Chemistry

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Third Law of Thermodynamics

The entropy of a substance varies directly with temperature. Lower the temperature, lower is the entropy. For example, water above 100° C at one atmosphere exists as a gas and has higher entropy (higher disorder). The water molecules are free to roam about in the entire container.

When the system is cooled, the water vapour condenses to form a liquid. The water molecules in liquid phase still can move about somewhat freely. Thus the entropy of the system has decreased. On further cooling, water freezes to form ice crystal.

The water molecules in the ice crystal are highly ordered and entropy of the system is very low. If we cool the solid crystal still further, the vibration of molecules held in the crystal lattice gets slower and they have very little freedom of movement (very little disorder) and hence very small entropy.

At absolute zero {0 K (or) – 273°C}, theoretically all modes of motion stop.

Absolute zero is a temperature that an object can get arbitrarily close to but absolute zero will remain unattainable.

Thus the third law of thermodynamics states that the entropy of pure crystalline substance at absolute zero is zero. Otherwise it can be stated as it is impossible to lower the temperature of an object to absolute zero in a finite number of steps. Mathematically,

= 0 for a perfectly ordered crystalline state.

Crystals with defects (imperfection) at absolute zero, have entropy greater than zero. Absolute entropy of a substance can never be negative.

In simple terms, the third law states that the entropy of a perfect crystal of a pure substance approaches zero as the temperature approaches zero. The alignment of a perfect crystal leaves no ambiguity as to the location and orientation of each part of the crystal.

The third law of thermodynamics states that the entropy of a system approaches a constant value as the temperature approaches absolute zero.

Third law of thermodynamics statement states that the entropy of a pure crystal at absolute zero is zero. An example that states the third law of thermodynamics is vapours of water are the gaseous forms of water at high temperature. The molecules within the steam move randomly.

The third law of thermodynamics has two important consequences: it defines the sign of the entropy of any substance at temperatures above absolute zero as positive, and it provides a fixed reference point that allows us to measure the absolute entropy of any substance at any temperature.

The second law of thermodynamics states that the entropy of any isolated system always increases. The third law of thermodynamics states that the entropy of a system approaches a constant value as the temperature approaches absolute zero.

Entropy is related to the number of possible microstates, and with only one microstate available at zero kelvin the entropy is exactly zero. The third law was developed by the chemist Walther Nernst during the years 1906-1912. It is often referred to as Nernst’s theorem or Nernst’s postulate.

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Gibbs Free Energy

One of the important applications of the second law of thermodynamics is to predict the spontaneity of a reaction under a specific set of conditions. A reaction that occurs under the given set of conditions without any external driving force is called a spontaneous reaction. Otherwise, it is said to be non-spontaneous. In our day today life, we observe many spontaneous physical and chemical processes, which includes the following examples.

1. A waterfall runs downhill, but never uphill, spontaneously.

2. A lump of sugar dissolves spontaneously in a cup of coffee, but never reappears in its original form spontaneously.

3. Heat flows from hotter object to a colder one, but never flows from colder to hotter object spontaneously.

4. The expansion of a gas into an evacuated bulb is a spontaneous process, the reverse process that is gathering of all molecules into one bulb is not spontaneous.

These examples show that the processes that occur spontaneously in one direction, cannot take place in opposite direction spontaneously. Similarly a large number of exothermic reactions are spontaneous. An example is combustion of methane.

CH₄ + 2O₂ → CO₂ + 2H₂O
∆H° = – 890.4 kJ mol^-1

Another example is acid-base neutralization reaction:

H⁺ + OH^– → H₂O
∆H° = – 57.32 kJ mol^-1

However, some endothermic processes are also spontaneous. For example ammonium nitrate dissolves in water spontaneously though this dissolution is endothermic.

From the above examples we can come to the conclusion that exothermicity favours the spontaneity but does not guarantee it. We cannot decide whether or not a chemical reaction will occur spontaneously solely on the basis of energy changes in the system. We know from second law of thermodynamics that in a spontaneous process, the entropy increases. But not all the processes which are accompanied by an increase in entropy are spontaneous.

In order to predict the spontaneity of a reaction, we need some other thermodynamic function. The second law of thermodynamics introduces another thermodynamic function called Gibbs free energy which finds useful in predicting the spontaneity of a reaction. The Gibbs free energy (G) was developed in the 1870’s by Josiah Willard Gibbs. He originally termed this energy as the “available energy” to do work in a system.

This quantity is the energy associated with a chemical reaction that can be used to do work.

