## February 2019

### Study the law of mass action for college courses

The law of mass action was first formulated by two Norwegian chemists, Guldberg and Waage in 1864. The study of the law of mass action was formulated by the observation of a huge deposit of sodium carbonate on Egyptian take shore.

A large amount of sodium chloride intake water and calcium carbonate on the take shore made the reverse reaction possible.

CaCl₂(aq) + Na₂CO₃(aq) → CaCO₃(s) + 2NaCl(aq)

They recognized that chemical equilibrium is a dynamic and not static equilibrium. The fact that the forward rate and reverse rate of reactions become the same.

The rate of a chemical reaction is proportional to the product of effective concentrations of reacting species, each raised to a power that is equal to the corresponding stoichiometric number of the substance in the balanced chemical equation.

The effective concentration is a thermodynamic quantity and it represents as,
• Molar concentration when the solution is dilute and the system behaves ideally. The unit of molar concentration moles/lit.
• Partial pressure in the atmosphere unit for the gaseous system and when the gas pressure of the system very low.
• Pure solid and pure liquid, active mass is assumed to be unity since their mass does not affect the rate of reaction.

#### Forward and reverse reaction rates at equilibrium

Let us apply the law of mass action to a general chemical reaction.

A + B ⇆ C + D

As the reaction proceeds in the forward direction, the concentration of reactants decreases and the forward reaction rate also decreases.

When the products are getting accumulated in the system, the reverse reaction also starts. Thus according to the law of mass action the forward and reverse reaction rates

Rf ∝ [A] × [B]
∴ Rf = Kf × [A] × [B].

Rb ∝ [C] × [D]
∴ Rb = Kb × [C] × [D].

Where [A], [B and [C], [D] are the concentration of forward and reverse reaction respectively in molar units. Kf and Kb are the rate constants of the forward and reverse reaction and they do not depend on concentration at a given temperature.

As the reactions proceed in the forward reaction rate decreasing but reverse reaction rate increasing and attained a state when they are equal.

The state where forward and reverse reaction rates are equal is called the chemical equilibrium. There will be no further change in the composition of the system.

At equilibrium of a chemical reaction,
Rf = Rb

or, Kf × [A] × [B] = Kb × [C] × [D]

∴ Kf/Kb = [C] × [D]/[A] × [B]

whare [A], [B], [C] and [D] are the equilibrium concentration of reacants and product.

Kf/Kb = Kc
where Kc = concentration equilibrium constant of a chemical reaction.

At a given temperature Kc is constant does not depend on the concentration of the reacting components.

Kc = [C] × [D]/[A] × [B]

#### How does concentration affect the rate of reaction?

Let us consider a general chemical equation

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 the reactants and product.

The values of the equilibrium constant of a chemical reaction depend on the mode of writing its stoichiometric balanced chemical equation.

#### The chemical equation for the formation of ammonia

The chemical reaction for the formation of ammonia from nitrogen and hydrogen, we can write the equation

N₂ + 3 H₂ ⇆ 2 NH₃.
The concentration equilibrium constant = Kc.
∴ Kc = [NH₃]²/[N₂] [H₂]².

If we write the chemical reaction
½ N₂ + 3/2 H₂ ⇆ NH₃
The contraction equilibrium constant = K՛c.
∴ K՛c = [NH₃]/[N₂]3/2[H₂]½

From the above studies, it is clear then Kc, and K՛c are not equal in magnitude and concentration equilibrium constant depends on the modes of writing a balanced chemical equation.

∴ Kc = (K՛c)½

If a chemical equation multiplied by n then the general rule for writing equilibrium concentration n will be raised to the power.
Kc = (K՛c)n
Online college chemistry courses

#### How do you find partial pressure at equilibrium?

When all the reactants and products are the gas-phase reacting system, the concentration expressed as partial pressure. Let the chemical equation for the above expression
Ɣ₁A₁ + Ɣ₂A₂ ⇆ Ɣ₃A₃ + Ɣ₄A₄

where Ɣ₁, Ɣ₂, Ɣ3, and Ɣ₄ are the stoichiometric coefficients.

∴ 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.

Problem
For the dissociation of N₂O₄ ⇆ 2NO₂, how to calculate the fraction of original N₂O₄ dissociated at equilibrium in terms of Kp and total pressure?

