Engineering Chemistry BCHY101L

Dr. S. Arockiasamy
Vellore Institute of Technology Chennai campus

1
Thermodynamics

2
 Laws of thermodynamics - entropy
Thermodynamics: Basic Terminologies change (selected processes) –
spontaneity of a chemical reaction and
Gibbs free energy - heat transfer;
Thermodynamic Systems: the quantity of matter or a region
System +Surroundings = Universe in space upon which attention is Properties of a system:
concentrated in the analysis of a
# Intensive Properties:
problem
Everything
external to
the system

# Extensive Properties

3
Thermodynamics: Basic Terminologies

# State function:
Depends on the initial state & final state; independent Internal energy (U) = Kinetic energy + Potential energy
of the path used to reach from. Example: T
(Temperature), P (Pressure), U (Internal energy), H
(Enthalpy) etc. • It’s a state function & an extensive property of the
system.
• Internal energy of a system changes when energy is
transferred into or outside the system in the form
heat or work

# Path function: Change in internal energy of the system is given

Depends on the path between the initial & final state by
Example: W (work done), q (heat transferred) etc. 𝜟𝑼𝒔𝒚𝒔𝒕𝒆𝒎 = 𝑼𝒇𝒊𝒏𝒂𝒍 𝒔𝒕𝒂𝒕𝒆 − 𝑼𝒊𝒏𝒊𝒕𝒊𝒂𝒍 𝒔𝒕𝒂𝒕𝒆
4
Thermodynamics: Basic Terminologies
State of a system Work done on/by the system
➢ The state of thermodynamic variables such
as P, T, V, composition which describes the W (Work) = F (force) x w (distance moved in the direction of force)
system is called state of the system.
➢ when one/more variables undergo change, Heat
the system is said to have undergone a Movable piston
(Q)
change of state w

System
• Adiabatic – No heat transferred Piston Rod
• Isothermal – Constant temperature
• Isobaric – Constant pressure Cylinder
• Isochoric – Cconstant volume System Surrounding
boundary

When Gas is heated  it will expand and push the piston, thereby doing work on the piston.

The work done (dw) when the system expands by dV against a pressure Pex: dw = −PexdV
𝒗
Total work done by the system to expand from volume Vi to Vf: W = -‫ 𝒙𝒆𝑷 𝒇 𝒗׬‬dV
𝒊

 This is an example of the system doing the work on the surrounding 5

6
0th Law of Thermodynamics

➢ According to 0th law:

If two systems (A and B) are in thermal equilibrium with
a third system (C), then those two systems are in thermal
equilibrium with each other.

Two physical systems are in thermal

equilibrium if there is no net flow of heat
(thermal energy) between them when
they are connected by a path permeable
to heat.

7
1st Law of Thermodynamics
It’s the law of conservation of energy

The energy of an isolated system remains constant. Whenever a quantity of energy

(some form) disappears, an exactly equivalent quantity of energy (some other form)
must make an appearance.

➢ Heat (q) and work (w) are equivalent ways of changing the internal energy of
a system
→ Example:
o If a weight has been raised/lowered in the surroundings, transfer of energy
happens by doing the work.
o If ice has melted in the surroundings, it indicated transfer of energy as heat.
➢ For a system, if w = work done on a system, q = 𝜟𝑼 = 𝒒 + 𝒘
energy transferred as heat to a system & ΔU =
resulting change in internal energy The sign of w and q:
✓ +ve if energy is transferred to the system as work/heat
✓ -ve if energy is lost from the system.
Therefore,
change in internal energy (𝛥𝑈 ) of a system = heat added to the system (q) - the work done by the system (w)

𝜟𝑼 = 𝒒 − 𝒘 8
 Application of 1st Law to the Expansion Work
➢ Isobaric process (constant pressure) ➢ Isochoric process (constant volume)

▪ the pressure is kept constant. ▪ the volume is kept constant

▪ The work done by the system in an isobaric process ▪ The work done is zero in an isochoric process
is simply the pressure multiplied by the change in
volume ✓ . Example of an isochoric system: A gas in a box
✓ Example of an isobaric system: A gas, being with fixed walls
slowly heated or cooled, confined by a piston in a
cylinder.

9
 Numerical from of 1st Law

❖ Example 1: Calculate w and ∆𝑈 for the conversion of 1 mole of water at 100℃ to steam at 1 atm pressure.
Heat of vaporisation of water at 100℃ is 40670 Jmol-1
Solution: w = P (Vf - Vi)
w = P (Vf - Vi) , = 101325 Nm-2 [0.0306 m3-(18 × 10-6 )m3]
= 101325 Nm-2 × 0.0306 m3 = 3100 Jmol-1
P = 1 atm = 101325 Nm-2

Vi = vol. of 1 mole of liquid water at 1 atm pressure Since, conversion of water to steam is
H2O = atomic mass of H2 * atomic mass of O2 accompanied by increase in volume, work is
= (2*1)+(1*16)
done by the system on the surroundings.
= 18 ml (1ml = 1*10-6m3)
= 18 × 10-6 m3
Hence, w = – 3100 Jmol-1

Vf = vol. of 1 mole of steam at 100℃ at 1 atm pressure According to First Law,

But w.kt, volume of one mole of gas at STP = 22.4 L, (STP = 273K, 1atm)
Therefore, at 100oC ..
ΔU= q+w
= (22.4 dm3 × 373 K)/273 K
=40670J+(−3100J)
= 30.60 dm3 = 0.0306 m3 =37570J.
10
❖ Example 2:
1 mole of an ideal gas expands against a constant external pressure of 1 atm from a volume
of 10 dm3 to a volume of 30 dm3. Calculate the work done by the gas in Joules.

