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Friday, 26 June 2015

THERMODYNAMICS

     
THERMODYNAMICS,enotropy,first law of THERMODYNAMICS,second law of THERMODYNAMICS,
THERMODYNAMICS
Chemical energy stored by molecules can be released as heat during chemical reactions
e.g.. when a fuel like methane, cooking gas or coal burns in air.
The chemical energy may also be used to do mechanical work
e.g. when a fuel burns in an engine
or to provide electrical energy
e.g. galvanic cell like dry cell. Thus, various forms of energy are interrelated and under certain conditions, these may be transformedfrom one form into another.
The study of the energy transformations is known as thermodynamics”.
The laws of thermodynamics deal with energy changes of macroscopic systems involving a large number of molecules.
Limitations of Thermodynamics---1. It is not concerned about how and at what rate the energy transformations are carried out, but is based on initial and final states of a system undergoing the change.
2. Laws of thermodynamics apply only when a system is in equilibrium or moves from one equilibrium state to another equilibrium state.
( Macroscopic properties like pressure and temperature do not change with time for a system in equilibrium state.)
THERMODYNAMIC STATE
The System and the Surroundings
A system is thepart of universe under observations and remaining universe constitutes the surroundings.
The universe = The system + The surroundings
Fig. 6.1 System and the surroundings
Note-The entire universe other than the system is not affected by the changes taking place in the system..
The wall that separates the system from the surroundings is called boundary. This is designed to allow us to control and keep track of all movements of matter and energy in or out of the system. It may be realor imaginary.
Types of the System – We classify the systems according to the movements of matter and energy in or out
of the system.
1. Open System -- In an open system, there is exchange of energy and matter between system and surroundings
[Fig. (a)]. The presence of hot water in open beaker.
(Here the boundary is an imaginary surface enclosing the beaker and reactants.)
2. Closed System-- In a closed system, there is no exchange of matter, but exchange of energy is possible
between system and the surroundings
[Fig. (b)]. The presence of hot water in a beaker covered with a lid.
Fi (a) Open, (b)closed (c) isolated systems
3. Isolated System-- In an isolated system, there is no exchange of energy or matter between the system and thesurroundings [Fig. (c)]. The presence of of hot water in a thermos flask.
The State of the System
The system must be described by specific quantitative properties such as pressure (p), volume (V), and temperature (T ) as well as the composition of the system.
(We need to describe the system by specifying it before and after the change)
We specify the state of the system by state functions or state variables.
The state of a thermodynamic system is described by its measurable or macroscopic (bulk) properties. We can describe the state of a gas by quoting its pressure (p), volume (V), temperature (T ), amount (n) etc.
Variables like p, V, T are called state variables or statefunctions because their values depend only on the state of the system and not on how it is reached.
The Internal Energy(U) as a State Function--
“The sum of all types of energy associated to the system i.e. the total energy of the system. It may be chemical, electrical, mechanical or any other type of energy you may think, it is called as the internal energy, Uof the system, which may change, when
1. heat passes into or out of the system,
2. work is done on or by the system,
3. matter enters or leaves the system.
(a) Work--- The system containing some quantity of matter in a thermos flask or in an insulated beaker. This would not allow exchange of heat between thesystem and surroundings through its boundary, this type of system called asadiabatic.
The manner in which the state of this system may be changed will be calledadiabatic process.
In Adiabatic process, there is no transfer of heatbetween the system and surroundings. Here, the wall separating the system and thesurroundings is called the adiabatic wall.
Fig. An adiabatic system which does not permit the transfer of heat through its boundary.
First law of thermodynamics-- states that-- The energy of an isolated system isconstant.It is commonly stated as the law of conservation of energy i.e., energy can neither be created nor be destroyed.
Change in the internal energy of the system
Let us call the initial state of the system as state ‘A’ and its temperature as TA. Let the internal energy of the system in state A be called UA. We can change the state of the system in different ways.
1. One way: We do some mechanical work, say 1 kJ, by rotating a set of small paddles and thereby churning water. Let the new state be called B state and its temperature, as TB. It is found that TB> TAand the change in temperature, ∆T = TBTA. Let the internal energy of the system in state B be UBand the change in internal energy, U =UBUA.
Second way: We now do an equal amount (i.e., 1kJ) electrical work with the help of an immersion rod and note down the temperature change. We find that the change in temperature is same as in the earlier case, say,
T = TBTA
(In fact, the experiments in the above manner were done by J. P. Joule between 1840–50 and he was able to show that a given amount of work done on the system, no matter how it was done (irrespective of path) produced the same change of state, as measured by the change in the temperature of the system.)
So, the internal energy Umay be define as a quantity, whose value is characteristic of the state of a system,
But the adiabatic work, wadrequired to bring about a change of state is equal to the difference between the value of U in one state and that in another state, ∆U i.e.,
U =U 2U1= wad
Therefore, internal energy, U, of the system is a state function.
The positive sign expresses that wadis positive when work is done on the system.
Similarly, if the work is done by the system, wadwill be negative.
Q. Can you name some other familiar state functions?
Ans. Some of other familiar state functions are V, p, and T.
