March 2019

Oxidation and reduction and redox reaction

    The electronic structure of elements and the chemical reaction has given the classical conceptions of oxidation and reduction. The study of electrochemical reactions provides quantitative basics of oxidation and reduction reactions.
    The reaction of magnesium metal burns in oxygen to produce magnesium oxide with the addition of oxygen. The reaction of magnesium metal with either oxygen or chlorine has led to the removal of two electrons from zerovalent magnesium metal forming magnesium(II) ion.
    On the other hand, the lost electrons have found new homes in oxygen or chlorine forming the oxide or chloride ion.
    It would thus appear the classical definition of oxidation and reduction has an intimate connection with redox definition.

What is an oxidation reaction?

    Classically oxidation has defined as the combination of oxygen or any other electronegative element with another element or compound or as the removal of hydrogen or any other electropositive element from a chemical compound.
What is oxidation reaction in redox reaction?
What is oxidation reaction?

Combination of oxygen to carbon, magnesium, and iron

  1. Carbon burns in oxygen to produce carbon dioxide. Here oxygen is combined with carbon to produce carbon dioxide and carbon is considered to have been oxidized.
    C + O₂ ⇆ CO₂
  2. Magnesium metal reacts with chlorine to produce magnesium chloride. In this reaction, magnesium is considered to have been oxidized because magnesium combines with oxygen to produced magnesium oxide.
    2Mg + O₂ ⇆ 2MgO
  3. Ferrous oxide heated in oxygen to produce ferric oxide. Here ferrous oxide is considered to have oxidized.
    4FeO + O₂ ⇆ 2Fe₂O₃

Combination of electronegative chlorine

    Chlorine gas passes through the colorless ferrous chloride solution it forms red ferric chloride. Ferrous chloride oxidized by the addition of electronegative element chlorine.
    2FeCl₂ + Cl₂ ⇆ 2FeCl₃

Removal of hydrogen oxidation

  1. Manganese dioxide reacts with concentrated hydrochloric acid to produce greenish-red chlorine gas.
    4HCl + MnO₂ ⇆ MnCl₂ + Cl₂ + H₂O
    Here hydrogen can remove from hydrogen chloride to form chlorine gas, hence the removal of hydrogen from hydrochloric acid indicated that hydrochloric acid can be oxidized.
  2. Hydrogen sulfide reacts with chlorine to produce sulfur.
    H₂S + Cl₂ ⇆ 2HCl + S
    In the above reaction, hydrogen can remove from hydrogen sulfide. Thus hydrogen sulfide oxidized to form sulfur.

Removal of electro-positive elements

    Potassium iodide reacts with hydrogen peroxide to produce iodine.
    2KI + H₂O₂ ⇆ 2KOH + I₂
    In the above reaction, electro-positive metal potassium can remove from potassium iodide to form iodine. Here potassium iodide oxidized.

What is the reduction reaction?

    Classically reduction has defined as the combination of hydrogen or any other electropositive element with another element or compound or as the removal of oxygen or any other electronegative element from a chemical compound.
What is the reduction reaction in redox reaction?
What is the reduction reaction?

Addition of hydrogen reduction

  1. Bromine can react with hydrogen to produce hydrogen bromide.
    H₂ + Br₂ ⇆ 2HBr
    In the above reaction, bromine can combine with hydrogen to produce hydrogen bromide. Thus bromine can be reduced.
  2. Hydrogen sulfide reacts with chlorine to produce hydrogen chloride and sulfur.
    H₂S + Cl₂ ⇆ HCl + S
    In the above reaction, chlorine can combine with hydrogen and reduced.

Addition of electro-positive elements

    2K₃[Fe(CN)₆ + 2KOH + H₂O₂ ⇆ 2K₄[Fe(CN)₆] + 2H₂O + O₂
    In the above reaction potassium ferricyanide combination with electro-positive element potassium to produce potassium ferrocyanide. Thus potassium ferricyanide can be reduced.

