Latest Post

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 with explanation
Atomic theory questions and answers
  • 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
    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,
ELi⁺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
    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.
Quantum numbers questions and answers
Quantum numbers
  • 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.

Ionization energy is supplied to an electron of a particular atom, it can be promoted to successively higher energy levels. If the supplied energy is large enough it may possible to remove the electron completely from the pull of the positive nucleus.
    The amount of energy required to remove the most loosely bound electron, that is outermost electron from an isolated gaseous atom of an element in its lowest energy state, that is ground state to produce cation is known as Ionization Potential or Ionization Energy of that element.
    M(g) + Ionization Energy M⁺(g) + e
    It is generally represented as I or IP and it is measured in electron volt (eV) or kilocalories (kcal) per gram atom. One electron volt (eV) being the energy gained by an electron falling through a potential difference of one volt.
∴ 1 eV = (Charge of an electron) × (1 volt)
= (1.6 × 10⁻¹⁹ Coulomb) × (1 Volt)
= 1.6 × 10⁻¹⁹ Joule
= 1.6 × 10⁻¹² erg

Successive Ionization Potentials

  1. The electrons are removed in stages one by one from an atom. The energy required to remove the first electron from a gaseous atom is called its first Ionization potential.
    M (g) + IP₁  M⁺ (g) + e
  2. The energy required to remove the second electron from the cation is called second Ionization potential.
    M⁺ (g) + IP₂ M⁺² (g) + e
  3. Similarly, we have third, fourth Ionization potentials.
    M⁺² (g) + IP₃ M⁺³ (g) + e
    M⁺ (g) + IP₄ M⁺⁴ (g) + e

Periodic trends ionization energy

    The following factors influence the magnitude of the Ionization Potentials:

Atomic radius ionization potential

    The greater the distance of an element from the positive charge nucleus of an element, the weaker will be the attraction and hence the lower the ionization potential of that element.
    If an atom is raised to an excited state by promoting one electron to a higher energy level, then that electron is further from the nucleus and it is observed that this excited electron is more easily detached than when the electron is in its ground state.
    Thus on moving from top to bottom in a group the ionization potential of the elements decreases with the increase of their nucleus electron separation.
    The successive increase in their values is due to the fact that it is relatively difficult to remove an electron from a cation having higher positive charge than from a cation having a lower positive charge or from a neutral atom.
Ionization Potential
The magnitude of Ionization Potential

Nuclear charge and ionization potential

    The higher the charge on the nucleus the more difficult it is to remove an electron and hence the higher the value of ionization potential.
    Thus the value of ionization potential generally increases in moving from left to right in a period, since the nuclear charge of the elements (atomic number) also increases in the same direction.
Ionization potential vs nuclear charge
    With increasing, atomic number electrons are added to orbitals of the same principal quantum number. These electrons add little to the size of the atom, the increasing nuclear charge brings about a contraction in size. In effect, therefore, ionization potential steadily increases along a period.

Filled and half-filled orbitals and ionization potential

    According to the Hund's rule atoms having half - filled or completely filled orbitals are comparatively more stable and hence more energy is needed to remove an electron from such atom.
    The ionization of such atoms is therefore relatively difficult than expected normally from their position in the periodic table. A few regulations that are seen in the increasing value of ionization potential along a period can be explained on the basis of the concept of the half-filled and completely filled orbitals.
    Be and N in the second period and Mg and P in the third period have a slightly higher value of ionization potentials than those normally expected.
    This is explained on the basis of extra stability of the completely - filled 2S - orbital in Be (2S²) and 3S - orbital in Mg (3S²) and of half-filled 2P - orbital in N (2S² 2P³) and 3P - orbital in P (3S² 3P³).

Screening Effect and ionization potential

    Electrons provide a screening effect on the nucleus. The outermost electrons are shielded from the nucleus by the inner electrons.
    The radial distribution functions of the S, P, d orbitals shows that for the same principal quantum number the S - orbital is the most penetrating., next is P - orbital and least penetrating is the d - orbital. Screening efficiency falls off in the order, S〉P〉d.
    As we move down a group, the number of inner - shells increases and hence the ionization potential tends to decreases.
Elements of II A Group: Be〉Mg〉Ca〉Sr〉Ba

Overall Charge on the Ionizing Species

    An increase in the overall charge on the ionizing species (M⁺, M⁺², M⁺³, etc) will enormously influence the ionization potential since electron withdrawal from a positively charged species is more difficult than from a neutral atom.
    The first ionization of the elements varies with their positions in the periodic table. In each of the table, the noble gas has the highest value and the alkali metals the lowest value for the ionization potential.

