IMPULSE PU SCIENCE COLLEGE, HUBLI
UNIT 7
THERMODYNAMICS
Thermodynamics: It is a branch of science, which deals with interrelationship between heat energy and other forms of
energy during physical and chemical processes. To understand thermodynamics, we need to know certain
thermodynamic terms.
1) System :The part of the Universe which is under study is called system. Ex : water in a open beaker
2) Surroundings : The portion of the universe excluding system is called surroundings.
Ex: Space around the beaker
3) Boundaries : A real or imaginary surface that separate the system from the surroundings is called boundary.
Types of system: There are 3 types of system
1) Open system: A system which can exchange both energy and matter with its surroundings is called open
system.
Ex : Water is heated in a open beaker.
2) Closed system:A system which can exchange only energy with its surroundings but not matter is called closed
system .
Ex : Water is in contact with its vapour in a closed vessel .
3) Isolated system: A system neither exchange energy nor matter with its surroundings is called isolated system.
Ex :Water is in contact with its vapour in a insulated closed vessel.
Types of processes :1)
2)
3)
4)
Isothermal process : A process is carried out at constant temperature is called isothermal process. (dT=0)
Isobaric process : A process is carried out at constant pressure is called isobaric process.dp=0
Isochoric process : A process is carried out at constant volume is called isochoric process. dv=0
Adiabatic process : A process in which no heat is exchange between system and its surroundings is called
adiabatic process. (dq=0)
5) Cyclic process : A process which brings back a system into its original state after a series of exchanges is called
cyclic process ex : Nitrogen cycle .∆H =0
6) Reversible process : A process which can reverse at any instant of time by changing the driving force by an
infinitesimal amount (small ) is called reversible process. Ex : Electrochemical reaction in a Daniel cell.
7) Irreversible process : A process which can occurs rapidly and the whole system is away from equilibrium
is called irreversible process.
Ex : All spontaneous processes .
Types of Properties : There are two types of properties . They are
1) Intensive property : These are properties, which doesnot depends on quantity of matter.
Ex : Temperature, pressure, density, specific heat, surface tension, m.p, b.petc .
2) Extensive properties : These are properties, which depends on quantity of matter.
Ex : mass, volume, heat, internal energy, enthalpy, entropy etc.
State of a system : It is a condition of system expressed by giving definite values for its properties such as
temperature, pressure, volume.
Ex : Water at O0C =Ice (solid state)
Water at 250C =liquid state
State variable or state functions : By specifying the values of certain properties such as temperature, pressure.
Volume etc, gives a clear idea of the state of a system. These properties are called state functions.
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Internal energy(U) : It is the sum of different kind of energy possessed by a system due to its chemical nature as well
as temperature, electronic , transitional etc.
Note :Types of energies are rotational, vibrational, electronic , transitional etc.
It is very difficult to measure absolute value of internal energy of the system. But change in internal energy , when a
system changes from state 1 to state 2 can be measured absolutely.
Internal energy may change, when
a) Heat absorbed or leaves the system
b) Work is done on the system or by the system
c) Matter enters or leaves the system
Change in internal energy by doing work:
It is a mode of energy transfer if there is difference in pressure of the system and surroundings till the pressure
equal.
Consider a system of state 1, its internal energy is U1 Let ‘w’ is the work done on the system. Therefore system changes
from state 1 to state 2, its internal energy be U2
∴ ΔU = U2 – U1
So internal energy is a state function in adiabatic work,Wad = U2 – U1 = ΔU∴ Wad = ΔU
w is +ve, when work is done on the system. w is -ve, when work is done by the system.
Change in internal energy by heat change :
The energy exchanged between a system and the surroundings when their temperatures are differentis called
heat (q)
If T1 is the temperature of a system and it absorbs heat from surroundings. So its temperature increases to T 2.
∴q = T2 – T1 = Δ U
∴ ΔU = qv, when no work is done at constant volume.
q is +ve, when heat is absorbed by system and q is -ve, when heat leaves the system
Change in internal energy by both work done and heat change:
(When change of state is brought by both doing work and transfer of head)
Consider a system of state 1 & its internal energy is U 1 + q. Let ‘ W ‘ is the amount of work done on the system. So
system changes from state 1 to state 2. Its energybecomes U1 +q+W. Let U2 is the internal energy of a system of state 2.
∴ U2 = U1 +q +w
∴ U2 - U1= q + w
ΔU = q + w
It is the mathematical expression of I law of thermodynamics.
It states that “energy can neither be created nor be destroyed, but it transformed from one form to another
form.”