For example, if we bring a change in temperature of the system from 25°C to 35°C, the change in temperature
is 35°C–25°C = +10°C, whether we go straight up to 35°C or we cool the system for a few degrees, then take
the system to the final temperature. Thus, T is a state function and the change in temperature is independent of
the route taken. Volume of water in a pond, for example, is a state function, because change in volume of its water is independent of the route by which water is filled in the pond, either by rain or by tube well or by both,
(b) Heat
We can also change the internal energy of a system by transfer of heat from the surroundings to the system or vice-versa without expenditure of work. This exchange of energy, which is a result of temperature
difference is called heat, q.
Let us consider a system with thermally conducting walls
Fig. A system which allows heat transferthrough its boundary.
We take water at temperature, TAin a container having thermally conducting walls, say made up of copper and enclose it in a huge heat reservoir at temperature, TB. The heat absorbed by the system (water), q can be
measured in terms of temperature difference , TBTA. In this case change in internal energy, ∆U= q, when no work is done at constant volume.
The q is positive, when heat is transferred from the surroundings to the system and q is negative when heat is transferred from system to the surroundings.
(c) The general case
Let us consider the general case in which a change of state is brought about both by doing work and by transfer of heat. We write change in internal energy for this case as:
U = q + w
For a given change in state, q and w can vary depending on how the change is carried out.
However, q +w = ∆U will depend only on initial and final state. It will be independent of the way the change is carried out. If there is no transfer of energy as heat or as work
(isolated system) i.e., if w = 0 and q = 0, then ∆U = 0.
The equation U = q + wis mathematical statement of the first law of thermodynamics..
Note: There is considerable difference between the character of the thermodynamic property energy and that of a mechanical property such as volume. We can specify an unambiguous (absolute) value for volume of a system in a particular state, but not the absolute value of the internal energy. However, we can measure only the changes in the internal energy, ∆U of the system.
APPLICATIONS
Many chemical reactions involve the generation of gases capable of doing mechanical work or the generation of heat. It is important for us to quantify these changes and relate them to the changes in the internal energy.
1.Work -- First we concentrate on the nature of work a system can do. We will consider only
mechanical worki.e., pressure-volume work.
For understanding pressure-volume work, let us consider a cylinder which contains one mole of an ideal gas fitted with a frictionless(no energy require to move the piston) piston. Total volume of the gas is Vi and pressure of the gas inside is p. If external pressure is pexwhich is greater than p, piston is moved inward till the pressure inside becomes equal to pex. As in fig (a)
Let this change be achieved in a single step and the final volume be Vf.
During this compression, suppose piston moves a distance, l and cross-sectional area of the piston is A
[Fig.(a)]. then, volume change = l × A = ∆V = (VfVi)
We also know, force/ area = pressurei.e. force per unit area is known as pressure.
Therefore, force on the piston = pex. A
If w is the work done on the system by movement of the piston then
w = force x distance = pex. A .l
= pex. (–∆V) = – pexV = – pex(VfVi)
Fig. (a) Work done on an ideal gas in a cylinder when it is compressed by a constant external pressure, pex
(in single step) is equal to the shaded area.
The negative sign of above expression is required to obtain conventional sign for w, which will be positive. It indicates that in case of compression work is done on the system. Here (VfVi)will be negative and negative
multiplied by negative will be positive. Hence the sign obtained for the work will be positive.
If the pressure is not constant at every stage of compression, but changes in number of finite steps, work done on the gas will be summed over all the steps and will be equal to −ΣpV [Fig (b)]
Fig. (b) pV-plot when pressure is not constant and changes in finite steps during compression from initial volume, Vito final volume, Vf. Work done on the gas is represented by the shaded area..
If the pressure is not constant but changes during the process such that it is always infinitesimally greater than the pressure of the gas, then, at each stage of compression, the volume decreases by an infinitesimal amount,
dV. In such a case we can calculate the work done on the gas by the relation
Here, pexat each stage is equal to (pin+ dp) in case of compression [Fig.(c)].
Fig. 6.5 (c) pV-plot when pressure is not constant and changes in infinite steps (reversible conditions) duringcompression from initial volume, Vi to final volume, Vf . Work done on the gas is represented by the shaded area.
In an expansion process under similar conditions, the external pressure is always less than the pressure of the system i.e., pex= (pin- dp).
In general case we can write, pex= (pin+ dp). Such processes are called reversible processes.
A process or change is said to be reversible, if a change is brought out in such a way that the process could, at any moment, be reversed by an infinitesimal change. A reversible process proceeds infinitely slowly by a series of equilibrium states such that system and the surroundings are always in near equilibrium with each other. Processes Fig.other than reversible processes are known as irreversible processes.
In chemistry, we relate the work term to the internal pressure of the system. We can relate work to internal pressure of the system under reversible conditions by writing above equation as follows:
(above equation)
Since dp × dV is very small we can write
Now, the pressure of the gas (pinwhich we can write as p now) can be expressed in terms of its volume through gas equation. For n mol of an ideal gas i.e., pV =nRT
Therefore, at constant temperature (isothermal process),
Free expansion: Expansion of a gas in vacuum (pex = 0) is called free expansion. No work is done during free expansion of an ideal gas whether the process is reversible or irreversible.
Now, we can write equation in number of ways depending on the type of processes.
Let us substitute w = – pexV in above equation , and we get
U = q −pexV
If a process is carried out at constant volume (∆V = 0), then
U = qV
the subscript V in qVdenotes that heat issupplied at constant volume.
Isothermal and free expansion of an ideal gas
For isothermal (T = constant) expansion of an ideal gas into vacuum ; w = 0 since pex = 0.
