February 2019

Law of mass action was first formulated by two Norwegian chemists, Guldberg and Waage in 1864.
The basis of their formulation is the observation of a huge deposit of sodium carbonate on Egyptian take shore. A large amount of NaCl intake water and CaCO₃ on the take shore made the reverse reaction possible.

CaCl₂(aq) + Na₂CO₃(aq) → CaCO₃(s) + 2NaCl(aq)
    The forward reaction occurs spontaneously in the laboratory. They state the law the rate of chemical reaction at a constant temperature is directly proportional to the active mass of the reactants. The active mass is thermodynamic quantity. We assumed active mass as,
  1. Molar concentration (moles/lit) when the solution is dilute, that is when the system behaves ideally.
  2. Partial pressure in the atmosphere unit for the gaseous system and when the pressure of the system is very low.
  3. For pure solid and pure liquid, active mass is assumed to be unity since their mass does not affect the rate of reaction.

Expression of Law of mass action

Let us Consider a reaction,
A + B ⇆ C + D
    Let the reacting system contains reactants only and [A] and [B] are their concentrations in molar units.
    According to the Law of mass action, the rate of the forward reaction,
Rf ∝ [A] × [B]

∴ Rf = Kf × [A] × [B]
    Though the rate equation is an experimental quantity and powers of concentration terms are determined experimentally, here we consider a special type of rate equation where the stoichiometric coefficients are raised in the of the concentration term(this is true for the elementary or one-step reaction).
    As the reaction proceeds in the forward direction, the concentration of A and B decreases and Rf also decreases.
    When the products are getting accumulated in the system, the backward reaction also starts and the rate of the backward reaction according to the law of mass action,
Rb ∝ [C] × [D]
∴ Rb = Kb × [C] × [D]
    Here Kf and Kb are the rate constants of the forward and backward reaction and they do not depend on concentration at a given temperature.
    As the reactions proceed in the forward reactions Rf is decreasing but Rb is increasing. A state is then attained when they are equal. This state is called the chemical equilibrium. There will be no further change in the composition of the system.
Thus at equilibrium,

Rf = Rb


or, Kf × [A] × [B] = Kb × [C] × [D]

∴ Kf/Kb = [C] × [D]/[A] × [B]
    [A], [B], [C] and [D] are the equilibrium concentration of A, B, C and D.
Again, Kf/Kb = Kc
where Kc is the concentration equilibrium constant of the reaction.
    At a given temperature for a reaction, Kc is constant does not depend on the concentration of the reacting components.
Kc = [C] × [D]/[A] × [B]

Equilibrium constant

Concentration equilibrium constant

    If we write the equation as,
Ɣ₁A₁ + Ɣ₂A₂ ⇆ Ɣ₃A₃ + Ɣ₄A₄
Where Ɣ₁, Ɣ₂, Ɣ₃, and Ɣ₄ are stoichiometric coefficients.
∴ Kc = [A₃]Ɣ3 × [A₄]Ɣ4/[A₁]Ɣ1 × [A₂]Ɣ2
    Where Kc is called the concentration equilibrium constant of the reaction and [A₁], [A₂], [A₃], and [A₄] is the equilibrium concentration of A, B, C, and D.
    However, the values of the equilibrium constant of a chemical reaction depend on the mode of writing its stoichiometric (balanced) equation.
    Thus, for the reaction of formation of NH₃ from N₂ and H₂, we can write the equation as,
N₂ + 3 H₂ ⇆ 2 NH₃

The equilibrium constant can be written as Kc = [NH₃]²/[N₂] [H₂]²

But if the equation is written as,
½ N₂ + 3/2 H₂ ⇆ NH₃

Then, K՛c = [NH₃]/[N₂]3/2[H₂]½

It is clear then Kc, and K՛c are not Equal in magnitude.

Thus, Kc = (K՛c)½
    The general rule is that if the equation is multiplied by n then the relation of Equilibrium concentration n will be raised to the power.
Kc = (K՛c)n

Pressure equilibrium constant

    When all the reactants and products are gases (that is, gas-phase reacting system), the expression of the equilibrium constant for the equation,
Ɣ₁A₁ + Ɣ₂A₂ ⇆ Ɣ₃A₃ + Ɣ₄A₄

Where Ɣ₁, Ɣ₂, Ɣ3, and Ɣ₄ are the stoichiometric coefficient
∴ Kp = (P₃Ɣ₃ × P₄Ɣ₄)/(P₁Ɣ₁ × P₂Ɣ₂)
    Where Kp is called the pressure equilibrium constant of the reaction and P₁, P₂, P3, and P₄ are the equilibrium partial pressure of reacting components.

