HSC: Chemistry Chapter no 4: Chemical Thermodynamics
CHAPTER 4 : CHEMICAL THERMODYNAMICS
4.1 Introduction
Summary
• Many physical and chemical changes involve energy.
Examples:
- Ice melts into water.
- Water changes into steam.
- Carbon burns to form carbon dioxide.
In all these changes, energy is either absorbed or released.
Energy can change from one form into another.
Examples:
- Dry cell: Chemical Energy → Electrical Energy
- Electroplating: Electrical Energy → Chemical Energy
- Hydroelectric dam: Potential Energy → Kinetic Energy → Electrical Energy
What is Thermodynamics?
Thermodynamics is the branch of chemistry that studies:
• Heat changes
• Work done
• Energy changes
during physical and chemical processes.
It only tells us how much energy changes.
It does NOT tell:
• How fast the reaction occurs.
• The mechanism (steps) of the reaction.
4.2 Terms Used in Thermodynamics
4.2.1 System and Surroundings
System
The part of the universe chosen for study is called the system.
Example:
Gas inside a cylinder is the system.
Surroundings
Everything outside the system is called the surroundings.
Examples:
• Cylinder
• Piston
• Air
• Room
Universe
Universe = System + Surroundings
Simple Formula
Universe = System + Surroundings
4.2.2 Types of System
There are three types of systems.
1. Open System
Definition
An open system exchanges both matter and energy with the surroundings.
Example
Hot coffee kept in an open cup.
What happens?
• Heat escapes.
• Water vapour escapes.
Therefore,
Matter ✔
Energy ✔
Example:
Tea in an open cup.
2. Closed System
Definition
A closed system exchanges only energy but not matter.
Example
Coffee covered with a lid.
What happens?
• Heat can escape.
• Water vapour cannot escape.
Therefore,
Matter ✘
Energy ✔
3. Isolated System
Definition
An isolated system exchanges neither matter nor energy with the surroundings.
Example
Coffee kept inside a thermos flask.
What happens?
• Heat cannot escape.
• Water vapour cannot escape.
Therefore,
Matter ✘
Energy ✘
Difference Between Types of Systems
| Type | Matter Exchange | Energy Exchange | Example |
|---|---|---|---|
| Open | Yes | Yes | Open coffee cup |
| Closed | No | Yes | Covered coffee cup |
| Isolated | No | No | Thermos flask |
4.2.3 Properties of System
Properties are characteristics of a system.
They are of two types.
A. Extensive Properties
Definition
Properties that depend on the amount of substance present.
Examples
• Mass
• Volume
• Number of moles
• Internal Energy (U)
• Heat Capacity
Simple Trick
More substance → More value
B. Intensive Properties
Definition
Properties that do not depend on the amount of substance.
Examples
• Pressure
• Temperature
• Density
• Melting point
• Boiling point
• Surface tension
• Viscosity
Simple Trick
Amount changes
Property remains same
Difference Between Extensive and Intensive Properties
| Extensive | Intensive |
|---|---|
| Depends on amount | Independent of amount |
| Mass | Pressure |
| Volume | Temperature |
| Internal Energy | Density |
| Number of moles | Boiling Point |
4.2.4 State Functions
Definition
A property whose value depends only on the present state of the system and not on the path followed is called a state function.
Examples
• Pressure (P)
• Volume (V)
• Temperature (T)
• Internal Energy (U)
• Enthalpy (H)
Initial State
Suppose a gas has:
Pressure = 1 bar
Volume = 1 dm³
Temperature = 300 K
This is called the initial state.
Final State
After compression,
Pressure = 2 bar
Volume = 0.5 dm³
Temperature = 300 K
This is the final state.
Only the initial and final states matter.
The method used to reach the final state does not matter.
Important Point
State functions depend only on:
✔ Initial State
✔ Final State
They do NOT depend on the path followed.
Process
Definition
The change of a system from one state to another is called a process.
Example
Gas expanding inside a cylinder.
Path
Definition
The route or method used to change the state of a system is called the path.
A process may have many different paths.