Gibbs free energy is defined as below

G = H – TS ………….. (7.35)

Gibbs free energy (G) is an extensive property and it is a single valued state function.

Let us consider a system which undergoes a change of state from state (1) to state (2) at constant temperature.

G₂ – G₁ = (H₂ – H₁) – T(S₂ – S₁)
∆G = ∆H – T∆S ……………. (7.36)

Now let us consider how ΔG is related to reaction spontaneity.

We know that
∆S_total = ∆S_sys + ∆S_surr

For a reversible process (equilibrium), the change in entropy of universe is zero.
ΔS_total = 0 [∵∆S_sys = ∆S_surr]

Similarly, for an equilibrium process ΔG = 0

For Spontaneous Process, ∆S_total > 0

hence for a spontaneous process, ΔG < 0

i.e. ΔH – T ΔS < 0 ……………. (7.37)

ΔH_sys is the enthalpy change of a reaction, TΔS_sys is the energy which is not available to do useful work. So ΔG is the net energy available to do useful work and is thus a measure of the ‘free energy’. For this reason, it is also known as the free energy of the reaction. For non spontaneous process, ΔG > 0

Gibbs free energy and the net work done by the system:

For any system at constant pressure and temperature
ΔG = ΔH – T ΔS …………. (7.36)
We know that,
ΔH = ΔU + PΔV
∴ ΔG = ΔU + PΔV – TΔS

from first law of thermodynamics
if work is done by the system
ΔU = q – w
from second law of thermodynamics
∆S = \(\frac{q}{T}\)
ΔG = q – w + pΔV – T(\(\frac{q}{T}\))
ΔG = – w + PΔV
– ΔG = w – PΔV ……………. (7.38)

Here, – PΔV represents the work done due to expansion against a constant external pressure. Therefore, it is clear that the decrease in free energy (- ΔG) accompanying a process taking place at constant temperature and pressure is equal to the maximum work obtainable from the system other than the work of expansion.

Criteria for Spontaneity of a Process

The spontaneity of any process depends on three different factors.

If the enthalpy change of a process is negative, then the process is exothermic and may be spontaneous. (ΔH is negative)

If the entropy change of a process is positive, then the process may occur spontaneously. (ΔS is positive)

The gibbs free energy which is the combination of the above two (ΔH – TΔS) should be negative for a reaction to occur spontaneously, i.e. the necessary condition for a reaction to be spontaneous is ΔH – TΔS < 0

The Table assumes ΔH and ΔS will remain the way indicated for all temperatures. It may not be necessary that way. The Spontaneity of a chemical reaction is only the potential for the reaction to proceed as written. The rate of such processes is determined by kinetic factors, outside of thermodynamical prediction.

Problem: 7. 8

Show that the reaction CO + (\(\frac{1}{2}\))O₂ → CO₂ at 300K is spontaneous. The standard Gibbs free energies of formation of CO₂ and CO are – 394.4 and – 137.2 kJ mole^-1 respectively.

ΔG_(reaction) of a reaction at a given temperature is negative hence the reaction is spontaneous.

Relationship between standard free energy change (ΔG°) and equilibrium constant (K_eq):

In a reversible process, the system is in perfect equilibrium with its surroundings at all times. A reversible chemical reaction can proceed in either direction simultaneously, so that a dynamic equilibrium is set up. This means that the reactions in both the directions should proceed with decrease in free energy, which is impossible. It is possible only if at equilibrium, the free energy of a system is minimum. Lets consider a general equilibrium reaction.

A+B ⇄ C + D

The free energy change of the above reaction in any state (ΔG) is related to the standard free energy change of the reaction (ΔG°) according to the following equation.

ΔG = ΔG° + RT ln Q …………. (7.39)

where Q is reaction quotient and is defined as the ratio of concentration of the products to the concentrations of the reactants under non equilibrium condition.

When equilibrium is attained, there is no further free energy change i.e. ΔG = 0 and Q becomes equal to equilibrium constant. Hence the above equation becomes.

