Adama Science and Technology

University

Pre-engineering general chemistry (Chem1011)

Thermodynamic Equilibrium and

Thermochemistry
Course instructor: Eneyew Tilahun(M.Sc.)

AA
 Principles of thermodynamics
 The word thermodynamics is combination of two words:

 thermes meaning heat and dynamics meaning the

inherent power to perform.
 Thus thermodynamics is concerned with the study of
heat, its effect and its properties.
 Branch of science that deals with energy levels and the transfer of
energy between systems and between different states of matter”
 Thermodynamics has two parts:-

 classical and

 statistical
 Classical thermodynamics concerned with macroscopic

observables such as P, T, V, 𝜌….

 Statistical thermodynamic deals with microscopic details like

dipole moment, molecular size, shape…...

 What are the basic principles of thermodynamics?

 The four laws of thermodynamics define fundamental physical

quantities (temperature, energy, and entropy) that characterize

thermodynamic systems at thermal equilibrium.

 Thermodynamic terms

 A thermodynamic system is the part of the universe which

is selected for thermodynamic consideration.

 It is separated from the rest of the universe (environment)

called surroundings by a definite boundary.

 So the environment rather than the system is surrounding.

 Example: If you have water in container, the water is your system,

the environment outside your container is the surrounding and

the wall of the container separating the water from the other
environment is boundary.
 Class Activity

1. Explain the basic Applications of thermodynamics.

2. What is the familiarity of thermodynamics?

Answer
1. To determine spontaneity of a rxn, in the design of a heating or
air conditioning system, used in vehicles, widely used in making
of thermal power plants, nuclear plants.

2. It never deal with rate of a reaction within the system.

 The remaining portion of the universe is called surrounding.

 There are three types of boundary (wall) separating system and

surrounding. These are
i. A permeable wall: a boundary that allows the passage of both
matter and energy.
ii. ii. A diathermal wall: a boundary that prevents the passage of
matter and allows the flow of energy.
iii. Adiabatic wall: a boundary neither allows the passage of energy
nor matter.
 State of a system

 It means a condition of a system described in terms of a certain

observable properties, such as T, P V etc. of the system.

 These are known as state variables or thermodynamic

variables.

 A system can be expressed by using three of the above variables.

 By specifying two of the three state variables, we can get the

other.
 Properties of a system

 It is know that each and every thing can be measured qualitatively

or quantitatively.

 In thermodynamics, measurable properties of a system can be

divided in to two; they are extensive and intensive properties.

A) Extensive Properties: properties which depend up on the

total amount of materials in the system.

Example: If you mix 25 mL of water with 10 mL of ethanol, the total

volume of the mixture is 35 mL which depend up on their sum.
mass,, internal energy, heat contents, free energy, entropy, heat
capacity.
B) Intensive properties (specific property ): properties which are

independent on the amount of material in the system.

Example: If you have 1L of water and 10L of water in two

containers, what do you think about the boiling point for water in

each container?

You know that the boiling point of water is 100℃. Whatever

the amount of water, its boiling point is the same in all amount or

volume. So boiling point of substance is intensive property.

 Other examples include density, molar properties(molar volume,

molar entropy, molar heat capacity, etc.), surface tension,

viscosity, specific heat, thermal conductivity, refractive index, P, T,
boiling and freezing points, vapor pressure of a liquid.

NB:- The ratio of two extensive properties is an intensive property.

Homogeneous vs heterogeneous properties

 A system is considered to be homogeneous if every intensive

property has the same value for every point of the system.
 A system is said to be heterogeneous if the intensive property

of one portion is different from the property of another portion.

Activity: Categorize the following as intensive or extensive
a) Chemical potential d) entropy
b) Density e) freezing point
c) Molar heat capacity f) Viscosity
a. Intensive
b. Intensive
c. Intensive
d. extensive b/c ratio of heat capacity and T (Q is extensive, T is intensive)
e. Intensive
f. Intensive
Thermodynamic equilibrium, Zero laws of thermodynamics

 A system is in thermodynamic equilibrium if it is in mechanical ,

thermal and chemical equilibrium.

