Kossel – Lewis Approach to Chemical Bonding

Kossel – Lewis Approach to Chemical Bonding

A logical explanation for chemical bonding was provided by Kossel and Lewis in 1916. Their approach to chemical bonding is based on the inertness of the noble gases which have little or no tendency to combine with other atoms. They proposed that the noble gases are stable due to their completely filled outer shell electronic configuration. Elements other than noble gases, try to attain the completely filled electronic configurations by losing, gaining or sharing one or more electrons from their outer shell.

For example, sodium loses one electron to form Na+ ion and chlorine accepts that electron to give chloride ion (Cl), enabling both atoms to attain the nearest noble gas configuration. The resultant ions, Na+ and Cl are held together by electrostatic attractive forces and the attractive force is called a chemical bond, more specifically an electrovalent bond.

Kossel – Lewis Approach to Chemical Bonding img 1

Kossel – Lewis Approach to Chemical Bonding

G. N. Lewis proposed that the attainment of stable electronic configuration in molecules such as diatomic nitrogen, oxygen etc… is achieved by mutual sharing of the electrons. He introduced a simple scheme to represent the chemical bond and the electrons present in the outer shell of an atom, called Lewis dot structure.

In this scheme, the valence electrons (outer shell electrons) of an element are represented as small dots around the symbol of the element. The first four valence electrons are denoted as single dots around the four sides of the atomic symbol and then the fit onwards, the electrons are denoted as pairs. For example, the electronic configuration of nitrogen is 1s2, 2s2, 2p3. It has 5 electrons in its outer shell (valence shell). The Lewis structure of nitrogen is as follows.

Kossel – Lewis Approach to Chemical Bonding img 2

Similarly, Lewis dot structure of carbon, oxygen can be drawn as shown below.

Kossel – Lewis Approach to Chemical Bonding img 3

Only exception to this is helium which has only two electrons in its valence shell which is represented as a pair of dots (duet).

Kossel – Lewis Approach to Chemical Bonding img 4

Kossel – Lewis Approach to Chemical Bonding

Octet rule

The idea of Kossel – Lewis approach to chemical bond lead to the octet rule, which states that “the atoms transfer or share electrons so that all atoms involved in chemical bonding obtain 8 electrons in their outer shell (valence shell)”.

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Types of Solutions – Different Types, Homogeneous

Types of Solutions – Different Types, Homogeneous

A solution is a homogeneous mixture of two or more substances, consisting of atoms, ions or molecules. The compound that is present in largest amount in a homogeneous mixture is called the solvent, and others are solutes. For example, when a small amount of NaCl is dissolved in water, a homogeneous solution is obtained.

In this solution, Na+ and Cl ions are uniformly distributed in water. Here water is the solvent as the amount of water is more compared to the amount of NaCl present in this solution, and NaCl is the solute.

Types of Solutions

The commonly used solutions are the solutions in which a solid solute is dissolved in a liquid solvent. However, solute or solvent can be in any of the three states of matter (solid, liquid, gas). If water is used as the solvent,the resultant solution is called as an aqueous solution. If solvents (Benzene, CCl4, ether etc.,) other than water is used, then the resultant solution is called as a nonaqueous solution.

The following table illustrates the different types of solutions based on the physical state of the solute and solvent.

Table 9.1 Types and Examples of Solutions

Types of Solutions img 1

There are three types of solutions that can occur in your body based on solute concentration: isotonic, hypotonic, and hypertonic.

Solid – liquid:

A solid solute in a liquid solvent. Examples will be salt (solute) dissolved in water (solvent) and sugar (solute) dissolved in water (solvent).

Liquid – liquid:

A liquid solute in a liquid solvent.

Types of Solutions

Gas – liquid:

A gas solute in a liquid solvent.

Bleach (sodium hypochlorite dissolved in water) dishwater (soap dissolved in water) carbonated beverages (carbon dioxide dissolved in water is what gives sodas their fizz) powdered drinks.

Aqueous solutions:

These solutions have water as the solvent. Examples of such solutions are sugar in water, carbon dioxide in water, etc.

Non-Aqueous Solutions:

These solutions have a solvent that is not water.

Types of Solution:

A solution of two substances is called binary solution. In solution, the component that present in small amount is known as solute and the component present in larger amount known as solvent. Nine kinds of solution are possible.