Also, Joule determined experimentally that
q = 0; therefore, ∆U = 0
Equation , ∆U = q + w can be expressed for isothermal irreversible and reversible changes as follows:
1. For isothermal irreversible change
q = – w = pex(VfVi)
2. For isothermal reversible change
3. For adiabatic change, q = 0,
U = wad


Enthalpy, H
(a) Another state function--- We know that the heat absorbed at constantvolume is equal to change in the internal energy i.e., U = qV
But most of chemical reactions are carried out at constant pressure, in flasks or test tubes under constant atmospheric pressure. We need to define another state function which may be suitable under these conditions.
We may write equation ∆U = q + w as U = qp- p∆V at constant pressure,
Where qpis heat absorbed by the system and –p∆V represent expansion work done by the system.
Let us represent the initial state by subscript 1 and final state by 2
We can rewrite the above equation as
U2U1= qpp (V2V1)
On rearranging, we get
qp= (U2 + pV2) – (U1+ pV1)
Now we can define another thermodynamic function, the enthalpy H [Greek word enthalpien, to warm or heat content] as :
H = U + pV
so, above equation becomes
qp= H2H1 = H
Although q is a path dependent function,
H is a state function because it depends on U, p and V, all of which are state functions.
Therefore, H is independent of path. Hence, qpis also independent of path.
For finite changes at constant pressure, we can write above equation as
H = U + pV
Since p is constant, we can write
H = U + pV
It is important to note that when heat is absorbed by the system at constant pressure, we are actually measuring changes in the enthalpy.
We knowH = qp, heat absorbed by the system at constant pressure.
H is negative for exothermic reactions which evolve heat during the reaction and ∆H is positive for endothermic reactions which absorb heat from the surroundings.
At constant volume (V = 0), U = qV,
therefore above equation becomes
H = U = qV
The difference between H and U is not usually significant for systems consisting of only solids and / or liquids because Solids and liquids do not suffer any significant volume changes upon heating.
The difference, however, becomes significant when gases are involved.
Let us consider a reaction involving gases. If VAis the total volume of the gaseous reactants, VBis the total volume of the gaseous products, nAis the number of moles of gaseous reactants and nBis the number of moles of gaseous products, all at constant pressure and temperature, then using the ideal gas law, we write,
pVA= nART and pVB= nBRT
Thus, pVBpVA= nBRT nART = (nBnA)RT
Or p (VBVA) = (nBnA) RT
Or p V = ngRT
Here, ngrefers to the number of moles of gaseous products minus the number of moles of gaseous reactants.
Substituting the value of pV in above equation we get
H = U + ngRT
This equation is used to calculate H from U and vice versa.
(b) Extensive and Intensive Properties-- An extensive property is a property whose value depends on the quantity or size of matter present in the system. For example, mass, volume, internal energy, enthalpy, heat capacity, etc.
Those properties which do not depend on the quantity or size of matter present are known as intensive properties. For example temperature, density, pressure etc.
A molar property is an intensive property, Xm, is the value of an extensive property X of the system for 1 mol of the substance. If n is the amount of matter, Xm = X/n is independent of the amount of matter. For examples -- molar volume, Vm and molar heat capacity, Cm.
Let us understand the distinction between extensive and intensive properties by considering a gas enclosed in a container of volume V and at temperature T [Fig.(a)]. Let us make a partition such that volume is halved, each part [Fig. (b)] now has one half of the original volume, V/2, but the temperature will still remain the same i.e., T. --- It is clear that volume is an extensive property and temperature is an intensive property.
Fig(a) A gas at volume V and temperature T Fig (b) Partition, each part having half the volume of the gas
(c) Heat Capacity
We know the increase of temperature is proportional to the heat transferred to the system.
q = coeff x ∆T
Here q is the heat supplied to the system and T is the increase in the temperature due to supplied heat.
The magnitude of the coefficient depends on the size, composition and nature of the system.
We can also write it as q = C T
The coefficient, C is called the heat capacity.
Thus, we can measure the heat supplied by calculating the temperature rise.
When C is large, a given amount of heat results in only a small temperature rise. Water has a large heat capacity i.e., a lot of energy is needed to raise its temperature. This is the reason it is a good coolant and used in water coolers.
C is directly proportional to amount of substance. The molar heat capacity of a substance, Cm = (C/n)
is the heat capacity for one mole of the substance and it is the quantity of heat needed to raise the temperature of one mole by one degree celsius (or one kelvin).
Specific heat, also called specific heat capacityis the quantity of heat required to raise the temperature of one unit mass of a substance by one degree celsius (or one kelvin).
Note—For finding out the heat, q, required to raise the temperatures of a sample, we multiply the specific heat of the substance, c, by the mass m, and temperatures change, ∆T as
q = c x m x ∆T = C T
(d) The relationship between Cp and CVfor an ideal gas------ At constant volume, the heat capacity, C is
denoted by CVand at constant pressure, this is denoted by Cp . Let us find the relationship between the two.
We can write equation for heat, q at constant volume as qV= Cv ∆T = ∆U
at constant pressure as qp= Cp ∆T = ∆H
The difference between Cp and CVcan be derived for an ideal gas as:
For a mole of an ideal gas, ∆H = ∆U + ∆ (pV )
= ∆U + ∆ (RT )
= ∆U + R∆T
H = U + R ∆T
On putting the values of ∆H and ∆U, we have
Cp ∆T = Cv ∆T + R ∆T
OR Cp = Cv + R
OR Cp - Cv = R
MEASUREMENT OF U AND H: CALORIMETRY---Anexperimental technique used to measure energy changes associatedwith chemical or physical processes is called calorimetry.