 Mechanical equilibrium: the pressure difference between the

system and its surroundings is infinitesimal (pressure is constant).

 Thermal equilibrium: is a type of thermodynamic equilibrium

in which temperature remains the same in all parts of a system.

 The concept behind thermal equilibrium holds one of the law of

thermodynamics which is called zeroth law of thermodynamics.

 If system A is in thermal equilibrium with system B and

system B is in thermal equilibrium with system C, then a system

A is in thermal equilibrium with a system C.

 Chemical equilibrium: is a type of thermodynamic equilibrium

in which the composition of the system remains constant

throughout the system.
 Thermodynamic process

 A process is a change in the state of the system over time, starting

with a definite initial state and ending with a definite final state.

 Process terminology

 Adiabatic – no heat transferred

 Isothermal – constant temperature

 Isobaric – constant pressure

 Isochoric – constant volume

a) Adiabatic process: a process in which there is no exchange of
heat between the system and the surrounding (Q = 0).

Consider the equation ΔU = Q -W

 When a system expands adiabatically, W is positive (the system

does work) so ΔU is negative.

 When a system compresses adiabatically, W is negative (work is

done on the system) so ΔU is positive.

b) Isothermal process: a process carried out under condition of

constant temperature.

• Any heat flow into or out of the system must be slow enough to

maintain thermal equilibrium.

c) Isobaric (Isopiestic) process: a process that is carried out
under condition of constant pressure.

Example: Water boiling in a saucepan.

d) Isochoric (Isovolumetric) Process: a process in which volume of a

system remains constant. When the volume of a system doesn’t
change, it will do no work on its surroundings.

W = 0 and as a result ΔU = Q

Example: Heating gas in a closed container.

e) Cyclic process: if a system after undergoing through series of

changes in its state come back to initial state. The path followed is
known as cyclic Path.
f) Reversible process: if at any instant during transformation, the
system does not deviate from equilibrium by more than an
infinitesimal (very small or values closed to zero but greater than
zero) amount.

 Change is takes place both forward and backward direction.

g) Irreversible process: process that occurs rapidly or

spontaneously such that it does not remain in equilibrium during
transformation.

Example:
 Examples:

 expansion of gas against zero applied pressure,

 dissolution of a solute in solvent, mixing of gases,

 flow of liquid from higher level to lower level.

 State Function

 State functions are thermodynamic functions in which change in

the values of these quantities do not depend on how change is carried

out, but depends on the initial and final state of the system.

 change of a state function during a process depends only on the initial

and final states of the system, not on the path of the process.

 Examples: U, S, H, G, chemical potential, CV and CP

 The finite change of state function ∆𝑋 in a process can be written

as X.

 So it can be represented as ∆𝑋 = ∆X = X2-X1 , where X1 is the value

of initial state and X2 is the value of final state of the process
respectively.
 Path function: is one which depends up on how change in the

system takes place. Example heat, q and work, w

q= path dq and w = path dw

The infinitesimals dq and dw are called inexact differentials.

First Laws of Thermodynamics

Heat and work

 Heat is a form of energy transfer to or from a system which

occurs from hot to cold. Heat gained (absorbed) is considered +;

heat lost by system to surroundings is −.

 Activity

 q = mass (in grams) x T x C

 T = change in temperature and C = Specific Heat,

 Units are either: J/(g oC) or cal/(g oC)

Examples:
 Class work

1. A student must use 225 mL of hot water in a lab procedure.

Calculate the amount of heat required to raise the temperature

of 225 mL of water from 20℃ to 100℃ (specific heat of water is

4.184 g/J.℃ )

2. Calculate the specific heat capacity of a new alloy if a 15.4 g

sample absorbs 393 J when it is heated from 0.0 ℃ to 37.6 ℃

  Calorimetry - the measurement of the heat into or out of a

system for chemical and physical processes.

 The device used to measure the absorption or release of heat in

chemical or physical processes is called a “Calorimeter”.