Ionization potential questions answers


  • Question
    Calculate the ionization potential of the Hydrogen atom in eV.
  • Answer
    We know that the energy corresponding to the transition from n = ∞ to n = 1 gives the ionization potential of the hydrogen atom.
Thus the IP of the Hydrogen atom,
EH = (2π²me⁴/h²)[(1/n₁²) - (1/n₂²)]
∴ EH = 2.179 × 10⁻¹¹ erg
= 2.179 × 10⁻¹⁸ Joule
= (2.179 × 10⁻¹⁸)/(1.6 × 10⁻¹⁹) eV
= 13.6 eV
  • Question
    Calculate the second ionization potential of Helium (given IP of Hydrogen = 13.6 eV)
  • Answer
    The ground state electronic configuration of helium is 1S². The second ionization potential means the removal of the second electron from the 1S orbital against the nuclear charge of +2.
So we have: IPHe = (2π²mZ²e⁴/h²)[(1/n₁²) - (1/n₂²)]
= Z² × IPH
∴ Second Ionization Potential of Helium, = 22 × 13.6
= 54.4 eV
  • Question
    In first transition series electron filling up processes begins in the 3d level below a filled 4S level. During ionization process will a 4S electron or a 3d electron be lost first? Explain with reference to Chromium.
  • Answer
    During ionization, the 4S electron lost first.
    For details follow the link
  • Question
    Arrange the following with the increasing order of Ionization Potentials Li, Be, B, C, N, O, F, Ne.
  • Answer
    LiㄑBㄑBeㄑCㄑOㄑN〈FㄑNe

Electron Affinity of an element is defined as the amount of energy released when an electron is added to a neutral gaseous atom in its lowest energy state that is, ground state to convert it into uni-negative gaseous ion.
    In simple words, the electron affinity of an atom is defined as the energy liberated when a gaseous atom captures an electron.
A(g) + ElectronA- (g) + Electron Affinity
    Evidently, electron affinity is equal in magnitude to the ionization energy of the species formed. The definition of electron affinity as the exothermicity of the electron capture reaction would lead to the negative sign according to our usual thermodynamic convention. But electron affinity is normally described with a positive sign.
    Negative electron affinities are known. These negative values indicate that the species concerned does not like to have any more electron.

Measurement of Electron Affinity

    Electron affinities are difficult to obtain. These are often obtained from indirect measurements, by analysis of Born-Haber energy cycles in which one step is electron capture or by the direct study of electron capture from heated filaments.

Factors Influencing Electron Affinity

    The magnitude of Electron Affinity (EA) is influenced by the following factors such as,
  1. Atomic Size.
  2. Effective Nuclear Charge.
  3. Electronic Configuration.

Electron affinity vs atomic Size

    Larger the atomic radius lesser the tendency of the atom to attract the additional electron towards itself and lesser is the force of attraction exerted by the nucleus on the extra electron being added to the valence shell of the atom.
    Thus Electron Affinity values decrease with increases atomic radius.

Electron affinity vs effective nuclear charge

    Higher the magnitude of effective nuclear charge (Zeff) towards the periphery of an atom greater is the tendency of the atom to attract the additional electron towards itself.
    Again greater the force of attraction exerted by the nucleus on the extra electron being added to the valence shell of the atom. As a result, higher the energy released when an extra electron is added to form an anion. Thus the magnitude of Electron Affinity of an atom increases with increasing Zeff value.

Electron affinity vs electronic configuration

    nS², nS², nP⁶, nS², nP⁶
    valence shell configuration posses very low electron affinity value due to having their stable valence shell configuration.
Electron Affinity
Electron affinity of  different elements

Electron affinity trend

Electron affinity across a group

    In a group when move from top to bottom the size of atom generally increases with increasing atomic number and consequently, the magnitude of electron affinity decreases in the same direction.
    There are some exceptions to this general rule as is evident from the following examples:
    Although the elements of the second period of the periodic table are relatively smaller in size than those of the third-period elements, the electron affinity values of elements of the second period are smaller than the electron affinity values of third-period elements.
    This unexpected behavior is explained by saying that the much smaller sizes of the second-period elements give a very much higher value of charge densities for the respective negative ions. A high value of electron density is opposed by the interelectronic repulsion forces.
    Electron Affinity of fluorine is lower than that of Chlorine. The lower values of electron affinity for Florine due to the electron-electron repulsion in relatively compact 2P-Orbital of Florine atom. Thus the electron affinity values of halogens are:
F Cl Br I
- 3.6 eV - 3.8 eV - 3.5 eV - 3.2 eV