Given data : P = 1atm

Solution: ❖ Example 3: 10 moles of an ideal gas expands

𝑤 = −𝑃𝑒𝑥𝛥𝑉 reversibly and isothermally from a pressure of 10
= − 1 𝑎𝑡𝑚 30 𝑑𝑚3 − 10𝑑𝑚3 atm to 2 atm at 300 K. Calculate the work done.
𝑃𝑖
= −20 𝑑𝑚3 𝑎𝑡𝑚 Solution: w = −𝑛𝑅𝑇 ln
𝑃𝑓
10 𝑎𝑡𝑚
= −(10 mol) 8.314 𝐽 𝑚𝑜𝑙 −1 𝐾 −1 300 𝐾 ln
R = 8.314 Jmol-1K-1 2 𝑎𝑡𝑚
= − 40.15 × 103J
= 0.08206 dm3 atm K-1 mol-1

8.314 𝐽 𝑚𝑜𝑙 −1 𝐾 −1
1 atm =
0.08206 𝑑𝑚3 𝑚𝑜𝑙 −1 𝐾 −1

Therefore,
8.314 𝐽 𝑚𝑜𝑙 −1 𝐾−1
𝑤 = −20 𝑑𝑚3 𝑎𝑡𝑚 = −20 𝑑𝑚3 × = – 2026 J
0.08206 𝑑𝑚3 𝑚𝑜𝑙 −1 𝐾 −1 11
 Numerical from of 1st Law

❖ Example 4: 1 mole of an ideal monoatomic gas at 27 C expands reversibly and adiabatically from a volume of
𝐶 3
10 dm3 to a volume of 20 dm3. Calculate q, ∆U, W and ∆H. Given 𝑣 =
𝑅 2

Solution: Since for an adiabatic process, q =0,

Since the process is adiabatic, q = 0 hence ΔU = w = – 1384 J
𝑇𝑓 𝑉𝑖 ΔH = ΔU + ΔPV = ΔU + ΔnRT
𝐶𝑣 ln = 𝑅 ln
𝑇𝑖 𝑉𝑓 = ΔU + nR(Tf- Ti)
𝑇𝑓 10 𝑑𝑚3

3
ln = ln( )  Tf = 189K → 𝑛𝑅 𝑇𝑓 − 𝑇𝑖
2 300 20 𝑑𝑚3
→ = (1 mol) 8.314 𝐽 𝑚𝑜𝑙 −1 𝐾 −1 189 − 300
𝜕𝑈 = −923 𝐽
𝐶𝑣 =  𝑑𝑈 = 𝐶𝑣 𝑑𝑇
𝜕𝑇 𝑣 ΔH = – 1384 J – 923 J = – 2307 J

For a finite change in n moles,

∆𝑈 = 𝑛𝐶𝑣 ∆𝑇 # Note: ΔH can also be calculated by using the
3 relation 𝒅𝑯 = 𝑪𝒑 𝒅𝑻
= (1 mol) × 8.314 𝐽 𝑚𝑜𝑙 −1 𝐾 −1 (189 − 300)𝐾
2
= – 1384 J

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 2nd Law of Thermodynamics
▪ Why we need for the 2nd law of thermodynamics?
➢ the 1st law uses the internal energy to identify
→ The 1st law of thermodynamics does not tell us anything
permissible changes
about the direction of change. The direction of
➢ the 2nd law uses the entropy to identify which
spontaneous change of a process is defined by the 2nd law
of these permissible changes are
of thermodynamics
spontaneous.
→ A spontaneous process points towards the
❑ 2nd law of thermodynamics direction in which the total entropy increases.
▪ Heat does not flow spontaneously from a cool body ➢ Entropy (S) is a state function
to a hotter body
➢ Thermodynamic definition of entropy
▪ The entropy (S) of an isolated system increases in
the course of a spontaneous change. → The thermodynamic definition of entropy
ΔStot > 0 concentrates on the change in entropy
(dS) that occurs as the result of a physical
Where, Stot = Ssys + Ssur or chemical process.
→ dqrev is the energy transferred as heat
S = the entropy of the system of interest, & reversibly to the system at the absolute
Ssur = the entropy of the surroundings temperature T.
# Note: when considering applications of the 2nd law – it is a statement
𝒇 𝒅𝒒𝒓𝒆𝒗
about the total entropy of the overall isolated system (the 𝒅𝑺 = 𝒅𝒒𝒓𝒆𝒗ൗ𝑻, 𝜟𝑺 = ‫𝒊׬‬
‘universe’), not just about the entropy of the system of interest. 𝑻
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 Entropy
▪ Process that lead to an increase in entropy (ΔS > 0) ➢ Entropy change for the system of an isothermal
expansion of a perfect gas

ΔU = 0, q = -w  𝑞𝑟𝑒𝑣 = −𝑤𝑟𝑒𝑣

as, wrev =― nRT ln (𝑉𝑓 /𝑉𝑖)

[from the expression of work done in a reversible

isothermal expression]

 𝑞𝑟𝑒𝑣 = 𝑛𝑅𝑇𝑙𝑛(𝑉𝑓/𝑉𝑖)

as 𝑑𝑆 = 𝑑𝑞𝑟𝑒𝑣ൗ𝑇

Δ𝑆 = 𝑛𝑅𝑙𝑛(𝑉𝑓/𝑉𝑖)

☻ Notice the increasing disorder in above processes 14

 Entropy
➢ Total Entropy change in irreversible (spontaneous) process of a perfect gas

