Thermodynamics notes

Class XI notes

1. 1 GURUKUL Sector-7, Rohini
THERMODYNAMICS
It is the branch of chemistry which deals with energy changes in the chemical reactions and their
feasibility.
System – Part of universe in which observations are made.
Surroundings - Part of universe excluding system = Universe – System
Types of System –
a) Open System - There is exchange of energy and matter between system and surroundings. e.g.
Open beaker.
b) Closed system - There is no exchange of matter but exchange of energy is possible between
system and surroundings. e.g. presence of reactants in a closed vessel made up of conducting
material.
c) Isolated system - There is no exchange of energy as well as matter between system and
surroundings. e.g. Presence of reactants ina thermos flask or any insulated closed vessel.
State of the system - The system must be described in order to make useful calculations by specifying
quantitatively each of the properties such as pressure (P), volume (V) and temperature (T) as well as
the composition of the system.
The state of the system is described by its measurable or macroscopic (bulk) properties.
State functions or state variables - Those properties of a system which depend only on the state
(initial and final state) of the system and do not depend upon the path of the system e.g. P.V.T.
Path functions - Those properties of a system which do not depend upon the state (initial and final
state) of the system but depend only upon the path of the system e.g. work done.
Types of processes :
a) Isothermal process – that occurs at constant temperature.
b) Isobaric process – that occurs at constant pressure
c) Isochoric process – that occurs at constant volume
d) Adiabatic process – that occurs at constant heat i.e., there is no heat exchange between system
and surroundings.
Internal energy (U or E) - It is the energy possessed by the system. It depends upon –
a) Heat passes into or out of the system
b) Work is done on or by the system
c) Matter enters or leaves the system
Internal energy is a state function as it depends upon initial and final state of the system
We cannot calculate the exact value of internal energy as it is the sum of many different kinds of
energy such as mechanical energy, chemical energy, electrical energy, etc. whose values are difficult to
calculate separately.
For adiabatic process, ΔU = U2 – U1 = Wad
FIRST LAW OF THERMODYNAMICS –
- According to it, energy of an isolated system is constant

2. 2 GURUKUL Sector-7, Rohini
- Energy can neither be created nor be destroyed
Mathematically, ΔU = q + w
Where U = change in internal energy, q = heat absorbed by the system, W = work done on the system.
Proof – Let initial internal energy = U1, final internal energy = U2. If ‘W’ work is done on the system
and ‘q’ heat is absorbed. Then,
U2 = U1 + q + W
U2 – U1 = q + W
ΔU = q + W
Sign Conventions –
Work done by the system = (.), Work done on the system = (+)
Heat absorbed → q = (+) and U = (+), Heat evolved → q = (–) and U = (–)
During expansion W = (–). During compression W = (+)
For adiabatic process, ΔU = – Wad
For thermally conducting process, ΔU = – q
For closed system, ΔU = q + w
If the work is done by the system, ΔU = q – W → ΔU = q – W
APPLICATIONS :
1. Work – (I = length, A = area) For 1 mole of a gas
Change in volume V = 1 × A
As P = Force / area
→ F on piston = Pext . A


Work done = F × distance = P × A × I = P × (–ΔV). Therefore, W = – P Δ V
2. 
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
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
 
 
 
nx
x
dx
&xdx
V
V
nnRT–
V
dV
nRT–WdV
V
nRT
–W
dVP–dW
2
V
1
V
2
V
1
V
1
2
W
o
2
V
1
V
.ext
l
l

W = – 2.303 nRT log V2 / V1
Also, As Boyle’s law : P1/P2 = V2/V1
→ W = – 2.303 nRT log P1/P2