Mole fraction equilibrium constant

If we write the equation as,

Ɣ₁A₁ + Ɣ₂A₂ ⇆ Ɣ₃A₃ + Ɣ₄A₄

where Ɣ₁, Ɣ₂, Ɣ3, and Ɣ₄ are the stoichiometric coefficient
∴ Kx = (x₃Ɣ₃ × x₄Ɣ₄)/(x₁Ɣ₁ × x₂Ɣ₂)
    Where Kx is called a mole fraction equilibrium constant of the reaction and x₁, x₂, x₃, and x₄ are the equilibrium mole fraction of reacting components.
Equilibrium constant from mass action law
Mass action law

Relation between Kp and Kc

The interrelations of these equilibrium constants are as follows,

Kp = (P₃Ɣ₃ × P₄Ɣ₄)/(P₁Ɣ₁ × P₂Ɣ₂)

The ideal gas Equation,

PV = nRT, may be written as,
P = (n/V)RT = CRT
Where C is the concentration of gas expressed as an amount per unit volume.

= {(C₃RT)Ɣ₃ × (C₄RT)Ɣ₄}/{(C₁RT)Ɣ₁ × (C₂RT)Ɣ₂}
∴ Kp = Kc(RT)ΔƔ
ΔƔ = (Ɣ₃ + Ɣ₄) - (Ɣ₁ + Ɣ₂)
  1. For the reaction in which total number of reactant molecules and of resultant molecules are same
  2. H₂ (g) + I₂ (g) ⇆ 2HI
    KP = (PHI)²/{(PH₂)(PI₂)}
    = (CHIRT)²/(CH₂RT) (CI₂RT)
    = [(CHI)²/{(CH₂) (CI₂)}] × [(RT)²/(RT)(RT)]
    = Kc
    Thus when (Ɣ₃ + Ɣ₄) = (Ɣ₁ + Ɣ₂), Kp = Kc
  3. For the reactions in which the number of molecules of reactants differ from that of the resultant
  4. 2SO₂(g) + O₂(g) ⇆ SO₃(g)
    Here, Kp = Kc×(RT){1 - (2+1)}
    = Kc×(RT)⁻²
    Thus when, (Ɣ₃ + Ɣ₄) ≠ (Ɣ₁ + Ɣ₂), Kp ≠ Kc

Equilibrium constant from Mass action law
Law mass action

Relation between Kp and Kx

∴ Kp = Kx(P)ΔƔ
ΔƔ = (Ɣ₃ + Ɣ₄) - (Ɣ₁ + Ɣ₂)

Problems solutions

Problem
    For the dissociation N₂O₄ ⇆ 2 NO₂, obtain an expression for the fraction of original N₂O₄ dissociated at equilibrium in terms of Kp and total pressure.
Solution

The reaction is, N₂O₄ ⇆ 2 NO₂
    Let a mole of N₂O₄ is taken initially and x moles of N₂O₄ is dissociated at equilibrium then mole number of N₂O₄ and NO₂ at equilibrium is (a - x) and 2x. Total moles number at equilibrium = (a -x + 2x) = (a + x). Equilibrium partial pressure of the components is {(a - x)/(a + x)}P and {2x/(a + x)}P where P = Total pressure of the reacting system.
The expression of the equilibrium constant,
Kp = (PNO₂)²/PN₂O₄
= {2x/(a + x)}P/{(a - x)/(a + x)}P
∴ Kp = (4x²P)/(a² - x²)
    The fraction of the original N₂O₄ dissociated at equilibrium ɑ = x/a. Replacing (x/a) by ɑ, we have,
Kp = (4α²P)/(1 - α²)
Problem
    Calculate the Kc value of the reaction N₂ + 3H₂ ⇆ 2NH₃ at 400⁰C, Given Kp at the same temperature 1.64 × 10⁻⁴.
Solution
    Kp = 0.5
Problem
    At 100⁰C the vapor density of N₂O₄ is 25 at 1 atm. Show that Kp = 9.6.
Solution

N₂O₄ (g) ⇆ 2NO₂ (g)
    Let 1 mole of N₂O₄ is taken initially (t = 0) and x mole of N₂O₄ has reacted at equilibrium.
    So the mole number of each component is (1-x) and 2x and total moles at equilibrium,
    (1-x+2x) = (1+x).
    So total moles have increased from 1 to (1+x). Let Volume increase from V₁ to V₂.
So, (1+x) = V₂/V₁
    As density and hence vapor density is inversely proportional to volume so vapor density will decrease from d₁ to d₂.
Hence, (1+x) = V₂/V₁= d₁/d₂.
The molecular weight of N₂O₄ is 92 and vapor density,
= 92/2
= 46

Due to dissociation, it is = 25.
∴ 1+x = 46/25
or, x = 0.84

Now partial pressure are,
PNO₂= {2x/(1+x)}P
= (2×0.84)/1.84
= 0.913 atm
PN₂O₄ = {(1-x)/(1+x)}P
= 0.16/1.84= 0.087

∴ Kp = (PNO₂)2/PN₂O₄
= (0.913)2/0.087

≃ 9.6

    If the thermal energy is much greater than the forces of attraction then we have the matter in a gaseous state.
    Molecules in the gaseous state move very large speeds and the forces of attraction between them are not sufficient to bind the molecules at one place with the result the molecules move practically independent of one another.
    There exist no boundary surface and therefore gases tend to fill completely any available space, that is they do not possess fixed volume.