Removal of oxygen reduction

    When hydrogen can pass through a black color heated cupric oxide, oxygen can remove from cupric oxide to form red color copper. Thus cupric oxide can be reduced.
CuO + H₂ ⇆ Cu +H₂O

Removal of electro-negative elements

    Sulfur dioxide gas passes through a red ferric sulfate solution it turns greenish ferrous sulfate solution.
Fe₂(SO₄)₃ + 2SO₂ + H₂O ⇆ 2FeSO₄ + 2H₂SO₄
    In the above reaction ferric sulfate removes sulfate to form ferrous sulfate. Thus ferric sulfate reduced to form ferrous sulfate.

Ideal gas law for gas molecules

The gases are characterized by the lack of definite volume or shape. In the gaseous state, the matter has the properties of filling completely any available space to a uniform density. Low density and high compressibility are also pronounced properties of gases.

Boyle's, Charles, and Avogadro's law gives the birth of the ideal gas law for gas molecules. Ideal gas law for n moles ideal gas,
How to measure pressure, density, molecular weight from ideal gas law?
Ideal gas law
At any given temperature and pressure the volume of a gm mole of any gas will be the same. Thus R is a universal constant of gas.

The equation gives us a connection between the pressure, volume, and temperature of a gas. Thus this equation is the equation of state for a gas molecule.

Problem
Why an ideal gas can not be liquefied?

Solution
The molecules to come close to each other for the liquefaction of gas molecules. The intermolecular attraction needed for the liquefaction of gas.
An ideal gas has no intermolecular attraction and hence ideal gas can not be liquefied.

Further putting the condition of the critical state to the ideal gas law.

PV = RT
or, (dP/dV)T = - RT/V²

At the critical state (dP/dV)T = 0

∴ (- RTc/Vc²) = 0
or, Tc = 0⁰ K
Thus for liquefying an ideal gas necessary to bring the temperature below 0⁰ K. Naturally this temperature not possible thus the gas can not be liquefied.

How to measure gas pressure?

Problem
How to measure pressure excreted on the walls of a 3 liter of the flask when 7 grams of nitrogen are introduced into the same at 27°C?

Solution
PV = nRT
or, PV = (g/M)RT
Molecular weight of nitrogen 28 gm mol⁻¹

∴ P = (7/28) × (0.082 × 300)/3
= 2.05 atm

How to measure the density of gas molecules?

Ideal gas law for n mole gas
PV = nRT.
or, P = (g/V) × (RT/M)

∴ P = dRT/M
where d = density.
Problem
How to measure the density of ammonia gas at 100⁰ C, when confined by the pressure of 1600 mm Hg?

Solution

P = dRT/M
or, d = PM/RT

Molecular weight of ammonia = 17 gm mol⁻¹
P = 1600 mm Hg = 1600/760 atm
R = 82.1 cc atm mol⁻¹ K⁻¹
T = (273 +100) K

∴ d = (17 × 1600)/(760 × 82.1 × 373)
Density of ammonia = 0.00117 gm/cc

The molecular weight of gas molecules

Ideal gas law
PV = nRT
or PV = (g/M) × RT
where g = weight of the gas
M = molecular weight.

∴ PV = gRT/M

Problem
Calculate the molecular weight of a gas, 12.8 gms of which occupy 10 liters at a pressure of 750 mm and at 27° C?

Solution
31.91 gm mol⁻¹

How to calculate the kinetic energy of the gas molecules?

Total kinetic energy
Etotal = N₀ × (Ē)
where Ē = ³/₂ R/N₀
= ³/₂ RT.

Problem

How to calculate the total kinetic energy of 0.5 mol of an ideal gas at 273 K?

Solution
Total kinetic energy
= n (3RT/2)
= 0.05 {(3/2) × 8.314 × 273}

= 1702 Joule

Chemical equilibrium in thermodynamics

Chemical equilibrium is an important article for study college courses and every student has common questions, why do chemical reactions occur?

