Latest Post

    Le-Chatelier principle predicts quantitatively the effect on the system at equilibrium when some of the variables such as temperature, pressure, and concentration of the equilibrium of a chemical reaction.

Le-Chatelier principle

    If a system at equilibrium is subjected to change, the system will react in such a way so as to oppose or reduce the change if this is possible that is the system tends to balance or counteract the effects of any imposed stress.

Effect of pressure

    According to Le-Chatelier principle with the increase of pressure, the reaction will shift in a direction where the no of moles is reduced thus the system will try to lower the pressure.
N₂ + 3 H₂ ⇆ 2 NH₃
    With the increase of pressure is to shift the reaction in a direction where the sum of the stoichiometric number of gaseous molecules is lowered thus lowering the pressure.
    In other words, an increase in pressure shifts the equilibrium to the low volume side of the rection whereas a decrease of pressure shifts it to the high volume side.

Effect of temperature

    According to Le-Chatelier principle with the increase of temperature, the equilibrium will shift in the endothermic direction that is shifted to the high enthalpy side.
    If the reaction proceeds from low enthalpy side to high enthalpy side heat is absorbed and it is for this reason this direction is known as endothermic direction.
N₂ + 3 H₂ ⇆ 2 NH₃ ΔH = -22 kcal
where ΔH = ΣHproduct - ΣHreactant
    Thus the enthalpy of the reactants in the above reaction higher than that of the products. Thus with the increases in temperature backward reaction favors and thus the equilibrium shifted to the higher enthalpy side and the production of ammonia is decreased.
    With the decrease of temperature, the equilibrium will shift in the exothermic direction that is shifted to the low enthalpy side.
    If the reaction proceeds from high enthalpy side to low enthalpy side heat is released and it is for this reason this direction is known as exothermic direction.
    Thus with the decreases in temperature forward reaction favors and thus the equilibrium shifted to the low enthalpy side and the production of ammonia is increased.
N₂ + O₂ ⇆ 2 NO ΔH = +44 kcal
    Enthalpy of the reactants in the above reaction lower than that of the products. Thus with the increases in temperature forward reaction favors and thus the equilibrium shifted to the lower enthalpy side and the production of NO- is increasing.
    Decreases in temperature for the above reaction backward reaction favors and the equilibrium shifted to the high enthalpy side and the production of NO is decreased.

Addition of inert gas

    Addition of inert gas(He, Ne, Ar, etc) at constant temperature by two way
Constant volume
    Addition of inert gas at constant volume can not affect the equilibrium since the concentration of the total reacting components remain unchanged.
Constant pressure
    When inert gas is added to the system at equilibrium at constant pressure the volume of the reacting system is increased and thereby total concentration is decreased.
    According to Le-Chatelier principle, the system will move in the direction in which no of moles is increases and thereby the concentration of the system is also increased.
Le-Chatelier principle and addition of inert gas on equilibrium
Le-Chatelier principle

Effect of catalyst on equilibrium

    Catalyst can speed up the reaction it does not affect the equilibrium of the reaction. A reversible reaction, the equilibrium state is one in which the forward and reverse reaction rates are equal.
    Presence of a catalyst, speed up both the forward and backward reaction, thereby allowing the system to reach equilibrium faster.
    This is very important that the addition of a catalyst has no effect on the final equilibrium position of the reaction. Thus we can not increases the production of the product.
    Catalysts can be lowering the transition state and the reaction proceeds faster rate. It can be lowering the energy of the transition state(rate-limiting step), catalysts reduce the required energy of activation to allow the reaction proceeds faster rate and reach equilibrium more rapidly.

Properties of the reacting system

    Le-Chatelier principle provides the reacting system some special features.
  1. For examples, if the volume of the nonreactive system is decreased by a specific amount the pressure rises correspondingly.
    In the reactive system, the equilibrium shifted to the low volume sides (if ΔV ≠ 0), so the pressure increases become less than in the non-reactive system.
    The response of the system is moderate in the shift in the equilibrium position makes the reactive system higher compressibility than the non-reactive one.
  2. Similarly, if the fixed quantity of the heat is supplied to the non-reacting system temperature is corresponding increases.
    In a reactive system, such amount of heat supplied does not increase the temperature so much as the non-reacting system. Since the equilibrium is a shift to the higher enthalpy side and the temperature is less increased.
    This shift of equilibrium makes the heat capacity much higher than the non-reactive system. This is useful since the reacting system is chosen as a heat storage medium.