Boyle’s law

    At constant temperature, the volume of a definite mass of a gas is inversely proportional to its pressure.
    That is the volume of a given quantity of gas, at a constant temperature decreases with the increase of pressure and increases with the decreasing pressure.
    Let, a Cylinder contains 10 ml of gas, at constant temperature and 1 atm pressure. if the pressure increases to 2 atm then the volume also decrees to 5 ml.

Representation of Boyle's law

V ∝ 1/P
∴ PV = K

where, K is a constant whose value depends upon the,
(i)Nature of the gas.
(ii)Mass of the gas.

For a given mass of a gas at a constant temperature,
P₁V₁ = P₂V₂

where V₁ and V₂ are the volumes at pressure P₁ and P₂ respectively.

Graphical representation of Boyle's law

    The relation between pressure and volume can be represented by an arm of a rectangular hyperbola given below.
    As the value of the constant in the equation will change with temperature, there will be a separate curve for each fixed temperature. These curves plotted at different fixed temperatures are called isotherms.
Graphical representation of Boyle's law
Boyle's law

    At constant temperature, a given mass of ideal gases the product of pressure and volume is always the same. If the product of pressure and volume represents in Y-axis and pressure represents X-axis a straight line curve is obtained parallel to X-axis. This graph shows that at a constant temperature the product of pressure and volume does not depend on its pressure.

Pressure density relationship of gases

    At constant temperature(T) a definite mass of gas has pressure P₁ at volume V₁ and pressure P₂ at volume V₂.
According to Boyle’s law,
P₁ V₁ = P₂ V₂
or, P₁/P₂ = V₂/V₁

Again, let the mass of the gas = M
Density D₁ at pressure P₁ and density D₂ at pressure P₂
Thus, D₁ = M/V₁ and D₂ = M/V₂
or, V₁ = M/D₁ and V₂ = M/D₂

Again, P₁/P₂ = (M/D₂) × (D₁/M) = D₁/D₂
or, P₁/P₂ = D₁/D₂
or, P/D = Constant(K)
or, P = K × D

∴ P ∝ D
    Thus, at constant temperature density of a definite mass of a gas is proportional to its pressure.

Volume temperature relationship of gases

    At constant pressure a definite mass of gas, with the increasing temperature, the volume also increases and with decreasing temperature, the volume also decreases. That is, the volume of a given mass of gas at constant pressure is directly proportional to its kelvin temperature.

Charl’s law

    At constant pressure, each degree rise in temperature of a definite mass of a gas, the volume of the gas expands 1/273.5 of its volume at 0⁰C.

Representation of Charl's law

    If V₀ is the volume of the gas at 0⁰C, then 1⁰C rise of temperature the volume of the gas rise V₀/273.5 ml.
∴ 1⁰C temperature the volume of the gas, (V₀ + V₀/273) ml = V₀ (1 + 1/273) ml.
At t⁰C temperature the volume of the gas,
Vt = V₀ (1+ t/273) ml

= V₀ (273 + t°C)/273 ml
    It is convenient to use the absolute temperature scale on which temperature is measured Kelvin(K). Reading on this scale is obtained by adding 273 to the celsius scale value.

The temperature on the Kelvin scale is,
T K = 273 + t°C
∴ Vt = (V₀ × T)/273 = (V₀/273) T
    Since V₀, the volume of a gas at 0°C, has a constant value at a given pressure, the above relation expressed as,
Vt = K₂ T
∴ V ∝ T
    Where K₂ is constant whose value depends on the nature, mass, and pressure of the gases.
    According to the above relation, Charl’s law states as, at constant pressure, the volume of a given mass of gas is directly proportional to its kelvin temperature.

Graphical representation of Charl's law

    A typical variation of volume of gas with a change in its kelvin temperature a straight line plot was obtained, called isobars. The general term isobar, which means at constant pressure, is assigned to these plots.
Properties of gases and gas laws
Char's law

Absolute temperature

    Since volume is directly proportional to its Kelvin temperature, the volume of the gas is theoretically zero at zero Kelvin or - 273⁰C.
    However, this is indeed hypothetical because all gases liquefy and then solidity before this low temperature reached. In fact, no substance exists as a gas at the temperature near kelvin zero, through the straight-line plots can be extra plotted to zero volume.
    The temperature corresponds to zero volume is -273⁰C.

Density temperature relationship of gases

From Charl’s law, V₁/V₂ = T₁/T₂

Again, the mass of the gas is M.

Density D₁ and D₂ at the volume V₁ and V₂ respectively.
Then, V₁ = M/D₁ and V₂ = M/D₂

Thus, (M/D₁ )/(M/D₂ ) = T₁/T₂
or, D₂/D₁ = T₁/T₂
∴ D ∝ 1/T 
    Thus at constant pressure, the density of a given mass of gases is inversely proportional to its temperature.

Combination of Boyle’s and Charl’s law

From Charl's law, V ∝ 1/P when T constant.