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Derivation for mechanical work done expression in irreversible process :
Consider a gas enclosed in a cylinder, it is fitted with weightless or friction less piston. If external pressure (Pex) is more
than internal pressure (Pin). The piston moves inwards till the pressure equal to Pex. ∴ The volume decreases from Vi to
Vf
Piston
During this compression, piston moves a distance ‘l’ and area of cross section ‘A’
l
∴Volume change = l× A = Δ V = (Vf – Vi)
WKT pressure =
∴
𝐹𝑜𝑟𝑐𝑒
𝐴𝑟𝑒𝑎
Force on piston = Pressure( ex)× area
=Pex×A
Workdone (W) = Force × displacement
∴ W = Pex× A ×l
W =Pex× - Δ V
W = -Pex× ΔV
W = - Pex (Vf– Vi)
If pressure is not constant at every stage of compression, but changes in number in number of finite steps, work done
on the gas will be summed over all the steps.
∴ W = - ∑ 𝑃𝑒𝑥 Δ V
Work done during isothermal reversible process
If pressure is not constant, but changes during the process such that it is always infinitesimally greater than pressure of
gas, then, at each stage of compression, the volume decreases by an infinitesimal amount, dV.
vf
W = -∫vi Pex dv
Pex= Pin + dp in case of compression and Pin – dp in expansion process. In general Pex = Pin±dp in reversible process.
We can relate work to internal pressure of the system under reversible conditions
vf
wrev = - ∫vi Pex × dv
vf
= - ∫vi ( Pin ± 𝑑𝑝) dv
vf
∴ w = - ∫vi Pin dv
For ‘n’ mole of gas PV = nRT∴P =
since dp× dv is very small
nRT
V
∴at constant temperature ( isothermal process),
vf nRT
×
V
W = - ∫vi
dv
W = - nRT log e(Vf – Vi)
W = -2.303nRT log10
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Vf
Vi
Or
P1
W = -2.303nRT × log10P2
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Note : Free expansion : Expansion of gas in vacuum is called free expansion. So Pex = 0& no workdone during free
expansion.∴ Δ U = q + w
Δ U = q – Pex× ΔV;Δ U = qv
Problem : 1) Calculate the maximum workdone, when 2 moles of an ideal gas expands isothermally and
reversibly from a volume of 10 L to 20 L at 270c. [R = 8.314J/mol]
𝑉2
W = -2.303RT log10𝑉1
20
W = -2.303 × 2 × 8.314 × 300× log1010
W = - 2.303 × 2× 8.314 × 300×0.3010
W = -3457.97 J/ mol
2) Calculate the maximum work done, when 3 mole of an ideal gas contract isothermally and reversibly
from a volume of 2 tm to 10 atm at 298K [R = 8.314 J / K/mol]
𝑃1
W =-2.303 nRT log10𝑃2
2
= -2.303 x 3x 8.314 x 298 x log1010 5
= - 2.303 x3x8.314x298x(0-0.6990)
= 2.303x3x8.314 x298x0.6990
= 11965.16 J / mol
Note :Work done in isothermal irreversible expansion of an ideal gas are of two types:
a . Expansion against constant external pressure
w =-pex ∆V
b. Expansion against zero pressure(in vaccum)(free expansion)
w=-pex ∆V = -0 x∆V = 0
No work done during free expansion of the ideal gas whether the process is reversible or irreversible.
Interpretation of first law of Thermodynamics :
1. In an isothermal process
∆U =0 (temperature remains constant and there is no change in internal energy)
putting this condition in ∆U = q + w
0=q+w
q = -w
ie work done in an isothermal process is equal to the heat absorbed by the system
2. In adiabatic change, q =0
∴ Δ U = q – w ;ΔU = 0 – w ; W = - Δ U
The work done in this system is equal to change in internal energy of the system.
3. Isochoric Process:∆V =0, ∴ w = 0 hence heat change at constant volume is called internal energy change( q v = ΔU)
Enthalpy or heat content [H] :
Total heat content of a system at constant pressure is called enthalpy.
Enthalpy of a system is defined as the sum of internal energy and the product of pressure and volume of the system.
Mathematically, it is defined as
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H = U + PV
where H=Enthalpy, U= Internal energy, P = Pressure
V=Volume of system
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Show that heat change at constant Pressure = ΔH
WKT, heat change at constant volume is called internal energy change.
i.e qv = ΔU
But most of the chemical reactions are conducted at constant pressure [in flask or test tubes]. So we need to define
another state function ΔH
According to first law of thermodynamics
ΔU =qp – P Δ V at constant pressure. It can rewrite at two states as
U2 - U1 = qP – P (V2 – V1)
U2 - U1 = qP – PV2 + PV1
∴ (U2 + PV2) - (U1 + PV1) = qp
H2 – H1 = qp
∴
ΔH = qp
Note :- If Hp is enthalpy of reactant, and HR is the enthalpy of product. Enthalpy change means “heat evolved” or “heat
absorbed” at constant pressure.
