^q-]^r = r.S mol/dm³

Hence K_sp = [q.S] ^q [r.S] ^r

= S^q+r. q^q.r^r

For example for the salt, calcium Phophate, Ca₃(PO₄)₂

Ca₃(PO₄)₂ ↔ 3Ca_aq²⁺ + 2PO_4(aq)^3-

K_sp = [Ca²⁺] ³ [PO₄^3-]²

= S³⁺².3³.2²
= 108S⁵

Past JEE Question

For a sparingly soluble salt A_pB_q, the relationship of its solubility product (K_sp) with its solubility (s) is

a. K_sp = s^p+q.p^p.q^q
b. K_sp = s^p+q.p^q.q^p
c. K_sp = s^pq.p^p.q^q
d. K_sp = s^pq.(pq)^p+q)

(2001)

Answer: a

Common Ion Effect

Common ion effect may be defined as the suppression of the degree of dissociation of a weak acid or a weak base by the addition of a strong electrolyte containing a common ion (ion common to the weak acid or base or the strong electrolyte).

Example:
The case of ionisation of a weak base NH4OH.
At equilibrium if we add solid NH4Cl to the solution,the concentration of NH4+ ions increases.

Due to increase in NH4+ ions, the equilbrium shifts to the left and ionisation of NH4OH reduces.

This principle is made use of in purification of sodium chloride

sodium chloride obtained from sea water or lakes is always impure.

The saturated solution of impure sodium chloride is prepared by dissolving in minimum quantity of water required.

Then HCL gas is passed through this solution (CL- is the common ion).

Due to increase of common chloride ions, the dissociation of sodium chloride is suppressed and sodium chloride is thrown down as precipitate. From the initial saturated solution, now sodium chloride is precipitated.

But the chlorides of impurities still remain in solution. So pure NaCl is obtained using the principle of common ion effect.

JEE Revision - pH and Buffer Solutions

Ionic Product of Water

The Ionic Product of Water, Kw, is the equilibrium constant for the reaction in which water undergoes an acid-base reaction with itself. That is, water is behaving simultaneously as both an acid and a base.

H2O(l) + H2O(l) = H3O+(aq) + OH-(aq)

Kw = [H3O+(aq)][OH-(aq)]

At 298 K, the value of Kw is 1 x 10-14 mol^2 dm^-6. This makes the concentration of H+ ions equal to 1 x 10-7 mol dm^-3, and therefore the pH is 7. This is defined as 'neutral'.

From the above equilibrium expression, taking -log10 throughout

pKw = pH + pOH = 14

Ionic Product does not apply only to water. It applies, for example, to the equilibrium in liquid ammonia:

NH3 + NH3 = NH2- + NH4+

pH

pH is related to hydrogen ion concentration. Since the H+ ion concentration in solution is often small, the concentration is generally expressed as the logarithm of its reciprocal, which is called a pH value. Therefore, pH is defined as

pH = -log[H+]

For a ten times increase in H+ ion concentration there is a decrease in the pH value of one unit.

Given the pH of a solution, its H+ concentration can be found:

[H+(aq)] = antilog -pH

or

[H+(aq)] = 10^-pH

JEE Revision - Acids and bases (Bronsted and Lewis concepts)

Acids and bases definitions (Arrhenius, Bronsted and Lewis)

Arrhenius defined acid as a hydrogen compound which in water solution give hydrogen ions.
He defined base as a hydroxide compound which in water solution gives hydroxide ions.

Lowry and Bronsted
An acid is defined as a substance having a tendency of lose or to donate one or more protons.
A base is defined as a substance having a tendency to accept or add a proton

Lewis Theory of Acids and Bases

Acid: An acid is any substance (molecule, ion or atom) that can accept a lone pair of electrons to form a coordinate bond (*Remember coordinate bond and lone pair topics in chapter on Bonding)

Base: Base is any species (molecule, ion or atom) that can donate a lone pair of electrons to form a co-ordinate bond.

JEE Revision Hydrolysis of salts.

Hydrolysis is the name for a substance chemically reacting with water.

Hydrolysis should be distinguished from solvation, which is the process of water molecules associating themselves with individual solute molecules or ions.

I. Salts of Weak Acids

In general, all salts of weak acids behave the same, therefore we can use a generic salt to represent all salts of weak acids.

Let NaA be a generic salt of a weak acid and A¯ its anion.