4.2.5 Path Functions
Definition
Properties whose values depend on the path followed are called path functions.
Examples
• Heat (Q)
• Work (W)
Unlike state functions, path functions change depending on how the process is carried out.
4.2.6 Thermodynamic Equilibrium
Definition
A system is said to be in thermodynamic equilibrium when all its state functions remain constant with time.
Examples of constant state functions:
• Pressure
• Volume
• Temperature
If these remain constant, the system is in equilibrium.
If they keep changing, the system is not in equilibrium.
4.2.7 Process and Its Types
A process is the change of a system from one equilibrium state to another.
There are five important types.
1. Isothermal Process
Definition
A process in which temperature remains constant.
Condition
ΔT = 0
Since temperature remains constant,
Internal Energy also remains constant.
ΔU = 0
Example
Slow expansion of gas in contact with surroundings.
2. Isobaric Process
Definition
A process in which pressure remains constant.
Condition
ΔP = 0
Example
Most reactions carried out in an open beaker.
3. Isochoric Process
Definition
A process in which volume remains constant.
Condition
ΔV = 0
Example
Reaction inside a closed rigid container.
4. Adiabatic Process
Definition
A process in which no heat is exchanged between the system and surroundings.
Condition
Q = 0
If reaction is exothermic
Temperature increases.
If reaction is endothermic
Temperature decreases.
Example
Gas compressed inside a perfectly insulated cylinder.
5. Reversible Process
Definition
A process that can be reversed by making a very small change in external conditions.
Characteristics
• Occurs very slowly.
• Can move in both directions.
• System remains in equilibrium throughout.
• Pressure difference is extremely small.
Features of Reversible Process
- Driving force and opposing force differ by only a very small amount.
- Can be reversed easily.
- Happens infinitely slowly.
- System remains in equilibrium at every stage.
Chapter Quick Revision
Thermodynamics → Study of energy changes.
System → Part under study.
Surroundings → Everything outside the system.
Universe = System + Surroundings.
Open System → Matter + Energy exchange.
Closed System → Only Energy exchange.
Isolated System → No exchange.
Extensive Property → Depends on amount.
Intensive Property → Independent of amount.
State Function → Depends only on initial and final state.
Path Function → Depends on path followed.
Thermodynamic Equilibrium → State functions remain constant.
Isothermal → ΔT = 0.
Isobaric → ΔP = 0.
Isochoric → ΔV = 0.
Adiabatic → Q = 0.
Reversible Process → Can be reversed by an infinitesimal change.
4.3 Nature of Heat and Work
Heat and work are two ways by which energy is transferred between a system and its surroundings.
They are not stored inside a system. They exist only when energy is transferred.
4.3.1 Nature of Work (W)
What is Work?
In physics,
Work is done when a force moves an object through a certain distance.
Formula
W = Force × Distance
W = F × d
where,
W = Work
F = Force
d = Distance moved
Work in Thermodynamics
In thermodynamics, work mainly occurs because of a change in the volume of a gas.
This is called Pressure-Volume Work (PV Work).
When a gas expands or contracts, it pushes or is pushed by a piston.
This causes work to be done.
Why is it called PV Work?
Pressure acts on the piston.
The piston moves, changing the volume.
Therefore,
Pressure × Volume Change = Work
Hence it is called Pressure-Volume (PV) Work.
Example 1: Decomposition of Hydrogen Peroxide
Reaction
2H₂O₂(l) → 2H₂O(l) + O₂(g)
What happens?
• Oxygen gas is produced.
• The amount of gas increases.
• Gas pushes the piston upward.
• The piston lifts the weight placed on it.
Therefore,
The system performs work on the surroundings.
Energy flows
System → Surroundings
Therefore,
Work is done by the system.
Example 2: Reaction between Ammonia and Hydrogen Chloride
Reaction
NH₃(g) + HCl(g) → NH₄Cl(s)
What happens?
• Two gases combine to form a solid.
• Gas volume decreases.
• Piston moves downward.
• Surroundings push the piston.
Therefore,
The surroundings perform work on the system.