✓ E.g: Isothermal expansion of an ideal gas at constant

temperature into vacuum ▪ In an isolated system, there is no heat transfer and
As, w = 0, 𝛥𝑈 = 0  𝑞 = 0 [from 1st law] 𝒅𝒒 = 𝟎 , thus dS ≥ 𝟎
 no heat is absorbed by or  all natural processes are spontaneous & irreversible.
removed from the surrounding,
Hence, Δ𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔 = 0 ▪ Reversible process:
Δ𝑆𝑚 = 𝑅𝑙𝑛(𝑉𝑓/𝑉𝑖) 𝒅𝑺𝒕𝒐𝒕𝒂𝒍 = 𝒅𝑺𝒔𝒚𝒔𝒕𝒆𝒎 + 𝒅𝒔𝒔𝒖𝒓𝒓𝒐𝒖𝒏𝒅𝒊𝒏𝒈 = 𝟎
As, 𝑆𝑡𝑜𝑡𝑎𝑙 = 𝑆𝑠𝑦𝑠𝑡𝑒𝑚 + 𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔  spontaneous in neither direction and is at equilibrium
 Δ 𝑆𝑡𝑜𝑡𝑎𝑙 = 𝑆𝑠𝑦𝑠𝑡𝑒𝑚 + 𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔
= 𝑅𝑙𝑛(𝑉𝑓/𝑉𝑖) + 0 ▪ Irreversible process
= 𝑅𝑙𝑛(𝑉𝑓/𝑉𝑖) 𝒅𝑺𝒕𝒐𝒕𝒂𝒍 > 𝒅𝑺𝒔𝒚𝒔𝒕𝒆𝒎 + 𝒅𝒔𝒔𝒖𝒓𝒓𝒐𝒖𝒏𝒅𝒊𝒏𝒈 > 𝟎

→ As Vf > 𝑉𝑖, this spontaneous (irreversible)  all spontaneous process occurring in Nature is
isothermal expansion of a gas is accompanied by irreversible, entropy of the universe is increasing
the increase in entropy. continuously
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Entropy
▪ The 1st law & the 2nd law of thermodynamics were summed up by German Physicist Rudolf Clausius as below:
The energy of the universe remains constant; the entropy of the universe tends towards a maximum

➢ Entropy change during different processes

▪ For an ideal gas (1 mole) with variable T & V

𝑇𝑓 𝑉𝑓
∆𝑆 = 𝐶𝑣 ln + 𝑅 ln
𝑇𝑖 𝑉𝑖
▪ For an ideal gas with variable P & T
𝑇𝑓 𝑃𝑓
∆𝑆 = 𝐶𝑝 ln − 𝑅 ln
𝑇𝑖 𝑃𝑖
▪ For an ideal gas in an isothermal process
𝑉𝑓 𝑃𝑓
∆𝑆𝑇 = 𝑅 ln = − 𝑅 ln
𝑉𝑖 𝑃𝑖
▪ For an ideal gas in an isobaric process
𝑇𝑓
∆𝑆𝑃 = 𝐶𝑝 ln
𝑇𝑖
▪ For an ideal gas in an isochoric process
𝑇𝑓
∆𝑆𝑣 = 𝐶𝑣 ln
𝑇𝑖 16
 Numerical from 2nd Law of Thermodynamics

❖ Example 4: 5 mole of an ideal gas expands ❖ Example 5: Calculate the change in entropy accompanying
reversibly from a volume of 8 dm3 to 80 the heating of 1 mole of Helium gas (assumed ideal) from
dm3 at a temperature of 27 ∘C. Calculate the a temperature of 298 K to a temperature of 1000 K at
change in entropy. constant pressure. Assume Cv =3/2 R
Solution: Solution: As 𝐶𝑝 − 𝐶𝑣 = 𝑅
Δ𝑆 = 𝑛𝑅𝑙𝑛(𝑉𝑓/𝑉𝑖)  Cp =𝐶𝑣 + 𝑅 = 2.5 𝑅
Given that 𝑛 = 2, 𝑉𝑓 = 80 dm3, 𝑉𝑖 = 8 dm3 𝑇𝑓 1000𝐾
∆𝑆𝑃 = 𝐶𝑝 ln = 2.5 × 8.314 𝐽𝑚𝑜𝑙 −1 𝐾 −1 ln
𝑇𝑖 298𝐾
Δ𝑆 = (5 mol) (8.314 J K-1mol-1) 𝑙𝑛(80 dm3/8 dm3) = 25.17 𝐽𝑚𝑜𝑙 −1 𝐾 −1
= 95.73 JK-1
❖ Example 6: 1 mol of an ideal gas expands reversibly from a volume of 10 dm3 and 298 K to a volume of 20 dm3
and temperature 250 K. Assuming Cv =3/2 R, calculate the entropy change for the process.
Solution: 𝑇𝑓 𝑉𝑓
∆𝑆 = 𝐶𝑣 ln + 𝑅 ln
𝑇𝑖 𝑉𝑖
250 𝐾
= 1.5 (8.314 𝐽𝑚𝑜𝑙 −1 𝐾 −1 ) ln + (8.314 𝐽𝑚𝑜𝑙 −1 𝐾 −1 ) ln (20 dm3/10
298𝐾
dm3)
= 3.57 JK-1mol-1
17
➢ Heat engine
▪ Its a device which transforms heat into work
▪ This happens in a cyclic process
▪ Heat engines require a hot reservoir to supply energy (QH) and a cold reservoir to take in the excess
energy (QC)
– QH is defined as positive, QC is defined as negative
➢ Carnot Cycle
▪ A Carnot cycle (named after the French engineer
Sadi Carnot) consists of four reversible stages in
which a gas (the working substance) is either
expanded/compressed in various ways
▪ To demonstrate the maximum convertibility of heat
into work
▪ The system consists of 1 mole of an ideal gas
which is subjected to four strokes

18
Carnot Cycle

A. 1st stroke:
Curve AB: A→B: Isothermal expansion at Th
➢ Four stages of Carnot Cycle: Work done by the gas
 The gas is placed in thermal contact with QH (at Th) and
undergoes reversible isothermal expansion from A to B.
 The entropy change is qh/Th ( qh = the energy supplied
to the system as heat from the hot source)
𝑽𝑩
𝒒𝒉 = −𝒘𝟏 = 𝑹𝑻𝒉 𝒍𝒏
𝑽𝑨
B. 2nd stroke:
Curve BC (B → C): Adiabatic expansion,
Work done by the gas
 Contact with QH is broken & the gas undergoes
reversible adiabatic expansion from B to C.
 No energy leaves the system as heat, ΔS = 0
 The expansion is carried on until the temperature of
the gas falls from Th to Tc (the temperature of Qc)