Two specific examples of salts of weak acids:

Substance ---Formula---The anion portion (A¯)
sodium acetate---NaC2H3O2---C2H3O2¯
sodium benzoate---C6H5COONa---C6H5COO¯

The generic chemical reaction may be written thusly:

A¯ + H2O --> HA + OH¯

This reaction is of a salt of a weak acid (NOT the acid) undergoing hydrolysis, the reaction with water.

Important things:

1) The Na+ IS NOT involved. Its source is the strong base (NaOH) that helped form the salt (called NaA in the generic example) and it DOES NOT affect the pH. Its presence in both writing the chemical reactions and doing the calculations is deleted. However, keep in mind that Na+ is present in the solution.

2) HA is the UNDISSOCIATED acid.
It is not the acid that makes the acidic pH of a solution, it is the amount of hydrogen ion (or hydronium ion, H3O+).
In order to produce the hydrogen ion, the acid must dissociate.

3) There is free hydroxide ion (OH¯) in the solution. This is the thing that makes the pH greater than 7.

If there is acid (HA) and base (OH¯), why don't they just react and give back the reactants on the left side?

The answer? This reaction is an equilibrium.

When a chemical reaction comes to equilibrium, there is a mixture of all involved substances in the reaction vessel. This mixture is characterized by a constant composition. (constant composition DOES NOT imply equal composition.)

The key point that makes a reaction come to equilibrium is that it is reversible. This means that both the forward reaction and the reverse reaction can happen,
The reaction comes to equilibrium when the rates of the two reactions (forward and reverse) become equal.

So, while it is true that the HA and OH¯ will react in the reverse direction, so can the A¯ and the H2O in the forward direction. The key point is that the reaction happens in such a way that a small amount of HA and OH¯ are present at equilibrium.

When calculations are done, the important points will be (1) how much OH¯ is formed and (2) what is the pH of the solution?

--------------------------------------------------------------------------------

II. Salts of Weak Bases

In general, all salts of weak bases behave the same, therefore we can use a generic salt to represent all salts of weak bases.
Let B be a generic base and HB+ its salt. (Compare how this is worded compared to the "salt of weak acid" discussion.)

HB+ is a cation, but that word is not used as much in discussions as is "anion" is above.

Two specific examples of salts of weak bases:

Substance---Formula---The cation portion (HB+)
ammonium chloride---NH4Cl---NH4+
methyl ammonium nitrate---CH3NH3NO3---CH3NH3+

The chloride ion, Cl¯, and the nitrate ion, NO3¯ tend to be used in examples.

The chemical reaction may be written as:

HB+ + H2O --> B + H3O+

This reaction is of a salt of a weak base (NOT the base) undergoing hydrolysis, the reaction with water.

It is very important that you notice several things:

1) There is an anion involved, but it is not usually written. For example Cl¯ could be the anion, but it IS NOT involved. Its source is a strong acid (HCl) that helped form the salt and it DOES NOT affect the pH. However, Cl¯ is present in the solution.

2) B is the UNPROTONATED base. Keep in mind that it is not the base that makes the basic pH of a solution, it is the amount of hydroxide ion (OH¯). In order to produce it, the base must protonated by the water.

3) There is free hydronium ion (H3O+) in the solution!! This is the thing that makes the pH less than 7.

If there is base (B) and acid (H3O+), why don't they just react and give back the reactants on the left side?

The answer, of course, is given in above in the discussion of salts of weak acids. It would be the same explanation here.

Of course, when calculations are done, the important points will be (1) how much H3O+ is formed and (2) what is the pH of the solution?

Saturday, January 19, 2008

IIt JEE Ch. 7. CHEMICAL EQUILIBRIA - Core Points for Revision

JEE Syllabus

Chemical equilibrium:
Law of mass action;
Equilibrium constant,
Le Chatelier's principle (effect of concentration, temperature and pressure); Significance of ΔG and ΔGo in chemical equilibrium;

Solubility product, common ion effect,
pH and buffer solutions;
Acids and bases (Bronsted and Lewis concepts);
Hydrolysis of salts.

The syllabus has two main components: Equilibrium among ions and equilibrium among compounds
---------

There are reactions where one can see the reverse action also to be active and see the equilibrium point. In this case

A + B → C + D and

C+D → A+B both reactions keep taking place.

At the point of equilibrium the rate of both reactions is same. Formation of A+B is equal to consumption of A+B.