Energy flows
Surroundings → System
Therefore,
Work is done on the system.
Conclusion
Work is simply one way of transferring energy between the system and the surroundings.
4.3.2 Nature of Heat (Q)
Definition
Heat is the energy transferred because of a temperature difference between the system and the surroundings.
Heat always flows
From Higher Temperature
↓
To Lower Temperature
until both reach the same temperature.
When is heat absorbed?
If the surroundings are hotter,
Heat enters the system.
Example
Ice absorbs heat and melts.
When is heat released?
If the system is hotter,
Heat leaves the system.
Example
Hot tea cools down.
4.3.3 Sign Convention of Heat and Work
In thermodynamics, signs (+ and –) are very important.
Always remember:
Positive means
Energy enters the system.
Negative means
Energy leaves the system.
Heat (Q)
+Q
Heat is absorbed by the system.
Energy enters the system.
Example
Melting of ice.
–Q
Heat is released by the system.
Energy leaves the system.
Example
Burning of fuel.
Work (W)
+W
Work is done on the system.
Energy enters the system.
Example
Compressing a gas.
–W
Work is done by the system.
Energy leaves the system.
Example
Expansion of gas.
Easy Trick to Remember
Positive (+)
Energy comes INTO the system.
Negative (–)
Energy goes OUT of the system.
Important Note
Heat (Q) and Work (W) are Path Functions.
They depend on the path followed.
4.4 Expression for Pressure–Volume (PV) Work
This is one of the most important derivations in this chapter.
Learn every step carefully.
Derivation of PV Work
Step 1
Suppose a gas is enclosed inside a cylinder fitted with a frictionless movable piston.
When the gas expands,
the piston moves upward by a small distance "d".
The external pressure opposes this movement.
Step 2
We know from mechanics,
Work = Force × Distance
W = F × d
This is our first equation.
Step 3
Pressure is defined as
Pressure = Force / Area
P = F/A
Rearranging,
Force = Pressure × Area
Since the external pressure opposes the expansion,
Force = –Pext × A
where,
Pext = External Pressure
A = Area of piston
Negative sign indicates that the force opposes the motion.
Step 4
Substitute the force into the work equation.
W = F × d
W = (–Pext × A) × d
Therefore,
W = –Pext × A × d
Step 5
Area × Distance gives Volume.
That is,
Volume Change
ΔV = A × d
Step 6
Substitute
ΔV = A × d
into the previous equation.
W = –Pext × ΔV
This is the required expression.
Final Formula
W = –PextΔV
or
W = –Pext(V₂ – V₁)
where,
W = Work
Pext = External Pressure
V₁ = Initial Volume
V₂ = Final Volume
ΔV = Change in Volume
Why is there a Negative Sign?
The negative sign comes from the sign convention.
During expansion,
Energy leaves the system.
Therefore,
Work is negative.
During compression,
Energy enters the system.
Therefore,
Work is positive.
Case 1: Expansion
Gas expands.
Final volume is greater than initial volume.
V₂ > V₁
Therefore,
ΔV is positive.
Hence,
W = –PΔV
becomes negative.
Meaning
Work is done by the system.
Case 2: Compression
Gas is compressed.
Final volume becomes smaller.
V₂ < V₁
Therefore,
ΔV is negative.
Hence,
W becomes positive.
Meaning
Work is done on the system.
Important Conclusions
Expansion
• Volume increases.
• Work is negative.
• System loses energy.
• Work is done by the system.
Compression
• Volume decreases.
• Work is positive.
• System gains energy.
• Work is done on the system.
Memory Shortcut
Expansion
Volume ↑
Work = Negative
System does work
Compression
Volume ↓
Work = Positive
Surroundings do work
4.4.1 Free Expansion
Definition
When a gas expands into a vacuum without any external pressure, the process is called free expansion.
In vacuum,
External Pressure = 0
Therefore,
W = –PextΔV
Putting
Pext = 0
W = 0
Why is Work Zero?
There is no external force opposing the gas.
Since nothing resists the expansion,
the gas does not perform any work.