Question answers

Problem
    Why does vapour pressure of a liquid decreases with the addition of a nonvolatile solid solute?
Answer
    The pure solve is the mole faction x1 = 1 but when the non-volatile solute is added to the solvent the mole fraction of the solvent is decreased from 1 that is x1 ㄑ1.
    To reduce the effect according to Le-Chatelier principle the solvent is less vapourised and the mole fraction of the solvent in a solution is thus improved. Thus there occurring lowering of vapour pressure.
Problem
    N₂ + 3 H₂ ⇆ 2 NH₃ ΔH = -22 kcal what would most likely to happen if increasing the pressure of the reaction?
Answer
    According to Le-Chatelier principle yield of NH₃ is increased if pressure is increasing.
Question
    N₂ + 3 H₂ ⇆ 2 NH₃ ΔH = -22 kcal equilibrium shifted to forward direction when
(A) concentration of NH3 increases
(B) pressure is decreasing
(C) concentration of N2 and H2 decrease
(D) pressure increases and temperature decreases
Answer
    (D) pressure increases and temperature decreases

Bohr's atomic theory

    According to Rutherford's atomic theory, the entire mass of an atom was concentrated in a tiny positively charged nucleus around which the extranuclear electrons are moving in constant motion in the orbit.
    But a moving charged particle emits radiation, will then loss kinetic energy and eventually hit the nucleus.
    To resolve the anomalous position Bohr's atomic theory was proposed which is based on several novel postulates.
    Following formulas will be widely used in solving the questions of Bohr's atomic theory.
Atomic theory questions and answers
Atomic theory
Question
    The energy of an electron in the first Bohr orbit of the hydrogen atom is -13.6 eV. What is the energy value of the electron in the excited state of Li⁺²?
Answer
Energy of an electron in nth orbit,
En = (E₁/n²) × Z²

For Li⁺², Z = 3 and in the excited state of Li⁺², n = 2

Putting the values on the above equation we have,

Energy of Li⁺² = (-13.6/4) × 3²
= - 30.6 eV
Question
    What is the radius of the second orbit of a hydrogen atom? Give radius of first orbit = 0.529 Å.
Answer

Radius of an electron in nth orbit,
rn = r₁ × n²

For second orbit n = 2

Thus the radius of the second orbit of a hydrogen atom,
r₂ = 0.529 × 2²
= 2.12 Å

Quantum number orbitals

    For size shape and orientation of an orbital, we need to know about the four quantum numbers and these quantum numbers are designed as, principal quantum number(n), azimuthal quantum number(l), magnetic quantum number(m) and spin quantum number(s).
    Following rules will be widely used in solving the questions of quantum number orbitals.
Principal quantum number is denoted by n

It can have integral values from 0 to ∞

Azimuthal quantum number denoted by l
For a given value of n

l = 0, 1, 2, 3, ........ (n -1)

Total number of different values of l equal to n

Magnetic quantum number is denoted by ml
For a given value of l

ml = +l to -l

Total number of different values of ml equal to (2l + 1)
Question
    Write the correct set of four quantum numbers for the valence electron of Rubidium.