Types of Solutions

Properties of a Solution:

Solutions are homogeneous mixtures of two or more substances whose components are uniformly distributed on a microscopic scale. The component present in the greatest amount is the solvent, and the components present in lesser amounts are the solute (s).

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Introduction of Solutions

Introduction of Solutions

There are many chemicals that play an important role in our daily life. All these chemicals are in different physical forms, viz solid, liquid and gas. If we do close examination on their composition, we could find that most of them are mixtures and rarely pure substances.

One more interesting aspect is that most of the mixtures are homogeneous irrespective of their physical state and such homogeneous mixtures are called as solutions.

Sea water is one of the naturally existing solutions which covers more than 70% of the earth’s surface. We cannot imagine life on earth without sea water. It contains many dissolved solids, mostly NaCl. Another important naturally occurring solution is air.

Introduction of Solutions

Air is a homogeneous mixture of nitrogen, oxygen, carbon dioxide, and other trace gases. Even solid material such as brass is a homogeneous mixture of copper and zinc. In the above examples the solutions are in different physical states viz… liquid (sea water), gas (air) and solid (alloys), and one common property of all the above is their homogeneity.

The homogeneity implies uniform distribution of their constituents or components throughout the mixture. In this chapter, we learn about the solutions and their properties.

A solution is a particular type of mixture. Mixtures in chemistry are combinations of different substances where each substance retains its chemical properties. Generally, mixtures can be separated by non-chemical means such as filtration, heating, or centrifugation.

Definition of a Solution

Solution, in chemistry, a homogenous mixture of two or more substances in relative amounts that can be varied continuously up to what is called the limit of solubility. The term solution is commonly applied to the liquid state of matter, but solutions of gases and solids are possible.

A solution is a homogeneous mixture of two or more components in which the particle size is smaller than 1 nm. Common examples of solutions are the sugar in water and salt in water solutions, soda water, etc. In a solution, all the components appear as a single phase.

Introduction of Solutions

A solution is a homogeneous type of mixture of two or more substances. A solution has two parts: a solute and a solvent. The solute is the substance that dissolves, and the solvent is the majority of the solution.

Solution is very important in the study of foods and human nutrition. Only substances which can be dissolved can be assimilated. Many substances which will not dissolve in pure water will dissolve in water which contains something else in solution.

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Van’t Hoff Equation – Chemical Equilibrium – Derivation, Formula

Van’t Hoff Equation – Chemical Equilibrium – Derivation, Formula

This equation gives the quantitative temperature dependence of equilibrium constant (K). The relation between standard free energy change (ΔG°) and equilibrium constant is

ΔG° = – RTln K ……………. (1)
We know that
ΔG° = ΔH° – TΔS° ………………. (2)
Substituting (1) in equation (2)
– RTln K = ΔH° – TΔS°

Rearranging

Van’t Hoff Equation img 1

Differentiating equation (3) with respect to temperature,

Van’t Hoff Equation img 2

Equation 4 is known as differential form of van’t Hoff equation.

Van’t Hoff Equation

On integrating the equation 4, between T1 and T2 with their respective equilibrium constants K1 and K2.

Van’t Hoff Equation img 3

Equation 5 is known as integrated form of van’t Hoff equation.

Problem:

For an equilibrium reaction Kp = 0.0260 at 25° C ΔH = 32.4 kJmol-1, calculate Kp at
37° C

Solution:

T1 = 25 + 273 = 298 K
T2 = 37 + 273 = 310 K
ΔH = 32.4 KJmol-1 = 32400 Jmol-1
R = 8.314 JK-1 mol-1
Kp1 = 0. 0260
Kp2 = ?

Van’t Hoff Equation img 4

Van’t Hoff Charle’s law at the constant concentration the osmotic pressure of a dilute solution is directly proportional to the absolute temperature(T) i.e. παT. παCT. v=CRT. The equation is called van’t Hoff general solution equation.

The Van’t Hoff equation gives the relationship between the standard gibbs free energy change and the equilibrium constant. It is represented by the equation -ΔG°=RTlogeKp.

The Van ‘t Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the Van ‘t Hoff factor is essentially 1.

Van’t Hoff Equation

The van’t Hoff theory describes that substances in dilute solution obey the ideal gas laws, resulting to the osmotic pressure formula π = (n/V)RT = [Ci]RT where R is the gas constant, T the absolute temperature, and [Ci] the molar concentration of solute i in dilute solution (1).

