Chemical Thermodynamics Class 11 Notes Chemistry Chapter 5

Basic Terminology

HEAT, ENERGY AND WORK
HEAT (Q):

• Energy is exchanged between system and surround in the form of heat when they are at different temperatures.
• Heat added to a system is given by a positive sign, whereas heat extracted from a  system is given negative sign.
• It is an extensive property.
• It is not a state function.

ENERGY:

• It is the capacity for doing work.
• Energy is an extensive property.
• Unit : Joule.

Work (W):

• Work = Force × Displacement i.e. dW = Fdx
• Work done on the system is given by positive sigh while work done by the system is given negative sign.
• Mechanical Work or Pressure-Volume Work: work associated with change in volume of a system against an external pressure.
•  Work done in reversible process: W=
   
  W = – 2.303 nRT log v₂/v₁  =  –2.303 nRT log p₁/p₂
• Wok done in isothermal reversible contraction of an ideal gas:
  ?W = – 2.303 nRT log v₂/v₁  =  –2.303 nRT log p₁/p₂              
•  Unit : Joule.

Internal Energy (E or U):

• Sum of all the possible types of energy present in the system.
• ΔE  = heat change for a reaction taking place at constant temperature and volume.
• ΔE is a state function.
• It is an extensive property.
• Value of ΔE is -ve for exothermic reactions while it is +ve for endothermic reactions.
  

First Law of Thermodynamics:
Energy can neither be created nor destroyed although it can be converted from one form to another.
or
Energy of an isolated system is constant.
Mathematical Expression
Heat observed by the system = its internal energy + work done by the system.
i.e. q = dE + w
For an infinitesimal process
dq = dE + dw
Where, q is the heat supplied to the system and w is the work done on the system.
For an ideal gas undergoing isothermal change ΔE =0.
so q= -w.

For an isolated system, dq=0

so, dE = - dw

For system involving mechanical work only

ΔE = q - pdV

At constant volume i.e. isochoric process

ΔE = qv  

For Isothermal Process

ΔE = 0

or

 q = - pdV  =-W

For adiabatic process

?q = 0
or
ΔE = W

Enthalpy (H):

H = E+PV
At constant pressure:
dH = dE + pdV
For system involving mechanical work only
dH = Q_P (At constant pressure)
For exothermic reactions:
dH = -ve
For endothermic reactions:
dH = +ve

Relation between dH and dE:
dH = dE + dn_g RT
Where,
dn_g = (Number of moles of gaseous products - Number of moles of gaseous reactants)

Heat capacity:

Amount of heat required to rise temperature of the system by one degree.   
C = q / dT
Specific heat capacity: Heat required to raise the temperature of 1 g of a substance by one degree.
C_s = Heat capacity / Mass in grams
Molar heat capacity: Heat required to raise the temperature of 1 g of a substance by one degree.
C_m = Heat capacity / Molar mass.
Heat capacity of system at constant volume:
 C_v  =  (dE/dT)_v
Heat capacity of system at constant pressure:
C_p  =  (dE/dT)p

C_p – C_v = R
Variation Of Heat Of Reaction With Temperature:

dC_P = (dH₂ - dH₁)/(T₂-T₁)   &    dC_V = (dE₂ - dE₁)/(T₂-T₁
Bomb Calorimeter

Heat exchange = Z × Δ
Z–Heat capacity of calorimeter system
ΔT– Rise in temp.
Heat changes at constant volumes are expressed in ΔE and Heat changes at constant
pressure are expressed in dH.

ENTHALPIES OF REACTIONS

(i) Enthalpy of Formation

• Definitions - Enthalpy change when one mole of a given compound is formed from its elements
• Example - H₂(g) + 1/2O₂(g) →  2H₂O(l), Δ_fH = –890.36 kJ / mol

(ii) Enthalpy of Combustion

• Definitions - Enthalpy change when one mole of a substance is burnt in oxygen.
• Example - CH₄ + 2O₂(g) →CO₂  + 2H₂O(l), Δ_combH  = –890.36 kJ / mol

(iii) Enthalpy of Neutralization

• Definitions - Enthalpy change when one equivalent of an acid is neutralized by a base in dilute solution.
• Example - H⁺ (aq) + OH^– (aq) → H₂O(l) Δ_neutH = –13.7 kcal

(iv) Enthalpy of Hydration

• Definitions - Enthalpy change when a salt combines with the required number of moles of water to form specific hydrate.
• Example - CuSO₄(s) + 5H₂O (l) → CuSO₄5H₂O, Δ_hydH° = –18.69 kcal

(v) Enthalpy of Transition

• Definitions - Enthalpy change when one mole of a substance is transformed from one allotropic form to another allotropic form.
• Example - C (graphite) → C(diamond), Δt_ransH^° = 1.9 kJ/mol

(vi) Enthalpy of Sublimation

• Definitions - Enthalpy change when one mole of a solid substance sublime at constant temp. and 1 bar pressure
• Example - CO₂(S) → CO₂(g) Δt_fusH° = 6.00 kJ/mol

(vii) Enthalpy of fusion

• Definitions - Enthalpy change when one mole of a solid melts
• Example - H₂O(S) → H₂O (l) Δt_subH° = 73.00 kJ/mol

Hess’s Law of constant heat summation:
The total enthalpy change of a reaction is the same, regardless of whether the reaction is completed in one step or in several steps.
According to Hess’s law: ΔH = ΔH₁ + ΔH₂

Born–Haber Cycle:
Applying Hess’s law we get
ΔH₁ + 1/2 ΔH₂ + ΔH₃ + ΔH₄ + ΔH₅ = ΔH_f (MX)  (Lattice energy)
Lattice energy: The change in enthalpy that occurs when 1 mole of a solid crystalline substance is formed from its gaseous ions.

SECOND LAW OF THERMODYNAMICS
Statement:
It is impossible to take heat from a hot reservoir and convert it completely into work by a cyclic process without transferring a part of it to a cold reservoirs.

Mathematically:

ΔS =  q_rev/T

Where,
ΔS is entropy change.
Entropy is the degree of randomness thus it increases with increase in randomness of particles of the system i.e.  ΔS is positive for melting of ice.
At equilibrium, ΔS = 0
For a spontaneous process, ΔS > 0
Entropy change in an isothermal reversible expansion of a gas
Spontaneous Processes: These type of physical and chemical changes occur of its own under specific circumstances or on proper initiations. For example: Flow of liquids from higher to lower level.

Gibbs free energy(ΔG):

ΔG = ΔH - TΔS
ΔG = nRT ln K_eq
ΔG = nFE_cell
At equilibrium, ΔG = 0
For spontaneous process, ΔG < 0

Bond Energies:

Average amount of energy required to break one mole bonds of that type in gaseous molecules.
H–OH(g) → 2H(g) + ½O(g)                
ΔH = 498 kJ
O–H(g) → H2(g) + ½O₂ (g)        
ΔH = 430 Kj
ΔH_O–H = (498 + 430)/2 = 464 kJ mol^–1

Efficiency of a heat engine (carnot cycle):

W = R (T₂ – T₁) ln v₂/v₁
q₂ = RT₂ ln v₂/v₁
W = q₂

Efficiency (h). h =