ΔG° = – RT ln K_eq

This equation is known as Van’t Hoff equation.
ΔG° = – 2.303 RT log K_eq …………….. (7.40)
We also know that
ΔG° = ΔH° – T ΔS° = – RT ln K_eq

Problem: 7.9

Calculate ΔG° for conversion of oxygen to ozone 3/2 O₂ ⇄ O_3(g) at 298K, if K_P for this conversion is 2.47 × 10^-29 in standard pressure units.
Solution:
ΔG° = – 2.303 RT log K_p
Where
R = 8.314 JK^-1mol^-1
K_p = 2.47 × 10^-29
T = 298K
ΔG° = – 2.303(8.314)(298)log(2.47 × 10^-29)
ΔG° = 163229 Jmol^-1
ΔG° = 163.229 KJ mol^-1

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Second Law of Thermodynamics

Need for the Second Law of Thermodynamics:

We know from the first law of thermodynamics, the energy of the universe is conserved. Let us consider the following processes:

1. A glass of hot water over time loses heat energy to the surrounding and becomes cold.

2. When you mix hydrochloric acid with sodium hydroxide, it forms sodium chloride and water with evolution of heat.

In both these processes, the total energy is conserved and are consistent with the first law of thermodynamics. However, the reverse process i.e. cold water becoming hot water by absorbing heat from surrounding on its own does not occur spontaneously even though the energy change involved in this process is also consistent with the first law.

However, if the heat energy is supplied to cold water, then it will become hot. i.e. the change that does not occur spontaneously and an be driven by supplying energy. Similarly, a solution of sodium chloride does not absorb heat energy on its own, to form hydrochloric acid and sodium hydroxide.

But, this process can not be driven even by supplying energy. From these kinds of our natural experiences, we have come to know that certain processes are spontaneous while the others are not, and some processes have a preferred direction. In order to explain the feasibility of a process, we need the second law of thermodynamics.

Various Statements of the Second Law of Thermodynamics

Entropy

The second law of thermodynamics introduces another state function called entropy. Entropy is a measure of the molecular disorder (randomness) of a system. But thermodynamic definition of entropy is concerned with the change in entropy that occurs as a result of a process.

It is defined as, dS = dq_rev / T

Entropy Statement:

The second law of thermodynamics can be expressed in terms of entropy. i.e “the entropy of an isolated system increases during a spontaneous process”.

For an irreversible process such as spontaneous expansion of a gas,
∆S_total > 0
∆S_total > ∆S_system + ∆S_surrounding
i.e; ∆S_total > ∆S_system + ∆S_surrounding

For a reversible process such as melting of ice,
ΔS_system = – ∆S_surrounding
∆S_total = 0

Kelvin-Planck Statement:

It is impossible to construct a machine that absorbs heat from a hot source and converts it completely into work by a cyclic process without transferring a part of heat to a cold sink. The second law of thermodynamics explains why even an ideal, frictionless engine cannot convert 100% of its input heat into work.

Carnot on his analysis of heat engines, found that the maximum efficiency of a heat engine which operates reversibly, depends only on the two temperatures between which it is operated.

Efficiency = work performed / heat absorbed

q_h – heat absorbed from the hot reservoir
q_c – heat transferred to cold reservoir

………….. (7.27)

For a reversible cyclic process

T_h >> T_c
Hence, η < 1

Efficiency in percentage can be expressed as

Clausius Statement:

It is impossible to transfer heat from a cold reservoir to a hot reservoir without doing some work.

Problem: 7.10

If an automobile engine burns petrol at a temperature of 816° C and if the surrounding temperature is 21° C, calculate its maximum possible efficiency.
Solution:
% Efficiency = \(\left[\frac{\mathrm{T}_{\mathrm{h}}-\mathrm{T}_{\mathrm{c}}}{\mathrm{T}_{\mathrm{h}}}\right]\) × 100
Here
T_h = 816 + 273 = 1089 K;
T_c = 21 + 273 = 294 K
% Efficiency = (\(\frac{1089-294}{1089}\)) × 100
% Efficiency = 73%

Unit of Entropy:

The entropy (S) is equal to heat energy exchanged (q) divided by the temperature (T) at which the exchange takes place. Therefore, The SI unit of entropy is JK^-1.

Spontaneity and Randomness

Careful examination shows that in each of the processes viz., melting of ice and evaporation of water, there is an increase in randomness or disorder of the system. The water molecules in ice are arranged in a highly organised crystal pattern which permits little movement. As the ice melts, the water molecules become disorganised and can move more freely.

The movement of molecules becomes freer in the liquid phase and even more free in the vapour phase. In other words, we can say that the randomness of the water molecules increases, as ice melts into water or water evaporates. Both are spontaneous processes which result in a increase in randomness (entropy).

Standard Entropy Change(ΔS°):

It is possible to calculate the actual entropy of a substance at any temperature above 0 K. The absolute entropy of a substance at 298 K and one bar pressure is called the standard entropy So. The third law of thermodynamics states, according to Nernst, that the absolute entropy of elements is zero only at 0 K in a perfect crystal, and standard entropies of all substances at any temperature above 0 K always have positive values. Once we know the entropies of different substances, we can calculate the standard entropy change (∆S°_r) for chemical reactions.