In calorimetry, the process is carried out in a vessel called calorimeter,which is immersed in a known volume of a liquid, with known heat capacity of the liquid in which calorimeter is immersed and the heat capacity of calorimeter, it is possible to determine the heat evolved in the process by measuring temperature changes. Measurements are made under two different conditions:
i) at constant volume, qV
ii) at constant pressure, qp
(a) Measurement of heat change at constant volume i.e. U measurements---
For chemical reactions, heat absorbed at constant volume, is measured in a bomb calorimeter (Fig.). Here, a steel vessel (the bomb) is immersed in a water bath. The whole device is called calorimeter. The steel vessel is immersed in water bath to ensure that no heat is lost to the surroundings. A combustible substance is burnt in pure dioxygen supplied in the steel bomb. Heat evolved during the reaction is transferred to the water
around the bomb and its temperature is monitored. Since the bomb calorimeter is sealed, its volume does not change i.e., the energy changes associated with reactions are measured at constant volume. Under these conditions, no work is done as the reaction is carried out at constant volume in the bomb calorimeter. Even for reactions involving gases, there is no work done as ∆V = 0. Temperature change of the calorimeter produced by the completed reaction is then converted to qV, by using the known heat capacity of the calorimeter with
the help of equation.
Fig. Bomb calorimeter
(b)Measurement of heat change at constant volume i.e.H measurements---
Measurement of heat change at constant pressure (generally under atmospheric pressure) can be done in a calorimeter shown in Fig.
Fig. Calorimeter for measuring heat changes at constant
pressure (atmospheric pressure).
We know that ∆H= qp (atconstant p) and, therefore, heat absorbed orevolved, qp at constant pressure is also calledthe heat of reaction or enthalpy of reaction, ∆rH.
In an exothermic reaction, heat is evolved, and system loses heat to the surroundings. Therefore, qp will be negative and ∆rH will also be negative.
Similarly in an endothermic reaction, heat is absorbed, qp is positive and ∆rH will be positive.
ENTHALPY CHANGE, rH OF A REACTION ñ REACTION ENTHALPY
In a chemical reaction, reactants are converted into products and is represented by,
Reactants →Products
The enthalpy change duing a reaction is called the reaction enthalpy. The enthalpy change of a chemical reaction, is given by the symbol ∆rH
rH = (sum of enthalpies of products) – (sum of enthalpies of reactants)
(Here symbol Σ (sigma) is used for summation and ai and bi are the stoichiometric coefficients of the products and reactants respectively in the balanced chemical equation. For example, for the reaction
CH4(g) + 2O2(g) →CO2 (g) + 2H2O (l)
= [Hm (CO2,g) + 2Hm (H2O, l)]– [Hm (CH4, g) + 2Hm (O2, g)]
where Hm is the molar enthalpy.
Significanse-- Knowledge of this quantity is required when one needs to plan the heating or cooling required to maintain an industrial chemical reaction at constant temperature. It is also required to calculate temperature dependence of equilibrium constant.
(a) Standard enthalpy of reactions--- The standard enthalpy of reaction is the enthalpy change for a reaction when all the participating substances are in their standard states.
The standard state of a substance at a specified temperature is its pure form at 1 bar.
Note---Enthalpy of a reaction depends on the conditions under which a reaction is carried out. It is, therefore, necessary that we must specify some standard conditions. For example, the standard state of liquid ethanol at 298 K is pure liquid ethanol at 1 bar; standard state of solid iron at 500 K is pure iron at 1 bar. Usually data are taken at 298 K.
Standard conditions are denoted by adding the superscript Ө to the symbol ∆H, e.g., ∆HӨ
(b) Enthalpy changes during phase transformations---
Phase transformations involve energy changes.
(i) enthalpy of fusion or molar enthalpy of fusion, fusHӨ
Ice, for example, requires heat for melting. Normally this melting takes place at constant pressure (atmospheric pressure) and during phase change, temperature remains constant i.e. melting point (at 273 K).
H2O(s) ---H2O(l ); ∆fusHӨ = 6.00 kJ mol-
Here ∆fusHӨ is enthalpy of fusion in standard state. If water freezes, then process is reversed and equal amount of heat is given off to the surroundings.
The enthalpy change that accompanies melting of one mole of a solid substance in standard state is called standard enthalpy of fusion or molar enthalpy of fusion, fusHӨ
Melting of a solid is endothermic, so all enthalpies of fusion are positive. Similarly freezing of a liquid is exothermic, So all enthalpies of freezing are negative.
(ii) enthalpy of vaporization or molar enthalpy of vaporization, vapHӨ
Water requires heat for evaporation. At constant temperature of its boiling point Tb and at constant pressure:
H2O(l) ---H2O(g ); ∆vapHӨ= 40.79 kJ mol-
vapHӨis the standard enthalpy of vaporization.
Amount of heat required to vaporize one mole of a liquid at constant temperature and under standard pressure (1bar) is called its standard enthalpy of vaporization or molar enthalpy of vaporization, vapHӨ.