ΔH = HP -HR
∴ It is energy equation in thermo chemistry
Types of reactions:Exothermic reaction : A reaction in which heat is liberated is called exothermic reaction.
C(s) + O2(g)
CO2(g)
: Δ H = -393.5kJ
or
C(s) + O2(g)
CO2(g)
+ 393.5kJ
In exothermic reactions, HP is less than HR ∴ ΔH = -ve and in exothermic reactions, products are more stable than
reactants.
Endothermic reactions : A reaction in which heat is absorbed is called endothermic reactions.
N2(g) + O2(g)
2NO(g) : ΔH = 180 kJ OR
N2(g)+ O2(g)
2NO(g)– 180kJ
In endothermic reactions, Hp> HR∴ Δ H = + Ve & in such reactions products are less stable than reactants .
Thermo chemistry:
It is a branch of chemistry which deals with the study of heat changes taking place in all physical and chemical changes.
Thermochemical equations :
These are balanced chemical equations, which indicates physical states of reactants and products and also indicates
heat liberated or absorbed in a reaction.
Ex : C(s) + O2(g)
CO2(g) : Δ H =-393.5kJ
N2(g)+ O2(g)
2NO(g) : ΔH = + 180kJ
Relation between𝚫H and 𝚫U
The difference betweenΔH and ΔU is not significant in solids and liquids because volume change (is very
small)is not much significant in solids and liquids. However the difference is significant in gases.
Consider a gaseous reaction, if VA and VB be the volume of gaseous reactants and gaseous products and nA and nB be
the number of moles of gaseous reactants and products respectively at constant pressure & temperature. According to
ideal gas law.
PVA = nART and PVB = nB RT
Thus PVB - PVA = nBRT -nART
P(VB - PVA) = (nB - nA) RT
∴ P Δ V = RT Δng
Δ H = Δ U + R T Δ ng
∴ Δ H = Δ U + P ΔV becomes
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Note : 1) The equation which relates enthalpy and internal energy system is H = U+PV’
2. The equation which relates enthalpy and internal energy of a reaction is ΔH = ΔU + RT Δng
Δng = number of moles of gaseous product - number of moles of gaseous reactant
1) Write Δ H and ΔU relationship for the following reactions.
a) N2(g) + 3H2(g)
2NH3(g)
Δng = 2 – 4 = -2
∴ Δ H = ΔU +RT x -2
∴
ΔH =Δ U – 2RT
b) C(s) + O2(g)
CO2(g)
Δn = 1-1 =0
ΔH = ΔU +RT x 0
c) PCl5(g)
PCl3(g) + Cl2(g)
ΔH = ΔU + RT x 1
Δ H= Δ U + RT
NOTE: If Δn = 0, Δ H = ΔU
Δng = -ve, ΔH <Δ U:Δng = +ve,
ΔH >Δ U
Problem : Molar enthalpy change for vapourisation of 1 mol of H2O at constant pressure is 41KJ / mole. Calculate heat
change at constant volume at 1000C [R = 8.314J/K / mol]
ΔH =ΔU + RT Δ ng
H2O(1)
H2O(g) : Δn = 1-0=1
ΔU = ΔH – RTΔn
H2O(1)
H2O(g) :
Δn= 1- 0= 1
ΔU = Δ H – RT x Δn
ΔU = 41000 – 1 x 8.314 x 373
= 41,000 – 3096
= 37904J / mol = 37, 904KJ / mol
Heat capacity (C ): It is defined as the quantity of heat required to raise the temperature of a substance by one Kelvin.
q α ∆T ; q = C∆T
where C= a proportional constant known as heat constant & ∆T = change of temperature
Heat capacity depends on size, composition, nature of substance. Higher the heat capacity, smaller is the temperature
rise & vice versa.
Molar heat capacity (cm)
cm=
C
n
It is defined as the quantity of heat required to raise the temperature of 1 mole of substance by one Kelvin or 1oC with
no change in physical state.
q = ncm∆T
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where n= no. of moles, cm = molar heat capacity
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Unit of molar heat capacity : J/mol/K
Specific heat capacity (c ) or (s): It is the quantity of heat required rise the temperature of 1g or one unit mass of a
substance by one Kevin or one degree Celsius with no change in physical state
Relationship between the heat capacity and specific heat capacity of a substance is C=mc, where m is mass of a
substance in grams and ‘c’ is specific heat of a substance.