From Boyl's law, V ∝ T when P constant.
    When all the variables are taken into account the variation rule states as, 
Then, V ∝ T/P
or, PV/T = K(constant)
∴ (P₁V₁)/T₁ = (P₂V₂)/T₂ = constant

∴ PV = KT
    Thus the product of the pressure and volume of a given mass of gas is proportional to its kelvin temperature.

Questions answers

Problem
    At 1 atm pressure and 300 K temperature, the volume of the gas is 2000 cm³, then calculate the volume of this gas at 600 K temperature and 2 atm pressure.
Answer
From the combination of Boyle's and Charl's law is,
(P₁V₁)/T₁ = (P₂V₂)/T₂
    Here, P₁ = 1 atm; V₁ = 2000 cm³ and T₁ = 300K and P₂ = 2 atm; V₂= ? and T₂ = 600 K
∴ 1×2000/300 = 2×V₂/600
or, V₂ = (1×2000×600)/(300×2)
= 2000 cm³

    Simply one equation can be used to distinguish between real gases vs ideal gases and this is,
PV = nRT
    The gas which obeys this equation under all conditions of temperature and pressure is called ideal gas and the gas which does not obey this equation under all conditions of temperature and pressure is called real gases. A number of points can be discussed to compare the two types of gases.

What is Ideal gas?

  1. The ideal gas cannot be liquefied since it has no inter-molecular attraction and so that molecule will not condense.
  2. The coefficient of thermal expansion(ɑ) depends on the temperature(T) of the gas and does not depends on the nature of the gas.
  3. The coefficient of compressibility(β) similarly depends on the pressure(P) of the gas and will be the same for all gases.
  4. When P is plotted against V, at constant temperature(T) a rectangular hyperbola curve is obtained as demanded by Boyle's Low,
  5. PV = Constant
    The hyperbola Curve at each temperature is called one isotherm and at a different temperature, we have different isotherms. Two isotherms will never intersect.
  6. When PV is plotted against P, at constant T a straight line parallel to P-axis is obtained. At different temperatures, there will be different parallel lines.
Comparison between ideal and real gases
Ideal gas graph

    When an ideal gas passes through a porous plug from higher pressure to lower pressure within the insulated enclosure, there will be no change in the temperature of the gas. This confirms that the ideal gas has no inter-molecular attraction.
Problem
    Show that the coefficient of thermal expansion of ideal gas depends on the temperature of the gas.
Answer
Coefficient of thermal expansion(α) is defined as,
α = (1/V)[dV/dT]P

Ideal gas equation for 1-mole gas is,
PV = RT

Hence [dV/dT]P = R/P
α = (1/V) × (R/P)
= (R/PV)

= 1/T
This means all the gases have the same coefficient of thermal expansion.
Problem
    Show that the coefficient of compressibility of an ideal gas depends on the pressure of the gas.
Answer
Coefficient of Compressibility(β) is defined as,
β = - (1/V)[dV/dP]T

The ideal gas equation for 1-mole gas is,
PV = RT

Hence [dV/dP]T = - (RT/P2)
β = (1/V) × (RT/P2)
= (RT/P2V)
= (RT/PV) × (1/P)

= (1/P)
This means the coefficient of compressibility depends on the pressure(P) of the gas.

What are the real gases?

  1. This gas could be liquefied since it has an intermolecular attraction which helps to coalesce the gas molecules.
  2. The coefficient of thermal expansion (ɑ) is found to vary from gas to gas that is α depends on the nature of the gas.
  3. The coefficient of compressibility (β) also is found to depend on the nature of the gas.
  4. When P is plotted against V, a rectangular hyperbola is obtained only at a high temperature (above the critical temperature).
  5. But a temperature below the critical temperature(C), the gas is liquefied after certain pressure depends on temperature. The point C is the critical point where the liquid and gas can be indistinguishable.
  6. When PV is plotted against P for real gas, the following plots, called Amagat curve are obtained.
  7. When real gases pass through porous plug from higher pressure to lower pressure within the insulated enclosure, there occurs a change of temperature. This is due to the fact that real gases have inter-molecular attraction and when the gas expands, the molecules have to spend energy to overcome inter-molecular attraction and so the temperature of the gas drops down.

Amagat curve

Comparison between Ideal and Real Gases
Amagat curves
    It shows that for most gases, the value of Z decreases attains minimum and then increases with the increase of P.
    Only Hydrogen(H₂) and Helium(He) baffle this trend and the curve rises with the increase of pressure(P) from the very beginning.
    For CO₂, there is a large dip in the beginning, In fact, for gases which are easily liquefied, Z dips sharply below the ideal line in the low-pressure region.