Therefore,
Work Done = 0
4.4.2 Units of Energy and Work
SI Unit
Joule (J)
Relationship
1 J = 1 kg m² s⁻²
Pressure Unit
Pascal (Pa)
1 Pa = 1 kg m⁻¹ s⁻²
Chemistry Unit
bar dm³
Conversion
1 bar = 10⁵ Pa
1 dm³ bar = 100 J
This conversion is very important for numericals.
Important Formula Summary
Work in Mechanics
W = F × d
Pressure
P = F/A
Force
F = PA
Volume Change
ΔV = A × d
Pressure–Volume Work
W = –PextΔV
General Form
W = –Pext(V₂ – V₁)
Free Expansion
Pext = 0
W = 0
Unit Conversion
1 dm³ bar = 100 J
Quick Revision
• Work is the transfer of energy due to expansion or compression.
• Heat is the transfer of energy due to temperature difference.
• Heat and Work are Path Functions.
• Positive Work → Work done on the system.
• Negative Work → Work done by the system.
• Expansion → W is negative.
• Compression → W is positive.
• PV Work Formula = W = –PextΔV.
• Free Expansion occurs in vacuum.
• During free expansion, W = 0.
• 1 dm³ bar = 100 J.
4.5 Concept of Maximum Work
This is one of the most important concepts in Thermodynamics.
What is Maximum Work?
Imagine pushing a heavy box.
If nobody resists,
very little effort is needed.
If someone pushes back,
more effort is needed.
Similarly,
a gas performs more work when it expands against a larger opposing pressure.
The larger the opposing pressure,
the greater the work done.
If external pressure becomes zero,
No work is done.
Example
Expansion in vacuum.
If external pressure becomes almost equal to gas pressure,
Maximum work is obtained.
Therefore,
Maximum work is obtained only during
Reversible Expansion.
Why?
Suppose
Gas Pressure
P
External Pressure
Pext
For expansion to occur,
Gas pressure must be slightly greater than external pressure.
That is
P > Pext
If both become equal,
Expansion stops.
If external pressure becomes larger,
Compression starts.
Therefore,
Maximum work is obtained when
Pext differs from P by only an extremely small amount.
Important Point
Maximum Work
↓
Reversible Process
↓
Very Slow Process
↓
Pressure Difference is Extremely Small
4.5.1 Derivation of Maximum Work
This is one of the most important derivations for HSC examinations.
We shall derive it step by step.
Step 1
During reversible expansion,
the piston moves by a very tiny distance.
The volume changes by
dV
instead of
ΔV
Here,
"d" means
an infinitely small change.
Step 2
Small work done
dW
is
dW = –Pext dV
This is exactly the same PV work equation,
except that the volume change is extremely small.
Step 3
During reversible expansion,
External Pressure is almost equal to Gas Pressure.
Difference is extremely small.
Therefore,
Pext = P – dP
where
dP
is an extremely small pressure difference.
Step 4
Substitute this into the equation.
dW
= –(P–dP)dV
Open the bracket.
dW
= –PdV +dPdV
Step 5
Now observe
dPdV
is the multiplication of two extremely tiny quantities.
Its value becomes almost zero.
Therefore,
Ignore it.
Hence,
dW = –PdV
This is a very important result.
Step 6
This equation gives work done only for one tiny step.
The whole expansion contains thousands of such tiny steps.
Therefore,
Add all of them.
In mathematics,
adding infinitely many tiny quantities is called
Integration.
Hence,
∫dW
= –∫PdV
Step 7
Using Ideal Gas Equation
PV = nRT
Pressure
P = nRT/V
Substitute into the equation.
Wmax
= –∫nRT/V dV
Since temperature remains constant,
nRT is constant.
Take it outside the integral.
Wmax
= –nRT∫dV/V
Step 8
Integration Formula
∫dV/V
=lnV
Apply limits
Initial Volume
V₁
Final Volume
V₂
Therefore,
Wmax
= –nRT[lnV₂–lnV₁]
Step 9
Using logarithm rule
lnA–lnB
=ln(A/B)
Therefore,
Wmax
= –nRT ln(V₂/V₁)
Step 10
Chemistry usually uses logarithm base 10.