Heat changes in a chemical reaction :- (ΔH) = q
Δ H = mass x specific heat x temperature change
∴ ΔH = q = m x c x ∆T
Or
q =mx c x ΔT
q=Cx ΔT
Derive Mayer equation CP – CV = R for an ideal gas:
At constant volume, heat capacity is denoted by Cv and at constant pressure, heat capacity is denoted by Cp∴ equation
for heat at constant volume is qv = CvΔT and for heat at constant pressure is qp = Cp∆T
We know that, For one mole of an ideal gas,PV = RT
ΔH =Δ U + ΔPV
ΔH =ΔU + ΔRT
because PV = RT
∴ ΔH =ΔU + R Δ T
∴ CPΔ T = CvΔT + R Δ T
CpΔT – CvΔT = R ΔT
(Cp – Cv)∆ T = R∆T
∴ Cp – Cv= R
Measurement of 𝚫H and 𝚫U : An instrument used to measure 𝚫U and 𝚫H is bomb calorimeter and
calorimeter.
Bomb Calorimeter-Measurement of ΔU :- Bomb calorimeter is used to determine the heat of combustion at constant
volume. Calorimeter is a device to measure the heat absorbed or released during a chemical or physical change.
Bomb calorimeter consists of sealed steel vessel, immersed in water bath to prevent heat is lost to surroundings.
Combustible substance placed in bomb and burnt in oxygen and the heat evolved is transferred to the water, so the
temperature increases. Bomb calorimeter is sealed, so its volume doesnot change ∴ ΔV = 0. By knowing heat capacity,
temperature change, heat changes can be calculated according to qv = Cv x∆ T
Heat produced during combustion= heat capacity of calorimeter x sp. Heat of water x ∆T
Measurement ofΔH :The heat change during a reaction under constant pressure is called enthalpy of reaction. The
enthalpy of reaction is determined by using the principle of calorimetry.
Calorimeter consists of a polythene bottle fitted with a stopper carrying thermometer and stirrer. A known volume of
reaction mixture is taken in a calorimeter, note down the temperature before reaction and after reaction. By knowing
heat capacity, temperature change, and mass of the solution, heat change can be calculated according to
qp = ΔH = mass x heat capacity x Δ T
Δ H= m x c x t J / mol
Problem : ‘1’ g of graphite is burnt in bomb calorimeter, at 298K and 1 atm pressure, temperature rises from 298 to
299K and heat capacity in 20.7KJ / K. Calculate enthalpy change for a reaction.
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C(s) + O2(g)
CO2(g)
qv= -CvΔT
=-20.7 x 1
=-20.7KJ / K
But for 1 mole (12g), qv = 12 x 20.7
qv= 2.48 x 102KJ / mol
In this case,
ΔH = ΔU, because
Δng=0
Enthalpy of a reaction (∆r H) :
It is the enthalpy change that occurs, when reactants involved in a reaction have completely reacted to form product
under given conditions.
C(s) + O2(g)
CO2 : ΔRH = -393.5KJ
∴enthalpyof a reaction is -393.5KJ
H2(g) + Cl2(g)
2HCl(g) : ΔRH = - 184KJ
∴enthalpy of a reaction is – 184KJ
Note :- In general
Δ H = Hp - Hr
It is the energy equation in thermochemistry
Standard enthalpy of reactions( ∆ r H0):
It is the enthalpy change produces, when reactants involved in a reaction have reacted completely at standard
conditions i.e 298K and 1 bar pressure in their standard states.
Note. : Standard State: It is the state of a pure substance at one bar pressure and 298K.
The standard conditions are indicated by a subscript degree sign (0 )
Standard enthalpy of free elements is considered as zero for example O2(g), H2(g), N2(g), Cgraphite, S(rhombic), Pwhite etc are zero.
Factors affecting enthalpy of a reaction :1)
2)
3)
4)
5)
quantities of reactant
Physical state of reactants and products
Temperature
Allotropic modification
ΔU and ΔH
Types of enthalpy of reaction :
1)
2)
3)
4)
5)
6)
Enthalpy of formation
Enthalpy of combustion
Enthalpy of phase change
Enthalpy of automization
Enthalpy of solution
Lattice enthalpy
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Standard Enthalpy of formation :∆f H0
It is the enthalpy change produces, when one mole of the compound is formed from its elements in their standard
states ie 1 bar pressure and 298K
Ex: C(s) + O2(g)
CO2:
ΔH = -393.5KJ
∴enthalpy of formation of CO2 is -393.5KJ
Note: exothermic compounds are more stable than endothermic compounds
Write formation reactions for the following compounds H2O, NH3, CH4, C2H6, CH3COOH, C2H5OH, C6H6, C6H12,O6.