Boyle temperature

    At some intermediate temperature TB called Boyle temperature, the initial slope is zero.
    At the Boyle temperature, the Z versus P line of an ideal gas is tangent to that of a real gas when P approaches zero. The latter rises above the ideal gas line only very slowly.
    Thus, at the Boyle temperature, the real gas behaves ideally over a wide range of pressure, because the effect of the size of molecules and intermolecular forces roughly compensate each other. Boyle Temperature(TB) is given by,
[d(PV)/dP]T when P → 0
    The Boyle temperature of some gases are given below
Gases TB
Hydrogen (H₂) -156⁰ C
Helium (He) -249⁰ C
Nitrogen (N₂) 59⁰ C
Methane (CH₄) 224⁰ C
Ammonia (NH₃) 587⁰ C
    Thus we can see that for H₂ and He, the temperature of 0⁰C is above their respective Boyle temperature and so they have Z values greater than unity.
    The other gases at 0⁰C are below their respective Boyle temperature and so hay has Z values less than unity in the low-pressure region.

Compressibility factor

    An important single parameter called the compressibility factor (Z) is used to measure the extent of deviation of the real gases from ideal behavior. It is defined as,
Z = PV/RT
    When Z=1, the gas is ideal or there is no deviation from ideal behavior.
    When Z ≠ 1, the gas is non-ideal and the departure of the value of Z from unity is a measure of the extent of non-ideality of the gas.
    When Zく1, the gas is more compressible then ideal gas and when Z 〉1, the gas is a less compressible then ideal gas.
Problem
    At 273 K and under pressure of 100 atm the compressibility factor of O₂ is 0.97. Calculate the mass of O₂ necessary to fill a gas cylinder of 108.5 lit capacity under the given conditions.
Answer
    Mass of O₂ = 1600 gm = 1.6 Kg

The wavelength of the hydrogen spectrum

The wavelength of the hydrogen spectrum was determined experimentally and it was possible to relate the frequencies of different spectral lines by a simple equation known as the Rydberg equation.
frequency and wavelength of the hydrogen spectrum
Wavelength of the hydrogen spectrum

Energy of an electron in the hydrogen atom

The explanation of the Rydberg equation is now simple. The most stable orbit must be the one with the lowest energy (ground state).

When n = 1 the energy of an electron is lowest and it the lowest energy state and the states with n 〉 1 are the higher energy excited states.

Provided the right amount of energy is supplied the electron may jump from n = 1 to another higher level. Conversely energy will be released in the form of light of definite frequency when the excited electron returns to its ground state.

Since the energy and frequency of the emitted light are connected by Plank relation,
ΔE = hν
or, ν = ΔE/h
The energy corresponding to a particular line in the emission spectrum of a hydrogen atom is the energy difference between the initial state 1 and the final state 2.

From the Bohr's model of hydrogen atom energy of an electron at a particular energy level.
En = - 2π²me⁴/n²h²

So that, ΔE= E₂ - E₁
= - (2π²me⁴/n²h²) - [- (2π²me⁴/n²h²)]
= (2π²me⁴/h²)[1/n² - 1/n²]

Then ν corresponding to the energy E.
ν = ΔE/h
= (2π²me⁴/h³)[(1/n²) - (1/n²)]

= R[(1/n²) - (1/n²)]
Where R is the Rydberg constant.
This relation tells us that as we go to higher and still higher orbits the energy gap between two immediate orbits continues to decreases.

Wavenumber and frequency of the hydrogen spectrum

Using the mass of an electron =9.108 × 10⁻²⁸ gm, the Bohr model predicts the following value of the Rydberg constant.

R = 2π²me⁴/h³
= {2 × (3.1416)² × (9.108 × 10⁻²⁸ gm) × (4.8 × 10⁻¹⁰ esu)4}/(6.627 × 10⁻²⁷ erg sec)³
= 3.2898 × 10¹⁵ cycles sec⁻¹.

Evaluate R in terms of wavenumber (⊽) instead of frequency. Wavenumber and frequency are connected with,

λν = c and ν = c/λ = c⊽
Thus, ⊽ = ν/c
where c is the velocity of light.

Recall the frequency (ν) indicates the number of waves passing a given point per second and is expresses as cycles per second. Wavenumber (⊽) on the other hand stands for the number of waves connected in unit length that is per centimeter (cm⁻¹) or per meter (m⁻¹).

Thus, R = (3.2898 × 1015 sec⁻¹)/(2.9979 × 1010 cm sec⁻¹).
= 109737 cm⁻¹
= 10973700 m⁻¹

The experimental values of R are 109677 cm⁻¹ (10967700 m⁻¹) showing a remarkable agreement between the experiment and Bohr's model.

Spectral series of the hydrogen atom

Putting n = 1, n = 2, n = 3, etc in the Rydberg equation we get the energies of the different stationary states for the hydrogen electron.

The transition energies can be calculated from the Rydberg equation. The experimental hydrogen spectrum of the atom also exhibited several series of lines.

Question
Why a large number of lines appear in the hydrogen spectrum although it contains only one electron?

Answer
This is because any given sample of hydrogen contains an almost infinite number of atoms.
Under normal conditions, the electron of each hydrogen atom remains in the ground state near the nucleus that is n = 1 (K - Shell).

When energy is supplied to this sample of hydrogen gas by passing an electron discharge through the gas in the discharge tube, individual atom absorbed different amounts of energy.