This leading question in study chemistry can not be answered in a simple sentence Details study reveal that the two primary factor responsible for the feasibility of a chemical reaction.
  1. The potential energy of the reacting system must be lowered by the reaction.
  2. Reactants posses potentially to react, they must find a suitable path to react at a participle rate under specific conditions.
The fast concern with the thermodynamics and study it under chemical equilibrium and the second one constitute the study of chemical kinetics.

What is heat change in chemical reactions?

All the natural process follow a general trend that they take place in a direction which results in an ultimate decrease in the chemical energy of the universe.
  • The release of energy in a chemical reaction is readily demonstrated by the exothermic reaction.
  • Endothermic reactions are accompanied by the absorption of energies.
Many exothermic reactions are reversible and the reverse process must be endothermic in nature. The synthesis of ammonia is an endothermic reversible reaction.
  • Forward reaction: N₂ + 3H₂ → 2NH₃ ΔH = -92.22 kJ mol⁻¹.
  • Backword reaction: 2NH₃ → N₂ + 3H₂ ΔH = +92.22 kJ mol⁻¹.
Evaluation of heat can not be considered the necessary driving property of a chemical reaction. To study the spontaneous process of a chemical reaction two factors control the ultimate energy change of the chemical process.
  1. Enthalpy change (ΔH).
  2. Entropy change (ΔS).
These two changes collectively determined by another fundamental property of the system known Gibbs free energy (ΔG).

Van't Hoff proposed the equilibrium of a chemical reaction is constant at a given temperature. The quantitative relation between chemical equilibrium and Gibbs free energy by using the Gibbs - Helmholtz equation.

Predicting products of chemical reactions

Predicting the product of the chemical reaction can not be cited if the favorable external condition not maintained. The reaction proceeds to some extent and they stop by converting some portion of reactant to product.

If hydrogen and iodine vapor kept at a constant temperature in a closed vessel, only a portion of hydrogen and iodine converted into hydrogen iodide and then the reaction is stopped.

H₂ + I₂ ⇆ 2HI

If at the same temperature some hydrogen iodide is taken in the closed vessel, a faction of hydrogen iodide converted into hydrogen and iodine and rest of hydrogen iodide remain unchanged.

The amount of hydrogen, iodine, and hydrogen iodide remail unchanged in both the experiment.
When a chemical reaction reached such a stage that no further action is apert is called the equilibrium of the chemical reaction.

The equilibrium point of the chemical reaction maintains the following criteria
  1. Approachability from both ends.
  2. Permanency of the equilibrium.
  3. The incompleteness of the reaction.
  4. Dynamic nature of the equilibrium point.
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Bismuth chloride and excess water reaction

Hydrolysis of bismuth chloride by added water, the milky-like solution appears due to the formation of product bismuth oxychloride.

BiCl₃ + H₂O ⇆ BiOCl + 2HCl

If hydrochloric acid solution added the milkiness disappears showing the reversible reaction occurs.
This shows the reversible nature of this chemical reaction.

Problem
Why heat of reaction is the same whether a catalyst used or not?

Solution

H is a state function hence heat of a reaction, ΔH does not change if the initial state and final state of a process are the same.

A catalyst cannot change the initial and final state of a reaction, hence ΔH remains the same whether a catalyst is used or not. Therefore the statement is correct.

What is the Law of mass action in chemistry?

Norwegian Physicists, Guldberg and Waage in 1867 developed the quantitative relation between the amount of the chemical product and reactant at equilibrium point is known as mass action law.

At constant temperature, the rate of a chemical reaction is proportional to the active masses of the reacting system.
  1. Molar concentration (moles/lit) when the solution is dilute.
  2. Partial pressure in the atmosphere unit for the gaseous system.
  3. For pure solid and pure liquid, active mass is assumed to be unity.
Chemical Equilibrium Questions and Answers
Chemical equilibrium
More details study online for the college course of Law of mass action.

Critical constants of gases definition

The critical constant of gases is an impotent article for students who want to study online college courses in chemistry.

