KINETICS OF ZERO ORDER REACTIONS:

Consider a general chemical reaction its follow zero order kinetics

[A] ₀ =initial concentration of reactant A

[A]_t = concentration of after time t

[B]_t = concentration of product B after time t

On Integration of above equation:

On rearrangement equation (1) give equation of straight line

Note: Concentration of reactant after regular interval of time will constitute an arithmetical progression (AP) 

Similarly concentration of product B after time t

Unit of rate constant:    K= Mol/litre second

Half life period:  Time required converting half of the reactant to product i.e. life of zero order reaction may give as

Imp note: In zero order reaction average rate of reaction is equal to instantaneous rate of reaction

100% completion of time: It is time in which reactant completely converted into product.

ILLUSTRATIVE EXAMPLE (1): The graph show decomposition of Ammonia  on Pt surface it initial concentration of ammonia is 0.1 M then calculate the time required for the 40% completion of reaction.

SOLUTION:   

ILLUSTRATIVE EXAMPLE (2): A certain zero order reaction has k=0.025 M /S for disappearance of A what will be the concentration of  A after 15 second if initial concentration is 0.5 M?

SOLUTION: concentration of A after time of t is [A]_t

KINETICS FOR nth ORDER REACTION:

Consider a chemical reaction have one reactant only

Differential rate expression may be written as:

Separating same variable same side

On integrating by rule

By equation (1) and (2)

Graphical representation:

Half life period:

Graphical representation:

Note : This formula do not valid for n=0 and n=1

ILLUSTRATIVE EXAMPLE (1) :  A graph plotted between log  t₅₀   Vs log conc. (a)  is a straight line what conclusion withdraw from the given graph?

SOLUTION: for a given reaction

ILLUSTRATIVE EXAMPLE (1) :  What  will be the order of reaction and rate constant for a chemical change having log  t₅₀   Vs log conc. (a)  curve as

SOLUTION: for a given reaction

Case (1) n=2 i.e.    2^nd order chemical kinetics:

Graphical representation: 

Half life period:

Graphical representation:

Case (2) n= 3 i.e.    3^nd order chemical kinetics:

Graphical representation:

Half life period:

Graphical representation:

Important kinetic expression for reaction of type as:

Graphs of various order:

W h e r e

[A]₀ = initial concentration

[A] = concentration at time t

t_1/2 = time taken for initial concentration of reactant to finish by 50%

t _3/4 = time taken for initial concentration of reactant to finish by 75%

ATOMIC MASS AND MOLECULAR MASS:

ATOM AND MOLECULES:

Atom: Atom is the ultimate electrically neutral, made up of fundamental particle (Electron, neutron, Proton) which shows the characteristic properties of the element and exist freely in a chemical reaction.

Molecules: A molecule is the smallest particle made up of one or more than one atom in a definite ratio having stable and independent existence e.g.H₂ ,O₂ He HCl , CaCO₃ etc. 

(1) Atomic Mass: As atoms are very tiny particles, their absolute masses are difficult to measure. However it is possible to determine the relative masses of different atoms if small unit of mass is taken as standard there is three conventions given as.

(1) This standard was mass of one atom of hydrogen and taken as unity.

(2)  And also mass of one atom is 1/16th part of oxygen atom

(3) And now mass of one atom is 1/12th part of C-12 atom.

The atomic mass of an element can be defined as the number which indicates how many times the mass of one atom of the element is heavier in comparison to the mass of one atom of hydrogen.

(2) Atomic Mass Unit:

(1) The quantity 1/12Th mass of an atom of carbon-12 is known as the atomic mass unit and is abbreviated as amu.

(2) The actual mass of one atom of carbon-12 is –

Determination of atomic mass:

(1)  Applying Dulong and Petit’s law.

(2)  Cannizzaro’s methods

(3)  By mitscherlich’s law of isomorphism.

(4)  By measurement of V.D. of volatile chloride or bromide.

(1) Dulong & Petits Law: The product of specific heat of pure element and atomic mass of the element is equal to 6.4.