Solution

N₂O₄ ⇆ 2 NO₂

Let one mole of N₂O₄ taken initially and x moles of N₂O₄ dissociated at equilibrium then mole number of N₂O₄ and NO₂ at equilibrium (a - x) and 2x.

Total moles number at equilibrium = (a -x + 2x) = (a + x). Equilibrium partial pressure of the components {(a - x)/(a + x)}P and {2x/(a + x)}P where P = total pressure of the reacting system.

∴ Kp = (PNO₂)²/PN₂O₄
= {2x/(a + x)}P/{(a - x)/(a + x)}P
= (4x²P)/(a² - x²)

The fraction of the original N₂O₄ dissociated at equilibrium ɑ = x/a.
∴ Kp = (4α²P)/(1 - α²)

#### Mole fraction equilibrium of a chemical reaction

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. Law of mass action

#### The relation between Kp and Kc in equilibrium

The interrelations between Kp and Kc can be expressed for the chemical reaction in equilibrium

Ɣ₁A₁ + Ɣ₂A₂ ⇆ Ɣ₃A₃ + Ɣ₄A₄

∴ Kp = (P₃Ɣ₃ × P₄Ɣ₄)/(P₁Ɣ₁ × P₂Ɣ₂)

The ideal gas law for n moles ideal gas
PV = nRT
or, P = (n/V)RT = CRT
where C = concentration of gas moles per unit volume.

∴ Kp = {(C₃RT)Ɣ₃ × (C₄RT)Ɣ₄}/{(C₁RT)Ɣ₁ × (C₂RT)Ɣ₂}

 Kp = Kc(RT)ΔƔ ΔƔ = (Ɣ₃ + Ɣ₄) - (Ɣ₁ + Ɣ₂)
1. A chemical reaction in which the total number of reactant molecules and of resultant molecules are the same.
2. For the reactions in which the number of molecules of reactants differs from the resultant molecules of the chemical reaction. Chemical reactions at equilibrium
Problem
How to calculate Kp form Kc for the chemical reaction, N₂ + 3H₂ ⇆ 2NH₃ at 400⁰C?
Kc at 400°C = 1.64 × 10⁻⁴.

Solution
Kp = 0.5

#### The relation between Kp and Kx in chemical equilibrium

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

The above study represents some chemical equations and chemical equilibrium and the existence of an equilibrium constant.

The rate of some reversible chemical reactions before the thermodynamic approach developed by Law of mass action.

### Study properties of gases

If the thermal energy is much greater than the forces of attraction then we have the matter in a gaseous state.

Gas molecules move very large speeds and the forces of attraction between them are not sufficient to bind the gas molecules at one place with the result the molecules move practically independent of one another.

There exist no boundary surface and therefore gas molecules tend to fill completely any available space, that is they do not possess fixed volume.

### Volume measurement or 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.

The 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.

#### Mathematical presentation of Boyle's law

V ∝ 1/P
∴ PV = K

where K = constant of gas.

The value of gas constant K depends on the nature of the gas and mass of the gas molecules.

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 presentation of Boyle's law

The relation between pressure and volume can be presented 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 presentation of gases
At constant temperature, a given mass of ideal gas 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.

#### Density measurement for gases

At constant temperature(T) a definite mass of gas has pressure P₁ at volume V₁ and pressure P₂ at volume V₂.

Boyle’s law for gases,
P₁ V₁ = P₂ V₂
or, P₁/P₂ = V₂/V₁.

Let the mass of the gas = M and
density = D₁ at pressure P₁
density = D₂ at pressure P₂.

∴ D₁ = M/V₁ and D₂ = M/V₂
or, V₁ = M/D₁ and V₂ = M/D₂

P₁/P₂ = (M/D₂) × (D₁/M) = D₁/D₂
or, P₁/P₂ = D₁/D₂
or, P/D = constant

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

### Temperature measurement 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.

#### Charles law for gas molecules

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.

#### Mathematical presentation of Charles 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, Charles law states as, at constant pressure, the volume of a given mass of gas is directly proportional to its kelvin temperature.

#### Graphical presentation of Charles 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. Graphical presentation of Charles law

#### Kelvin scale for gas molecules

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 = -273⁰C.

#### Density measurement from Charles law

Charles law, V₁/V₂ = T₁/T₂

Mass of the gas = M.