−𝒘𝟐 = −𝑪𝒗 (𝑻𝒉 − 𝑻𝒄 ) 19

Carnot Cycle
C. 3rd stroke:
Curve CD (C → D): Isothermal compression at TC,
Work done on the gas.
▪ The gas is placed in contact with the cold sink (Qc) and undergoes a
reversible isothermal compression from C to D at Tc.
▪ Energy is released as heat to the cold sink; the entropy change of
the system = qc/Tc, where qc is negative.
𝑉𝐷
−𝑞𝑐 = 𝑤3 = 𝑅𝑇𝑐 ln
𝑉𝐶

D. 4th stroke:
Curve DA (D → A): Adiabatic compression
Work done on the gas
▪ Contact with Qc is broken and the gas undergoes
reversible adiabatic compression from D to A such that
the final temperature is Th.
▪ No energy enters the system as heat, so the change in
entropy is zero.
specific heat (Cv), the quantity of heat required to raise the temperature of one gram of a 𝑤4 = 𝐶𝑣 (𝑇ℎ − 𝑇𝑐 )
20
substance by one Celsius degree at constant volume
 Carnot Cycle

▪ The area enclosed by the four curves represents the net

work done by the engine in one cycle
▪ The total change in entropy around the cycle is the sum
of the changes in each of these four steps:
𝑞ℎ 𝑞𝑐
ර 𝑑𝑆 = +
𝑇ℎ 𝑇𝑐
𝑞ℎ 𝑇ℎ
❖ For an ideal gas, ‫ = 𝑆𝑑 ׯ‬0 and =−
𝑞𝑐 𝑇𝑐

Efficiency of a heat engine

• The thermal efficiency of a heat engine is
𝑤𝑜𝑟𝑘 𝑝𝑒𝑟𝑓𝑜𝑟𝑚𝑒𝑑 𝑊
𝜂= =
ℎ𝑒𝑎𝑡 𝑎𝑏𝑜𝑠𝑟𝑏𝑒𝑑 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 ℎ𝑜𝑡 𝑠𝑜𝑢𝑟𝑐𝑒 𝑞ℎ
= 1 − 𝑞𝑐 /𝑞ℎ = 1 − 𝑇𝑐 /𝑇ℎ
[Kelvin’s statement]
➢ The “engine” statement of the 2nd Law:
– it is impossible for any system to have an
efficiency of 100% (𝜂 = 1) 21
 Application of the Carnot Cycle

• Energy efficiency of the

Carnot cycle is independent
of its working substance.
• Any cyclic process that
absorbs heat at one
temperature and rejects
heat at another
temperature and is
reversible has the energy
efficiency of a Carnot cycle.

→ Thermal devices or thermal machines are one of the

applications of this cycle.
→ Refrigeration, Air conditioners & Heat pumps
✓ The heat pumps to produce heating,
✓ the refrigerators to produce cooling,
● These appliances are heat engines operating
✓ the steam turbines used in the ships, in reverse.
✓ the combustion engines of the combustion vehicles ● By doing work, heat is extracted from the cold
✓ the reaction turbines of the aircraft reservoir & exhausted to the hot reservoir
22
 Numerical from 2nd Law of Thermodynamics
❖ Example 1: Calculate the efficiency of a ❖ Example 2: Heat supplied to a Carnot engine is 1897.8 kJ. How
certain power station operates with much useful work can be done by the engine which works
superheated steam at 300 ℃ (Th = 573 K) between 0 ∘C and 100 ∘C.
and discharges the waste heat into the
Solution:
environment at 20 ℃ (T = 293 K).
Tc​= 0 + 273 = 273 K, Th ​= 100 + 273 = 373 K
Solution: Tc= (1 − η)Th qh = 1897.8 kJ
 η = 1 – (Tc /Th) w = 𝑞ℎ × (1 − 𝑐 )
𝑇
𝑇ℎ
Theoretical efficiency (η) = 1 – (293K/573K) = 1897.8 kJ × (1 −
273 𝐾
)
= 0.489 373 𝐾
= 508.7 kJ
= 48.9 %
 Work done by the engine is 508.7 kJ
# In practice, there are other losses due to
mechanical friction and the fact that
turbines do not operate reversibly.

23
❖ Example3 : Calculate the maximum efficiency of a heat engine operating between 100 ∘C and 25 ∘C

Solution: For engines, we have

24

Tc​ = 25 + 273 = 298K

Th​ = 100 + 273 = 373 K

Tc= (1 − η)Th
 η = 1 – (Tc /Th)

𝜂 = 1 – (298 K/373 K)
= 0.201
= 20.1 %
Free Energy
▪ The criterion of spontaneity in terms of Gibbs Free energy change (dG) and Helmholtz energy (dA)

𝑑𝐺 ≤ 0 & 𝑑𝐴 ≤ 0

➢ In an endothermic reaction: ▪ as dG = dH − TdS

it is possible for dG to be negative

provided that the entropy of the
system increases so much that TdS
outweighs dH.
 Endothermic reactions are
therefore driven by the increase of
entropy of the system

dH > 0
but if such a reaction is to be spontaneous at constant
temperature and pressure, G must decrease.
25
 Free Energy

➢ In an exothermic reactions ➢ At chemical equilibrium,

▪ commonly spontaneous
because dH < 0 & then dG < 0 provided TdS is not so
negative that it outweighs the decrease in enthalpy. ➢ dG = 0
▪ Free energy change with temperature
and pressure:
𝑃2
∆𝐺 = 𝑛𝑅𝑇 ln
𝑃1
𝑉
= 𝑛𝑅𝑇 ln 1
𝑉2

26
Spontaneity of a Chemical Reaction

• A spontaneous reaction is a reaction that favors the formation of products at the conditions under
which the reaction is occurring. Spontaneous processes may be fast or slow, but they occur without
outside intervention.
Ex. i) Conversion of graphite to diamond is slow;
ii) A burning fire is relatively a fast reaction.
• "In any spontaneous process there is always an increase in the entropy of the universe“
• For a given change to be spontaneous, ΔSuniverse must be positive.