Rate of forward reaction = k-f[A][B]

Rate of reverse reaction or backward reaction = k-r[C][D]

Therefore k-f[A][B] = k-r[C]{D]

This gives k-f/k-r = [C]{D]/[A][B]

The equilibrium constant is always written as products by reactants.

For the a general reaction

aA + bB ↔ cC+dD (Normal two arrows are used for reversible reaction. Only one arrow with heads on both sides is used here to tide over the inability to show two arrows.)

k-eq = [C]^c[D]^d/[A]^a[B]^b

The equilibrium constant may or may not have units.

In the case of 2A ↔ 2B +C

The units of equilibrium constant are going to be: (mol/l)^2(mol/l)/(mol/l)^2

= mol/l

Ionic Equilibrium – Introduction

Acids, basess and salts when dissolved n water dissociate to some extent and form ions. In the ion formation, an equilibrium is established between ionized and unionized (whole) molecules as this ionization is a reversible reaction. Such an equilibrium that involves ions is called ionic equilibrium.

Acids and bases definitions (Arrhenius, Bronsted and Lewis)

Arrhenius defined acid as a hydrogen compound which in water solution give hydrogen ions.
He defined base as a hydroxide compound which in water solution gives hydroxide ions.

Lowry and Bronsted
An acid is defined as a substance having a tendency of lose or to donate one or more protons.
A base is defined as a substance having a tendency to accept or add a proton

Lewis Theory of Acids and Bases

Acid: An acid is any substance (molecule, ion or atom) that can accept a lone pair of electrons to form a coordinate bond (*Remember coordinate bond and lone pair topics in chapter on Bonding)

Base: Base is any species (molecule, ion or atom) that can donate a lone pair of electrons to form a co-ordinate bond.

Ostwald’s Dilution law
α = SQRT(K/C)

α = Total mole of acid or base dissociated/Total mole of acid or base present in the solution

Sunday, October 21, 2007

Study Guide Ch. 7. CHEMICAL EQUILIBRIA

Post updated on 29 August 2009

JEE Syllabus

Chemical equilibrium:
Law of mass action;
Equilibrium constant,
Le Chatelier's principle (effect of concentration, temperature and pressure); Significance of DG and DGo in chemical equilibrium;
Solubility product, common ion effect,
pH and buffer solutions;
Acids and bases (Bronsted and Lewis concepts);
Hydrolysis of salts.

The syllabus has two main components: Equilibrium among ions and equilibrium among compounds

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Main topics in TMH Book Chapter

Section I CHEMICAL EQUILIBRIUM
Section II IONIC EQUILIBRIUM IN ACQUEOUS SOLUTIONS
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Chemical equilibrium

Some chemical reactions appear to go only in one direction and they are said to go to completion. The reactants are totally consumed or some of the reactants are totally consumed and because of that reaction comes to a stop.

But there are many reactions, wherein a reaction among products takes place and reactants are formed due to it. In such reactions, initially in the reaction products keep forming and their concentration increases to certain level and then an equilibrium is reached from which point there is no change in the concentration of products or reactants.

Reversible reactions

All chemical reactions are theoretically reversible. But in some reactions, the reverse reaction is so slight and takes place at such a low rate that for all practical purposes the reaction is considered to be irreversible.

Example: Formation of water.

But there are reactions where one can see the reverse action also to be active and see the equilibrium point. In this case

A + B → C + D and

C+D → A+B both reactions keep taking place.

At the point of equilibrium the rate of both reactions is same. Formation of A+B is equal to consumption of A+B.

From the rate law (Refer Chemical Kinetics chapter), we can write

Rate of forward reaction = k-f[A][B]

Rate of reverse reaction or backward reaction = k-r{C]{D]

Therefore k-f[A][B] = k-r[C]{D]

This gives k-f/k-r = [C]{D]/[A][B]

This is called as equilibrium constant. When the concentrations of C and D and A and B reach this proportion equilibrium is reached in the reaction.

The equilibrium constant is always written as products by reactants.

For the a general reaction

aA + bB ↔ cC+dD (Normal two arrows are used for reversible reaction. Only one arrow with heads on both sides is used here to tide over the inability to show two arrows.)

k-eq = [C]^c[D]^d/[A]^a[B]^b

The equilibrium constant may or may not have units. As we know concentrations are in mol/liter.