A gas can be liquefied by lowering temperature and increasing pressure. But the influence of temperature is more important.

Most of the real gases are liquefied at ordinary pressure by the suitable lowering of the temperature. But gas cannot be liquefied unless its temperature below a certain value depending upon the nature of the gas whatever high the pressure may be applied.

The temperature at which the gas can be liquified is called its critical temperature. Gas can only be liquefied when the temperature below the critical temperature.

Conditions for liquefaction of gases

The essential condition for the liquefaction of gases is to maintain critical temperature, pressure, and volume of the gas molecules. The critical temperature, pressure, and volume can simply represent as Tc, Pc, and Vc respectively.
  1. The critical temperature is the maximum temperature at which a gas can be liquefied and the temperature above which a liquid cannot exist.
  2. Critical pressure is the maximum pressure required to cause liquefaction at the critical temperature.
  3. Critical volume is the volume occupied by one mole of gas molecules at critical temperate and critical pressure.

Liquefaction of CO₂ Andrews isotherms

In 1869, Thomas Andrews carried out an experiment in which P - V relations for liquefication of CO₂ at different temperatures.
Liquefaction of CO₂ and critical constants of gas
Critical constant of gas
  1. High temperatures the isotherms for liquefication of CO₂ look like those of an Ideal gas.
  2. Low temperatures, the curves have altogether different appearances. For curve a as the pressures increase, the volume of the gas decreases in curve A to B.
  3. At the point, B liquefaction commences and the volume decreases rapidly as the gas converted to a liquid with much higher density. This conversion takes place at constant pressure.
  4. Point C, liquefaction of CO₂ is complete and thus the CD is evidence of the fact that the liquid cannot be easily compressed. Thus, we note that AB represents the gaseous state, BC, liquid, and vapor in equilibrium, and CD shows the liquid state only.
  5. Still higher temperatures we get a similar type of curve as discussed in the above point, except that the horizontal portion is reduced. The pressure corresponding to this portion is higher than at lower temperatures.
  6. At temperatures, Tc the horizontal portion is reduced to a mere point. At temperatures higher then Tc there is no indication of qualification at all.
For every gas can have a limit of temperature above which the gas can not be liquefied, no matter what the pressure is.

Continuity of state in the gaseous state

It appears from the Amagat curve at T there is discontinuity or break during the transformation of gas to liquid. But it is not so. The continuity of the states from the gas to liquid can be explained from the following Andrews isotherm ABCD at T₁.

The gas at A is heated to B at constant volume along AB. Then the gas is gradually cooled at constant P along BC, the volume is reduced considerably. The gas is again cooled at constant volume until point D is reached. Nowhere in the process liquid would appear.

At D, the system is a highly compressed gas. But from the curve, this point is the representation of the liquid state. Hence there is hardly a distinction between the liquid state and the gaseous state.

There is no line of separation between the two phases. This is known as the principle of continuity of the state.

Explanation of critical phenomenon in chemistry

Van der Waals equation for 1-mole real gas
(P + a/V2)(V - b) = RT
or, V3 - (b + RT/P)V2 + (a/P)V - ab/P = 0

This equation has three roots in volume for given values of a, b, P and T. It found that either all the three roots are real or one is real and the other two are imaginary.

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Critical constants of gas and Van der Waals equation
Critical phenomena
  1. At higher temperatures and higher volume regions, the isotherms look much like the isotherms from Ideal gas law.
  2. At the temperature lower than Tc the isotherm exhibits a maximum and a minimum. For certain values of pressure, the equation gives three roots of volume as V1, V2, and V3 at pressure.
    The section AB and ED of the Van der Waals curve at T1 can be realized experimentally. ED represents supersaturated or super-cooled vapor and AB represents super-heated liquid. Both of these states are meta-stable.
    These states are unstable in the seance that slight disturbances are sufficient to cause the system to revert spontaneously into the stable state with the two phases present in equilibrium.
  3. The section BCD of the Van der Waals isotherms cannot be realized experimentally. In this region the slope of the P - V curve is positive. Increasing or decreasing the volume of such a system would increase or decrease in pressure. The line BCD represents the metastable state.