NOTE:

(1) But this law is not applicable to lighter element like boron, carbon, silicon. To obtain correct atomic mass of element first of all equivalent mass of the element is known by any other method and their atomic mass = eq. weight ´ valency

(2) In which valency has whole number value which can be deduced by dividing approximate by equivalent mass.  Dulongs and Petit’s Law:  Atomic mass ´ specific heat = 6.4

ILLUSTRATIVE EXAMPLE(1): The specific heat of a metal of atomic mass 32 is likely to be:

                                (A)  0.25                                                                (B)  0.24

                                (C)  0.20                                                                (D)  0.1

SOLUTION:

ILLUSTRATIVE EXAMPLE(2): On dissolving 6 gm of metal in sulphuric acid, 13.53g of the metal sulphate was formed. The specific heat of metal is 0.057 Cal/g. What is equivalent mass of metal, valency and exact atomic mass?

SOLUTION:

(2)  Cannizzaro’s methods:

If an element has several compound with other same or different elements of known atomic mass then the compound that has minimum presence of former element indicate the atomic mass of former element.

Procedure:

(1)  First of all the molecular mass of all compounds known by applying

      V.D ´ 2 = mol. weight

(2)  By analysis the presence of the desired element in each compound is known.

(3)  The mass that is lowest among all the compound indicate the atomic mass

of that element

ILLUSTRATIVE EXAMPLE (3): Estimate the atomic mass of nitrogen given that vapour density of NH₃ = 8.5, Nitrous oxide = 22, Nitric oxide = 15, Nitrogen peroxide = 23, Nitrogen
trioxide = 38.

SOLUTION:

(3)  Law of Isomorphism

When two or more compound forms similar type of crystals or able to form mixed crystals, they are known as isomorphs. For examples: MgSO₄.7H₂O, ZnSO₄.7H₂O and FeSO₄.7H₂O are isomorphs of each other as their crystals posses same shape.

According to Mitscherlich [year 1819]. The valency of elements that are similarly placed to that of other elements in their isomorphs are always same.

In the above example Fe, Zn and Mg have same valency [2] and equal ratio of water molecule in each isomorphs.

If equivalent mass of one element is known then atomic mass can be calculated by knowing the valency of other isomorphs key element.

ILLUSTRATIVE EXAMPLE(4): Which pair of the following substances is said to be isomorphous?

                        (A) White vitriol and blue vitriol        (B) Epsom salt and Glauber’s salt

                        (C) Blue vitriol and Glauber’s salt      (D) White vitriol and Epsom salt

SOLUTION:   Epsom salt (MgSO₄.7H₂O) and White vitriol (ZnSO₄.7H₂O) contains divalent cation Mg²⁺ and Zn²⁺ and same number of water molecules as water of crystallization which hold criteria for isomorphism. Hence (D) is correct.

(4)  Atomic mass from vapour density of a chloride:

The following steps are involved in this method

(1) Vapour density of chloride of the element is determined

(2) Equivalent mass of the element is determined

(3) Let the valency of the element be x. The formula of its chloride will be MCl_X

ILLUSTRATIVE EXAMPLE(5): One gram of chloride was found to contain 0.835g of chlorine. Its vapour density is 85. Calculate its molecular formula

SOLUTION:

TRY YOURSELF:

Exercise (1): Two oxides of a metal contain 63.2% and 69.62% of the metal. The specific heat of the metal is 0.117. What is the formula of the two oxides?

Exercise (2): 1 g of a metal which has specific heat of 0.06 combines with oxygen to form 1.08 g of oxide. What is the mass of M?

Exercise (3): The chloride of a solid metallic element contains 57.89% by mass of the element. The specific heat of the element is 0.0324 cal deg^-1g^-1. Calculate the exact atomic mass of the element.

Exercise (4): White vitriol (hydrated sulphite) is isomorphous with MgSO₄.7H₂O. White vitriol contains 22.95% zinc and 43.9% of water of crystallization. Find the atomic mass of zinc.

Exercise (5): Two oxides of metals A and B are isomorphous. The metal A whose atomic mass is 52, forms a chloride whose vapour density is 79. The oxide of the metal B contains 47.1% oxygen. Calculate the atomic mass of B.