Some of atoms absorbed such energy to shift their electron to third energy level (n = 3) while some others may absorb a large amount of energy to shift their electron to the fourth (n = 4), fifth (n = 5), sixth (n = 6) and seventh (n = 7) energy level.

The electrons in the higher energy level are relatively unstable and hence drop back to the lower energy level with the emission of energy in the form of line spectrum containing various lines in of particular frequency and wavelength.

What is the spectrum of a hydrogen atom?

Emission hydrogen spectrum
Emission hydrogen spectrum

Lyman series of the hydrogen atom

Transition to the ground state (n = 1) from n = 2, n = 3, n = 4, etc exited states constitute the Lyman series in the ultraviolet region. The wavelength of these spectral lines can be calculated by the following equation.

⊽ = 1/λ = R[(1/n²) - (1/n²)]
For Lyman series, we have n = 1 and n = 2, 3, 4...

When n = 2,
1/λ = 109677{1 - (1/4)}
or, 1/λ = {(109677 × 3)/4} cm⁻¹
∴ λ = {4/(109677 × 3)}
= 1215 × 10⁻⁸ cm
= 1215 Å.

When n = ∞,
1/λ = 109677{1 - (1/∞2)}
or, 1/λ = 109677 cm⁻¹
∴ λ = (1/(109677)
= 912 × 10⁻⁸ cm
= 912 Å.

Question
The wavenumber of the get spectrum of hydrogen in the Lyman series is 82200 cm⁻¹. Show that the transition occurs to the ground-state (n =1) from n = 2. (R = 109600 cm⁻¹).

Answer
We know that wavenumber,⊽ = 1/λ = R[(1/n²) - (1/n²)]
For Lyman series n = 1 and wave Number (⊽) = 82200 cm⁻¹
Thus, 82200 = 109600[(1/1²) - (1/n²)]
or, 1/n22 = 1 - (822/1096)
= 274/1096
=1/4
∴ n = 2.

Balmer series of the hydrogen atom

Transition to the n = 2 level from n = 3, n = 4, n = 5, etc exited states constitute the Balmer series in the visible region. The wavelength of these spectral lines can be calculated by the following equation.

⊽ = 1/λ = R[(1/n²) - (1/n²)]
For Balmer series, we have n = 2 and n = 3, 4, 5, ...

Question
Which of the Balmer lines fall in the visible region of the spectrum wavelength 4000 to 7000 Å?(Given R = 109737 cm⁻¹).

Answer
The Balmer series of the hydrogen spectrum get comprises the transition from n 〉2 levels to the n = 2 level.
4000 Å = 4000 × 10⁻⁸ cm = 4 × 10⁻⁵ cm and 7000 Å = 7 × 10⁻⁵ cm. Therefore wave number, ⊽ = 1/λ = (1/4) × 10⁵ to (1/7) × 10⁵ cm⁻¹.

Transition energy of the Balmer lines, ⊽ = 1/λ = 109677[(1/2²) - (1/n²)]
Taking (⊽) = (1/4) × 10⁵ cm⁻¹
We have,(1/4) × 10⁵ cm⁻¹ = 109737{(1/4) - (1/n²)}
= 1.1 × 10⁵{(1/4) - (1/n²)} cm⁻¹
∴ 1/4 ⋍ 1.1{(1/4) - (1/n²)}
or, n² ⋍ 44
So n = 7 ( nearest whole number).

Taking (⊽) = (1/7) × 10⁵ cm⁻¹
We have,(1/7) × 10⁵ cm⁻¹ = 109737{(1/4) - (1/n²)}
= 1.1 × 10⁵{(1/4) - (1/n²)} cm⁻¹
∴ 1/7 ⋍ 1.1{(1/4) - (1/n²)}
or, n² ⋍ 9
So n = 3 ( nearest whole number).

So all transition from n =7 to n = 3 while falling to n = 2 will generate the Balmer series.

Question
Calculate the wavelength of Hɑ and Hβ of the Balmer series.

Answer
Transition energy of the Balmer lines, ⊽ = 1/λ = 109677[(1/2²) - (1/n²)]
Again for Hɑ line n2 = 3 and for Hβ line n2 = 4
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
= 6564 Å.

Again the wavelength for Hβ - line,
1/λ = 109677 [(1/2²) - (1/4²)]
= 109677 × (3/16)
∴ λ = 16/(3 × 109677)
= 4.863 × 10⁻⁵ cm
= 4863 Å.

Paschen series spectrum

Transition to the n = 3 level from n = 4, n = 5, n = 6, etc exited states constitute the Paschen series in the near-infrared region. The wavelength of these spectral lines can be calculated by the following equation.

⊽ = 1/λ = R[(1/n²) - (1/n²)]

For Paschen series, we have n = 3 and n = 4, 5, 6, ...

Bracket series spectrum

Transition to the n = 4 level from n = 5, n = 6, n = 7, etc exited states constitute the Bracket series in the far-infrared region. The wavelength of these spectral lines can be calculated by the following equation.