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Le Chateliers Principle on Equilibrium

Le Chateliers Principle on Equilibrium

Many chemical reactions that have industrial importance such as synthesis of ammonia are reversible in nature. It is important to know the reaction conditions to produce maximum yield. If a system at equilibrium is disturbed by modifying the reaction conditions, then the system adjusts itself the new conditions to re-establish the equilibrium.

The effect of change in reaction conditions such as temperature, pressure and concentration etc. on a system at equilibrium can be predicted by Le Chatelier-Braun principle.

It states that “If a system at equilibrium is disturbed, then the system shift itself in a direction that nullifis the effect of that disturbance.”

Le-Chatelier's Principle

Effect of Concentration

At equilibrium, the concentration of the reactants and the products does not change. The addition of more reactants or products to the reacting system at equilibrium causes an increase in their respective concentrations.

According to Le Chatelier’s principle, the effect of increase in concentration of a substance is to shift the equilibrium in a direction that consumes the added substance.

Let us consider the reaction

H2(g) + I2(g) ⇄ 2HI(g)

The addition of H2 or I2 to the equilibrium mixture, disturbs the equilibrium. In order to minimize the stress, the system shift the reaction in a direction where H2 and I2 are consumed. i.e., the formation of additional HI would balance the effect of added reactant. Hence, the equilibrium shift to the right (forward direction) i.e. the forward reaction takes place until the equilibrium is re-established. Similarly, removal of HI (product) also favours the forward reaction.

If HI is added to the equilibrium mixture, the concentration HI is increased, and system proceeds in the reverse direction to nullify the effect of increase in concentration of HI.

Le-Chatelier's Principle

Let us explain the effect change in concentration by considering the formation of HI from H2 and I2. At equilibrium, the concentrations of HI, H2 and I2 are 1 M, 0.2 M and 0.1 M respectively.

Le-Chatelier's Principle img 1

The equilibrium was disturbed by adding 0.1 M iodine to the reaction mixture. After sometime, the concentration of HI is found to be 1.092 M. Let us verify whether the system proceeds to re-establish the equilibrium according to Le Chatelier’s principle.

Le-Chatelier's Principle img 2

Now,
Concentration of HI = 1 + 2x = 1.092 M
2x = 0.092
x = 0.046 M

Therefore, the concentration of hydrogen and iodine at this stage,

[H2] = 0.2 – X = 0.2 – 0.046 = 0.154
[I2] = 0.2 – X – 0.2 – 0.046 = 0.154

The reaction quotient Q at this stage,

Le-Chatelier's Principle img 3

The Q value is equal to Kc value.

So, we can conclude that the system has reestablished the equilibrium at this stage and the addition of iodine results in the increase in the HI concentration.

Le-Chatelier's Principle img 4

Effect of Pressure

The change in pressure has significant effect only on equilibrium systems with gaseous components. When the pressure on the system is increased, the volume decreases proportionately and the system responds by shifting the equilibrium in a direction that has fewer moles of gaseous molecules. Let us consider the synthesis of ammonia from nitrogen and hydrogen.

N2(g) + 3H2(g) ⇄ 2NH3(g)

Let the system be allowed to attain equilibrium in a cylinder with a piston. If we press the piston down to increase the pressure, the volume decreases. The system responds to this effect by reducing the number of gas molecules. i.e. it favours the formation of ammonia. If we pull the piston upwards to reduce the pressure, the volume increases. It favours the decomposition of ammonia

Le-Chatelier's Principle img 5

However, when the total number of the moles of the gaseous reactants and the gaseous products are equal, the change in pressure has no effect on system at equilibrium.

Let us consider the following reaction

H2(g) + I2(g) ⇄ 2HI(g)
2 moles of reactants ⇒ 2 moles of products

Here, the number of moles of reactants and products are equal. So, the pressure has no effect on such equilibrium with Δng = 0.

Le-Chatelier's Principle

Effect of Temperature

If the temperature of a system at equilibrium is changed, the system responds by shiftng the equilibrium in the direction that attempts to nullify the effect of temperature.

Let us consider the formation of ammonia.

Le-Chatelier's Principle img 6

In this equilibrium, the forward reaction is exothermic i.e. heat is liberated while the reverse reaction is endothermic i.e. heat is absorbed.