Answer
    The atomic number of Rubidium atom is 37. Thus the electronic configuration of rubidium is,
1S² 2S² 2P⁶ 3S² 3P⁶ 4S² 3d¹⁰4P⁶5S¹
or, [Kr]³⁶ 5S¹
    So the valence electron of rubidium means 5S¹ electron of a rubidium atom. Thus for 5S¹ electron, principal quantum number = 5, azimuthal quantum number = 0, magnetic quantum number = 0, and spin quantum number = +½.
    So the correct set of four quantum numbers for the valence electron of Rubidium atom are,
5, 0, 0, +½
Question
    What are the four quantum numbers of the 19th electron of Cr atom?
Answer
    The four quantum numbers of the 19th electron of Cr atom are 4, 0, 0, +½
Question
    How many electrons in an atom can have the following quantum numbers n=4 and l=1?
Answer
    n = 4 means principal quantum number 4 and l = 1 means the azimuthal quantum number is 1, that is P orbital.
    Thus when n =1 and l = 0 we can describe 4P orbital. For P orbital magnetic quantum numbers = -1, 0, +1 and each magnetic quantum number has two spin quantum numbers.
    So (2 × 3) = 6 electrons in an atom can have the following quantum numbers n =4 and l =1.
    Question
    How many possible orbitals are there for n = 4?
    Answer
    n = 4 means principal quantum number 4.
    Thus when n =4, l = 0, 1, 2, 3. That is 4S, 4P, 4d, and 4f orbitals. S orbital has one orientation of space, p orbitals have three orientation in space, d orbitals have five orientation in space, and f orbitals have seven orientation in space. 
    Thus the number of possible orbitals when n = 4 is (1 + 3 + 5 + 7) = 16
    Question
    How many possible orbitals are there for n = 3, l =1, and ml =0 ?
    Answer
    n = 3 means principal quantum number 3, l = 1 means azimuthal quantum number = 0, and ml = 0 means magnetic quantum number zero that is one orientation in space.
    Thus the number of orbitals is 1, 3S orbital.
    Question
    An electron with the highest value of the magnetic quantum number as 3, what is its principal quantum number?
    Answer
    If the highest value of the magnetic quantum number is 3, then the highest value of the azimuthal quantum number is 3.
    Thus the principal quantum number of this orbital is (l+1) = 3+1 = 4.
    Question
    What is the wavelength of the Hɑ line of the Balmer series of hydrogen?
    Answer
    The wavenumber of the Balmer lines, ⊽ = 1/λ = 109677[(1/2²) - (1/n²)]
Again for Hɑ line n2 = 3
Thus the wavelength for Hɑ - line, 1/λ = 109677 [(1/2²) - (1/3²)]
= 109677 × (5/36)
∴ λ = 36/(5 × 109677)
= 6.564 × 10⁻⁵ cm
= 656.5 nm
    Question
    The lowest wavelength of Lyman series of the hydrogen atom is x. What is the highest wavelength of Paschen series of the He⁺² ion?
    Answer
    The highest wavenumber of the Lyman lines, ⊽max = 1/λmin = R[(1/1²) - (1/∞²)]
or, 1/x = R
or, R = 1/x
    The lowest wavenumber of the Paschen series of He⁺² ion is,
⊽min = 1/λmax = R Z² [(1/3²) - (1/4²)]
or, λmax = (9 × 16)/(4 × 7R)
∴ λmax = 36x/7
    Question
    How many photons of light having the wavenumber a is necessary to provide 3 J of energy?
    Answer
    From the plank theory,
E = nhν
Where n is the number of photon and E is the energy of the photon source.
3 = n h c ⊽ where ⊽ = wave number
∴ n = 3/hca
    Question
    What is De-Broglie wavelength of an electron orbiting in the n = 6 state of a hydrogen atom? (r₀ = Bohr's radius)
    Answer
We know that 2πr = nλ

or, λ = 2πr/n
where r = 6² × r₀ =36 r₀
∴ λ = (2π ×36 r₀)/6 = 12πr₀
    Question
    Find out the number of unpaired electrons of an ion M+x(Z of M = 25) with the magnetic moment √15 B. M. and also find out the value of x?
    Answer
We know that Magnetic moment = √n(n+2)
Thus √15 = √n(n+2)
or, √3(3+2) = √n(n+2)
∴ n = 3
Again the electronic configuration of M is [Ar]¹⁸ 4S² 3d⁵
    We find out that the number of unpaired electrons of M+x ion is 3. For this electronic configuration, we need to remove four electrons from its valence shell.
Thus the electronic configuration of M is
[Ar]¹⁸ 3d³
Thus x = 4

Alkenes

Unsaturated hydrocarbon alkenes

    The compounds contain at least one pair of adjacent carbon atoms linked by multiple bonds, then that compound is said to be unsaturated.

Ethylene

    Ethylene contains a double bond. There are only four univalent hydrogen atoms present in ethylene, therefore ethylene said to be unsaturated compound.
H₂C = CH₂

Acetylene

    Acetylene contains a triple bond and there are only two univalent hydrogen atoms.
HC ☰ CH

Alkenes

    The alkenes are the unsaturated hydrocarbons that contain one double bond. They have the general formula CnH2n, as they contain two hydrogen atoms less than the alkanes, alkenes are called unsaturated hydrocarbons.
    The double bond is called 'olefinic bond or 'ethylenic bond'. The name olefin arose from the fact that ethylene was called 'olefiant gas'( oil-forming gas) since it forms oily liquids when treated with chlorine or bromine.
    The original name given to this homologous series was olefine, but it was later decided to reserve the suffix - ine for basic substances only.