 Changes in enthalpy = H and q = H

 Thus, q = H = m x C x T, where

 H is negative for an exothermic reaction and positive for an

endothermic reaction.
Specific heats of substances at 298.15 K
 An exothermic process is a process that releases energy as

heat in to its surroundings.

 Example: All combustion reactions are exothermic. Heat is

negative in exothermic process.

 An endothermic process is a process in which energy is

acquired from its surroundings as heat. Heat is positive in this

process.

 Example: Vaporization of water is endothermic process.

 The specific heat (s/c) of a substance is the amount of heat (q)

required to raise the temperature of one gram of the substance

by one degree Celsius.

 The heat capacity (C) of a substance is the amount of heat (q)

required to raise the temperature of a given quantity (m) of the

substance by one degree Celsius. C = m x s and q = m x s x t

q = C x t and t = tfinal - tinitial

 How much heat is given off when an 869 g iron bar cools from

94oC to 5oC? Solution: and s of Fe = 0.444 J/g • oC

t = tfinal – tinitial = 5oC – 94oC = -89oC, q = mst
= 869 g x 0.444 J/g • oC x –89oC = -34,000 J
 Work
 Work is the motion against an opposing force

 If an object is displaced through a distance dx against a force

f(x), then the amount of work which has to be done is

x

x
= =
 Fdx
x0
w   Fdx  ( PA)( x  x0 )   PdV
x0

 Gravitational work=-mgh
 Electrical work = -nFE
 Mechanical (pressure-volume work)=-PdV
Example: The capacity to do work is-----------------

A) Expansion Work B) Heat C) Energy D) All

 Work don the gas:

W  PV

B/c work is done by the gas


 Activity
1. A sample of nitrogen gas expands in volume from 1.6 L to 5.4 L at
constant temperature. What is the work done in joules if the gas
expands (a) against a vacuum and (b) against a constant pressure of
3.7 atm? Solution w = -P V
(a) V = 5.4 L – 1.6 L = 3.8 L P = 0 atm
W = -0 atm x 3.8 L = 0 L.atm = 0 joules
(b) V = 5.4 L – 1.6 L = 3.8 L P = 3.7 atm

w = -3.7 atm x 3.8 L = -14.1 L•atm

101.3 J = -1430 J
w = -14.1 L•atm x
1L•atm
 Consider the equation:

 Negative sign is in accord with convention that work done by a

system is taken to be a negative quantity.

 If we divide Mg by A, the area of the piston, and multiply by h

 But Mg/A is the external pressure exerted on the gas and Ah is

the change in volume experienced by the gas:

 After compression, Pext = Pf , work is positive because work is done

on the (gas) system when it is compressed.



Summery
 The first law of thermodynamics is known as the law of

conservation of energy.

 It states that energy can be neither created nor destroyed.

 Mathematical formulation in terms of change of internal energy,

heat and work for first law is: ∆U = q + w.

 Assume that U1 and U2 is the initial and final value of internal

energy after w is done on the system and heat,

 , ∆U = q

∆U = W

 If there is no change in internal energy of the system, ∆U the first

law may be written as q = -W.

 Thus, when the energy of the system is kept constant then heat

absorbed by the system is equal to the work done by the

system W = -q.

 Thus, the energy of the system kept constant and work is done

on the system, then heat must flow from the system to the
surroundings.
 Example: If an electric motor produces 15 KJ of energy each second
as mechanical work and lost 2 KJ as heat to the surrounding, what is
the change in internal energy?

∆U = -2 K J-15 KJ = -17 KJ.

-15 indicate that the energy (the capacity to do work) is produced

by the electric motor; mean work is done by the motor.

2) Suppose that, when the spring was wound, 100KJ of work was
done on it. But 15KJ escaped to the surroundings as heat. What is
the change in internal energy of the spring?
 Thermochemistry

 Is chemistry of reaction of heat.

 Thermochemistry is the branch of physical chemistry deals with

the heat change during chemical or physical process.

 Thermochemistry is the study of heat change in chemical

reactions.

 Since heat is depends on the path of the reaction, it is more

convenient to express the energy of the system in terms of

thermodynamic quantities as they are independent of path.