P (P )istonexternal
area
Mole of ideal gas

3. 3 GURUKUL Sector-7, Rohini
Now cases :
a) For isothermal process (T = constant) i.e., for isothermal expansion of ideal gas into vacuum,
W = 0 and q = 0 → ΔU = 0
b) For isothermal irreversible change, q = W = P (V2 – V1)
c) For isothermal reversible change, q = – W = nRT InV2 / V1
d) For adiabatic change, q = 0 → ΔU = Wad
Enthalpy (H) :
It is the heat absorbed or heat evolved by the system.
As U = qp – P Δ V (expansion)
qp = ΔH = ΔU + P Δ V
Proof : ΔU = q – PΔV
U2 – U1 = qP – P (V2 – V1) at constant P
qp = (U2 + PV2) – (U1 + PV1)
Therefore, H = U + PV
qp = H2 – H1 = ΔH ( it is also known as state function)
Now, if H = U + PV
ΔH = ΔU + ΔpV + pΔV
As H = qp (at constant Pressure) i.e. ΔP = 0
Therefore, ΔH = ΔU + PΔV
ΔH = (+) for endothermic reaction, ΔH = (.) for exothermic reaction.
Derivation ΔH = ΔU + ΔngRT
Proof ΔH = ΔU + PΔV = ΔU + P (V2 – V1)
Now, if PV = nRT. So, PV1 = n1RT and PV2 = n2RT
ΔH = ΔU + n2RT – n1RT = ΔU + (n2 – n1) RT
ΔH = ΔU + ΔngRT
Where Δng = n2 – n1 for gaseous state = np – nr
– Extensive Property – Property whose value depends upon the quantity of matter contained in
the system. e.g. mass, volume, internal energy, enthalapy, heat capacity. etc.
– Intensive Property - Property whose value does not depend on the quantity or size of matter
present in the system. e.g. temperature, density, pressure
– Heat Capacity (C) – It is the amount of heat required to raise the temperature of a substance
by IC.
Specific Heat Capacity (CS) –Amount of heat required to raise the temperature of 1gram of a
substance by 1°C (1K).
Molar Heat Capacity (Cm) –Amount of heat required to raise the temperature of 1 mole of a
substance by 1°C.
Q = m C ΔT where ΔT = T2 – T1

4. 4 GURUKUL Sector-7, Rohini
Relationship between CP and CV for ideal gases –
For constant volume, qV = CVΔT = ΔU and at constant P, qP = CPΔT = ΔH
For 1 mole of a gas, ΔH = ΔU + R ΔT
On putting the values of ΔH and ΔU, we get CPΔT = CVΔT + R ΔT
VP = VC + R Therefore, CP – CV = R
We can measure energy changes (ΔH and ΔU) associated with chemical or physical processes by an
experimental technique called calorimetry. The process is carried out in calorimeter immersed in a
known volume of a liquid.
Knowing that heat capacity of liquid in which calorimeter is immersed and heat capacity of
calorimeter, it is possible to determine the heat evolved in the process of measuring temperature
changes under two different values, a) at constant volume, b) at constant pressure.
ΔH = HP – HR
Enthalpy changes - Reactants → Products
The enthalpy changes accompanying the reaction is known as reaction enthalpy (ΔH)
Standard enthalpy of reaction - The reaction enthalpy changes for a reaction when all the
participating substances are in their standard states [standard temperature and pressure (1 bar) [Δ,H
0
]
DEFINITIONS
1. Standard enthalpy of fusion (ΔfusH
0
) - It is the heat evolved or absorbed by the system when
one mole of a solid substance melts in standard state.
2. Standard enthalpy of vapourisation (vapH
0
) - Amount of heat require to vapourize mole of a
liquid at constant temperature and under standard pressure (1 bar)
3. Standard enthalpy of sublimation (ΔsubH
0
) - Amount of heat absorbed or evolved when
sublimes at constant temperature and standard pressure (1 bar)
ΔH is directly proportionsal to the intermolecular interactions in substance
4. Standard enthalpy of formation (ΔtH
0
) - Amount of heat absorbed or evolved when 1 mole
of compound is formed from its elements in their most stable states of aggregation (reference
state) at 25°C and 1 bar.
Example - H2 (g) + 1/2O2 (g) → H2O (1) ΔtH
0
= 285.8KJ/mol; C (graphite) + 2H2 (g) → CH4 (g)
5. Standard enthalpy of combustion – (ΔCH
0
) Enthalpy change when 1 mole of a substance is
burnt in presence of air completely.
6. Enthalpy of atomization (ΔaH
0
) – It is enthalpy change on breaking 1 mole of bonds
completely to obtain atoms in the gas phase.
Example - CH4 (g) → C (g) + 4H (g) Na (s) → Na (g)
7. Bond enthalpy (ΔbondH
0
) - Energy is required break a bond and released to form a bond.
The amount of heat absorbed or released due to break or form one mole of bonds of reactants is
known as bond enthalpy. ΔrH
0
= BER – BEP

5. 5 GURUKUL Sector-7, Rohini
8. Enthalpy of solution (ΔsolH
0
) – It is the enthalpy change
when 1 mole of a substance is dissolved in specified
amount of solvent.