1
2
1) H2 + O2
H2O
2)
3)
4)
5)
2NH3
CH4
C2H6
CH3COOH
N2 + 2H2
C + 2H2
2C + 3H2
2C +2H2
1
6) 2C + 2 O2
C2H5OH
7) 6C + 3H2
C6H6
8) 6C + 6H2 + 3O2
C6H12,O6
Enthalpy of combustion(∆c H) : It is the enthalpy change produces, when one mole of a substance is completely burnt in
air or oxygen.
Ex: 1) C(s) + O2(g)
CO2(g) :
Δ H = -393.5KJ
∴enthalpy of combustion of carbon solid is – 393.5KJ
EX :2) CH4(g) + 2O2(g)
CO2(g) + 2H2O(g) : ΔH = -890KJ
∴ Enthalpy of combustion of CH4gas is–890KJ
Note:1) Enthalpy of formation may be +ve, Enthalpy of combustion is always –ve
2) Write combustion reaction for the following substance H2, C2H6, C2H5OH, CH3COOH. C6H12O6, CS2, CH3OH
& C6H6,
1)
2)
3)
4)
5)
6)
7)
8)
H2 + ½ O2
H2O
C2H6 + 7/2 O2
2CO2 + 3H2O
C2H6OH + 3O2
2CO2 + 3H2O
CH3COOH + 2O2
2CO2 + 2H2O
C6H12O6 + 6O2
6CO2 + 6H2O
CS2 + 3O2
CO2 + 2SO2
CH3OH + 3/2 O2
CO2 + 2H2O
C6H6 + 15/2 O2
6CO2 + 3H2O
Enthalpy of solution(𝚫sol H) : It is the enthalpy change produces, when one mole of a solute is completely dissolved in
specified amount of solvent.
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Enthalpy of solution at infinite dilution( Integral enthalpy of solution): It is the enthalpy change produces, when one
mole of a solute is completely dissolved in such a large quantity of solvent so that further addition of solvent does not
cause any change in enthalpy.
Ex :NaCl(s) + aqua
NaCl(aq) :
ΔH = +5.5KJ
∴enthalpy of solution of NaCl(s) is + 5.5KJ
Note : It may be exothermic or endothermic.
Enthalpy of Hydration of ions:It is defined as the energy released by hydration of ions furnished by one mole of ionic
compound.
When an ionic salt is dissolved in water it involves two energy changes:
1. Lattice enthalpy and 2. Enthalpy of hydration
∆solH = ∆latticeH0 + ∆hydH0
Enthalpy of atomization:
It is the enthalpy change on breaking one mole of bonds completely to obtain gaseous atoms.
Ex : 1) H2(g)
2H(g) :
2) CH4(g)
ΔaH = 435KJ
C(g) + 4H(g)ΔaH = 1665KJ
∴ ΔC-H H = ¼ × 1665KJ = 416KJ / mol
Note :ΔRH = ∑ bond enthalpy of products - ∑ bond enthalpy of reactants.
Bond enthalpy (∆bondH)
1. Bond dissociation enthalpy: It is the enthalpy change when one mole of covalent bonds of gaseous covalent
compounds is broken to form products in the gas phase.
It is same as the enthalpy of atomization for all Diatomic molecules
Ex;
H2(g)
→
2H(g) :
ΔbondH = 435 kJ/mol
Cl2(g) → 2Cl(g) :ΔbondH = 242 kJ/mol
Mean bond enthalpy: in case of polyatomic molecules, all the bonds between similar atoms are identical in bond length
but differ in their strength. So this case, the mean of bond dissociation enthalpies of all the bonds present in the
compound is taken. Thus, in such cases bond enthalpy is called the mean bond enthalpy.
Ex: In water, there are two O-H bonds. The bond dissociation enthalpies are
H2O(g) → H(g) + OH(g)
∆H = 497.8 kJ
OH(g) → H(g) + O(g)
∆H = 428.5 kJ
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The bond enthalpy is the average of these two bond dissociation enthalpies
Bond enthalpy of O-H bond = 497.8 + 428.5 = 463.15 kJ
2
Enthalpy of phase transformation : There are 3 important types of enthalpy of phase change.
They are
1) Enthalpy of fusion
2) Enthalpy of vaporization
3) Enthalpy of sublimation
1) Standard or Molar enthalpy of fusion( 𝚫 fus H0): It is the enthalpy change that accompanies when one
mole of solid substance melts in their standard state.
Ex :H2O(g)
H2O(1) : Δ fus H0 = 6.00KJ / mol
2) Standard or Molar Enthalpy of vaporization ( vapH0): It is the enthalpy change produces, when one mole
of a liquid vaporizes at constant temperature and one bar pressure.
Ex: H2O(1)
H2O(g) : Δvap H0 = +40.79Kj /mol
3) Enthalpy of sublimation: (𝚫 sub H0): It is the enthalpy change produces, when one mole of a solid
substance sublimes at constant temperature & 1 bar pressure.