(iii) enthalpy of sublimation or molar enthalpy of sublimation subHӨ
Sublimation is direct conversion of a solid into its vapour. Solid CO2or ‘dry ice’ sublimes at 195K with ∆subHӨ=25.2 kJ mol–1; naphthalene sublimes slowly and for this 73.0 kJ mol1-subHӨ.
Standard enthalpy of sublimation, subHӨis the change in enthalpy when one mole of a solid substance sublimes at a constant temperature and under standard pressure (1bar).
---The magnitude of the enthalpy change dependson the strength of the intermolecularinteractions in the substance undergoing the phase transfomations. For example, the stronghydrogen bonds between water molecules hold them tightly in liquid phase. For an organicliquid, such as acetone, the intermolecular dipole-dipole interactions are significantlyweaker. Thus, it requires less heat to vaporize 1 mol of acetone than it does to vaporize 1 molof water.
(c) Standard enthalpy of formation---- The standard enthalpy change for theformation of one mole of a compound from its elements in their most stable states ofaggregation (also known as reference states) is called Standard Molar Enthalpyof Formation. Its symbol is fHӨ,
wherethe subscript ‘ f ’ indicates that one mole ofthe compound in question has been formed in its standard state from its elements in theirmost stable states of aggregation. The reference state of an element is its most stable state ofaggregation at 25°C and 1 bar pressure.
For example, the reference state of dihydrogen is H2gas and those of dioxygen, carbon and sulphur are O2gas, Cgraphite and Srhombicrespectively. Some reactions with standard molar enthalpies of formation are given below.
Note---It is important to note that a standard molar enthalpy of formation, ∆fHӨ, is just a special case of ∆rHӨ,, where one mole of a compound is formed from its constituent elements, as in the above three equations, where 1 mol of each, water, methane and ethanol is formed.
--In contrast, the enthalpy change for an exothermic reaction:
is not an enthalpy of formation of calcium carbonate, since calcium carbonate has been formed from other compounds, and not from its constituent elements.
--Also, for the reaction given below, enthalpy change is not standard enthalpy of formation,
Here two moles, instead of one mole of the product is formed from the elements, i.e.,
Therefore, by dividing all coefficients in the balanced equation by 2, expression for enthalpy of formation of HBr (g) is written as
Importance—If we want to know how much heat is required to decompose calcium carbonate to lime and
carbon dioxide, with all the substances in their standard state.
Here, we can make use of standard enthalpy of formation and calculate the enthalpy change for the reaction. The following general equation can be used for the enthalpy change calculation.
where a and b represent the coefficients of the products and reactants in the balanced equation. Let us apply the above equation for decomposition of calcium carbonate. Here, coefficients ‘a’ and ‘b’ are 1 each.
Thus, the decomposition of CaCO3(s) is an endothermic process and you have to heat it for getting the desired products.
(d) Thermochemical equations--- A balanced chemical equation together withthe value of its ∆rH is called a thermochemicalequation. We specify the physical state (alongwith allotropic state) of the substance inan equation. For example:
The above equation describes the combustion of liquid ethanol at constant temperature and pressure. The negative sign of enthalpy change indicates that this is an exothermic reaction.
Note---It is necessary to remember the following conventions regarding thermochemical equations.
1. The coefficients in a balanced thermochemical equation refer to the number of moles (never molecules) of reactants and products involved in the reaction.
2. The numerical value of ∆rHӨ,refers to the number of moles of substances specified by an equation. Standard enthalpy change ∆rHӨ,will have units as kJ mol–1.
To illustrate the concept, let us consider the calculation of heat of reaction for the following reaction :

From the Table of standard enthalpy of formation ∆fHӨ), we find :

Then--
Note-- that the coefficients used in these calculations are pure numbers, which are equal to the respective stoichiometric coefficients. The unit for ∆rHӨis kJ mol–1, which means per mole of reaction.
Once we balance the chemical equation in a particular way, as above, this defines the moleof reaction.
If we had balanced the equation differently, for example,

then this amount of reaction would be one mole of reaction and rHӨwould be
It shows that enthalpy is an extensive quantity.
3. When a chemical equation is reversed, the value of ∆rHӨis reversed in sign. For example

(e) Hessís Law of Constant Heat Summation---We know that enthalpy is a state function,therefore the change in enthalpy isindependent of the path between initial state (reactants) and final state (products). In otherwords, enthalpy change for a reaction is the same whether it occurs in one step or in aseries of steps. This may be stated as follows in the form of Hess’s Law.
If a reaction takes place in several steps then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.
Importance of this law---1. We can calculate the enthalpy of those reactions, whose enthalpy determination is not possible practically, for example we want to calculate enthalpy change for the reaction
Here CO(g) is the major product but some CO2gas is always produced in this reaction. Therefore, we cannot measure enthalpy change for the above reaction directly. However, if we can find some other reactions involving related species, it is possible to calculate the enthalpy change for the above reaction.
Let us consider the following reactions:
(i)
 (ii)
We can combine the above two reactions in such a way so as to obtain the desired reaction. To get one mole of CO(g) on the right, we reverse equation (ii). In this, heat is absorbed instead of being released, so we change sign of ∆rHӨvalue
 (iii)
Adding equation (i) and (iii), we get the desired equation,
In general, if enthalpy of an overall reaction A→B along one route is ∆rHӨand ∆rHӨ1, ∆rHӨ2,
rHӨ3..... representing enthalpies of reactions leading to same product, B along another route,then we have
rHӨ= ∆rHӨ1+ ∆rHӨ2+ ∆rHӨ3...