Electron affinity across a period

    In a period, when we move from left to right Zeff value towards the new coming electron gradually increases with the increasing atomic number and hence the electron affinity gradually increases in the same direction.
    A plot of electron affinities of elements up-to chlorine against atomic number shown as,
Electron affinities as functions of atomic number
Electron affinities as functions of atomic number

Electron affinity questions answers

  • Question
    Calculate the electron affinity of chlorine from the Born - Haber cycle, given the following date : Lattice Energy = - 774 kJ mol⁻¹ , Ionization Potential of Na = 495 kJ mol⁻¹, Heat of Sublimation of Na = 108 kJ mol⁻¹, Energy for Bond Dissociation of chlorine (Cl2) = 240 kJ mol⁻¹ and Heat of Formation of NaCl = 410 kJ mol⁻¹.
  • Answer
    Born - Haber Cycle for the formation of NaCl (S) is:
Born Haber Cycle for the calculation of electron affinity
Born Haber Cycle
From the above Born - Haber cycle we can write as,

-UNaCl - INa + ECl - SNa - ½DCl - ΔHf = 0 

or, ECl = UNaCl+INa+ SNa +½DCl + ΔHf 

∴ ECl = -774 + 495 + 108 + 120 + 410 

= 359 kJ mol-1
  • Question
    Account for the large decrees in electron affinity from lithium to beryllium despite the increase in nuclear charge.
  • Answer
    Atomic number and the electronic distribution of lithium and beryllium are:
Li    3    1S² 2S¹

Be    4    1S²2S²
    Lithium has an incompletely filled 2S sub-shell while beryllium has the subshell filled. Thus lithium can receive an electron in its 2S sub-shell but for beryllium, a still higher energy 2P level has to be made of. 
    A filled shell or sub-shell leads to some extra stability. Hence beryllium resists an extra electron more than does lithium.
  • Question
    Why electron affinity of Nitrogen is - 0.10 eV while that of Phosphorus is + 0.70 eV?
  • Answer
    Atomic number and the electronic configuration of Nitrogen and Phosphorus are
N    7    1S²2S²2P³

P    15    1S²2S²2P⁶3S²3P³
    Due to the smaller size of Nitrogen atom when an extra electron is added to the stable half - filled 2Porbital some amount of energy is required and hence electron affinity of nitrogen is negative. 
    On the other hand, due to the bigger size of P atom in compare to nitrogen less amount of energy released when the extra electron is added to the stable half - filled 3P orbitals and thus electron affinity of Phosphorus is expressed with a positive sign.
  • Question
    Explain why the electron affinity of chlorine is more than fluorine?
  • Answer
    The halogen possesses large electron affinity values indicating their strong tendency to form anions. This is easily understandable because their electronic configurations are only one electron short of the next noble gas element.
    The electron affinity of chlorine is greater than that of F. This is probably due to the very small size of F atom which results in a very high charge density and consequently strong repulsion between existing valence shell electrons and the entering electron (new coming electron).
    On the other hand, Cl being a bigger size, charge density is small and thus such repulsion is not strong enough. Hence the electron affinity Cl is greater than that of F.
  • Question
    Why electron gain enthalpy of Be and Mg are almost zero?
  • Answer
    Be and Mg have their electron affinity values equal to almost zero. Since Be and Mg have completely filled S orbitals.
Be      4   1S² 2S²

Mg   12   1S² 2S² 2P⁶ 3S²
    The additional electron will be entering the 2P-orbital in the case of Be and 3P-orbitals in the case of Mg which are of considerably higher energy than the 2S- Orbitals respectively.
  • Problem 6:
    Why the electron affinity of noble gases is zero?
  • Answer:
    Inert gases in which the nS and nP orbitals are completely filled (nS² nP⁶ configuration) the incoming electron must go into an electron shall have the larger values for the principal quantum number, n. Thus inert gas has its electron affinity values equal to zero.
    More Questions answer on inorganic chemistry

Slater's Rules are some set of empirical rules to calculate the screening constant (σ) of various electrons present in different orbitals of an atom or an ion. Once we get the value of screening constant it is easy enough to find the effective nuclear charge (Zeff).