We know
lnx
=2.303 logx
Therefore,
Multiply by
2.303
Final Equation
Wmax
= –2.303 nRT log(V₂/V₁)
This is the required derivation.
Final Formula
Maximum Work
Wmax
= –2.303 nRT log(V₂/V₁)
Another Important Formula
Using
P₁V₁=P₂V₂
V₂/V₁
=P₁/P₂
Substitute this.
Wmax
= –2.303 nRT log(P₁/P₂)
This formula is also frequently asked in examinations.
Conditions for Maximum Work
- Process must be reversible.
- Expansion should occur very slowly.
- Temperature must remain constant.
- Pressure difference should be infinitesimally small.
- Gas should behave ideally.
Easy Memory Trick
Free Expansion
↓
No resistance
↓
No Work
W=0
Normal Expansion
↓
Some resistance
↓
Some Work
Reversible Expansion
↓
Maximum resistance possible
↓
Maximum Work
Formula Summary
Normal Work
W = –PextΔV
Maximum Work
Wmax = –∫PdV
Ideal Gas
PV = nRT
Final Formula
Wmax = –2.303 nRT log(V₂/V₁)
Pressure Form
Wmax = –2.303 nRT log(P₁/P₂)
Quick Revision
• Maximum work is obtained only in a reversible process.
• Reversible expansion occurs very slowly.
• External pressure is only infinitesimally smaller than the gas pressure.
• The derivation uses:
- PV work equation
- Ideal gas equation (PV = nRT)
- Integration
- Logarithm conversion (ln x = 2.303 log x)
These derivation steps are commonly asked in HSC board exams, so understanding the reason behind each mathematical step is more important than memorizing the equations.
4.6 Internal Energy (U)
What is Internal Energy?
Every substance contains some energy stored inside it.
This stored energy is called Internal Energy.
It is represented by the symbol:
U
Internal energy is present because the particles (atoms, ions or molecules) are always moving and attracting or repelling each other.
Internal Energy Consists of Two Types of Energy
1. Kinetic Energy
It is the energy due to the motion of particles.
Examples
• Rotation
• Vibration
• Translation (movement from one place to another)
The faster the particles move, the greater their kinetic energy.
2. Potential Energy
It is the energy due to the attractive or repulsive forces between particles.
Examples
• Attraction between molecules
• Attraction between ions
• Chemical bonds
Therefore,
Internal Energy
=
Kinetic Energy
Potential Energy
Change in Internal Energy
When a system absorbs or releases heat or work, its internal energy changes.
Formula
ΔU = U₂ − U₁
where
U₁ = Initial Internal Energy
U₂ = Final Internal Energy
ΔU = Change in Internal Energy
Nature of Internal Energy
Internal Energy is
✔ State Function
✔ Extensive Property
When Does Internal Energy Increase?
Internal energy increases whenever energy enters the system.
Examples
Heat absorbed
Suppose
30 kJ heat is supplied.
Then
ΔU = +30 kJ
Reason
Energy enters the system.
Work done on the system
Suppose
20 kJ work is done on the system.
Then
ΔU = +20 kJ
Reason
Energy enters the system.
When Does Internal Energy Decrease?
Whenever energy leaves the system.
Example
System releases
10 kJ heat
and performs
15 kJ work
Total energy leaving
=10+15
=25 kJ
Therefore
ΔU = −25 kJ
Memory Trick
Energy enters
↓
Internal Energy increases
(+)
Energy leaves
↓
Internal Energy decreases
(−)
4.7 First Law of Thermodynamics
This is one of the most important laws in Chemistry.
Almost every competitive exam asks this topic.
Statement
Energy can neither be created nor destroyed.
It can only change from one form into another.
Other Ways of Stating the First Law
The textbook gives three equivalent statements.
Statement 1
Energy of the universe remains constant.
Statement 2
The total internal energy of an isolated system remains constant.
Statement 3
Energy can neither be created nor destroyed.
It only changes from one form to another.