It can be represented as---
2.ENTHALPIES FOR DIFFERENT TYPES OF REACTIONS
Here we give name to enthalpies specifying the types of reactions.
(a) Standard enthalpy of combustion (symbol : cHӨ)
Combustion reactions are exothermic in nature. These are important in industry, rocketry, and other walks of life. Standard enthalpy of combustion is defined as theenthalpy change per mole (or per unit amount)
of a substance, when it undergoes combustion and all the reactants and products being in their standard states at the specified temperature.
Cooking gas in cylinders contains mostly butane (C4H10). During complete combustion of one mole of butane, 2658 kJ/mol of heat is released. We can write the thermochemical reactions for this as:
Similarly, combustion of glucose gives out 2802.0 kJ/mol of heat, for which the overall equation is :
Our body also generates energy from food by the same overall process as combustion, although the final products are produced after a series of complex bio-chemical reactions involving enzymes.
(b) Enthalpy of atomization (symbol: aHӨ)
Consider the following example of atomization of dihydrogen
H2(g) →2H(g); ∆aHӨ= 435.0 kJ /mol
Here H atoms are formed by breaking H–H bonds in dihydrogen. The enthalpy change in this process is known as enthalpy of atomization, ∆aHӨ. It is the enthalpy change on breaking one mole of bonds completely to obtain atoms in the gasphase.
In case of diatomic molecules, like dihydrogen (given above), the enthalpy of atomization is also the bond dissociation enthalpy.
The other examples of enthalpy of atomization can be
CH4(g) →C(g) + 4H(g); ∆aHӨ= 1665 kJ/ mol
Note-- that the products are only atoms of C and H in gaseous phase. Now see the following reaction:
Na(s) →Na(g) ; ∆aHӨ= 108.4 kJ/ mol
In this case, the enthalpy of atomization is same as the enthalpy of sublimation.
(c) Bond Enthalpy (symbol: bondHӨ)
Chemical reactions involve the breaking and making of chemical bonds. Energy is required to break a bond and energy is released when a bond is formed. It is possible to relate heat of reaction to changes in energy associated with breaking and making of chemical bonds. With reference to the enthalpy changes
associated with chemical bonds, two different terms are used in thermodynamics. (i) Bond dissociation enthalpy (ii) Mean bond enthalpy
Let us discuss these terms with reference to diatomic and polyatomic molecules.
Diatomic Molecules: Consider the following process in which the bonds in one mole of dihydrogen gas (H2) are broken: H2(g) →2H(g) ; ∆H-HHӨ= 435.0 kJ /mol
The enthalpy change involved in this process is the bond dissociation enthalpy of H–H bond.
The bond dissociation enthalpy is the change in enthalpy when one mole of covalent bonds of a gaseous covalent compound is broken to form products in the gas phase.
Note--- that it is the same as the enthalpy of atomization of dihydrogen. This is true for all diatomic molecules. For example: Cl2(g) →2Cl(g) ; ∆Cl-ClHӨ= 242 kJ/ mol
O2(g) →2O(g) ; ∆O-OHӨ= 428 kJ/ mol
In the case of polyatomic molecules, bond dissociation enthalpy is different for different bonds within the same molecule.
Polyatomic Molecules: Let us now consider a polyatomic molecule like methane, CH4. The overall thermochemical equation for its atomization reaction is given below:
In methane, all the four C – H bonds are identical in bond length and energy. However, the energies required to break the individual C – H bonds in each successive step differ :
Therfore--
In such cases we use mean bond enthalpy of C and H bond.
For example in CH4, ∆C-HHӨis calculated as:
= 416 kJ/ mol
Note-- It has been found that mean C–H bond enthalpies differ slightly from compound to compound, as in
CH3CH2Cl, CH3NO2etc, but it does not differ in a great deal*. Using Hess’s law, bond enthalpies can be calculated. The standard enthalpy of reaction, ∆rHӨis related to bond enthalpies of the reactants and
products in gas phase reactions as:
The net enthalpy change of a reaction is the amount of energy required to break all the bonds in the reactant molecules minus the amount of energy required to break all the bonds in the product molecules.
Note-- This relationship is approximate and is valid when all substances (reactants and products) in the reaction are in gaseous state.
(d) Enthalpy of Solution (symbol : ∆solHӨ)
Enthalpy of solution of a substance is the enthalpy change when one mole of it dissolves in a specified amount of solvent.
Note--The enthalpy of solution at infinite dilution will be equal to the enthalpychange observed on dissolving the substance in an infinite amount of solvent when theinteractions between the ions (or solute molecules) are negligible.
When an ionic compound dissolves in a solvent, the ions leave their ordered positions on the crystal lattice. These are now more free in solution. But solvation of these ions (hydration in case solvent is water) also occurs at the same time. This is shown diagrammatically, for an ionic compound, AB (s)

The enthalpy of solution of AB(s), ∆solHӨ,in water is, therefore, determined by the selective values of the lattice enthalpy, ∆latticeHӨand enthalpy of hydration of ions, ∆hydHӨas
solHӨ = ∆latticeHӨ + ∆hydHӨ
For most of the ionic compounds, ∆solHӨis positive and the dissociation process is endothermic. Therefore the solubility of most salts in water increases with rise of temperature. If the lattice enthalpy is very high, the dissolution of the compound may not take place at all.
