CHAPTER 5

Thermodynamic:

            It is the branch of chemistry which deals with flow of heat during the chemical reaction.

Types of Thermochemical Reactions
Thermo-chemical reactions are of two types.
1. Exothermic Reactions
2. Endothermic Reactions

Endothermic Reaction:

            The reactions in which heat energy is absorb is called Endothermic Reaction. (∆H=117.6 KJ/mol_­^-1­)

Reactants + Heat —-> Products

Exothermic Reaction:

            The reactions in which heat energy is evolved is called exothermic reaction. (∆H=-394 KJ/mol_­^-1­)

Reactants —-> Products + Heat

Thermodynamic Terms:

1) System:

            The material portion of the universe which is under observation or under consideration is called system.

2) Surrounding:

            The environment of a system is called surrounding.

3) State:

            The description of the properties of a system at a particular moment is called state.

Types of State:

1) Initial State:

            The properties of a system at the start of experiment are called initial state.

2) Final State:

            The properties of a system at the end of experiment are called Final State.

Change in State:

            The difference in the properties of a system from initial state to final state is called change in state.

Change=Final state – Initial state.

Example:

            Change in Temperature ∆T = T₂ – T₁

Properties of System:

1) Intensive Properties:

            The physical properties of a system which are independent of the quantity of matter are called Intensive Properties.

Example: Temperature, density, pressure, specific gravity , melting point, boiling point, refractive index.

2) Extensive Properties:

            The physical properties of a system which depend to the quantity of matter are called Extensive Properties.

Example: Mass, volume, Mole numbers, Enthalpy, entropy, Internal Energy etc.

Macroscopic Properties:

            The properties of a system in a bulk rather than individual are called Macroscopic properties.

First Law of Thermodynamics:

            This law was given by Helmholtz in 1847, it states that

“Energy can neither be created nor destroyed but it can transform from one form to another”

In other words the total energy of a system and its surrounding must remain constant.

Derivation:

            Consider an ideal gas in a cylinder, having an internal energy E₁. The volume of the gas be V₁ when ‘q’ quantity of heat energy is supplied to the system then part of the energy is utilized in doing work ‘w’, by pushing the piston in the upward direction and the remaining energy is absorb by the gas. Hence the internal energy becomes ‘E₂’.

            Now, the change in the internal energy is

∆E=E₂ – E₁

According to the first law of thermodynamic, the supplied energy must be equal to the sum of change in internal energy and the work done i.e;

∆E=q - W

OR

q=∆E + W ----- (1)

Pressure-Volume Work:

(fig 5.1 from book)

            Let a gas is enclosed in a cylinder having moveable piston at a height ‘h₁’. The area of cylinder is ‘A’. When heat energy is supplied to the system from a piston moves upward at height ‘h₂’

            We know that,

Work done = force x displacement

OR

W = F x d

But,

P = F/A

Or,

            F = P x A

            .:. W = P x A x d

As,

            A = L x B

And d=∆h

            .:. W = P x L x B x ∆h

Again;

            L x B x ∆h = ∆V

            .:. W = P∆V

Substituting the value of ‘W’ in eq(1), we get,

            q = ∆E + P∆V ----- (2)

Process at Constant Volume:

            Let, an ideal gas is enclosed in a cylinder having a fixed piston. The internal energy of the system is ‘E₁’. When ‘q’ quantity or heat energy is supplied to the system then its internal becomes ‘E₂’.

Now, change in internal energy

∆E = E₂ – E₁

Change in volume

            ∆V = 0

Applying 1^st law of thermodynamics

            q = ∆E + P∆V

Substituting the value of ∆V

            q = ∆E + P(0)

            .:. q_v = ∆E

            Hence, the heat energy supplied at constant volume is equal to change in internal.

Process at Constant Pressure:

            Let, an ideal gas is enclosed in a cylinder fitted with moveable piston. Its internal energy is ‘E₁’ and initial volume is ‘V₁’. When ‘q’ quantity of heat energy is supplied to the system then its internal energy becomes ‘E₂’ and ‘V₂’, respectively

            Now; the change in internal energy

            ∆E = E₂ – E₁

and the change in volume

            ∆V = V₂ – V₁

Applying 1^st law of thermodynamics

            q = ∆E + P∆V

or

            q_p = ∆E + P∆V ----- (1)

`.` the pressure is constant

Substituting the value of ∆E and ∆V in eq(1),

We get

            q_p = (E₂ – E₁) + P(V₂ – V₁)

or

            q_p = E₂ – E₁ + PV₂ – PV₁

or,

            q_p = (E₂ + PV₂) – (E₁ + PV₁)

as (E + PV) = H (enthalpy)

.:. q_p = H₂ – H₁

Hence;

            q_p = ∆H

Thermochemistry:

It is a branch of chemistry which deals with the measurement of heat evolved or absorbed during a chemical reaction.
The unit of heat energy which are generally used are Calorie and kilo Calorie or Joules and kilo Joules.
1 Cal = 4.184 J
OR
1 Joule = 0.239 Cal

FORMULA’S

q / ∆Q = ∆E + W

W = P∆V        .:. W = Work Done => w.d by the system = +ve … => w.d on the system = -ve

∆Q = ∆E + P∆V        .:. q = Heat Energy => Heat energy is supplied to the system = +ve … => Heat energy is released = -ve

∆Qp = ∆H

∆Qv = ∆E       .:. Internal Energy => Increase +ve / decrease –ve

Hess’s Law of Constant Heat of summation:

Statement:

            If a chemical reaction is completed in a single step or in several steps the total enthalpy change for the reaction is always constant.
OR

The amount of heat absorbed or evolved during a chemical reaction must be independent of the particular manner in which the reaction takes place.

e.g:

            A à D , ∆H   Single step

            A à B , ∆H₁   1^st Step

                B à C , ∆H₂    2^nd Step

            C à D , ∆H₃   3^rd Step

.:. ∆H₁ + ∆H₂ + ∆H₃ = ∆H

Verification:

            Formation of Na₂CO₃ form NaOh:-

2NaOH + CO₂ à NaHCO₃ + H₂0 , ∆H=-90kj/mole       (single step)

NaOH + CO₂ à NaHCO₃ , ∆H₁=-49 Kj/mole                 (1^st step)

NaHCO₃ + NaOH à Na₂CO₃ + H₂0, ∆H₂=-41 Kj/mole            (2^nd Step)

Total change in enthalpy : ∆H₁ + ∆H₂

=(-49) + (-41)

=-90Kj/mole

Application of Hess’s Law:

1)      It help in calculating the heat of reaction

2)      It help in calculating the heat of formation, when direct measurement is not possible.

Heat of Formation:

            The changes in enthalpy when one mole of a compound is obtain form its element is called heat of formation.

Standard Heat of Formation:

            The change in enthalpy when one mole of a compound is obtain form its element at 25 Degree C and 1atm pressure is called Standard Heat of Formation.