Nomenclature of alkenes

Common naming of alkenes

    In the common naming system of an olefin is named according to the following rules,
    The total number of carbon atom in the olefin is counted and the name of the corresponding alkane is determined.
H2C=CH2
corresponding alkane is ethane
    By changing the name of the corresponding alkane, the suffix -ane of the latter into - ylene.
For ethane changing the suffix -ane of the latter into - ylene.
    The position of the double bond is indicated by number 1, 2, 3, 4...., or Greek Letters α, β, ⋎, ઠ, ...., the end carbon atom nearest to the double bond is denoted by 1, next 2,and so on or α, next β, and so on.
    These letters are then known as locants. The locants of the double bond carbon atom are then placed before the name of the olefin as obtain from rule 1 and 2.
    A hyphen is written in between the locants and the name. The locants are used only to name alkenes containing more than three carbon atom. alkenes of low molecular weights only have common names.
    Problem
    Write down the common names of the following compounds: (i) CH3CH2CH2CH=CH2 (ii) CH3CH=CHCH2CH3.
    Answer
(i) CH₃ - CH₂ - CH₂ - CH=CH2
Parent alkane is pentane

Thus the name of the compound is,
1 - pentylene or α - pentylene

(ii) CH3 - CH = CH - CH2 - CH3
Parent alkane is pentane

Thus the name of the compound is
2 -pentylene or β - pentylene

Substituted or derived naming of alkenes

    Another method of nomenclature is to consider ethylene as the parent substance and higher member is derivatives of ethylene.
    If the compound is mono-substituted then no difficulty arises in naming. But the compound is a disubstituted derivative of ethylene isomerism is possible. Since the alkyl groups are of attached same or different carbon atoms.
    When the groups are attached to the same carbon atom of the olefins named as the asymmetrical compound.
    When the groups are attached to the different carbon atom of the olefins named as the symmetrical compound.
CH3 ㄧCH = CH2
Methylethylene
CH3ㄧCH2ㄧCH = CH2
Ethylethylene
CH3 ㄧ(H3C)C = CH2
as - dimethyl ethylene
CH3ㄧCH = CHㄧCH3
sym - dimethyl ethylene

I.U.P.A.C. naming of alkenes

    According to the I.U.P.A.C. system of nomenclature, the class suffix of the olefins is - ene, and so the series becomes the alkene series.
  1. The longest carbon chain containing the double bond is chosen as the parent alkene.
    CH₃C(CH₃)₂CH₂CH(CH₃)CH=C(CH₃)CH₂CH₃
    The parent part is here base chain, it consists of 8 carbons. The basename, therefore, is to be derived from octane.
  2. The position of the double bond and side chains are indicated by numbers, the lowest number possible being given to the double bond, and this is placed before the suffix.
  3. IUPAC naming alkenes or Olifins
    IUPAC naming alkenes
  4. The name of which is obtained by changing the suffix - ane of the corresponding alkanes into - alkenes.
    To give the lowest number of possible double-bonded carbon is numbered 3.
  5. There are four branches, One methyl branch on carbon atom number 3, three methyl branches on 5th and 7th carbon atoms.
    These are to be indicated as prefixes to the base name. Their names with their locants are 3, 5, 7, 7 - tetramethyl.
Hence the full name is,
3, 5, 7, 7 - tetramethyl - 3 - octene
    Problem
    What are the names of the following compounds in the IUPAC system ? (i) CH3 - CH2 - CH = CH2 (ii) C(CH3)2 = CH2 (iii) CH3 - CH = C(CH3) - CH2 - CH3 (iv) CH2 = C(C2H5) - CH(CH3)2
    Answer
(i) CH₃ - CH₂ - CH = CH₂
but-1-ene

(ii) (CH₃)₂C = CH₂
2-methylprop-1-ene

(iii) CH₃ - CH = C(CH₃) - CH₂ - CH₃
3-methylpent-2-ene

(iv) CH₂ = C(C₂H₅) - CH(CH₃)₂
2-ethyl-3-methylbut-1-ene
    Problem
    Write out the (ignoring stereochemistry) of the isomeric pentanes, and name them by the IUPAC system. Give the structures of the products formed from each on ozonolysis.
    Answer
    The molecular formula of the pentene is C5H12.
    Now take each one in turn and introduce one double bond, starting at the least substituted end and shifting the double bond inwards.
CH₃CH₂CH₂CH=CH₂
pent-1-ene

CH₃CH₂CH=CHCH₃
pent-2-ene

CH₃CH(CH₃)CH=CH₂
3-methylbut-1-ene

CH₃C(CH₃)=CHCH₃
2-methylbut-2-ene

CH₃CH₂C(CH₃)=CH₂
2-methylbut-1-ene
    The product obtained from the ozonide depends on the nature of the reagents used. Here we small use of Zn and acid to give aldehyde and/or ketones.
CH₃CH₂CH₂CH=CH₂

CH₃CH₂CH₂CHO + HCHO

CH3CH2CH=CHCH3

CH3CH2CHO + CH3CHO

CH3CH(CH3)CH=CH2

CH3(CH3)CHCHO + HCHO

CH3CH2C(CH3)=CH2

HCHO + CH3COCH2CH3

CH3C(CH3)=CHCH3

CH3(CH3)C=O + CH3CHO

    Chemical kinetics differential rate low shows the dependence of the rate with the concentration of the reacting species.
    But the integrated rate law of the chemical kinetics provides the concentration of these species at any time from the start of the reaction.