Ex: CO2(s)
CO2(g) : Δsub H0 = +73.00KJ /mol
Lattice enthalpy : It is the enthalpy change produces, when one mole of ionic compound dissociates into gaseous
ions.
Ex: NaCl(s)
Na+(g) + Cl-(g) : ΔlatticeH = + 788 kJ / mol
Since it is impossible to determine lattice enthalpies directly by experiment, an indirect method i.e., Born –Haber
method is used, in which the principle of Hess’s law is applied.
Calculation of Lattice enthalpy of NaCl solution by Born Haber cycle:
Formation of Sodium chloride crystal
Na(s) + ½ Cl2(g)
NaCl(s): ΔfH0 = - 411.2 kJ/mol
It can also occurs in five steps:
1) Sublimation of sodium metal to get gaseous sodium atom
Na(s)
Na(g): ΔsubH0 = -108.4kJ /mol
2) Ionization sodium atoms to sodium ion
Na(g)
Na+(g) : Δi H0 = + 495.6kJ /mol
3) Dissociation of Chlorine molecule into chlorine atom
½ CL2(g)
Cl(g) : Δbond H0 = 121.5 kJ/mol
4) Conversion of gaseous chlorine atom to chlorine ion
Cl(g) + eCl-(g): Δeg H0 = -348.6kJ/mol
5) Gaseous Na+ combines with gaseous Cl- forming NaCl crystals
Na+(g) + Cl-(g)
NaCl(s) : ΔLattice H0= ?
Applying Hess’s law:
∆fH0= ∆subH0 +∆iH0 + ∆bondH0+∆egH0 +∆latticeH0
ΔLattice H0 = -411.2- 108.4- 495.6-121+348.6=- 788kJ /mol
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It can be summarized as
∆subH0
∆iH0
Na(s)
Na(g)
+
∆bondH0
∆egH0
1/2 Cl2(g)
Cl(g)
NaCl
Na+(g)
+
Cl-(g)
∆latticeH0
Laws of Thermochemistry : There are two important laws of Thermochemistry .
1) Laplace and Lavoisier law(First law of Thermochemistry): It states that “ If the forward reaction is
exothermic, the backward reaction is equally endothermic.” Or The Enthalpy of reaction is exactly equal but
opposite in sign for the reverse reaction.
C(s)+ O2(g)
CO2(g) : ΔH= - 393.5 KJ
C2(g) + O2(g)
C (g) + O2: ΔH = + 393.5 KJ
2) Hess’s Law of constant Heat Summation:
“ The Enthalpy change involved in a given reaction is the same, whether the reaction occurs in single step or in
several steps”. ie
Δ H = Δ H1 + Δ H2+ …....…..
Illustration : Carbon can be converted to carbon dioxide by two different methods
I) Carbon burns in excess of air or Oxygen to give CO2 in one step .
Single step : C(s) + O2(g)
CO2 : ΔH = -393.5kJ
II) Carbon burns in limited amount of air or Oxygen to give CO. The CO is then oxidized to form CO2
C(s) + ½ O2(g)
CO(g) : ΔH1 = -110.5kJ
CO(g) + ½ O2(g)
CO2(g) : ΔH2 = -283.0kJ
According to Hess law
ΔH= ΔH1 +ΔH2
-393.5 =-110.5 + -283
Application of Hess law :
1) Using Hess law, thermo chemical equation can be added or subtracted or multiplies or divided like algebraic
equations.
2) Using Hess law, enthalpy of many reactions e.g heat of combustion, heat of transformation etc can be calculated
where direct determination is not possible.
Problems :
1) For a reaction, H2 + Cl2
2HCl, ΔH =-184KJ. What is the enthalpy of formation of HCl.
Enthalpy of formation =
−184
=
2
92KJ
2) For a reaction 2Al2O3
4Al +3O2 : ΔH = +3340KJ
What is the enthalpy of formation of Al2O3? Given reaction is decomposition.
∴ Formation is 3Al +3O2
2Al2O3 :ΔH = -3340KJ
∴enthalpy of formation of Al2O3 is
−3340
= -1670KJ
2
3) Enthalpy of formation of A and B are -92KJ and -180KJ respectively. Which is more stable.
Stability is inversely proportional to enthalpy so between A and B, B is more stable because enthalpy is less.
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4) Enthalpy of combustion of C2H5OH(l) is -1367KJ. What amount of heat liberated by burning ‘23’ g of C2H5OH(l)
Combustion of C2H5OH can be represented by a equation
C2H5OH + 3O2
2CO2 + 3H2O :ΔH = -1367KJ.