Determination of critical constants of gas molecules?

Again with the increase of temperature the minimum and maximum points come close to each other and at the critical point (C) both maximum and minimum coalesce. The slope and curvature both become zero at this point.
Mathematical condition of the critical point
(dP/dV)T = 0
(d²P/dV²)T = 0.

Van der Waals equation for 1 mole gas
(P + a/V²)(V - b) = RT or, P = {RT/(V - b)} - a/V².
Differentiating with respect to the volume at constant temperature gives the slope of the curve.

∴ Slope = (dP/dV)T = - {RT/(V - b)²} + 2a/V³.
Curvature = (d²P/dV²)T = {2RT/(V - b)³} - 6a/V⁴.

Volume measurement at critical point

When P = Pc, V = Vc, and T = Tc.
- {RTc/(Vc - b)2} + 2a/Vc³ = 0
or, RTc/(Vc - b)² = 2a/Vc³.

{2RTc/(Vc - b)³} - 6a/Vc⁴ = 0
or, 2RTc/(Vc - b)3 = 6a/Vc⁴
or, (Vc - b)/2 = Vc/3
∴ Vc = 3b.

Temperature measurement at critical point

RTc/(Vc - b)² = 2a/Vc³.
Vc =3b.

RTc/4b² = 2a/27b³
∴ Tc = 8a/27Rb

Pressure measurement at critical point

Van der Walls equation at the critical state,
Pc = RTc/(Vc - b) - a/Vc²
Again Vc = 3b and Tc = 8a/27Rb
∴ Pc = a/27b²
Problem
Calculate Van der Waals constants for ethylene. (Tc = 280.8 K and Pc = 50 atm).

Answer
a = 0.057 lit mol⁻¹ and b = 4.47 lit² atm mol⁻²

Compressibility factor for real gas molecule

Z = PV/RT
At critical point of gases
Zc = RTc/PcVc = {(a/27b²) × 3b}/{R × (8a/27Rb)} = 3/8
= 0.375

Critical coefficient = RTc/PcVc = 8/3
= 1.66

Derivation of Van der Waals constant from critical constants

a and b for real gas can be determined from critical constants of gas, Tc and Pc. Vc in the expression is avoided due to difficulty in its determination.

b = Vc/3 PcVc/RTc = 3/8
or, Vc = (3/8) × (RTc/Pc)
∴ b = (1/8)(RTc/Pc)

a = Pc × 27b² = 3 × Pc × (3b)²
= 3 PcVc²
= 3Pc × (3RTc/8Pc)²
∴ a = (27/64)(R²Tc²/Pc)
Problem
The critical constants for water are 647 K, 22.09 MPa and 0.0566 dm³ mol⁻¹. What is the value of a and b?
Solution
Tc = 647 K
Pc = 22.09 Mpa = 22.09 × 10³ kPa
Vc = 0.0566 dm³ mol⁻¹
b = Vc/3 = (0.0566 dm³ mol⁻¹)/3

∴ b = 0.0189 dm³ mol⁻¹

a = 3 Pc Vc²
= 3 (22.09 × 10³) × (0.0566)²
∴ a = 213.3 kPa mol⁻²

Problem
Argon has Tc = - 122°C, Pc = 48 atm. What is the radius of the argon atom?

Answer
Radius of argon atom = 1.47 × 10⁻⁸ cm

Value of critical temperature of the ideal gas

For an ideal gas a = 0, since there exist no forces of attraction between the gas molecules. Thus the critical temperature of the ideal gas equal to zero. Cool up to below the critical temperature is the essential condition for an ideal gas to be liquified.

It is obvious that an ideal gas cannot be liquefied if the gas can not attain temperature below zero kelvin.

Chemistry 1

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