Zero-order kinetics questions

    Rate of the zero-order chemical kinetics reaction does not depend on the concentration of the reactants.
Zero-order kinetics questions
Zero-order chemical kinetics
    Question
    The rate constant of a chemical reaction is 5× 10⁻⁸  mol lit⁻¹sec⁻¹. What is the order of this reaction? How many secs need to change concentration from 4 × 10⁻⁴ moles lit⁻¹ to 2 × 10⁻² moles lit⁻¹?

    Answer
In chemical Kinetics unit of the rate constant in nth order reaction
= (unit of concentration)1-n (unit of time)⁻¹
Given unit of the rate constant = mol lit⁻¹sec⁻¹
= (unit of concentration)(unit of time)⁻¹
    Compare the above two equation We have 1 -n = 1 or, n = 0 Thus the reaction is a zero-order reaction.
    And the integration rate equation at two times (x₂ - x₁) = k (t₂ - t₁)
    Here at the time t₂, x₂ = 2 × 10⁻² moles lit⁻¹ and at time t₁, x₁ = 4 × 10⁻⁴.
Hence the time required to change the above concentration(t₂ - t₁) = (x₂ - x₁)/t
= (2 × 10⁻² - 4 × 10⁻⁴)/5× 10⁻⁸ sec
= 3.92 × 10⁵ Sec
    Questions
    The half-life of a zero-order reaction is x and the reaction is completed on t₁ time. What is the relation between x and t₁?
    Answer
Form the Zero-order kinetics half-life(t½) = [A]₀/2k
or, x = [A]₀/2k
or, [A]₀ = 2kx
    Again for zero order chemical kinetics, [A]₀ - [A] = kt when the reaction completed concentration of [A] = 0.
Thus [A]₀ = kt₁
Compare the above two equation we have,
kt₁ = 2kx
or, t₁ = 2x
    Question
    When the rate of the reaction is equal to the rate constant. What is the order of the reaction?
    Answer
    For zero-order chemical kinetics, the rate of the reaction is proportional to zero power of the reactant. 
That means r ∝ [A]⁰
or, r = k
Thus the reaction is a zero-order reaction.
    Question
    For a reaction, N₂ + 3 H₂ → 2NH₃, if d[NH₃]/dt = 2 × 10⁻⁴ mol lit⁻¹sec⁻¹, What is the order and the value of - d[H₂]/dt of this reaction?
    Answer
    From the unit of the rate constant, we can easily find out the order of this reaction. Here the unit of the rate constant is mol lit⁻¹sec⁻¹, this is the unit of zero-order reaction thus the reaction is zero-order chemical kinetics.
Rate of reaction of zero-order chemical kinetics is
- d[N₂]/dt = - ⅓ d[H₂]/dt
= ½ d[NH₃]/dt

Thus form the above equation, - ⅓ d[H₂]/dt = ½ d[NH₃]/dt
or, - d[H₂]/dt = (3/2) × d[NH₃]

Given d[NH₃]/dt = 2 × 10⁻⁴ mol lit⁻¹sec⁻¹

-d[H₂]/dt = (3 × 2 × 10⁻⁴ mol lit⁻¹sec⁻¹)/2
= 3 × 10⁻⁴ mol lit⁻¹sec⁻¹
    Question
    For a zero-order reaction N₂O₅ → 2NO₂ + ½ O₂, the rate of disappearance of N₂O₅ is 6.25 × 10⁻³  mol lit⁻¹sec⁻¹, what is the rate of formation of NO₂ and O₂ respectively?
    Answer
Rate of reaction in zero-order chemical kinetics is
- d[N₂O₅]/dt = ½ d[NO₂]/dt
= 2d[NH₃]/dt

Rate of disappearance of N₂O₅ is,
6.25 × 10⁻³  mol lit⁻¹sec⁻¹ = - d[N₂O₅]/dt

Thus the rate of formation of NO₂
= (2 × 6.25 × 10⁻³  mol lit⁻¹sec⁻¹)
= 1.25 × 10⁻²  mol lit⁻¹sec⁻¹.