∆S°_r = ΣS°_products – ΣS°_reactants …………… (7.30)

Standard Entropy of Formation:

Standard entropy of formation is defined as “the entropy of formation of 1 mole of a compound from the elements under standard conditions”. It is denoted as ∆S°_f. We can calculate the value of entropy of a given compound from the values of S° of elements.

Problem: 7.6

Calculate the standard entropy change for the following reaction (∆S°_f), given the standard entropies of CO₂(g), C(s), O₂(g) as 213.6 , 5.740 and 205 JK^-1 respectively.

C(g) + O₂(g) → CO₂(g)

Entropy Change Accompanying Change of Phase

When there is a change of state from solid to liquid (melting), liquid to vapour (evaporation) or solid to vapour (sublimation) there is a change in entropy. This change may be carried out at constant temperature reversibly as two phases are in equilibrium during the change.

…………… (7.31)

Entropy of Fusion:

The heat absorbed, when one mole of a solid melts at its melting point reversibly, is called molar heat of fusion. The entropy change for this process is given by

…………. (7.32)

where ∆H_fusion is Molar heat of fusion. T_f is the melting point.

Entropy of Vapourisation:

The heat absorbed, when one mole of liquid is boiled at its boiling point reversibly, is called molar heat of vapourisation. The entropy change is given by

…………… (7.33)

where ΔH_v is Molar heat of vaporisation. T_b is the boiling point.

Entropy of Transition:

The heat change, when one mole of a solid changes reversibly from one allotropic form to another at its transition temperature is called enthalpy of transition. The entropy change is given

…………….. (7.34)

where ΔH_t is the molar heat of transition, T_t is the transition temperature.

Problem: 7.7

Calculate the entropy change during the melting of one mole of ice into water at 0° C and 1 atm pressure. Enthalpy of fusion of
ice is 6008 J mol^-1.

Given:

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Latice Energy (ΔH_lattice)

Lattice energy is defined as the amount of energy required to completely remove the constituent ions from its crystal lattice to an infinite distance from one mole of crystal. It is also referred as lattice enthalpy.

NaCl(s) → Na⁺(g) + Cl^–(g)
ΔH_lattice = + 788 kJ mol^-1

From the above equation it is clear that 788 kJ of energy is required to separate Na⁺ and Cl^– ions from 1 mole of NaCl.

Born – Haber cycle

The Born-Haber cycle is an approach to analyse reaction energies. It was named after two German scientists Max Born and Fritz Haber who developed this cycle. The cycle is concerned with the formation of an ionic compound from the reaction of a metal with a halogen or other non-metallic element such as oxygen.

Born-Haber cycle is primarily used in calculating lattice energy, which cannot be measured directly. The Born-Haber cycle applies Hess’s law to calculate the lattice enthalpy. For example consider the formation of a simple ionic solid such as an alkali metal halide MX, the following steps are considered.

∆H₁ – Enthalpy change for the sublimation M(s) to M(g)
ΔH₂ – Enthalpy change \(\frac{1}{2}\)X₂(g) to X(g) for the dissociation of
ΔH₃ – Ionisation energy for M(g) to M⁺(g)
ΔH₄ – Electron affinity for the conversion of X(g) to X^–(g)
U – the lattice enthalpy for the formation of solid MX
ΔH_f – enthalpy change for the formation of solid MX directly form elements

According to Hess’s law of heat summation

∆H_f = ∆H₁ + ∆H₂ + ∆H₃ + ∆H₄ + U

Let us use the Born – Haber cycle for determining the lattice enthalpy of NaCl as follows:

Since the reaction is carried out with reactants in elemental forms and products in their standard states, at 1 bar, the overall enthalpy change of the reaction is also the enthalpy of formation for NaCl. Also, the formation of NaCl can be considered in 5 steps. The sum of the enthalpy changes of these steps is equal to the enthalpy change for the overall reaction from which the lattice enthalpy of NaCl is calculated.