It shows that 1 mole (46g) of C2H5OH burnt and produces 1367 KJ of heat
∴23g of C2H5OH burnt and produces
23 ×1367
=
46
683.5KJ of heat
5) Explain thermo chemical equation
C6H6(1)+
15
O2
2
6CO2 + 3H2O :
ΔH = -3368KJ.
It shows that one mole of C6H6(1) is completely bourns with
15
mole
2
of oxygen and give 6 mole of CO2 and 3
mole of water with the liberation of 3368KJ of heat.
6) When one mole of CH4 gas burnt in 2 mole of oxygen gas, gives one mole of CO2 and 2 mole of water with the
liberation of 890KJ of heat. Write thermo chemical equation.
CH4(g) + 2O2(g) + 2H2O(1) :
ΔH = -890KJ
7) Enthalpy of combustion of C(s), H2(g) and CH4 gas are -393.5KJ, -285KJ and -890KJ respectively. Calculate
enthalpy of formation of CH4.
Solution.: - The given data is
C(s) + O2(g)
CO2(g) : ΔH = -393.5KJ-------(1)
H2(g) + ½ O2(g)
H2O(1) : ΔH = -285KJ -----------(2)
CH4(g) + 2O2(g)
CO2(g) + 2H2O(1)∴ ΔH = -890KJ--------(3)
Required equation is C(s) + 2H2(g) :
CH4(g) : ΔH =?
To get required equation, 1+ 2 x 2-3
∴ C(s) + 2H2(g)
CH4(1) : ΔH = - 73.5KJ
∴ Enthalpy of formation of CH4(g) is - 73.5KJ
8) Enthalpy of formation of CO2(g) , H2O(1) are -393.5KJ, -285KJ and -274KJ respectively. Calculate enthalpy of
combustion of C2H5OH.
Given data is
C(s) + O2(g)
CO2(g) : ΔH = -393.5KJ-------(1)
H2(g) + ½ O2(g)
H2O(1) : ΔH = -285KJ -----------(2)
2C(s) + 3H2(g) + ½ O2
C2H5OH(1) :
ΔH = -274KJ--------(3)
Required equation is
C2H5OH + 3O2
2CO2 + 2H2O :Δ H = ?
To get required equation
1x2+2x3–3
2C + 2O2
2CO2 :
ΔH = -787KJ
3H2 + 3/2 O2
3H2O :ΔH = -855KJ
C2H5OH
2C + ½ O2 : O2 : ΔH = -274KJ
∴C2H5OH + 3O2
2CO2 + 3H2O : Δ H = - 1368KJ
∴ Enthalpy of combustion of C2H5OH is -1368KJ
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Spontaneous Process: A process which can occurs on its own , without any external support is called spontaneous
process.
Ex: Physical spontaneous process
1) Melting of ice
2) Evaporation of water
Chemical spontaneous process
1) Rusting of iron
2) Burning of white phosphorus
3) Zn + H2SO4 → ZnSO4 + H2
Non Spontaneous process : A process which does not occurs on its own, but they need external support is called
non spontaneous process.
Ex :
1) De composition of H2O into hydrogen and oxygen
2) Water flow out of well by pumping.
Criteria for spontaneity of a reaction:
1. Tendency of system to acquire a state of minimum energy
2, Tendency of system to acquire a state of maximum disorder
Entropy (S) : It is a property, which is measure of degree of randomness or disorder of a system.
More is the randomness, the larger is value of entropy
In general, entropy of solids are less than liquids, which is less than gases.
i,eSsolids<Sliquids<Sgases
The equation used to calculate change in entropy in a reversible chemical reaction is
ΔS =Sproducts - S reactants
q
But the equation used to calculate small entropy change is ∆s = Trev =
∆H
T
∴ The unit of entropy is J / K / mol
Spontaneity and Entropy:
 For an isolated system:
The process in isolated system will be spontaneous, if the entropy change of the system is positive (entropy increases
during the change) i.e ∆S > 0, But ∆U = 0
 For non-isolated system: (open and closed system)
For predicting the spontaneity of the reaction, the total change in entropy of the system and entropy of the surrounding
should be considered.
The total entropy change (∆Stotal) for the system and surroundings of spontaneous process is given by
∆Stotal =∆Ssystem + ∆Ssurr should be positive
ie Total entropy change should be positive ∆Stotal> 0
 The entropy of a system at equilibrium is maximum and change in entropy at equilibrium is zero.
Thus the entropy criterion about the spontaneity of a process is as follows
∆Stotal> 0, The process is spontaneous
∆Stotal< 0, The process is non spontaneous
∆Stotal = 0, The process is at equilibrium
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Criteria or driving forces of spontaneous process:
There are two driving forces of a spontaneous process. They are
1) Tendency of a reaction to attain a state of minimum energy(enthalpy).