If the temperature of the system is increased (by supplying heat energy), the system responds by decomposing some of the ammonia molecules to nitrogen and hydrogen by absorbing the supplied heat energy. Similarly, the system responds to a drop in the temperature by forming more ammonia molecules from nitrogen and hydrogen, which releases heat energy.

We have already learnt that the change in pressure or concentration causes a change in the equilibrium concentration such that the equilibrium constant remains the same. However, in case of change in temperature, the equilibrium is reestablished with a different equilibrium constant.

Le-Chatelier's Principle

Effect of a Catalyst

Addition of a catalyst does not affect the state of the equilibrium. The catalyst increases the rate of both the forward and the reverse reactions to the same extent. Hence, it does not change the equilibrium composition of the reaction mixture. However, it speeds up the attainment of equilibrium by providing a new pathway having a lower activation energy. For example, in the synthesis of NH3 by the Haber’s process iron is used as a catalyst. Similarly, in the contact process of manufacturing SO3, platinum or V2O5 is used as a catalyst.

Effect of Inert Gas

When an inert gas (i.e, a gas which does not react with any other species involved in equilibrium) is added to an equilibrium system at constant volume, the total number of moles of gases present in the container increases, that is, the total pressure of gases increases the partial pressure of the reactants and the products are unchanged. Hence at constant volume, addition of inert gas has no effect on equilibrium.

Table 8.3: Effect of concentration, pressure, temperature, catalyst and Inert gas on equilibrium

Le-Chatelier's Principle img 7

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Application of Equilibrium Constant

Application of Equilibrium Constant

The knowledge of equilibrium constant helps us to

  1. predict the direction in which the net reaction will take place
  2. predict the extent of the reaction and
  3. calculate the equilibrium concentrations of the reactants and products.

It is to be noted that these constants do not provide any information regarding the rates of the forward or reverse reactions.

Predicting the Extent of a Reaction

The value of equilibrium constant, Kc tells us the extent of a reaction, i.e., it indicates how far the reaction has proceeded towards product formation at a given temperature. A large value of Kc indicates that the reaction reaches equilibrium with high product yield.

On the other hand, a low value of Kc indicates that the reaction reaches equilibrium with low product yield. In general, if the Kc is greater than the 103, the reaction proceeds nearly to completion. If it is less than 10-3, the reaction rarely proceeds. If Kc is in the range 10-3 to 103, significant amount of both reactants and products are present at equilibrium.

Application of Equilibrium Constant

Dependence of extent of reaction on KC

Application of Equilibrium Constant img 1

Example

Consider the following equilibrium reactions and relate their equilibrium constants

  • N2 + O2 ⇄ 2NO; K1
  • 2NO + O2 ⇄ 2NO2; K2
  • N2 + 2O2 ⇄ 2NO2; K3

Application of Equilibrium Constant img 2

Predicting the Direction of a Reaction

From the knowledge of equilibrium constant, it is possible to predict the direction in which the net reaction is taking place for a given concentration or partial pressure of reactants and products.

Consider a general homogeneous reversible reaction,

xA + yB ⇄ lC + mD

For the above reaction under nonequilibrium conditions, reaction quotient ‘Q’ is defined as the ratio of the product of active masses of reaction products raised to the respective stoichiometric coefficients in the balanced chemical equation to that of
the reactants.

Application of Equilibrium Constant

Under non-equilibrium conditions, the reaction quotient Q can be calculated using the following expression.

Application of Equilibrium Constant img 3

As the reaction proceeds, there is a continuous change in the concentration of reactants and products and also the Q value until the reaction reaches the equilibrium. At equilibrium Q is equal to Kc at a particular temperature. Once the equilibrium is attained, there is no change in the Q value. By knowing the Q value, we can predict the direction of the reaction by comparing it with Kc.

If Q = Kc, the reaction is in equilibrium state.
If Q > Kc, the reaction will proceed in the reverse direction i.e., formation of reactants.
If Q < Kc, the reaction will proceed in the forward direction i.e., formation of products.

Application of Equilibrium Constant img 4

Example 1

The value of Kc for the following reaction at 717 K is 48.

H2(g) + I2(g) ⇄ 2HI(g)

At a particular instant, the concentration of H2, I2 and HI are found to be 0.2 mol L-1, 0.2 mol L-1 and 0.6 mol L-1 respectively. From the above information we can predict the direction of reaction as follows.

Application of Equilibrium Constant img 5

Since Q < Kc, the reaction will proceed in the forward reaction.