Simple Meaning
Suppose
Chemical Energy
↓
Heat Energy
↓
Mechanical Energy
↓
Electrical Energy
The form changes,
but the total amount of energy remains the same.
Formulation of First Law
Suppose
The system absorbs heat.
Heat supplied = Q
Suppose
Work is also done on the system.
Work = W
Both increase internal energy.
Therefore,
Increase in Internal Energy
=
Heat Supplied
Work Done
Mathematical Equation
ΔU = Q + W
This is called the Mathematical Form of the First Law of Thermodynamics.
Meaning of Symbols
ΔU
=
Change in Internal Energy
Q
=
Heat
W
=
Work
Very Important Point
Positive Q
Heat absorbed
Positive W
Work done on system
Negative Q
Heat released
Negative W
Work done by system
First Law for Different Processes
1. Isothermal Process
Temperature remains constant.
Therefore
ΔT = 0
For an ideal gas,
Internal energy depends only on temperature.
Therefore
ΔU = 0
Substitute into
ΔU = Q + W
0 = Q + W
Hence
W = −Q
Meaning
Whatever heat is absorbed
is completely converted into work.
No energy is stored inside the system.
Formula
W = −Q
2. Adiabatic Process
Definition
No heat exchange.
Therefore
Q = 0
Substitute
ΔU = Q + W
ΔU = W
Meaning
Only work changes internal energy.
No heat enters or leaves.
Formula
ΔU = W
Cases
Compression
↓
Work done on system
↓
Internal energy increases
Expansion
↓
System performs work
↓
Internal energy decreases
3. Isochoric Process
Definition
Volume remains constant.
Therefore
ΔV = 0
PV Work Formula
W = −PΔV
Since
ΔV = 0
Therefore
W = 0
No work is done.
Substitute
ΔU = Q + W
ΔU = Q
Since heat is supplied at constant volume,
we write
Qv
instead of Q.
Hence
ΔU = Qv
Important Formula
ΔU = Qv
Meaning
At constant volume,
all supplied heat increases internal energy.
No work is done because volume does not change.
4. Isobaric Process
Definition
Pressure remains constant.
Most laboratory reactions occur at constant atmospheric pressure.
Again,
First Law
ΔU = Q + W
Replace
W
by
−PΔV
Therefore
ΔU = Q − PΔV
Since pressure is constant,
heat is represented by
Qp
Hence
Qp = ΔU + PΔV
Important Formula
Qp = ΔU + PΔV
Meaning
At constant pressure,
part of the heat increases internal energy,
while the remaining heat performs expansion work.
Comparison of Important Processes
| Process | Constant Quantity | Formula |
|---|---|---|
| Isothermal | Temperature | W = −Q |
| Adiabatic | Heat | ΔU = W |
| Isochoric | Volume | ΔU = Qv |
| Isobaric | Pressure | Qp = ΔU + PΔV |
Important Points to Remember
Internal Energy
↓
State Function
Heat
↓
Path Function
Work
↓
Path Function
First Law
↓
ΔU = Q + W
Isothermal
↓
ΔU = 0
W = −Q
Adiabatic
↓
Q = 0
ΔU = W
Isochoric
↓
ΔV = 0
W = 0
ΔU = Qv
Isobaric
↓
ΔP = 0
Qp = ΔU + PΔV
Chapter Quick Revision
• Internal energy is the total energy stored inside a system.
• It consists of the kinetic energy and potential energy of particles.
• Internal energy is a state function and an extensive property.
• The First Law of Thermodynamics is based on the law of conservation of energy.
• Mathematical expression:
ΔU = Q + W
• In an isothermal process:
ΔU = 0, so W = −Q.
• In an adiabatic process:
Q = 0, so ΔU = W.
• In an isochoric process:
ΔV = 0, so W = 0 and ΔU = Qv.
• In an isobaric process:
Qp = ΔU + PΔV.
4.8 Enthalpy (H)
What is Enthalpy?
Enthalpy is the total heat content (total energy) of a system at constant pressure.