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

(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.

#### Charles law and Boyle's law for gas molecules

Charles law, V ∝ 1/P when T constant.

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

∴ PV = KT

The product of the pressure and volume of a given mass of gas is proportional to its kelvin temperature.

### Study ideal and real gases in chemistry

A comparison of ideal and real gases can be described by the ideal gas law for gas molecules. Ideal gas law for 1-mole gas

PV = nRT.

The gas which obeys this law under all conditions of temperature and pressure is called ideal gas and the gas which does not obey this law under all conditions of temperature and pressure is called real gases.

A gas, really hypothetical one, which follows the ideal gas law rigorously under all circumstances, has been named as ideal gas or perfect gas as distinct properties of gases.

### Ideal gas or perfect gas molecules

1. An ideal gas can not be liquefied because ideal gas molecules have no inter-molecular attraction. The gas molecules will not condense.
2. The coefficient of thermal expansion(ɑ) depends on the temperature of the gas and does not depends on the nature of the gas.
3. The coefficient of compressibility(β) similarly depends on the pressure of the gas and will be the same for all gases.
4. When pressure is plotted against volume at a constant temperature a rectangular hyperbola curve obtained.
5. The hyperbola curve at each temperature called one isotherm and at a different temperature, we have different isotherms. Two isotherms will never intersect.
6. When PV is plotted against pressure at a constant temperature a straight line plot obtained parallel. At different temperatures, there will be different parallel lines obtained.
7. 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. PV graph for an ideal gas

#### Thermal expansion for ideal gas molecules

Coefficient of thermal expansion(α) of a gas
α = (1/V)[dV/dT]P.

Ideal gas law for 1-mole gas
PV = RT.

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

This means thermal expansion will be independent of the nature of the gas and will be a function of temperature only. The values of thermal expansion for different gases are found to be different.

The coefficient of thermal expansion for hydrogen and carbon dioxide 2.78 × 10⁻⁷ and 3.49 × 10⁻⁷ respectively at 0°C and 500 atmospheres.

#### Coefficient of compressibility of an ideal gas

Coefficient of compressibility defined
β = - (1/V)[dV/dP]T

Ideal gas law for 1-mole gas
PV = RT.

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

= (1/P)

This means β should be a function of pressure only and should be the same for all gases. Experimentally the coefficient of compressibility has been found to be individual property.

### What are the real gases?

1. Real gas could be liquefied because gas molecules have an intermolecular attraction which helps to coalesce the gas molecules.
2. Thermal expansion (ɑ) found to vary from gas to gas. The coefficient of thermal expansion 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 pressure plotted against volume a rectangular hyperbola curve obtained only at a high temperature above the critical temperature.
5. But a temperature below the critical temperature(C), the gas can be liquefied after certain pressure depends on temperature. Liquid and gas can be indistinguishable in the critical point of the gases.
6. When PV is plotted against pressure for real or Van der Waals gases Amagat curve obtained.
7. Real gases pass through porous plug from higher pressure to lower pressure within the insulated enclosure, there occurs a change of temperature.
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.

#### Z vs P graph for real gases Z vs P graph for real gases
Most gases, the value of Z decreases attains minimum and then increases with the increased pressure of the gas.

Hydrogen and helium gas-only baffle this trend and the curve rise with the increased pressure of the gas from the very beginning.

Carbon dioxide gas can be easily liquified and Z dips sharply below the ideal gas line in the low-pressure region.

TB called Boyle temperature, the initial slope at TB zero. At TB, the Z vs P line of a gas tangent to that of a real gas when pressure approaches zero. Latter rises above the ideal gas line only very slowly.

Thus, at TB real gas behaves ideally over a wide range of pressure, because the effect of the size of gas molecules and intermolecular forces roughly compensate each other.

#### TB for hydrogen, helium, nitrogen, methane, ammonia

 Gases TB Hydrogen (H₂) -156⁰C Helium (He) -249⁰C Nitrogen (N₂) 59⁰C Methane (CH₄) 224⁰C Ammonia (NH₃) 587⁰C
For hydrogen and helium, TB lowers then 0⁰C temperature so Z values greater than unity.
For nitrogen, methane, and ammonia TB greater then 0⁰C so Z values less than unity in the low-pressure region.