⊽ = 1/λ = R[(1/n²) - (1/n²)]

For Bracket series, we have n = 4 and n = 5, 6, 7, ...

Pfund series spectrum

Transition to the n = 5 level from n = 6, n = 7, n = 8, etc exited states constitute the Pfund series in the far-infrared region. The wavelength of these spectral lines can be calculated by the following equation,

⊽ = 1/λ = R[(1/n²) - (1/n²)]
For Pfund series, we have n = 5 and n = 6, 7, 8, ...

Elementary particles of an atom
Elementary particles of an atom

History of an atom dalton model

  1. All the matter is made of atoms are indivisible and indestructible.
  2. All the atoms of a given element are identical mass and properties.
  3. Compounds are formed by the combination of two or more same or different kinds of atoms.
  4. A chemical reaction is a rearrangement of atoms.

History of the atomic particle

Rutherford has remarked that it is not in the nature of things for any one man to make the sudden violet discovery.
Science goes step by step and every man depends on the work of his predecessor.

The journey from Dalton's model of the atom to the modern structure of the atom was a long and arduous one. At the turn, this century much valuable information was being compiled.

This clearly indicates that Dalton's atomic theory no longer enjoyed the exalted position to grant it. Today an atom is considered to made up of a tiny nucleus carrying neutrons and protons.

This tiny nucleus has around itself a certain number of the negatively charged elementary particles carrying negligible mass, called electrons, arranged in a definite order.

Who discovers the electron of an atom?

Gases at low pressures, when subjected to high potential, becomes conducting and the various luminous effect was observed. When the pressure is quite low(0.01 mm), the tube remains dark (Crooks dark space) but a streak of rays, named cathode rays, emanating from the cathode.

This is confirmed by the fact that fluorescence is produced on the opposite wall where the rays impinge. The cathode rays have been very carefully studied for their many definite characteristics.
Elementary particles of an atom
Cathode Rays Experiment

Characteristic of cathode rays of an atom

  1. Cathode rays excite fluorescence on the glass walls where they impinge.
  2. The rays have traveled in straight lines, confirmed by the shadows of objects placed in their path. 
  3. The rays have penetrating power and can pass through thin metal foils. 
  4. They also possess considerable momentum, small paddle wheels placed in their path rotate from the impact with the rays.
  5. The cathode rays are deflected from their path by the application of a magnetic or electrostatic field. From the direction of deflection, the charge accompanying the rays is a negative one.
  6. When the rays impinge on a metal target, called anticathode, and placed on the path, a different type of radiation, the X-rays are produced. this new radiation not deflected in an electric or magnetic field. X-rays are really electromagnetic radiation of very short wavelength.

Charge of an electron of an atom

The electron carrying negatively charged and the charge of an electron(e),
4.8 × 10⁻¹⁰ esu
= 1.60 × 10⁻¹⁹ coulombs

Mass of an electron of an atom

Let the mass of an electron = m and charge = e, then e/m = 1.76 × 10⁸ Coulomb/gram.
∴ Mass of an electron, = (1.60 × 10⁻¹⁹)/(1.76 × 10⁸) gram
= 9.11 × 10⁻²⁸ gram

How to measure the electric charge?

The electrodeposition of silver from an aqueous solution of silver salt is a suitable experiment for the determination of the electronic charge.

Faraday's Low are readily interpreted by reference to the electrolysis of silver nitrate. The change at the cathode requires one electron for every Silver ion reduced.
Ag⁺ + e → Ag
If the electrons consumed at this electrode is equal to Avogadro number (6.023 × 10²³ mol⁻¹), 1 mole of Silver metal (107.9 gm) is produced. At the same time, 1 mole of electrons is removed from the anode and 1 mole of nitrate ions is discharged.

Thus, 96500 coulomb of electricity which is necessary to produce 1 equivalent mass of a substance at the electrode will be total charge carried by 1 mole of electrons.

Hence Charge carried by each electron is given by,
e = (96500 C mol⁻¹)/(6.023 × 10²³ mol⁻¹)
= 1.60 × 10⁻¹⁹ C

Who discovers proton of an atom?

Since the electrons contribute negligibly to the total mass of the atom and the atom is electrically neutral. That the nucleus must carry atomic particles which will account both for the mass and positive charge of the atom.

We have so far deliberately restricted the discussion on the discharge phenomena at low gas pressure. The production of cathode rays in discharge tubes inspired physicists to look for the oppositely charged ions, namely positive ions.

Goldstein added a new feature to the discharge tubes by using holes in the cathode. With this modification, it is observed that on operating such discharge tubes there appeared not only cathode rays traveling from the cathode to anode but also a beam of positively charged ions traveling from around anode to cathode.

Some of the positively charged atomic particles passed through the hole in the cathode and produce a spot on the far end of the discharge tube.
Elementary particles of an atom
Goldstein Experiment
The nature of these positive rays is extensively investigated by Thomson. It proved much more difficult to analyze the beam of the positive rays than to analyze a beam of electrons.