It includes:
• Internal Energy (U)
• Pressure–Volume (PV) Energy
Therefore,
Enthalpy = Internal Energy + PV Energy
Formula
H = U + PV
where,
H = Enthalpy
U = Internal Energy
P = Pressure
V = Volume
Change in Enthalpy
When a reaction takes place, the enthalpy changes.
Formula
ΔH = H₂ − H₁
where,
H₁ = Initial Enthalpy
H₂ = Final Enthalpy
Nature of Enthalpy
Enthalpy is:
✔ State Function
✔ Extensive Property
Derivation of ΔH = ΔU + PΔV
This is one of the most important derivations.
Step 1
We know,
H = U + PV
Step 2
For initial state,
H₁ = U₁ + P₁V₁
For final state,
H₂ = U₂ + P₂V₂
Step 3
Subtract the two equations.
ΔH = H₂ − H₁
= (U₂ + P₂V₂) − (U₁ + P₁V₁)
Step 4
Separate the terms.
ΔH = (U₂ − U₁) + (P₂V₂ − P₁V₁)
Step 5
We know,
U₂ − U₁ = ΔU
Therefore,
ΔH = ΔU + Δ(PV)
Step 6
For constant pressure,
P₁ = P₂ = P
Therefore,
Δ(PV) = P(V₂ − V₁)
= PΔV
Hence,
Final Equation
ΔH = ΔU + PΔV
This is the required derivation.
Heat at Constant Pressure
At constant pressure,
Heat exchanged is represented by
Qp
From the first law,
Qp = ΔU + PΔV
But
ΔH = ΔU + PΔV
Therefore,
Final Result
ΔH = Qp
Meaning
At constant pressure,
Heat supplied = Change in Enthalpy
4.8.1 Relationship Between ΔH and ΔU
For reactions involving gases,
Volume changes cannot be ignored.
We know,
ΔH = ΔU + PΔV
Using the ideal gas equation,
PV = nRT
For reactants,
PV₁ = n₁RT
For products,
PV₂ = n₂RT
Substitute into the equation.
ΔH = ΔU + (n₂RT − n₁RT)
Take RT common.
Final Equation
ΔH = ΔU + (n₂ − n₁)RT
Since,
Δng = n₂ − n₁
Therefore,
Final Formula
ΔH = ΔU + ΔngRT
Meaning of Symbols
ΔH = Enthalpy Change
ΔU = Internal Energy Change
Δng = Change in moles of gaseous substances
R = Gas Constant
T = Absolute Temperature (Kelvin)
When is ΔH = ΔU?
If only solids and liquids are involved,
their volume change is almost zero.
Therefore,
PΔV ≈ 0
Hence,
ΔH = ΔU
4.8.2 Work Done in Chemical Reactions
For gaseous reactions,
PV work can be calculated directly.
Formula
W = −ΔngRT
where,
Δng = Moles of gaseous products − Moles of gaseous reactants
Case 1
Products have more gas moles.
Δng is positive.
W is negative.
Meaning
System performs work.
Case 2
Reactants have more gas moles.
Δng is negative.
W is positive.
Meaning
Surroundings perform work.
Case 3
Equal gas moles.
Δng = 0
Therefore,
W = 0
No PV work is done.
4.9 Enthalpies of Physical Transformations
Physical transformations are changes of state without changing chemical composition.
4.9.1 Enthalpy of Fusion (ΔfusH)
Definition
Heat required to convert one mole of a solid into liquid at constant temperature and pressure.
Example
Ice → Water
H₂O(s) → H₂O(l)
ΔfusH = +6.01 kJ mol⁻¹
Positive sign means heat is absorbed.
Reverse Process
Water → Ice
ΔH = −6.01 kJ mol⁻¹
Heat is released.
4.9.2 Enthalpy of Vaporization (ΔvapH)
Definition
Heat required to convert one mole of liquid into vapour at constant temperature and pressure.
Example
Water → Steam
H₂O(l) → H₂O(g)
ΔvapH = +40.7 kJ mol⁻¹ (at 100°C)
Positive means heat absorbed.
