August 2019

    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 definition

    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 the 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
    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 the 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
    The 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 the 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.
    The presence of a catalyst, speeds 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

    The le-chatelier principle provides the reacting system with some special features.
  1. For example, 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

Question
    Why does vapour pressure of a liquid decrease 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 the Le-Chatelier principle the solvent is less vapourised and the mole fraction of the solvent in a solution is thus improved. Thus there occurring a lowering of vapour pressure.
Question
    N₂ + 3 H₂ ⇆ 2 NH₃ ΔH = -22 kcal what would most likely happen if increasing the pressure of the reaction?
Answer
    According to the 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) the concentration of NH3 increases
(B) pressure is decreasing
(C) the concentration of N2 and H2 decrease
(D) pressure increases and temperature decreases
Answer
    (D) pressure increases and temperature decreases

Bohr's model of an atom

    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.
Energy of an electron
Energy of an electron
    The following formulas will be widely used in solving the questions of Bohr's atomic theory.
Question
    The energy of an electron in the first Bohr orbital of the hydrogen atom is -13.6 eV. What is the energy value of the electron in the excited state of Li⁺²?
Answer
The 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

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

Second orbit n = 2

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

Quantum level of an atom

    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).
    The 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 1 to ∞

The 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

The 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
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 Chromium atom?
Answer
    The four quantum numbers of the 19th electron of chromium 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.
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 the Lyman series of the hydrogen atom is x. What is the highest wavelength of the 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 photons and E is the energy of the photon source.
3 = n h c ⊽ where ⊽ = wave number
∴ n = 3/hca
Question
    What is the 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.
Electronic configuration of M is
[Ar]¹⁸ 3d³
Thus x = 4

Alkenes nomenclature
Alkenes nomenclature

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 an unsaturated compound.
H₂C = CH₂

Acetylene

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

Alkenes

    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
the 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 rules 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 atoms. 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 the 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

    The rate of the zero-order chemical kinetics reaction does not depend on the concentration of the reactants.
Chemical kinetics questions answers
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
Question
    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 the 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 terms of A is ɑ and in terms 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.

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