ΔSuniv = Δ Ssys + Δ Ssurr

Gibbs Free Energy

• Free energy is energy that is available to do work. The free energy change of a reaction is a
mathematical combination of the enthalpy change and the entropy change.
ΔGo = ΔHo - TΔSo

NOTE : The change in enthalpy, change in entropy and change in free energy of a reaction are the driving forces behind all
chemical reactions. 27
Conditions for Spontaneity of a Chemical Reaction , (Changes in Enthalpy (ΔH), Entropy (ΔS), and Free Energy (ΔG))

• A spontaneous reaction is one that releases free energy, and so the sign of ΔG must be negative. Since
both ΔH and ΔS can be either positive or negative, depending on the characteristics of the particular
reaction, there are four different possible combinations as shown in the table below.
ΔGo = ΔHo - TΔSo

ΔHo ΔSo ΔGo

Negative Positive Always negative

Negative at higher temperatures, positive at lower

Positive Positive
temperatures

Negative at lower temperatures, positive at higher

Negative Negative
temperatures

Positive Negative Always positive

28
Change in Free Energy at different conditions
⮚ In an exothermic reactions
▪ commonly spontaneous ⮚ In an endothermic reaction:
because dH < 0 & then dG < 0 provided TdS is
not so negative that it outweighs the decrease in dH > 0
enthalpy. but if such a reaction is to be spontaneous at
▪ as dG = dH − TdS constant temperature and pressure, G must
it is possible for dG to be negative decrease.
provided that the entropy of the system
increases so much that TdS outweighs
dH.
⇒ Endothermic reactions are
therefore driven by the increase of
entropy of the system

⮚ At chemical equilibrium,

⮚ dG = 0

29
Problem 1
Q. How can you say that the following reaction is spontaneous or not? Justify the answer using the
standard entropy values given in the table.

Standard Entropy Values at

25°C

Substance So (J/K. mol)

H2 (g) 131.0 Solution
O2 (g) 205.0 From the absolute entropies of substances, we can calculate the entropy change by
H2O (l) 69.9
So of Products – So of Reactants

The entropy change for this reaction is highly negative because three gaseous molecules are being
converted into two liquid molecules. According to the drive towards higher entropy, the formation of
water from hydrogen and oxygen is an unfavorable reaction. In this case, the reaction is highly
exothermic and the drive towards a decrease in energy allows the reaction to occur. 30
Problem-2
Q.2. What is ΔG for the melting of ice at -10◦C if the H = 6.01 kJ/mol and S = 0.022 kJ K-1mol-1
T= −10∘C + 273 = 263K

Thus, we see that at −10∘C, the Gibbs free energy change ΔG is positive for
the melting of water. Therefore, we would predict that the reaction
is not spontaneous at −10 ∘C. 31
3rd Law of Thermodynamics

▪ At T = 0, all energy of thermal motion has been

quenched and in a perfect crystal all the
atoms/ions are in a regular, uniform array.
▪ The localization of matter and the absence of
thermal motion suggest that such materials also
have zero entropy.

▪ Statistical or microscopic definition of entropy:

S = k ln W

where, When T = 0, W = 1
S = the entropy, S = k lnW = 0
k = Boltzmann constant,
W = the number of microstates or the total  if the value zero is ascribed to the entropies of elements in
number of ways a molecular state can be their perfect crystalline form at T = 0, then all perfect
distributed over the energy states for a
crystalline compounds also have zero entropy at T = 0
specific value of total energy.

➢ Third law of thermodynamics: The entropy of all perfect crystalline substances is zero at T = 0.32
Engineering Chemistry

Module -1-Catalysis

Module:1 Chemical thermodynamics and kinetics (continues)

Kinetics – Order of Reactions, Concept of activation energy and energy barrier - Arrhenius equation-homo and
heterogeneous and Enzyme catalysis (Lock and Key Mechanism).

33
Chemical Kinetics

➢ Chemical kinetics is the branch of chemistry which deals with the study of rates
(or how fast) of chemical reactions, the factors affecting it and the mechanism
by which the reactions proceed.

The rate of a chemical reaction might depend on variables such as

❖ Pressure,
❖ Temperature, and
❖ Catalyst

It is to be contrasted with thermodynamics, which deals with the direction in which

a process occurs but in itself tells nothing about its rate

What is Chemical kinetics?

34
The rates of reactions Products conc. increases and Reactant conc. decreases

Products
The change in concentration of reactants or products per unit time Reactants

is called rate.

Consider the following reaction A → B

Here, the instantaneous rate of disappearance of the reactants (A ) at a The rates of reactions
given time, t (at constant volume) is

-d[A] +d[B]
dt dt
The rate is invariably a positive quantity. The minus sign given is
indicating the decreasing of concentration of reactants. Similarly the

Molar Concentration
positive sign implies the increase in concentration of products

Factors affecting the rate of the reactions :

1. Temperature : High T, reactants will posses Energy > Ea, more
collisions- Rate
2. Concentration : Higher the C of reactants-higher collisions- rate Reactant
increases
3. Surface area of reactants : Higher the surface area higher will be
the rate
4. Catalyst : Lowers activations energy and enhances the rate Time 35
The rates of reactions

Consider a general reaction,

A+B→C 1

The rate of reaction will be: Sample Questions:

Write rate expressions for the following reactions:
𝒅𝑨 𝒅𝑩 𝒅𝑪
− =− =
𝒅𝒕 𝒅𝒕 𝒅𝒕 1. NO2(g) + CO(g) → NO(g) + CO2(g)
2. 2HI(g) → H2(g) + I2(g)
Consider another general reaction:

aA + bB → cC + dD 2 Solutions:
where a, b, c d are stoichiometric coefficients
𝒅 𝑵𝑶𝟐 𝒅 𝑪𝑶 𝒅 𝑵𝑶 𝒅 𝑪𝑶𝟐
− =− = =
𝒅𝒕 𝒅𝒕 𝒅𝒕 𝒅𝒕
The rate of reaction will be:
𝟏 𝒅 𝑯𝑰 𝒅 𝑯𝟐 𝒅 𝑰𝟐
− = =
𝟐 𝒅𝒕 𝒅𝒕 𝒅𝒕
−1 𝑑 𝐴 1𝑑 𝐵 1𝑑 𝐶 1𝑑 𝐷
=− = =
𝑎 𝑑𝑡 𝑏 𝑑𝑡 𝑐 𝑑𝑡 𝑑 𝑑𝑡

36
Rate laws and rate constants
The rate law is the relationship between the rate and the concentration, which are related by a proportionality
constant k, known as rate constant.

Consider this reaction, aA + bB → cC + dD

and the rate law can be written as follows

Rate = k [A]m[B]n This equation is called the

rate law of the reaction.
where k = rate constant
m = Order of the reaction with respect to the concentration of A
n = Order of the reaction with respect to the concentration of B

➢ Important points about rate laws and rate constant:

✓ Rate law is a result of experimental observation. One can not look at the stoichiometry of the reaction
and predict the rate law (unless the reaction is an elementary reaction).

✓ The rate law is not limited to reactants. It can have a product term, For example: rate = k[A]m[B]n[C]c

✓ The rate constant is independent of the concentrations but depends on the temperature.
37
Order of the reaction
It is defined as “ The total number of reacting species whose concentration actually changes during the course
of the chemical reaction”
Briefly- “ Number of concentration terms which determines the dependence of rate of reaction”
Consider the following reaction
A+B→C
The total order of the
Rate = k [A]m[B]n reactions can be
determined
where k = rate constant experimentally and
m = Order of the reaction with respect to the concentration of A chemical equations
n = Order of the reaction with respect to the concentration of B fails to supply
The overall reaction order is the sum of the exponents in the rate law. complete
information about
the order
✓ m = 0 = Zero order k[A]0
✓ m = 1 = First order k[A]1
✓ m = 2 = Second order k[A]2
Examples:
H2 + Cl2 → 2HCl. Rate = k[H2]0 [Cl2]0 (Zero order)
SO2Cl2(g) → SO2(g) + Cl2(g) Rate = k[SO2Cl2]1 (First order)
NO2 → 2 NO + O2 Rate = k[NO2]2 (Second order)
2NO(g) + 2H2(g) → 2N2 (g) + 2H2O (g) Rate = k[NO]2 [H2]1 (Third order)
CH3COOC2H5 + H2O → CH3COOH + C2H5OH Rate = [CH3COOC2H5]1[H2O]0 (pseudo-first-order)
38
Inegrated rate Law
Since rate laws are in differential form, we must integrate them to find out the concentration as a function of time.

Integrated first-order rate law: consider the following unimolecular reaction where A gives B
A→B 1
The rate law equation can be written as follows If we plot ln [A]t Vs time, then we will get a straight line having
negative slope (-k). Rate constants can be determined from
𝒅𝑨
Rate = − =𝒌𝑨 experiment by plotting data in this manner.
𝒅𝒕

Separate concentration and time terms, we get

𝒅𝑨
= −𝒌𝒅𝒕
𝑨

Integrating over the limits [A]0 to [A]t and 0 to t,

𝒅𝑨
න = −𝒌 න𝒅𝒕
𝑨

[𝑨]𝒕 𝒕
𝒅𝑨
න = −𝒌 න 𝒅𝒕
𝑨
[𝑨]𝟎 𝟎

𝐥𝐧 𝑨 𝒕 − 𝐥𝐧 𝑨 𝟎 = −𝒌𝒕 39
Inegrated rate Law
Other forms:

ln
At = −kt
A0
𝐥𝐧 𝑨 𝒕 − 𝐥𝐧 𝑨 𝟎 = −𝒌𝒕

At = e −kt
A0
At = A0 e − kt
Integrated first order rate law

The exponential decay of the reactant in a

first-order reaction. The larger the rate
constant, the more rapid the decay.

40
Inegrated rate Law-calculation of rate constant

Question 1: The variation in the partial pressure of azomethane with time was followed at 600 K, with the results
given below. Confirm that the decomposition is first-order in azomethane, and find the rate constant at 600 K.
t/s p/pa lnp
0 10.9 2.388763
CH3N2CH3 (g) → CH3CH3 (g) + N2 (g) 1000 7.63 2.032088
2000 5.32 1.671473
3000 3.71 1.311032
4000 2.59 0.951658
5000 1.88 0.631272

Solution: 2.5
lnp
y = -0.0004x + 2.3981
To confirm that a reaction is first-order, R² = 0.9996
2
Rate = k [CH3N2CH3]
➢ plot ln([A]t/[A]0) against time and expect a straight line. 1.5

➢ Because the partial pressure of a gas is proportional to its 1

concentration, an equivalent procedure is to plot ln(p/p0) against t. 0.5

➢ First, convert p into ln(p/p0) and draw the graph. 0

➢ The plot is straight, confirming a first-order reaction, and its 0 1000 2000 3000 4000 5000 6000

slope is =3.610-4.
Therefore, rate constant k = 3.6 10-4 s-1.