Application of Equilibrium Constant

Example 2

The value of Kc for the reaction

N2O4(g) ⇄ 2NO2(g)

Kc = 0.21 at 373 K. The concentrations N2O4 and NO2 are found to be 0.125 mol dm-3 and 0.5 mol dm-3 respectively at a given time. From the above information we can predict the direction of reaction as follows.

Application of Equilibrium Constant img 6

The Q value is greater than Kc. Hence, the reaction will proceed in the reverse direction until the Q value reaches 0.21

Calculation of Concentration of Reactants and Products at Equilibrium

If the equilibrium concentrations of reactants and products are known for a reaction, then the equilibrium constant can be calculated and vice versa.

Let us consider the formation of HI in which, ‘a’ moles of hydrogen and ‘b’ moles of iodine gas are allowed to react in a container of volume V. Let ‘x’ moles of each of H2 and I2 react in a container of volume. Let ‘x’ moles of each of H2 and I2 react together to form 2x moles of HI.

H2(g) + I2(g) ⇄ 2HI(g)

Application of Equilibrium Constant img 7

Applying law of mass action,

Application of Equilibrium Constant img 8

The equilibrium constant Kp can also be calculated as follows:

We know the relationship between the Kc and Kp

Kp = Kc(RT)∆ng

Here the

∆ng = np – nr = 2 – 2 = 0

Hence Kp = Kc

Application of Equilibrium Constant img 9

Application of Equilibrium Constant

Solved Problem

One mole of H2 and one mole of I2 are allowed to attain equilibrium in 1 lit container. If the equilibrium mixture contains 0.4 mole of HI. Calculate the equilibrium constant.

Given data:

[H2] = 1 mol L-1 [I2] = 1 mol L-1
At equilibrium, [HI] = 0.4 mol L-1 Kc = ?

Solution:

Application of Equilibrium Constant img 10

Dissociation of PCl5:

Consider that ‘a’ moles of PCl5 is taken in a container of volume V. Let ‘x’ moles of PCl5 be dissociated into x moles of PCl3 and x moles of Cl2.

Application of Equilibrium Constant img 11

Applying law of mass action,

Application of Equilibrium Constant img 12

The equilibrium constant Kp can also be calculated as follows:

We know the relationship between the Kc and Kp

KP = KC(RT)∆ng

Here the

∆ng = np – nr = 2 – 1 = 1
Hence Kp = Kc (RT)
We know that PV = nRT
RT = \(\frac{PV}{n}\)
Where n is the total number of moles at equilibrium.
n = (a-x) + x + x = (a+x)

Application of Equilibrium Constant img 13

Synthesis of Ammonia:

Let us consider the formation of ammonia in which, ‘a’ moles nitrogen and ‘b’ moles hydrogen gas are allowed to react in a container of volume V. Let ‘x’ moles of nitrogen react with 3x moles of hydrogen to give 2x moles of ammonia.

N2(g) + 3H2(g) ⇄ 2NH3(g)

Application of Equilibrium Constant img 14

Applying law of mass action,

Application of Equilibrium Constant img 15

The equilibrium constant Kp can also be calculated as follows:

KP = KC(RT)>∆ng

∆ng = np – nr = 2 – 4 = – 2

Application of Equilibrium Constant img 16

Total number of moles at equilibrium,
n= a – x + b – 3x + 2x = a + b – 2x

Application of Equilibrium Constant img 17

Application of Equilibrium Constant

Solved Problems:

1. The equilibrium concentrations of NH3, N2 are 1.8 × 10-2 M, 1.2 × 10-2 M and 3 × 10-2 M respectively. Calculate the equilibrium constant for the formation of NH3 from N2 and H2. [Hint: M = mol lit-1]

Given Data:

[NH3] = 1.8 × 10-2M
[N2] = 1.2 × 10-2M
[H2] = 3 × 10-2M
Kc = ?

Solution:

Application of Equilibrium Constant img 18

2. The equilibrium constant at 298 K for a reaction is 100.

A + B ⇄ C + D

If the initial concentration of all the four species is 1 M, the equilibrium concentration of D (in mol lit-1) will be

Given data:

[A] = [B] = [C] = [D] = 1 M
Kc = 100
[D]eq = ?

Solution:

Let x be the no moles of reactants reacted

Application of Equilibrium Constant img 19

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