Q. Why do many fluorides tend to be less soluble than the corresponding chlorides?
Lattice Enthalpy-- The lattice enthalpy of an ionic compound is the enthalpy change which occurs when one mole of an ionic compound dissociates into its ions in gaseous state.
Since it is impossible to determine lattice enthalpies directly by experiment, we use an indirect method where we construct an enthalpy diagram called a Born-Haber Cycle
Let us now calculate the lattice enthalpy of Na+Cl(s) by following steps given below :
1. Na(s)→Na(g) , sublimation of sodium metal, ∆subHӨ=108.4 kJ/ mol
2. Na(g)→Na+(g)+ e−1(g) , the ionization of sodium atoms, ionization enthalpy ∆iHӨ= 496 kJ/ mol
3. 1/2Cl 2(g) → Cl(g), the dissociation ofchlorine, the reaction enthalpy is half the bond dissociation
enthalpy. 1/2bondHӨ= 121 kJ/ mol
4. Cl(g) + e−1(g)→Cl(g) electron gained by chlorine atoms. The electron gain enthalpy,
egHӨ= –348.6 kJ/ mol.
5. Na+ (g) + Cl(g)→ Na+Cl (s)
The sequence of steps is shown in Fig. 6.9, and is known as a Born-Haber cycle. The importance of the
cycle is that, the sum of the enthalpy changes round a cycle is zero.
Applying Hess’s law, we get,
latticeHӨ = 411.2 +108.4 +121+ 496 − 348.6
latticeHӨ = +788kJ
for NaCl(s)→ Na+ (g) + Cl(g)

Fig Enthalpy diagram for lattice enthalpy of NaCl
Internal energy is smaller by 2RT ( because ∆ng = 2) and is equal to + 783 kJ/ mol.
Now we use the value of lattice enthalpy to calculate enthalpy of solution from the expression:
solHӨ = ∆latticeHӨ + ∆hydHӨ
For one mole of NaCl(s), lattice enthalpy = + 788 kJ/ mol and ∆hydHӨ= – 784 kJ/ mol( from the literature)
solHӨ = + 788 kJ/ mol – 784 kJ/ mol = + 4 kJ/ mol
The dissolution of NaCl(s) is accompanied by very little heat change.
Note-- The first law of thermodynamics tells us about the relationship between the heat absorbed
and the work performed on or by a system.
Limitations of the first law of thermodynamics ---. It puts no restrictions on the direction of heat flow. However, the flow of heat is unidirectional from higher temperature to lower temperature.
Second law of thermodynamics OR SPONTANEITY of the system.--
In fact, all naturally occurring processes whether chemical or physical will tend to proceed spontaneously (by their own force) in one direction only.
For example,--1. a gas expanding to fill the available volume, 2. burning carbon in dioxygen giving carbon dioxide. These and manyother spontaneously occurring changes show unidirectional change
But heat will not flow from colder body to warmer body on its own, the gas in a container will not spontaneously contract into one corneror carbon dioxide will not form carbon and dioxygen spontaneously.. Q. ‘what is the driving force of spontaneously occurring changes ?
Q. What determines the direction of a spontaneous change ?
Here we study about the criterion for these processes whether these will take place or not and meaning of
the spontaneous reaction or change.
We think by our common observation that spontaneous reaction is one which occurs immediately when contact is made between the reactants. e.g. combination of hydrogen and oxygen. These gases may be
mixed at room temperature and left for many years without observing any perceptible change. Although the reaction is taking place between them, it is at an extremely slow rate. It is still called spontaneous reaction.
So spontaneity means having the potential toproceed without the assistance(help) of external
agencyí.
Limitation--it does not tell about the rate of the reaction or process.
Another aspect of spontaneous reaction or process, cannot reverse their direction on their own.We may summarise it as follows:
A spontaneous process is an irreversible process and may only be reversed by some external agency.
(a) Is decrease in enthalpy a criterion for spontaneity ?
e.g. flow of water down hill or fall of a stone on to the ground, we find that there is a net decrease in potential energy in the direction of change. Similarly we may conclude that a chemical reaction is spontaneous in a given direction, because decrease in energy has taken place, as in the case of exothermic reactions.
For example:
The decrease in enthalpy in passing from reactants to products may be shown for any exothermic reaction on an enthalpy diagram as shown in Fig.(a).
Fig. (a) Enthalpy diagram for exothermic reactions
Thus, the postulate that driving force for a chemical reaction may be due to decrease in energy is ‘reasonable’ as the basis of evidence.
Now let us examine the following reactions:
These reactions though endothermic, are spontaneous. The increase in enthalpy may be represented on an enthalpy diagram as shown in Fig.(b).
Fig (b) Enthalpy diagram for endothermicreactions
It means decrease in enthalpy alone can not be criterion for spontaneity of all the reactions.
(b) Entropy and spontaneity---Let us examine such a case in which H = 0 i.e., there is no change in
enthalpy, but still the process is spontaneous. E.g. diffusion of two gases into each other in a closed container which isolated from the surroundings as shown in Fig.
.
  1. (b) Fig. Diffusion of two gases
The two gases, say, gas A and gas B arerepresented by black dots and white dots respectively and separated by a movablepartition [Fig. (a)]. When the partition is withdrawn [Fig.( b)], the gases begin todiffuse into each other and after a period of time, diffusion will be complete.
We say that the system (b) has become less predictable.