 The energy changes in chemical reactions are due to bond

breaking and formation of new bonds.

 Internal energy and enthalpy of reaction

 Enthalpy of reaction: is the amount of heat absorbed or evolved

in the transformation of reactant at a given temperature and

pressure in to products at the same temperature and pressure.

 Example:

Conditions at which enthalpy of a reaction depends on

 Constant volume: when the reaction carried in closed and rigid

container. The first law of thermodynamics becomes:-

(*)


 from equation (*), the above equation
becomes

 Collect like terms (p with p and R with R), the equation becomes:

Since H =U+PV , the equation above is becomes:

 The difference in the enthalpy of the product and the reactant is

the change in enthalpy of the system. but

 For the system involving liquid and solid the volume of reactant

equal to the volume of product, so

 For an ideal gas, PV = nRT, then

take constants as common from both sides of the equation & becom
Class work
 Activity
3. How much heat is evolved when 266 g of white phosphorus (P4)
burn in air? (Mwt = 124 g/mol)
P4 (s) + 5O2 (g) P4O10 (s) , ∆H = -3013 kJ/mol
1 mol P4 3013 kJ
266 g P4 x x = 6470 kJ
123.9 g P4 1 mol P4

mass of asubstance
i.e heat energy evolved = x∆H
molar mass of a substance
Laws of Thermochemistry
s
 Adding the above reactions together gives:

Class Activity

g
 Class Activity
1. C + O2  CO2, find H
C + ½O2  CO H = – 110.5 kJ
CO + ½O2  CO2 H = – 283.0 kJ
C + CO + O2  CO + CO2 sH = – 393.5 kJ
I.e. C+ O2  CO2 H = – 393.5 kJ

Homework
Consider the reaction given below in the formation of CO2 and water as
shown:
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)

C3H8 (g)  3 C + 4 H2 (g)

3 C + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)
Use the standard enthalpy of formation for each species, calculate change of
enthalpy.
 Second Laws of Thermodynamics

 The first law is not capable of predicting the direction of a process.

 So the second law was developed for prediction of the direction of

the process (reaction).

statements of the 2nd law

 Entropy increases in irreversible process.

 Heat cannot pass from a colder body to a warmer body.

Entropy

 Measure of the randomness or disorderness in a molecular

system. and given as:

 Activity

a. ∆S<0, more ordered

b. ∆S>0, less ordered

c. ∆S>0, order decrease

d. ∆S<0, as temperature decrease, ∆S decrease

 Enthalpy and entropy relation

But so the above equation becomes:

 Entropy of universe

 Since then
 Mathematically second law of thermodynamics described as:

 Consider the chemical reaction given below:

 Class Activity
For the reaction given below, determine the change in entropy.
 Homework

1. Two gaseous pollutants that form in auto exhaust are CO and NO.
An environmental chemist is studying ways to convert them to less
harmful gases through the following equation:
 Third Law of Thermodynamics

 The entropy of all the perfect crystalline solids is zeros at

absolute zero temperature.

 The third law of thermodynamics is also referred to as Nernst

law. It provides the basis for the calculation of absolute

entropies of the substances, lim S  0
T 0
 It explains the behavior of solids at very low temperature.

 It helps in analyzing chemical and phase equilibrium.

 Entropy and molecular arrangement is given by Boltzmann

equation: where

W is the number of ways of molecular arrangements/microstates.

= – Gibbs Free Energy
 = – Gibbs Free Energy
Make this equation nicer:
solution
 Exercise on thermodynamics
1. What is the heat required to convert 25 grams of ice at -10 °C into steam at
150 °C?
2. What quantity of heat is necessary to convert 50 g of ice at 0 degrees C into
steam at 100 degrees C?

3. How much heat energy is required to warm 168 g copper from -12.2
°C to 25.6 °C?
4. If 187 J raises the temperature of 93.45 g of Ag from 18.5 to 27.0°C,
what is the specific heat capacity of silver?
5. Find heat energy to warm 168 g copper from -22.2 °C to 25.6 °C?