Let us calculate the lattice energy of sodium chloride using Born-Haber cycle

∆H_f = heat of formation of sodium chloride = – 411.3 kJ mol^-1
∆H₁ = heat of sublimation of Na(g) = 108.7 kJ mol^-1
∆H₂ = ionisation energy of Na(g) = 495.0 kJ mol^-1
∆H₃ = dissociation energy of Cl₂(g) = 244 kJ mol^-1
∆H₄ = Electron affity of Cl(S) = – 349.0 kJ mol^-1

U = lattice energy of NaCl
∆H_f = ∆H₁ + ∆H₂ + \(\frac{1}{2}\)∆H₃ + ∆H₄ + U
∴ U = (∆H_f) – (∆H₁ + ∆H₂ + \(\frac{1}{2}\)∆H₃ + ∆H₄)
⇒ U = (∆H_f) – (∆H₁ + ∆H₂ + \(\frac{1}{2}\)∆H₃ + ∆H₄)
U = (-411.3) – (108.7 + 495.0 + 122 – 349)
U = (–411.3) – (376.7)
∴ U = – 788 kJ mol^-1

This negative sign in lattice energy indicates that the energy is released when sodium is formed from its constituent gaseous ions Na⁺ and Cl^–.

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Hess’s Law of Constant Heat Summation

We have already seen that the heat changes in chemical reactions are equal to the difference in internal energy (ΔU) or heat content (ΔH) of the products and reactants, depending upon whether the reaction is studied at constant volume or constant pressure.

Since ΔU and ΔH are functions of the state of the system, the heat evolved or absorbed in a given reaction depends only on the initial state and final state of the system and not on the path or the steps by which the change takes place.

This generalisation is known as Hess’s law and stated as:

The enthalpy change of a reaction either at constant volume or constant pressure is the same whether it takes place in a single or multiple steps provided the initial and final states are same.

Application of Hess’s Law:

Hess’s law can be applied to calculate enthalpies of reactions that are difficult to measure. For example, it is very difficult to measure the heat of combustion of graphite to give pure CO.

However, enthalpy for the oxidation of graphite to CO₂ and CO to CO₂ can easily be measured. For these conversions, the heat of combustion values are – 393.5 kJ and – 283.5 kJ respectively. From these data the enthalpy of combustion of graphite to CO can be calculated by applying Hess’s law.

The reactions involved in this process can be expressed as follows

According to Hess law,

ΔH₁ = ΔH₂ + ΔH₃
– 393.5 kJ = X – 283.5 kJ
X = – 110.5 kJ

Hess’s law, also called Hess’s law of constant heat summation or Hess’s law of heat summation, rule first enunciated by Germain Henri Hess, a Swiss-born Russian chemist, in 1840, stating that the heat absorbed or evolved (or the change in enthalpy) in any chemical reaction is a fixed quantity and is independent of the law.

The Hess’s law states that the total enthalpy change during a complete chemical reaction is the same regardless of the path taken by the chemical reaction. Hess’s law can be seen as an application of the principle of conservation of energy.

Hess’s law is useful to calculate heats of many reactions which do not take place directly. It is useful to find out heats of extremely slow reaction. It is useful to find out the heat of formation, neutralization, etc.

Hess’ law can be used to determine the overall energy required for a chemical reaction, when it can be divided into synthetic steps that are individually easier to characterize. This affords the compilation of standard enthalpies of formation, that may be used as a basis to design complex syntheses.

Hess law of constant heat summation states that the total enthalpy change during a reaction is the same whether the reaction takes place in one step or in several steps.

Application:

Hess law is useful to calculate heats of many reactions which do not take place directly.

To experimentally measure the ΔH values of two reactions using the technique of constant pressure calorimetry. To apply these ΔH values in a Hess’s Law calculation to determine the enthalpy of combustion of a metal.

By converting the methanol to formaldehyde and hydrogen the enthalpy of the fuel has been increased. When the formaldehyde and the hydrogen are burned 86 kJ more energy will be released than when methonol is burned H is per mole. means that for each mole of methanol burned 677 kJ heat are released.

Hess’s law is due to enthalpy being a state function, which allows us to calculate the overall change in enthalpy by simply summing up the changes for each step of the way, until product is formed. All steps have to proceed at the same temperature and the equations for the individual steps must balance out.

Germain Henri Hess is noted today for two fundamental principles of thermochemistry: the law of constant summation of heat (known simply as Hess’s law) and the law of thermoneutrality.

What is the connection between Hess’s Law and the fact that H is a state function? Hess’s Law is a consequence of the fact that enthalpy is a state function. Since ΔH is independent of path, we can describe a process by any series of steps that add up to the overall process.

Hess’s law states that the energy change in an overall chemical reaction is equal to the sum of the energy changes in the individual reactions comprising it.

Hess’s Law of Constant HeatSummation (or just Hess’s Law) states that regardless of the multiple stages or steps of a reaction, the total enthalpy change for the reaction is the sum of all changes. This law is a manifestation that enthalpy is a state function.