In the case of 2A ↔ 2B +C

The units of equilibrium constant are going to be: (mol/l)^2(mol/l)/(mol/l)^2

= mol/l

For any reaction mechanism at equilibrium, the equilibrium constant is equal to the concentration of products over the concentration of the reactants, all raised to the power of the coefficients from the balanced equation.
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IONIC EQUILIBRIUM PORTION

Syllabus

Solubility product, common ion effect, pH and buffer solutions; Acids and bases (Bronsted and Lewis concepts); Hydrolysis of salts.

Syllabus rearranged

Acids and bases (Bronsted and Lewis concepts);
pH
buffer solutions;
common ion effect,
Solubility product,
Hydrolysis of salts.

Ionic Equilibrium – Introduction

Acids, basess and salts when dissolved n water dissociate to some extent and form ions. In the ion formation, an equilibrium is established between ionized and unionized (whole) molecules as this ionization is a reversible reaction. Such an equilibrium that involves ions is called ionic equilibrium.

Acids and bases definitions (Arrhenius, Bronsted and Lewis)

Arrhenius defined acid as a hydrogen compound which in water solution give hydrogen ions.
He defined base as a hydroxide compound which in water solution gives hydroxide ions.

Lowry and Bronsted came out with a different concept to broaden the definition of Arrhenius.

An acid is defined as a substance having a tendency of lose or to donate one or more protons.
A base is defined as a substance having a tendency to accept or add a proton

Lewis Theory of Acids and Bases

Acid: An acid is any substance (molecule, ion or atom) that can accept a lone pair of electrons to form a coordinate bond (*Remember coordinate bond and lone pair topics in chapter on Bonding)

Base: Base is any species (molecule, ion or atom) that can donate a lone pair of electrons to form a co-ordinate bond.

According to this concept acid-based reaction involves the formation of a co-ordinate covalent bond.

Dissociation of acids and bases into ions

Degree of dissociation: It is defined as the fraction of the total number of moles of an acid or base or electrolyte that dissociates into ions in acqueous solution when the equilibrium is attained. It is represented by α.

α = Total mole of acid or base dissociated/Total mole of acid or base present in the solution

The greater the degree of dissociation, the stronger the acid.

Strong acids HCl, H2SO4, HNO3
Weak acids H2CO3, H2S, CH3COOH

Strong bases NaOH, KOH
Weak bases NH4OH, Mg(OH)2, Ca(OH)2

Dissociation constants of weak acids

Represent acid by HA (H is hydrogen and A remaining portion of the acid)

HA + H2O ↔ H+ + A- (HA dissociates into H+ and A-)

Equilibrium constant from law of mass action = K_a = [H+][A-]/[HA]

K_a is also termed as dissociation constant of acids

Dissociation of bases

Bases are represented as BOH and they dissociate into B+ and OH- ions.

Equilibrium constant for dissociatin of bases from law of mass action =
K_b = [B+][OH-]/[BOH]

K_b is also termed as dissociation constant of bases.

Ostwald’s Dilution law

This law is the relationship between degree of dissociation and the dissociation constant.

If one mole of a weak electrolyte be dissolved in v litres of a solution. At equilibrium let α be the degree of dissociation.

Hence BA will be 1- α moles
B will be α moles and A will be α moles

As K = [B][A]/[BA] = (α/v)( α/v)/{(1- α)/v}
= α²/(1- α)v

But the concentration C = 1/v

Therefore K = α²C/(1- α)

If α is very small, 1- α can be assumed as equal to one and K becomes equal to α²C.

And α = SQRT(K/C)

In case of acids it can be written as

α = SQRT(K_a/C)

In case of bases it can be written as

α = SQRT(K_b/C)

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web sites
Reversible Reactions, Chemical Equilibrium,
http://www.docbrown.info/page04/4_74revNH3.htm

http://www.science.uwaterloo.ca/~cchieh/cact/c123/massacti.html

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JEE Question 2007 Paper II

23. Consider a reaction aG+bH--> Products. When concentration of both the reactants G and H is doubled, the rate increases by eight times. However, when concentration of G is doubled keeping the concentration of H fixed, the rate is doubled. The overall order of the reaction is
(A) 0
(B) 1
(C) 2
(D) 3

Solution: D

This is a question of chapter chemical kinetics.
Reason: When G alone is doubled, rate is doubled. That is the exponent of [G] is one in rate law. When both are doubled, rate increases by 8 times, telling us that exponent of [H] is 2 in rate law. So overall order of reaction is sum of the two exponents that is 3.
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