Screening Effect and Slater's rule

    The Valence electron in a multi-electron atom is attracted by the nucleus and repelled by the electrons of inner-shells.
what values do you estimate for zeff using slater's rules?
Screening Effect
    The combined effect of this attractive and repulsive force acting on the valence electron is that the Valence - electron experiences less attraction from the nucleus. This is known as the screening effect.

Slater's Rules and its application

    Slater's rules are applicable for calculating screening constant and Zeff for atom, ions or molecules and the rules are

Slater's Rules for an electron in the nS, nP level

  1. The first to do is to write out the Electronic Configuration of Elements of the atom or the ion in the following order and grouping.
    (1S) (2S, 2P) (3S, 3P) (3d) (4S, 4P) (4d) (4f) (5S, 5P) etc. 
    It may be noted that so far as the screening effect is concerned the S and P electrons belonging to the same principal quantum shell have the same effect as advocated by Slater.
    (1S)²(2S, 2P)⁸(3S)¹

  2. An electron in a certain (nS, nP) level is screened only by electrons in the same level and by the electrons of lower energy level.Electrons lying above (nS, nP) level do not screen any (nS, nP) electron to any extent.

  3. Higher energy electrons have no screening effect on any lower energy electrons.
    Thus the screening effect of the valence electron of the Sodium atom, electronic configuration of Na atom according to slater's rule,
    (1S)² (2S, 2P)⁸ (3S)¹
    For calculation of the screening effect, Valence electron will be excluded from our Calculation.
    And for calculation of screening effect of the 2P electron of the Sodium atom, one 2P electron, and 3S electrons will be excluded from our Calculation.

  4. Electrons of an (nS, nP) level shield valence electrons in the same group by 0.35 each. This is also true for the electrons of the nd or nf that is for the electrons in the same group.

  5. Electrons belonging to one lower quantum shell, that is (n-1) shell shield the valence electrons by 0.85 each.

  6. Electrons belonging to (n-2) or still lower quantum shell shields the valence electron by 1.0 each. This means the screening effect is complete.
Slater's Rules
Slater rule for S or P electron
    Zeff of Sodium atom
    Screening Constant (σ) = (2 × 1) + (8 × 0.85) + (0 × 0.35) = 8.8
    ∴ Zeff of Sodium atom = (11 - 8.8) = 2.2

Slater's Rules for an electron in the nd, nf level

    The above rules are quite well for estimating the screening constant of S and P orbitals. However when d or f electron being shielded the (iv) and (v) rule replaced by new rules for estimation of screening constant.
    The replaced rules are, all electrons below the nd or nf level contribute 1.0 each towards the screening constant.
Slater's Rules
Slater rule for d and f  electron
    Zeff for the 4S electron of Vanadium atom
    Vanadium has atomic number 23 and the electronic configuration according to the Slater's Rules is,
    (1S)² (2S, 2P)⁸ (3S, 3P)⁸ (3d)3 (4S)² 
    We have considered only one electron of the two 4S electrons.
    ∴ Screening Constant (σ) = (2 ×1.0) + (8×1.0) + (8×0.85) + (3×0.85) + (1×0.35) = 19.7
    ∴ Zeff for Vanadium atom = (23 - 19.7) = 3.3