Thus the rate of formation of O₂
= (6.25 × 10⁻³  mol lit⁻¹sec⁻¹)/2
= 3.125 × 10⁻²  mol lit⁻¹sec⁻¹.
    Question
    For the reaction H₂ + Cl₂ → 2HCl on sunlight and taking place on the water. What is the order of the reaction?
    Answer
    This is a zero-order reaction in chemical kinetics.
    Question
    A reaction carried out within A and B. When the concentration of B doubled then the rate of the reaction is one-fourth of the initial rate. What is the order of B in this reaction?
    Answer
    Let the order of the reaction in term of A is ɑ and in term of B is β.
Thus the rate of the reaction(r) = k [A]ɑ [B]β
where k is the rate constant of the reaction.

The initial concentration of A = [A]₀ and B = [B]₀

Thus the initial rate of the reaction(r₀) = k [A]₀ɑ [B]₀β
    When the concentration of B doubled then the rate of the reaction is one-fourth of the initial rate.
Thus, (r₀/4) = k [A]₀ɑ [2B]₀β

Compare these two equations we have, r₀/(r₀/4) = (k [A]₀ɑ [B]₀β)/(k [A]₀ɑ [2B]₀β)
or, 4 = 2
or, β = -2

Half-life in chemical kinetics

    Question
    In a chemical reaction, the rate constant of this reaction is 2.5 × 10⁻³  mol lit⁻¹sec⁻¹. If the initial concentration of the reactant is one  Find out the half-life this reaction?
    Answer
    From the unit of the rate constant, we can easily find out the order of this reaction. Here the unit of the rate constant is mol lit⁻¹sec⁻¹, this is the unit of zero-order reaction thus the reaction is zero-order chemical kinetics.
Thus for the Zero-order kinetics half-life(t½)
= [A]₀/2k
or, t½ = 1/(2.5 × 10⁻³) sec
= 0.4 × 10³ sec

First-order chemical kinetics

    Question
    In a radioactive reaction, the rate constant of this reaction is 2.5 × 10⁻³ sec⁻¹. What is the order of this reaction?
    Answer
In chemical Kinetics unit of the rate constant in nth order reaction
= (unit of concentration)1-n(unit of time)⁻¹

Given unit of the rate constant, = sec⁻¹
= (unit of concentration)⁰(unit of time)⁻¹

Compare the above two-equation,
we have 1 -n = 0
or, n = 1
    Thus the reaction is the first-order reaction.

Zero-order kinetics, the rate of these reactions does not depend on the concentration of the reactants.

zero-order kinetics

    Let us take a reaction represented as
      Product
    Let the initial concentration of the reactant a and product is zero. After the time interval t, the concentration of the reactant is (a-x) and concentration of the product is x. Thus x is decreases of concentration in zero-order reaction.

Zero-order kinetics in terms of product

    Thus the mathematical equation of zero-order kinetics in terms of product,
dx/dt = k₀
Where k₀ is the rate constant of the zero-order reaction.
or, dx = k₀dt

Integrating the above reaction,
dx = k₀ dt
or, x = k₀t + c
where c is the integration constant of the reaction.

When t = o, x is also zero thus, C = o Thus the above equation is,
x = k₀ t
    This is the relationship between decreases of concentration of the reactant(x) within time(t).

Zero-order kinetics in terms of reactant

Rate equation in terms of reactant,
-d[A]/dt = k₀ [A]⁰ = k₀
Where [A] is the concentration of the reactant at the time t.
or, - d[A] = k₀dt
Integrating the above equation,
We have - d[A] = k₀ ∫ dt
or, - [A] = k₀t + c
where c is the integration constant of the reaction.
If initial at the time t = 0 concentration of the reactant [A]₀ Then from the above equation,
- [A]₀ = 0 + c
or, c = -[A]₀
    Putting the value on the above equation,
- [A] = kt - [A]₀
    This is another form of the rate equation in zero-order kinetics.

The half-life of zero-order kinetics

    The time required for half of the reaction to be completed is known as the half-life of the zero-order reaction. It means 50% of reactants disappear in that time interval.

Half-life in zero-order kinetics

    If in a chemical reaction initial concentration is [A]₀ and after t time interval the concentration of the reactant is [A].
    Then, [A]₀ - [A] = kt
    Thus when t = t½, that is the half-life of the reaction, the concentration of the reactant [A] = [A]₀/2. Putting the value on the above equation,
We have [A]₀ - [A]₀/2 = k t½
or, k t½ = [A]₀/2
t½ = [A]₀/2k
    Thus for the zero-order kinetics the half-life of the reaction proportional to its initial concentration.

Examples of the zero-order kinetics

    The only heterogeneous catalyzed reactions may have zero-order kinetics.
Zero order kinetics integrated equation with examples
Zero-order kinetics reaction

Characteristics of zero-order kinetics

  1. The rate of the reaction is independent of concentration.
  2. Half-life is proportional to the initial concentration of the reactant.
  3. The rate of the reaction is always equal to the rate constant of the reaction at all concentration.