ANSWER KEYS:

Exercise (1): MO₂ and M₂O₃, Exercise (2): Molar mass of metal 100 g, Exercise (3): 195.21, Exercise (4): 65.3 (atomic mass of zinc) Exercise (5): Atomic mass of B = 27 amu

DALTON’S ATOMIC THEORY:

(1) Matter is made up of extremely small, indivisible particles called atoms.

(2) Atom of same substance are identical in all respect i.e. they posses same size, shape, mass, chemical properties etc.

(3) Atom of different substances are different in all respect i.e. they posses different shape, size, mass and chemical properties etc.

(4) Atom is the smallest particle that takes part in chemical reactions.

(5) Atom of different elements may combine with each other in a fixed, simple, whole number ratio to form compound atoms.

(6) Atom can neither be created nor destroyed i.e. atoms are indestructible.

Limitations of Dalton’s theory:

The main failures of Dalton’s atomic theory are:

(1) Atom was no more indivisible. It is made up of various sub-atomic particles like electrons, proton and neutron etc.

(2) It failed to explain how atoms of different elements differ from each other.

(3) It failed to explain how and why atoms of elements combine with each other to form compound atoms or molecules.

(4) It failed to explain the nature of forces that bind together different atoms in molecules.

(5) It failed to explain Gay Lussac’s law of combining volumes.

(6) It did not make any distinction between ultimate particle of an element that takes part in reaction (atoms) and the ultimate particle that has independent existence (molecules).

Modern Atomic theory:

(1) Atom is no longer supposed to be indivisible. Atom has a complex structure and is composed of sub-atomic particles such as electrons protons and neutrons.

(2) Atom of the same element may not be similar in all respects e.g. isotopes.

(3) Atom of different elements may be similar in one or more respects e.g. isobars.

(4) Atom is the smallest unit which takes part in chemical reactions.

(5) The ratio in which atoms unite may be fixed and integral but may not be simple. e.g. In sugar molecules C₁₂H₂₂O₁₁ the ratio of C, H and O atoms is 12:22:11 which is not simple.

(6) Atom of one element can be changed into atoms of other element for e.g. transmutation.

(7) Mass of atom can be changed in energy. 

     (E=MC²) According to Einstein mass energy relationship, mass and energy are inter-convertible. Thus atom is no longer indestructible.

ILLUSTRATIVE EXAMPLE: An important postulate of Dalton’s atomic theory is:

                        (A) an atom contains electrons, protons and neutrons

                        (B) atom can neither be created nor destroyed nor divisible

                        (C) all the atoms of an element are not identical

                        (D) all the elements are available in nature in the form of atoms

SOLUTION: The statement written in (B) is about law of mass conservation which is true for all chemical reaction. Hence (B) is correct.

PERCENTAGE COMPOSITION OF COMPOUNDS:

Law of Definite Proportions:  Compounds are consistent chemical combinations of atoms that can be expressed as:

(i)  Ratio of masses

(ii) Ratio of atoms

(iii) Ratio of moles of atoms

(iv) Percent composition

ILLUSTRATIVE EXAMPLE (1): Calculate the percent Nitrogen in Dinitrogen (N₂O) Monoxide?

SOLUTION:

ILLUSTRATIVE EXAMPLE (2): What is the % composition of each element in (Mg(OH)₂) magnesium hydroxide?

SOLUTION:

ILLUSTRATIVE EXAMPLE (3): Haemoglobin contains 0.33% of Iron by weight. The molecular weight of it is approx. 67200. The numbers of iron atoms (Atomic wt of Fe=56 u) present in one molecule of Haemoglobin are?

SOLUTION:

ILLUSTRATIVE EXAMPLE (4): The hydrated salt, Na₂SO4.nH₂O undergoes 55.9% loss in weight on heating and becomes anhydrous. The value of n will be?

SOLUTION:

ILLUSTRATIVE EXAMPLE (5): Air contain 20 % O₂ by volume. How many cm³ of air will be required for oxidation of 100 cm³(ml) of acetylene?

SOLUTION:

Since air contain 20 % oxygen by volume then amount of air required to react with 100cc/ml of C₂H₂

Try yourself:

Exercise (4): Two oxides of metal contain 72.4 and 70 of metal respectively if formula of 2nd oxide is M₂O₃ find that of the first.

Ans Key : M₃O_{4
Determination of Empirical and Molecular Formula:}