9. Lattice enthalpy – It is the enthalpy change which occurs when 1 mole of an ionic compound
dissociates into its ios in gaseous state (ΔLattice H or U)
10. Heat of hydration - Amount of enthalpy change when 1 mole of the anhydrous salt combines
with required number of moles of water so as to change into the hydrated salt.
CuSO4 + aq → CuSO4.5H2O
11. Heat of neutralization of an acid by a base - It is the heat change when 1 gram equivalent of
the acid is neutralized by a base, the reaction is carried out in dilute aqueous solution.
When 1 gram equivalent of HCI is neutralized by NaOH or vice-versa, 57.1 kJ of heat is
produced
– Heat of neutralization is taken for 1 gram equivalent of acid and base because neutralization
involves combination of 1 mole of H
+
ions with 1 mole of
–
OH ions. 1 gram of any acid on
complete dissociation gives 1 mole of H
+
ions but 1 mole of an acid may not give 1 mole of H
+

ions.
Example - 1 mole H2SO4 → 2 moles of H
+
ions on complete dissociation,
1 gram equivalent H2SO4 → 1 mole of H
+
ion
Hess’s law of constant heat summiation –
According to it, if a reaction takes place in several steps then its standard enthalpy is the sum of the
standard enthalpies of the intermediate reaction into which the overall reaction may be divided at same
temperature. OR
According to it, if a reaction takes place in one step or in many number of steps, the amount of energy
released or absorbed (enthalpy change) always remain constant at constant temperature.
Example - C (graphite, s) + O2 (g) → CO2
Step 1 : C (s) + ½ O2 → CO (g); Step 2 : CO (g) + ½ O2 → CO2 (g) i.e. ΔH = H1 + H2
Spontaneity - means having the potential to proceed without the assistance of external agency.
A spontaneous process is an irreversible process and may only be reversed by some external agency.
Entropy (S) – It is the measure of degree of randomness or disorder in the system.
Solid → Liquid → Gas [ΔS = (+); randomness increase]
Gas → Liquid → Solid [ΔS = (–) ; randomness decrease]
→ T
q
= S
→ For isolated system, ΔU = 0, ΔS = 0
→ For spontaneous process, Stotal = Ssystem + Ssurrounding > 0
→ When a system is in equilibrium, entropy is maximum but ΔS = 0. It is a state function.
Gibbs free energy - It is the amount of energy of a system that can be converted into useful work
(free E) Gibbs function, G = H – TS
ΔG = ΔH – TΔS – SΔT
At constant T, ΔG = ΔH – TΔS Δ
sol
H =
lattice
H +
hyd
HΔΔ
– – –
AB(s) A
+
(aq) + B
–
(aq)
Δ
sol
H
–
Δ
lattice
H
–
Δ
hyd
H
–
A
+
(g) + B
–
(g)

6. 6 GURUKUL Sector-7, Rohini
At equilibrium, ΔG = 0
Derivation : For spontaneous process, ΔG = (.)
As we know that, Strotal = Ssystem + Ssurrounding
If the system is in thermal equilibrium with surroundings, increase in enthalpy of surroundings =
decrease of enthalpy of system.
OG
]ST–HG[OG–
O)T–(
OH–STOStates,processeoustansponFor
H–STST
T
H
–SS
T
H
–HS
syssys
syssys
syssystotal
sys
systotal
sys
surrsurr
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