We may now formulate another postulate: in an isolated system, there is always a tendency for the systems’ energy to become more disordered or chaotic and this could be a criterion for spontaneous change.
Thus we introduce another thermodynamic function, entropydenoted asS. The above mentioned disorder is themanifestation of entropy. Hence entropy as a measure of the degree of randomness or disorder in the system. The greater the disorder in an isolatedsystem, the higher is the entropy.
In a chemical reaction--- 1. If the structure of the products is very much disordered than that of the reactants, there will be a resultant increase in entropy.
2. For a given substance, the crystalline solid state is the state of lowest entropy (most ordered), The gaseous state is state of highest entropy.
3. Whenever heat is added to the system, it increases molecular motions causing increased randomness in the system. Thus heat (q) has randomising influence on the system.
Entropy, like any other thermodynamic property such as internal energy U and enthalpy H is a state
function and ∆S is independent of path.
Q. Can we then equate ∆S with q ?
Ans. A system at higher temperature has greater randomness in it than one at lower temperature. Thus, temperature is the measure of average chaotic motion of particles in the system.
Note-- Heat added to a system at lower temperature causes greater randomness than when the same quantity of heat is added to it at higher temperature.
This suggests that the entropy change is inversely proportional to the temperature. S is related with q and T for a reversible reaction as :
The total entropy change (∆Stotal) for the system and surroundings of a spontaneous process is given by
-When a system is in equilibrium, the entropy is maximum, and the change in entropy, ∆S = 0.
-We can say that entropy for a spontaneous process increases till it reaches maximum and at equilibrium the change in entropy is zero.
Since entropy is a state property, we can calculate the change in entropy of a reversible process by
We find that both for reversible and irreversible expansion for an ideal gas, under isothermal conditions,
U = 0, but ∆Stotali.e., is not zero for irreversible process. Thus, ∆U does not discriminate
between reversible and irreversible process, whereas ∆S does.
(c) Gibbs energy and spontaneity
Hence for a system, it is the total entropy change, ∆Stotalwhich decides the spontaneity of the process. But most of the chemical reactions fall into the category of either closed systems or open systems. Therefore, for most of the chemical reactions there are changes in both enthalpy and entropy.
It is clear that neither decrease in enthalpy nor increase in entropy alone can determine the direction of spontaneous change for these systems. For this purpose, we define a new thermodynamic function the Gibbs energy or Gibbs function, G, as
G = H - TS
Gibbs function, G is an extensive property and a state function. The change in Gibbs energy for the system,
Gsyscan be written as
Gsys= ∆HsysTSsysSsysT
At constant temperature, ∆T = 0
Gsys= ∆HsysTSsys
OR ∆G = ∆H T S
Thus, Gibbs energy change = enthalpy change – temperature × entropy change,
And is referred to as the Gibbs equation,
G has units of energy because, both H and the TS are energy terms, since
TS = (K) (J/K) = J.
Now let us consider how G is related to reaction spontaneity.
We know, ∆Stotal= ∆Ssys+ ∆Ssurr
If the system is in thermal equilibrium with the surrounding, then the temperature of the surrounding is same as that of the system. Also, increase in enthalpy of the surrounding is equal to decrease in the enthalpy of the
system. Therefore, entropy change of surroundings,

Rearranging the above equation:
TStotal= TSsys– ∆Hsys
For spontanious process, S total=0, so
By using Gibbs equation, the above equation can be written as
Hsys is the enthalpy change of a reaction, TSsys is the energy which is not available to do useful work. So G is the net energy available to do useful work and is thus a measure of the ‘free energy’. Hence it
is also known as the free energy of the reaction.
G gives a criteria for spontaneity at constant pressure and temperature.
(i) If G is negative (< 0), the process is spontaneous.
(ii) If G is positive (> 0), the process is non spontaneous.
EFFECT OF TEMPERATURE ON SPONTANEITY OF REACTION
GIBBS ENERGY CHANGE AND EQUILIBRIUM
The sign and magnitude of the free energy change of a chemical reaction shows:
(i) Prediction of the spontaneity of the chemical reaction.
(ii) Prediction of the useful work that could be extracted from it.
Let us now examine the free energy changes in reversible reactions.
‘Reversible’ means a special way of carrying out a process such that system is at all times in perfect
equilibrium with its surroundings. It means a 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 a decrease in free energy,) It is possible only if at equilibrium the free energy of the system is minimum. If it is not, the system would spontaneously change to configuration of lower free energy. So, the criterion for equilibrium
Gibbs energy for a reaction in which all reactants and products are in standard state, is related to the equilibrium constant of the reaction as follows:
We also know that
For strongly endothermic reactions, the value of ∆rHӨ will be large and positive. In such a case, value of K will be much smaller than 1 and the reaction is unlikely to form much product. In case of exothermic reactions, ∆rHӨ is large and negative, and ∆rGӨ is likely to be large and negative too. In such cases, K will
be much larger than 1. We may expect strongly exothermic reactions to have a large K, and hence can go to near completion. ∆rGӨ also depends upon ∆rSӨ , if the changes in the entropy of reaction is also taken into account, the value of K or extent of chemical reaction will also be affected, depending upon whether ∆rSӨ is positive or negative. Using above equation ,
(i) It is possible to obtain an estimate of GӨ from the measurement of HӨ and SӨ , and then calculate K at any temperature for economic yields of the products.

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