On deflection by a magnetic and electric field, the positive ray beam produced a large defuse spot indicating that the e/m ratio of the constituents of the beam was not the same and that the atomic particles moved with different velocities.

Thomson further demonstrated that each different gas placed in the apparatus gave a different assortment of e/m. Since an H⁺ is produced from a hydrogen atom by the loss of one electron, which has but a negligible mass, it follows that the mass of an H⁺ is the same as that of the hydrogen atom (=1).

The particle represented by H⁺ is called a Proton and is considered an elementary particle that accounts for the positive charge of the nucleus.

Charge of a proton of an atom

The Proton carrying Positively charged and the charge of a proton(p),
4.8 × 10⁻¹⁰ esu = 1.60 × 10⁻¹⁹ coulomb

Mass of a proton of an atom

Let the mass of an Proton = m and charge = e,then e/m = 9.3 × 10⁴ Coulomb/gram.
∴ Mass of a Proton = 1.6725 × 10⁻²⁴ gm

Who discovers the neutron of an atom?

Attempts were now directed towards a correlation of atomic mass number ( = integer nearest to the atomic weight) and nuclear charge (= atomic number).

If A stands for the mass number and Z for the nuclear charge of an element, then Z units of nuclear charge means Z number of a proton inside the nucleus. But Z protons can contribute only Z mass units. The shortfall of the (A - Z) mass units bothered chemists and physicists for the quite same time.

Rutherford then suggested this shortfall must be made up by another elementary particle. This elementary particle has electrically neutral, and mass equal to that of the proton, namely 1.
He named this elementary particle in advance as a neutron. The glory of discovering the neutron went to Chadwick, one of Rutherford students.

An interpretation of the atomic nuclei on the basis of neutrons and protons is now a simple affair. Taking oxygen of mass number 16, for example, and recalling that the atomic number of the element is 8, we have an atomic nucleus composed of 8 protons and 8 neutrons.

Since neutrons contribute only to the mass of the element but do nothing towards charge it follows that there may exist species with the same number of protons but varying numbers of neutrons inside the nucleus. Such species must belong to the same element and must vary only in their mass numbers. They are called isotopes.

Thus ₁H¹, ₁H², ₁H³ are three isotopes of hydrogen, and ₈O¹⁶, ₈O¹⁷, and ₈O¹⁸ are three isotopes of oxygen.

Deflection of the alpha particle

Rutherford and his students describe some alpha particle scattering experiments. The alpha particles were already established by the group He⁺² from their behavior in electric and magnetic fields.

A beam of alpha particles obtained from spontaneously disintegrating polonium was directed on to a very thin foil of platinum or gold.

With the help of fluorescent zinc sulfide screen around the platinum or gold foil, any deflection of the alpha particle was observed.
Rutherford nucleus of an atom
Nucleus of an atom
The vast majority of the alpha particles passed the straight line through the foil. But a very limited few were found to be deflected back from the foil, some even appearing on the side of incidence.

Rutherford concluded that since most of the alpha particles passed straight through there must be a very large volume of empty space in the atom of the platinum or gold.

A very small part of the platinum or gold atom must be responsible for the large scattering of the few alpha particles, the central part was called the atomic nucleus.

Rutherford experiment nucleus of an atom

Conclusions and an atomic model emerged from Rutherford's experiment.
  1. All the positively charged and almost entirely mass of the atom was concentrated in a very small part of the atom and these central core called the atomic nucleus.
  2. The large deflection of an alpha particle from its original path was due to Coulombic repulsion between the alpha particle and the positive nucleus of an atom. The simple impact between the two such massive particles can lead to a scattering of the order of only 10.
  3. An alpha particle suffers little deflection while passing by an electron.
  4. The Radius of the atomic nucleus is ∼ 10-13 being the same as that of an electron. Since the radius of an atom is ∼ 10-8 it is obvious that an atom must have a very empty structure. From the above conclusions, an atomic structure is proposed by Rutherford.

Rutherford's atomic model

According to Rutherford's model, the entire mass of the atom was concentrated in a tiny, positively charged nucleus around which the extranuclear electrons were moving in a circular orbital of an atom.
Rutherford divided the atom into two-part,
  1. The nucleus of an atom
    Almost the entire mass of the atom is concentrated in a very small, central core called the Atomic nucleus.
    Since the extranuclear electrons contribute negligibly to the total mass of the atom and since the atom is electrically neutral it follows that the nucleus must carry particles which will account both for the mass and positive charge of the atom.
  2. The extranuclear electrons
    A very small positive nucleus was considered surrounded by electrons. Such a system cannot be stable if the election were in rest.
    Therefore it is proposed that the electron moving in circular orbits around the nucleus so that the Coulombic attraction between the nucleus and the electron was equal to the centrifugal force of attraction.

Defects of Rutherford's model

  1. The Rutherford model is, however, not in conformity with the classical model of electromagnetic radiation. A moving charged particle will emit radiation, will then loss kinetic energy and eventually will hit the nucleus.
  2. If the electrons lose energy continuously, the observed atomic spectra should be continuous, consisting of broad bands merging one into the other.

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