→ For non-spontaneous process, ΔG = (+)
→ Two driving processes / forces for a spontaneous process –
a) Enthalpy change (ΔH) → ΔH = (.)
b) Entropy change (ΔS) → ΔS = (+)
c) ΔS = (.) then T should be too low. Hence, ΔG = ΔH – TΔS would be (.)
Effect of temperature on spontaneity of reaction –






a) ΔG
0
= FE° where, n = number of electrons, F = Faraday’s Constant = 96500 C,
E
0
= Standard electrode potential = E°cathode E°amode
b) ΔG° = – 2.303 R T log Ke = equilibrium constant = [products] / [reactants]
Second law of thermodynamics –
According to it, entropy of universe is continuously increasing → Stotal > 0 for spontaneous process.
Third law of thermodynamics – Given by Planck
According to it, entropy of a pure and perfectly crystalline substance is zero at the absolute zero
of temperature.
Born-Haber cycle – Enthalpy diagram – for NaCl, MgCl2, CaO, CaF2, AlCl3
1. (–) (+) (–) Spontaneous
2. (–) (–) (–) at low T Spontaneous
3. (–) (–) (+) at high T non-spontaneous
4. (+) (+) (+) at low T non-spontaneous
5. (+) (+) (–) at high T spontaneous
6. (+) (–) (+) at all T non-spontaneous
Δ
r
H
–
Δ
r
S
–
Δ
r
G
–
Discription

7. 7 GURUKUL Sector-7, Rohini
THERMODYNAMICS QUESTIONS
1. The heat of combustion of ethyl alcohol (C2H5OH) is 1380.7 kJ/mol. If the heat of formation of
CO2 and H2O are 394.5 and 286.6 kJ/mol respectively, calculate the heat of formation of ethyl
alcohol. [–268.1 kJ/mol]
2. Calculate heat of formation of sucrose (C12H22O11) from the following data –
i) C12H22O11 + 12O2 → 12CO2 + 11H2O, ΔH = – 5200.7 kJ
ii) C + O2 → CO2 , ΔH = – 394.5 kJ
iii) H2 + ½O2 → H2O + 285.8 kJ [–2677.1 kJ/mol]
3. Calculate heat of hydrogenation of ethylene, given that heat of combustion of ethylene,
hydrogen and ethane are –1410.0, –286.2 and – 1560.0 kJ/mol respectively at 298 K.
4. ΔH for the reaction :
H – CN (g) + 2H2 (g) → (g) is – 150 kJ

Calculate the bond energy of C ≡ N bond.
[Given bond energies of C – H = 414 kJ/mol. H – H = 435 kJ / mol, C – N = 293 kJ/ mol
N – H = 396 kJ / mol] [ 839 kJ/ mol]
5. Calculate ΔH° for reaction :
CH2 = CH2 + 3O2 → 2CO2 + 2H2O
Given that the average bond energies of different bonds are -
Bond : C – H O = O C = O O – H C = C [–964 kJ/mol]
Bond E : 414 499 724 460 619 kJ / mol
6. Enthalpy and entropy changes of a reaction at 27°C are 40.63 kJ/mol and 108.8 J/mol/k
respectively. Predict the feasibility of the reaction at 27°C.
7. The heat of combustion of benzene in a bomb calorimeter (i.e. constant volume) was found to
be 3263.9 kJ/mol at 25°C. Calculate heat of combustion of benzene at constant pressure.
[ – 3267.6 kJ/mol]
8. Calculate the maximum work obtained when 0.75 mol of an ideal gas expands isothermally
and reversibly at 27°C from a volume of 15L to 25L.
9. The equilibrium constant for a reaction is 10. What will be the value of ΔG°? T = 300K.
10. What are the signs of ΔH and ΔS for the reaction – I2 (g) → 2I (g) ?
11. Enthalpies of formation of CO (g), CO2 (g), N2O (g) & N2O4 (g) are – 110, – 393, 81 and 9.7
kJ/mol respectively. Find the value of ΔrH for the reaction –
N2O4 (g) + 3 CO(g) → N2O (g) + 3CO2 (g)
12. Calculate the enthalpy of reaction for : (Cl4(g) → (g) + 4Cl (g) & calculate bond enthalpy of J
C – Cl in (Cl4(g)). Given : Δvap H
θ
= 30.5 kJ/mol, Δf H
θ
(Cl4) = – 135.5 kJ/mol, ΔaH
θ
(C)
= 715.0 kJ /mol & ΔaH
θ
(Cl2) = 242 kJ/mol. H H
| |
H – C– N – H
|
H