Slater's rules examples

  • Slater's rules for Sodium ion
    Electronic configuration of Na⁺ ion (1S)²(2S, 2P)⁸
    Screening Constant (σ) = (2 × 1) + (8 × 0.85) + (0 × 0.35) = 8.8
    Zeff for sodium ion = (11 - 8.8) = 2.2
  • Slater's rules for 2P electron of Sodium
    Electronic configuration of Na⁺¹ ion (1S)²(2S, 2P)⁸(3S)¹
    Screening Constant (σ) = (2 × 0.85) + (7 × 0.85) = 4.15
    Zeff for 2P electron of sodium = (11 - 4.15) = 6.85
  • Slater's rules for Magnesium ion
    Electronic configuration of Mg⁺² ion (1S)²(2S, 2P)⁸
    Screening Constant (σ) = (2 × 0.85) + (8 × 0.85) = 4.50
    Zeff for Magnesium ion = (12 - 4.50) = 7.50
  • Slater's rules for Potassium ion
    Electronic configuration of K⁺¹ ion (1S)²(2S, 2P)⁸(3S, 3P)⁸
    Screening Constant (σ) = (2× 1) + (8 × 0.85) + (8 × 0.35) = 11.6
    Zeff for Pottassium ion = (19 - 11.6) = 7.40
  • Slater's rules for valence electron of Fluorine
    Electronic configuration of K⁺¹ ion (1S)²(2S, 2P)⁷
    Screening Constant (σ) = (2 × 0.85) + (6 × 0.35) = 3.80
    Zeff for for valence electron of fluorine = (9 - 3.8) = 5.20
  • Slater's rules for Fluoride ion
    Electronic configuration of F⁻ ion (1S)²(2S, 2P)⁸
    Screening Constant (σ) = (2 × 0.85) + (8 × 0.35) = 4.50
    Zeff for for valence electron of fluorine = (9 - 4.50) = 4.50
  • Slater's rules for 3d electron of Vanadium
    Electronic configuration of Vanadum (1S)²(2S, 2P)⁸(3S, 3P)⁸(3d)³(4S)²
    Screening Constant (σ) = (2 ×1.0) + (8×1.0) + (8×1.0) + (2 ×0.35) = 18.70
    Zeff for for vanadium = (23 - 18.70) = 4.30
  • Slater's rules for Vanadium(II) ion
    Electronic configuration of Vanadum (1S)²(2S, 2P)⁸(3S, 3P)⁸(3d)³
    Screening Constant (σ) = (2 ×1.0) + (8×1.0) +(8×1.0) + (3×0.35)= 19.05
    Zeff for for vanadium = (23 - 19.05) = 3.95

Slater's Rules Questions Answers

  • Questions
    In the first transition series electron filling up process begins in the 3d level below a filled 4S² level. During ionization process will a 4S electron or a 3d electron be lost first? Explain with reference to chromium.
  • Answer
    Chromium has atomic number 24 and the electronic configuration according to the Slater's Rule is
    (1S)² (2S, 2P)⁸ (3S, 3P)⁸ (3d)⁵ (4S)¹
    ∴ Screening Constant (σ) for the 4S electron of Chromium atom
    = (2×1.0) + (8×1.0) + (8×0.85) + (5×0.85) + (0×0.35) = 21.05
    Thus Zeff for the 4S electron of chromium atom = (24 - 21.05) = 2.95
    ∴ Screening Constant (σ) for the 3d electron of Chromium atom
    = (2×1.0) + (8×1.0) + (8×1.0) + (3×0.85) + (4×0.35) = 19.40
    Thus Zeff for the 4S electron of chromium atom = (24 - 19.40) = 4.60
    Hence 3d electron is more tightly held than a 4S electron. So during ionisation, the 4S electron will be lost.
  • Question
    Calculate the effective nuclear charge of the hydrogen atom.
  • Answer
    The hydrogen atom has a single 1S valence electron. There is no other electron to screen it from the nuclear charge of a single proton.
    Thus, σ = 0 and Zeff = 1.0 - 0 = 1.0
    Thus hydrogen electron sees the full nuclear charge of the nucleus, that is the electron is totally exposed to the proton.
  • Question
    Comment on the variation in effective nuclear charge for a 2P electron from carbon to oxygen.
  • Answer
    Electronic distribution according to the Slater's Rule is:
C (1S)² (2S, 2P)⁴
N (1S)² (2S, 2P)⁵
O (1S)² (2S, 2P)⁶
    In carbon, the 2P electron is screened by 1S² 2S² 2P¹ electrons while in Nitrogen and Oxygen this is done by 1S² 2S² 2P² and 1S² 2S² 2P³ electrons respectively.
Thus Zeff for Nitrogen
= Zeff for carbon + (1 nuclear charge) - Screening due to one 2P electron
= Zeff for Carbon + 1 - 0.35
= Zeff for Carbon + 0.65 and Zeff for Oxygen
= Zeff for Nitrogen + 0.65
    Thus effective nuclear charge will go up by the same amount from carbon to Nitrogen and then to Oxygen.
    For more questions answers on Slater's Rules

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

Mathematical derivation of 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.
  • Mathematical derivation of 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 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).
  • Mathematical derivation of 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 thereactant 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
Examples of zero order kinetics

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 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.
    The Solution of the above question follow the link

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 chemical kinetics questions and answers
Rate laws of zero order chemical kinetics
  • Question
    The rate constant of a chemical reaction is 5× 10⁻⁸  mol lit⁻¹sec⁻¹. Find out 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 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 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

The Half-life of zero order 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 questions

  • 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.

Inorganic Chemistry

[Inorganic chemistry][column1]

Contact Form

Name

Email *

Message *

Powered by Blogger.