41
First-order Half-life
Half-life is the time taken for the original concentration to be reduced by half.
From the above discussions

ln
At = −kt
A0
 A0 
 
ln  2 
= −ktt / 2
A0
1
ln = −ktt / 2
2
− 0.693 = − kt t / 2
𝑡𝑡/2 = 0.693/𝑘

It is clear from the result that the half-life of a reactant is independent of its initial concentration for a
first-order reaction.
Therefore, if the concentration of A at some arbitrary stage of the reaction is [A], then it will have fallen to
1/2[A] after a further interval of (ln 2)/k.
42
 First-order Half-life
2. If 3.0 g of substance A decomposes for 36 minutes the
Another important point to note is the time constant,  (tau), mass of unreacted A remaining is found to be 0.375 g.
the time required for the concentration of a reactant to fall to What is the half-life of this reaction if it follows first-
1/e of its initial value. order kinetics?

 A / e  1 Solution:
k = − ln 0  = − ln = 1
 A0  e
At
ln = −kt
Thus, the time constant of a first-order reaction is the A0
reciprocal of the rate constant.
ln
At / t = −k
A0
Question: 0.375 g
ln
1. The half-life of a first-order reaction was found to be k =−
3g
10 min at a certain temperature. What is its rate 36 min
constant?
Solution:
k =0.0578 min-1
t t / 2 = 0.693 / k
k = 0.693/600s ln 2 0.693
t1 / 2 = = = 12 min
= 0.00115s-1 k 0.0578
43
 Integrated second-order rate law Pseudo First Order Reaction
➢ A pseudo first-order reaction can be defined as a second-order
or bimolecular reaction that is made to behave like a first-
Let us consider the reaction A→B
order reaction.
➢ This reaction occurs when one reacting material is present in
d A
− = k A great excess or is maintained at a constant concentration
2
Rate law can be written as
dt compared with the other substance.
A+B→C
d A
= −kdt So, if component B is in large excess and the concentration of B is
A 2
very high as compared to that of A, the reaction is considered to be
𝟏 𝟏 a pseudo-first-order reaction with respect to A and if component
𝑨𝒕
−
𝑨𝟎
= 𝒌𝒕 A is in large excess and the concentration of A is very high as
compared to that of B, the reaction is considered to be pseudo-first
order with respect to B.
Integrated zero-order rate law
For example:
CH3COOC2H5 + H2O → CH3COOH + C2H5OH
Let us consider the reaction A→B
Rate = k [CH3COOC2H5]
d A
− = k A
0
Rate law can be written as
dt The concentration of water is very high and thus does
not change much during the course of the reaction.
𝑨 𝒕− 𝑨 𝟎 = −𝒌𝒕
44
Arrhenius equation - Temperature dependence of reaction rates
Questions
The rate constant of most reactions increases with increase in the 1. The rate of the second-order decomposition of
temperature. acetaldehyde (CH3CHO) was measured over the
𝐸 temperature range 700–1000 K, and the rate constants
− 𝑎 E
Arrhenius equation 𝑘= 𝐴𝑒 𝑅𝑇 ln k = ln A − a are reported below. Find Ea and A.
RT
where, A is the pre-exponential factor and Ea is the activation energy.

❖ A plot of ln k against 1/T is a straight line when the reaction follows the Solution: First convert T in (103) and k into ln k. Now,
behavior described by the Arrhenius equation. plot ln k against 1/T. You will obtain the following
❖ The higher the activation energy, the stronger the temperature dependence graph having slope = -22.7 and intercept 27.7.
of the rate constant (i.e., the steeper the slope).
❖ If a reaction has zero activation energy, its rate is independent of
temperature.

If we plot the graph ln k vs 1/T, we will get

the value of A from the intercept at infinite
T (i.e.; 1/T=0) and the value of Ea from the
slope.
Now, Ea = 22.7 × (8.3145 J K−1 mol−1) × 103 K = 189 kJ mol−1
A = e27.7 dm3 mol−1s−1 = 1.1 × 1012 dm3 mol−1 s−145
Problem 2: The rate of the second-order decomposition of acetaldehyde (CH3CHO) was measured over the temperature range
700–1000 K, and the rate constants are reported below. Find Ea and A.

Solution: First convert T in (103) and k into ln k. Now, plot ln k

against 1/T. You will obtain the following graph having slope = -
22.7 and intercept 27.7.

Ln k = lnA-(Ea/R)T

Now, Ea = 22.7 × (8.3145 J K−1 mol−1) × 103 K

𝐸𝑎 = 189 kJ mol−1
−𝑅𝑇
𝑘= 𝐴𝑒
A = e27.7 dm3 mol−1s−1
= 1.1 × 1012 dm3 mol−1 s−1 46
Temperature dependence of reaction rates

2. The values of rate constants for a reaction are 9.51x10-9 L/mol.s and 1.10x10-5 L/mol.s at temperatures 500K and 600 K respectively.
Calculate the activation energy.

Solution:
Ea
ln k = ln A −
RT
k2 E 1 1 After substituting the given
ln =− a  − 
k1 R  T2 T1  values and calculating,
−1
k 1 1
Ea = − R ln 2  − 
k1  T2 T1  Ea= 176 kJ/mol

47
Energy of Activation (Ea) Energy barrier:
Let us consider the reaction A→B
𝐸𝑎 𝐸
Minimum amount of energy required to carryout a ln 𝑘 = ln 𝐴 − One can rewrite the
𝑘= − 𝑎
𝐴𝑒 𝑅𝑇
chemical reaction 𝑅𝑇 equation as:

To interpret Ea, lets us consider the collision between

molecules of reactants A and B:

48
Why Catalysis is important?
Catalysis is of great significance as the four major sectors of the world economy depends on
catalytic processes for their daily production:
A) petroleum and energy production, B) Pollution control, C) Pharma and Food industry, D)
Chemicals and polymer production,

Catalysis The economic

significance of the
catalyst industry is
enormous. The
catalytic processes
Petroleum and Pharmaceutical Chemicals and contribute greater
Industries that Pollution polymer
uses catalysts
Energy
control
and Food than 30-40% of
production Industry production
global GDP. The
global catalyst
market size was
Fabric, Cosmetics, estimated at USD
Gasoline, SOx, COx, Nox, Medicines,
Products
Diesel, Fuel oil hydrocarbon Diaries,
plastic, Fibers, 34.0 billion in 2019
required for Coatings and
day today life are produced are removed Beverages and is expected to
Adhesives
reach USD 35.1
49
billion in 2020.
What is a catalyst?