Unit of the rate constant in zero-order kinetics

    The rate equation in terms of product for the nth-order reaction is,
d[A]/dt = k [A]n
or, k = (d[A]/dt) × (1/[A]n)
    Thus the unit of rate constant(k) = (unit of concentration)/{unit of time × (unit of concentration)n}
    = (unit of concentration)1-n/unit of time
    Thus if zero-order kinetics the concentration is expressed in lit mole⁻¹ and time in sec
Then the rate constant = (lit mol⁻¹)/sec
= mol lit⁻¹sec⁻¹

Questions and answers of zero-order kinetics


    Question
    The rate constant of a chemical reaction is 5 × 10⁻⁸ mol lit⁻¹sec⁻¹. What is the order of this reaction? How long does it take to change concentration from 4 × 10⁻⁴ moles lit⁻¹ to 2 × 10⁻² moles lit⁻¹?
    Answer
    The reaction is a zero-order reaction and 3.92 × 10⁵ Sec take to change concentration from 4 × 10⁻⁴ moles lit⁻¹ to 2 × 10⁻² moles lit⁻¹.
    Questions
    The half-life of a zero-order reaction is x and the reaction is completed on t₁ time. What is the relation between x and t₁?
    Answer
    2x = t₁
    Question
    If the rate of the reaction is equal to the rate constant. What is the order of the reaction?
     Answer
    Zero-order reaction.
    Question
    For a reaction, N₂ + 3 H₂ → 2NH₃, if d[NH₃]/dt = 2 × 10⁻⁴mol lit⁻¹sec⁻¹, What is the order and the value of - d[H₂]/dt of this reaction?
    Answer
    Zero-order reaction and the value of d[H₂]/dt = 3 × 10⁻⁴ mol lit⁻¹sec⁻¹.
    Question
    For the reaction N₂O₅ → 2NO₂ + ½ O₂, the rate of disappearance of N₂O₅ is 6.25 × 10⁻³ mol lit⁻¹sec⁻¹, what is the rate of formation of NO₂ and O₂ respectively?
    Answer
    1.25 × 10⁻² and 3.125 × 10⁻³ mol lit⁻¹sec⁻¹
    Question
    For the reaction H₂ + Cl₂ → 2HCl on sunlight and taking place on the water. What is the order of the reaction?
    Answer
    This is a zero-order reaction.

Diffusion is the movement of gas molecules from high concentration to the low concentration. Diffusion is occurred on the gases because of the random movement of the gas molecules.
Grahams Law of diffusion or effusion
Graham's Law of diffusion or effusion
    The phenomenon of diffusion may be described as the tendency for any substance to spread uniformly throughout the space available to it.
    The passing out of gas through the pinhole in the wall of the container is called effusion.
    The rate of diffusion and effusion of the gas passing out depends on the density, pressure, and temperature of the gases.

Graham's law of diffusion

    At constant temperature and pressure, the rates of diffusion or effusion of different gases vary inversely as the square root of their densities.
    At constant temperature(T) and pressure(P) if the rate of diffusion or effusion = r and density = d.
r ∝ 1/√d
or, r = k/√d where k is a constant.

Rates of diffusion and densities

    If r₁ and r₂ be the rates of diffusion or effusion of two gases having densities d₁ and d₂ at constant temperature(T) and pressure (P).
∴ r₁ ∝ 1/√d₁ and r₂ ∝ 1/√d₂
Thus r₁/r₂ = √d₂/√d₁
Again density(d) ∝ Vapour density (D)
Thus r₁/r₂ = √D₂/√D₁

Rates of diffusion and molecular weight

    If r₁ and r₂ be the rates of diffusion or effusion of two gases having molecular weight M₁ and M₂ at constant temperature(T) and pressure (P).
We know that Molecular weight(M) = 2 × Vapour density (D)
Thus,  r₁/r₂ = √D₂/√D₁ = √M₂/√M₁

Rates of effusion and volume

    Let at constant temperature and pressure V₁ and V₂ be the volume of the two gases passing through the same hole with the time t.
Then V₁/t = r₁ and V₂/t = r₂
∴ r₁/r₂ = V₂/V₁ = √M₂/√M₁

Graham's law from the kinetic gas equation

    The rate of diffusion or effusion can be assumed to be directly proportional to the root mean square speed or any other average speed.
    Thus, r₁/r₂ = √u₁²/√u₂² where u₁ and u₂ are the root mean square speed of the gas molecules.
From the kinetic gas equation
PV = ⅓ m N u²
or, u² = 3PV/m N

For 1-mole ideal gas PV = RT and m N = M = molar mass of the gas.