#### Compressibility factor for gases

An important single parameter called the compressibility factor used to measure the extent of deviation of the real gases from ideal behavior.

Z = PV/RT

1. Z=1, the gas is ideal gas or there is no deviation from ideal behavior.
2. 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.
3. When Zく1, the gas is more compressible then ideal gas and when Z 〉1, the gas has less compressible then ideal gas.

Problem
Z for oxygen in 273 K temperature and 100 atm pressure 0.97. Calculate the weight of this oxygen gas necessary to fill a gas cylinder of capacity 108.5 liters.

Weight of oxygen = 1600 gram = 1.6 kilogram

### Hydrogen emission spectrum energy levels

Measurements in the emission hydrogen spectrum are feasible because each hydrogen atom has definite quantum energy in which electron resides. Ordinarily, they are in the ground state with the lowest energy level of an atom.

On the addition of energy by thermal or electrical one electron is removed to the higher energy level of the hydrogen atom.

However, this exited electron returns to the ground state and in this process extra energy emitted in the form of a photon.

The source of excitation influence the form and intensity of emission in such measurement. Through we get a line spectra of the hydrogen atom for spectroscopic study.

Photon energy and frequency of the emitted light from the hydrogen spectrum are connected by Plank relation.

ΔE = hν
or, ν = ΔE/h

#### Hydrogen emission spectrum wavelengths

The energy corresponding to a particular line in the emission spectrum of a hydrogen atom is the energy difference between the ground level and the exited level.

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

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

Frequency of emitted light from the spectrum related to energy by plank equation
ν = ΔE/h
= (2π²me⁴/h³)[(1/n²) - (1/n²)].

∴ ν = R[(1/n²) - (1/n²)]
where R is the Rydberg constant.

This is the Rydberg equation for the measurement of the emission spectral region of the hydrogen atom. Wavelength of the hydrogen spectrum
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.

λν = c and ν = c/λ = c⊽
Wavenumber, ⊽ = 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).

∴ 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.

### What is the emission spectrum of hydrogen?

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 exhibited several series of lines.

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

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 thermal or electrical 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.

Online college chemistry courses

### What is the spectrum region of a hydrogen atom? Emission hydrogen spectrum region

#### Lyman series of the hydrogen spectrum

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).

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 emission spectrum

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)]
n = 2 and n = 3, 4, 5, ...

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

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.

⊽ = 1/λ = 109677[(1/2²) - (1/n²)]
(⊽) = (1/4) × 10⁵ cm⁻¹.

∴ (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.
∴ n = 7 ( nearest whole number).

(⊽) = (1/7) × 10⁵ cm⁻¹
∴ (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.
n = 3 ( nearest whole number).

The 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.

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 lies in which region?

Transition to the n = 3 level from n = 4, n = 5, n = 6, etc exited states constitute the Paschen series lies 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)]

n = 3 and n = 4, 5, 6, ...

#### Brackett series of hydrogen spectral lines

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)]

n = 4 and n = 5, 6, 7, ...

#### Pfund series lies in which region

The wavelength of these spectral lines can be calculated by the following equation,

⊽ = 1/λ = R[(1/n²) - (1/n²)]
n = 5 and n = 6, 7, 8, ...
Transition to the n = 5 level from n = 6, n = 7, n = 8, etc exited states constitute the Pfund series lies in far-infrared spectrum region.

### Classification of elementary particles

Dalton's atomic model should as primary basic of chemistry and it grows up in the last century. The chemical reaction in chemistry clearly explained by this model.
• All the matter made of atoms is indivisible and indestructible.
• All the atoms of a given element are identical mass and properties.
• Compounds are formed by the combination of two or more same or different kinds of atoms.
• A chemical reaction is a rearrangement of atoms.
But Dalton's model of indivisibility of atom discarded by the discovery of various list elementary particles like proton, neutrons, and electrons.

#### The modern structure of an atom

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 to the modern structure of an atom was a long and arduous one. This century much valuable information was being compiled.