“A catalyst is a chemical substance which increases or decreases the rate of the reaction,
itself remaining unchanged in chemical properties or mass at the end of a reaction.”
The phenomenon of alteration of the rate of a reaction by a catalyst is known catalysis.

Catalytic reactions Transition state

TS*
➢ Use of catalysts forms a
TS** intermediate complex namely
Transition state (TS*) of lower
energy.
➢ Thereby, the activation energy*
of the catalytic reaction is
reduced compared to the
uncatalyzed reaction as shown in
Figure which leads to quick product
formation 50
Types of catalysis
Gaseous
Following are the main types of catalysis:

1. Heterogeneous catalysis

2. Homogeneous catalysis
Solid
3. Enzyme catalysis
In exhaust system
If the catalyst is present in a different phase than the
reactants then it is called heterogeneous catalyst and
the phenomenon is known heterogeneous catalysis.

In heterogeneous catalysis the reactions take place at the

interface of two phases. The catalyst is, often a solid and
adsorbs a liquid or a gas. This type of catalysis is of great
importance in many industrial processes. 51
Examples of Heterogeneous catalyst
(I) Heterogeneous catalysis with gaseous reactants (Contact catalysis)

(A) Manufacture of ammonia by the Haber process. Iron (Fe) acts as catalyst.

N2 (g) + 3H3 (g) + Fe(solid) → 2NH3 (g) + Fe

(B) Manufacture of sulphuric acid by the Contact process. Vanadium pentoxide (V2O5) or platinum are catalysts for the
production of SO3 (g) from SO2 (g) and O2 (g).

2SO2 (g) + O2 (g) + Pt (solid) → 2SO3 (g) + Pt

(II) Heterogeneous catalysis with liquid reactants

Catalysts used in many reactions in the petroleum and polymer industries. There are cases of heterogeneous catalysis
where a reaction in the liquid phase is catalysed by a substance in the solid state. An example is the decomposition of
H2O2 (aqueous) by MnO2 (s).

2H2O2 (aq) +MnO2 (s) → 2 H2O (l) + O2 (g)

(II) Heterogeneous catalysis with solid reactants

(D) Examples of reactions in which both the reactant and the catalyst are in the solid phase. The decomposition of
KClO3 is catalysed by solid MnO2.
2 KClO3 (s) + MnO2 (s)→ 2KCl (s) + 3O2 (g)
52
2. Homogeneous catalysis
In a reaction, if the catalyst is present in the same phase as the reactants, it is called a
homogeneous catalyst and the phenomenon is homogeneous catalysis.

Such catalysis can take place in gaseous reaction or reactions in solution.

These chemicals help in attaining the equilibrium more quickly by increasing the rates of both the
forward and reverse reactions to an extent.

Examples of homogeneous catalysis in the gas phase

(a) Oxidation of sulphur dioxide, SO2, by oxygen to sulphur trioxide, SO3, in presence of nitric oxide, NO,
in the Chamber Process for sulphuric acid manufacture.
2SO2 (g) + O2 (g) +NO (nitric oxide) → 2SO3 (g)

(b) The following reaction in the gas phase is catalyzed by traces of chlorine gas, particularly in presence
of light.
2N2O (g) + Cl2 (g) → 2N2 (g) + O2 (g)
In presence of light chlorine forms chlorine radicals, which react with N2O forming the intermediate
radical ClO*. The proposed mechanism is:
Step 1: N2O (g) + Cl* (g) → N2 (g) + ClO*(g)
Step 2: 2ClO*(g) → Cl2 (g) + O2 (g) 53
Examples of homogeneous catalysis in the solution phase

(a) Hydrolysis of ester in the presence of acid and alkali: (H2SO4, KOH)

CH3COOC2H5 (l) + H2O (l) → CH3COOH (aq) + C2H5OH (aq)

(b) Hydrolysis of sucrose (cane sugar) into glucose and fructose in

presence of minerals acids acting as catalysts:

C12H22O11 (aq) + H2O (l) → C6H12O6 (aq) + C6H12O6 (aq)

(cane sugar) (glucose) (fructose)

54
Enzyme catalysis – Michaelis menton mechanism

➢ Numerous organic reactions are taking place in the body of animals and plants to maintain the life
process.
➢ These reactions being slow remarkably catalysed by the organic compounds known as Enzymes.
➢ All enzymes have been found to be complex protein molecules.
➢ DEFINITION : Enzymes are protein molecules which act as catalysts to speed up organic reactions
in living cells. The catalysis brought about by enzymes is known as Enzyme Catalysis.

Example: A best example is the conversion of hydrogen peroxide in

to water and oxygen by catalase. It is an important enzyme
protecting the cell from oxidative damage by reactive oxygen species
(ROS = *O2- superoxide)

Enzymes lower the activation energy for reactions. The

lower the activation energy, the faster the rate of the
reactions.
H2O2 +catalase → H2O + O2
Enzymes are the best catalyst known so far.
Ea = 18 kcal/moles – without catalyst Enzyme’s effect on the activation energy
Ea = 11.7 kcal/moles – with colloidal Pt
Ea = 2 kcal/mole – with catalase enzyme 55
Molecular structure of Catalase
Mechanism of enzyme-catalyzed reactions

❑ Enzyme-catalyzed reactions work

in a lock and key fashion.

❑ The substrate uniquely fits like a

key into the active site of the
enzyme, forming a lock-key
complex.
ES-complex
Enzyme ❑ The substrate is converted into the
product by the enzyme at the
active site.

❑ The product is then released from

the active site.

56