Thus u² = 3RT/M

∴ r₁/r₂ = √u₁²/√u₂² = √(3RT/M₁)/√(3RT/M₂)
or,  r₁/r₂ = √M₂/√M₁
which is Graham's law of diffusion.

Practical application of Graham's law

    Application of Graham's law is in the partial separation of the components in a gas mixture. If the mixture is led out a tube made of porous walls, in a given time the lighter components will diffuse out more than the heavier ones.  By repeating the process with each sperate fraction from diffusion, the concentration of one component is considerable increases compared with that of the other. This is called the atmolysis.
    Argon has been concentrated with nitrogen in this way. The isotopes of neon, chlorine, bromine, etc have been partially separated by this method.
    Graham's law also used on for detecting the marsh gas in mines.

Problems solutions

    Problem
    At constant temperature and pressure 432 ml and 288 ml be the volume of the two gases passing through the same hole with the time 36 min and 48 min respectively. If the molecular weight of one gas is 64 gm mol⁻¹, what is the molecular weight of another gas?
    Solution
According to Graham's law of diffusion,
r₁/r₂ = √M₂/√M₁
    Here, r₁ = 432 ml/36 min = 12 ml min⁻¹,  r₂ = 288 ml/48 min = 6 ml min⁻¹ and M₂ = 64 gm mol⁻¹.
∴ 12/6 = √64/√M₁
or, M₁ = 64/4 = 16 gm mol⁻¹
    Problem
    A certain temperature the time required to the complete diffusion of 200 ml of hydrogen gas is 30 min. How many time required for the complete diffusion of 50 ml of oxygen gas at the same temperature?
    Solution
    Let time required for the complete diffusion of 50 ml of oxygen gas is t min.
According to Graham's law of diffusion,
r₁/r₂ = √M₂/√M₁
    Here, rate of diffusion of hydrogen (r₁) = 200 ml/30 min = 20/3 ml min⁻¹,  rate of diffusion of oxygen (r₂) = 50 ml/t min = 50/t ml min⁻¹,  molecular weight of hydrogen (M₁) = 2 gm mol⁻¹ and molecular weight of oxygen (M₂) = 32 gm mol⁻¹
∴ (200/30)/(50/t) = √(32/2)
or, t₂ = 30 min
    Problem
    Which of the two gases ammonia and hydrogen chloride diffuse faster and by which factor?
    Solution
According to Graham's law of diffusion,
rNH₃/rHCl = √MHCl/√MNH₃
    Here molecular weight of ammonia (NH₃) = 17 gm mol⁻¹ and molecular weight of hydrogen chloride (M₂) = 36.5 gm mol⁻¹
∴ rNH₃ = rHCl × (√MHCl/√MNH₃) = rHCl × (√36.5/√17) = 1.46 rHCl
    Thus ammonia will diffuse 1.46 times faseter then hydrogen chloride gas.
    Problem
    Landenberg found that a sample of ozonized oxygen containing 86.16% of ozone by weight required 430 seconds to diffuse under conditions where pure oxygen required 367.5 seconds. Determine the vapor density of ozone.
    Solution
    Let V is the volume diffusing out in each case. Let dm, do and d be the densities of the mixture, pure oxygen, and ozone respectively.
    Here, rate of diffusion of mixure (rm) = V ml/430 seconds = V/430 ml seconds⁻¹,  rate of diffusion of oxygen (ro) = V ml/367.5 seconds = 50/t ml seconds⁻¹
According to Graham's law of diffusion,
rm/ro = √do/√dm = √16/√dm
or, dm = (430/367)² × 16
≈ 21.91

Again for 100 gms, the volume = 100/dm = (86.16/d) + (13.84/do)
or, (100/21.91) = (86.16/d) + (13.84/16)
or, d = 23.3
    Problem
    Equal moles of hydrogen and oxygen is placed in a container with a pinhole through which escape. What fraction of oxygen escape in the time required for one half of the hydrogen to escape?
    Solution
    Let the time required for this escape is t, and the fraction of oxygen and hydrogen escape is nO₂ and nH₂ respectively.
Thus according to Graham's law of diffusion,
rO₂/rH₂ = √MH₂√MO₂
or, (nO₂/t)/(nH₂/t) = √2/√32
or, (nO₂/t)/(0.5/t) = 1/4
or, nO₂ = 1/8
[Chemical kinetics] [column1]

Contact Form

Name

Email *

Message *

Theme images by Jason Morrow. Powered by Blogger.