2) Tendency of a reaction to attain state of maximum randomness (entropy).
Thus spontaneous reaction depends on two factors called enthalpy and entropy. Generally in a spontaneous process,
enthalpy decreases and entropy increases. ieΔS = + ve and ΔH = -ve. But the term enthalpy and entropy cannot alone
predict whether the process is 100% spontaneous or non spontaneous. Because in all the spontaneous process
enthalpy does not decreases (melting of ice) and entropy doesnot increases (condensation of vapours)
Free energy : (G) It is an important factor, which can predict whether the process is 100% spontaneous or non
spontaneous.
It is the amount of energy available for doing work or Free energy is equal to the difference between enthalpy and the
product of absolute temperature and entropy.
i.e This equation is known as Gibbs Free energy equation G = H - TS
Where G = Free energy, H = Enthalpy of the reaction, S =Entropy and T = temperature in
Kelvin.
The change in free energy at constant temperature and pressure is given as
This equation is called Gibbs-Helmholtz equation.
∆G = ΔH- TΔS
Free energy change in a chemical reaction is the difference between free energy of products and free energy reactants.
ie
ΔG = GP - GR
Standard Free energy Change ∆G0: It is defined as free energy change of reaction, when the reactants and products are in
their standard state.
∆G0 = ΔH0 – TΔS0
Free energy change provides a criteria for spontaneity at constant pressure and temperature.
If ∆G = -ve (< 0), the reaction is spontaneous (feasible)
If ∆G = +ve (> 0), the reaction is non spontaneous.
If ∆G = 0 , the reaction is at equilibrium
Note :1) Δ G = 0 at equilibrium
2) Generally in a spontaneous process, enthalpy decreases, entropy increases and free energy decreases.
3) Standard free energy of formation of any elements is taken as zero.
Effect of temperature on spontaneity of reactions
From the knowledge of free energy change. One can predict, whether the process is spontaneous or non spontaneous.
1) If ∆H = -Ve =,
∆S = +ve,
ΔG = + ΔH – T . ΔS
ΔG = - Δ H – T (+ ΔS)
ΔG = - Δ H – T ΔS
ΔG = - Ve
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The process is spontaneous at all temperature.
2) If ∆ H = + Ve&∆S = -Ve,
∴ ΔG =Δ H - T ΔS becomes
ΔG = + ΔH – T (- ΔS)
ΔG = + ΔH + T ΔS
∴
ΔG = + ve
+Ve
The process is non spontaneous at all temperature.
3) If ∆ H = + Ve&∆S = +Ve
∴ ΔG =ΔH - TΔS becomes
ΔG = + ΔH – T (+ Δ S)
ΔG = + ∆ H – T ΔS
At high temperature TΔS >ΔH
∴ Δ G = -Ve and the process is spontaneous only at high temperature.
At low temperature TΔS <Δ H
∴ Δ G = +Ve and the process is non spontaneous at low temperature.
4) If
H = - ve and ΔS = -Ve
∴ ΔG =ΔH - TΔS becomes
∴ ΔG =Δ H – (- ΔS)
ΔG = - ΔH + T Δ S
At low temperature, T
S < H
∴ ΔG = - Ve and the process is spontaneous only at low temperature.
At high temperature, T S > H
∴ ΔG = + Ve and the process is non spontaneous at high temperature
Relationship between standard free energy change and equilibrium constant:
∆G0= - 2.303RT log 10 KP
Where
ΔG0 = standard free energy change ,R = gas constant
T = Absolute temperature and Kp = equilibrium constant
Second law of Thermodynamics: “All spontaneous process are irreversible and tend to attain equilibrium” Or
Total energy of the universe remains constant but total entropy increases. or
“The entropy of the universe is continuously increasing”.
Third law of Thermodynamics: “Entropy of perfect crystalline solid at absolute zero temperature is zero (Absolute zero
temperature = 0K or – 2730C)”
Problem :
Calculate standard free energy change , when equilibrium constant of a reaction is 50 at 270C [R = 8.314J / K/mol]
ΔGo = -2.303RT log10Kp
= -2.303 x 8.314 x 300 x log10 50
= -2.303 x 8.314 x 300 x 1.6990
= - 9759.29J / mol
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3) Standard free energy change of a 62KJ calculate equilibrium constant at 298K. [R = 8.314J /K/mol]
ΔG0 = -2.303RTlog10kp
62000 = -2.303 x 8.314 x 298 x log10kp
−62000
∴log10Kp =2.303 ×8.314×298
Log10Kp = -10.86
∴Kp
Kp
= antilog -10.86
= antilog 11.14
Kp= 1.461 x 10-11
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