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

The nucleus of an atom has around itself a certain number of the negatively charged particles carrying negligible mass, called electrons, arranged in a definite order. Discovery of electron, proton, and neutron

### JJ Thomson cathode ray experiment

Gases at low pressures, when subjected to high potential, becomes conducting and the various luminous effect was observed. When the pressure quite low (0.01 mm), the tube remains dark (Crooks dark space) but a streak of rays, named cathode rays, emanating from the cathode.
The cathode rays have many definite characteristics.
1. Production of fluorescence on the opposite wall where the rays impinge.
2. The rays travel in straight lines confirmed by the shadows of an object placed on their path.
3. The cathode rays defected from the path they travel by electric or magnetic field. The direction of deflection suggested that cathode rays are a negative charge particle.
4. When these rays impinge on the metal targets placed on their path x-ray is produced.

#### Charge and mass of an electron

The electron carrying negatively charged and the charge of an electron
= - 4.8 × 10⁻¹⁰ esu
= - 1.60 × 10⁻¹⁹ coulombs.

Let the mass of an electron = m and charge = e
∴ e/m = 1.76 × 10⁸ coulomb/gram.

Mass of an electron = (1.60 × 10⁻¹⁹)/(1.76 × 10⁸) gram
= 9.11 × 10⁻²⁸ gram.

#### Determination of charge of an electron

The electrodeposition of silver atom from an aqueous solution of silver salt. This is a suitable experiment for the measurement of the charge of an electron.

Faraday's Law is 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 equal to Avogadro number, 1 mole of a silver atom produced. At the same time, 1 mole of electrons removed from the anode and 1 mole of nitrate ions discharged.

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

∴ The charge carried by each electron
e = (96500-coulomb mol⁻¹)/(6.023 × 10²³ mol⁻¹)
= 1.60 × 10⁻¹⁹ coulomb

### Discovery of protons experiment

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

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.
The nature of these positive rays 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.

The hydrogen atom is the simplest atom with one electron revolving around the nucleus. The nucleus of the hydrogen atom carry unit positive charge. When the only electron of the hydrogen atom has removed the nucleus of hydrogen atom contains unit charge and mass.

The particle represented by hydrogen ion called a proton considered as a list elementary particle that accounts for the positive charge of the nucleus.

#### Relative charge and mass of a proton

The proton carrying positively charged and the charge of a proton
= +4.8 × 10⁻¹⁰ esu
= +1.60 × 10⁻¹⁹ coulomb

Let the mass of an proton = m and charge = +e
∴ e/m = 9.3 × 10⁴ coulomb/gram.

∴ Mass of a proton = 1.6725 × 10⁻²⁴ gm

Online college chemistry courses

### Discovery of the neutron by Chadwick

We have study the entire mass of an atom is concentrated in the nucleus and the weight of electrons being negligible.

Atomic number and mass number of hydrogen = 1 and protons alone account for the total mass of hydrogen atom. But except hydrogen, the proton alone cannot account for the total mass of the nucleus.

Helium atom 4 times heavy as an atom of hydrogen, hence helium nucleus must be 4 times heavier than a proton. But helium atom contains 2 protons, which account for two units of mass. The other two units must have been due to other particles.

Let the mass number of an atom = A, nuclear charge or number of protons of the atom = Z.
∴ (A - Z) shortfall of mass number due to other particles.

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. Rutherford named this particle in advance as a neutron. The glory of discovering the neutron went to Chadwick, one of Rutherford students.

The study of the atomic nucleus on the basis of neutrons and protons now a simple affair. Mass number and the atomic number of oxygen 16 and 8 respectively suggested that the atomic nucleus of oxygen composed of 8 protons and 8 neutrons.

Since neutrons contribute only to the mass of the element but do nothing towards the charge. Some of
the 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.

Protium, deuterium, and tritium are three isotopes of hydrogen with zero, one and two protons in the nucleus of the hydrogen atom.

Oxygen-16, oxygen-17, and oxygen-18 are three isotopes of oxygen with 8, 9, 10 number of protons on the nucleus.

### Who discovered the nucleus and with what experiment?

Rutherford and his students study some alpha particle scattering experiments. The alpha particles were already established by the helium ion from their behavior in electric and magnetic fields.

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

With the help of fluorescent zinc sulfide screen around the platinum or gold foil, any deflection of the alpha particle was observed. The 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.

#### Gold foil experiment rutherford nucleus

Conclusions and an atomic theory emerged from the gold foil 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.
Online college chemistry courses

#### What did rutherford expect from the gold foil experiment?

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.

#### Limitations of the Rutherford model of an atom

1. The Rutherford model is 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.

Name

Email *

Message *