Reverse Process
Steam → Water
Condensation
ΔH becomes negative.
4.9.3 Enthalpy of Sublimation (ΔsubH)
Definition
Heat required to convert one mole of solid directly into vapour.
Example
Dry Ice
Solid CO₂ → CO₂ Gas
Relationship
Sublimation occurs in two steps.
Solid
↓
Liquid
↓
Gas
Therefore,
Formula
ΔsubH = ΔfusH + ΔvapH
This equation is frequently asked in board examinations.
4.9.2 Enthalpy for Atomic and Molecular Changes
1. Enthalpy of Ionization (ΔionH)
Definition
Energy required to remove one electron from one mole of gaseous atoms.
Example
Na(g)
↓
Na⁺(g) + e⁻
Heat is absorbed.
Therefore,
ΔionH is always positive.
2. Electron Gain Enthalpy (ΔegH)
Definition
Energy change when one mole of gaseous atoms gains one electron.
Example
Cl(g) + e⁻
↓
Cl⁻(g)
Usually,
Heat is released.
Therefore,
Electron gain enthalpy is usually negative.
3. Enthalpy of Atomization (ΔatomH)
Definition
Heat required to convert one mole of a substance into gaseous atoms.
Example
Cl₂(g)
↓
2Cl(g)
Heat must be supplied.
Therefore,
ΔatomH is positive.
4. Enthalpy of Solution (ΔsolnH)
Definition
Heat change when one mole of a substance dissolves in a solvent.
Example
NaCl(s)
↓
NaCl(aq)
Dissolution Occurs in Two Steps
Step 1
Crystal lattice breaks.
Energy is absorbed.
This is called
Crystal Lattice Enthalpy
ΔLH
Always positive.
Step 2
Water molecules surround the ions.
Energy is released.
This is called
Hydration Enthalpy
ΔhydH
Always negative.
Relationship
Formula
ΔsolnH = ΔLH + ΔhydH
4.10 Thermochemistry
Definition
Thermochemistry is the branch of chemistry that studies heat changes during chemical reactions.
4.10.1 Enthalpy of Chemical Reaction (ΔrH)
Definition
The difference between the total enthalpy of products and reactants.
Formula
ΔrH = ΣH(products) − ΣH(reactants)
where,
Σ means "sum of."
Interpretation
If products have higher enthalpy than reactants,
ΔrH is positive.
Reaction absorbs heat.
It is Endothermic.
If reactants have higher enthalpy than products,
ΔrH is negative.
Reaction releases heat.
It is Exothermic.
4.10.2 Exothermic and Endothermic Reactions
Exothermic Reaction
Definition
A reaction that releases heat to the surroundings.
Characteristics
• Heat released
• Temperature of surroundings increases
• ΔH is negative
Examples
• Burning of coal
• Combustion of LPG
• Neutralization of acid and base
Endothermic Reaction
Definition
A reaction that absorbs heat from the surroundings.
Characteristics
• Heat absorbed
• Temperature of surroundings decreases
• ΔH is positive
Examples
• Photosynthesis
• Melting of ice
• Thermal decomposition of calcium carbonate
Difference Between Exothermic and Endothermic Reactions
| Exothermic | Endothermic |
|---|---|
| Heat is released | Heat is absorbed |
| ΔH is negative | ΔH is positive |
| Products have lower enthalpy | Products have higher enthalpy |
| Surroundings become warmer | Surroundings become cooler |
Complete Chapter Formula Sheet
- Work
W = −PextΔV
- Maximum Work
Wmax = −2.303 nRT log(V₂/V₁)
- First Law
ΔU = Q + W
- Isothermal Process
ΔU = 0
W = −Q
- Adiabatic Process
Q = 0
ΔU = W
- Isochoric Process
ΔU = Qv
- Isobaric Process
Qp = ΔU + PΔV
- Enthalpy
H = U + PV
- Enthalpy Change
ΔH = ΔU + PΔV
- Heat at Constant Pressure
ΔH = Qp
- Relation Between ΔH and ΔU
ΔH = ΔU + ΔngRT
- Work in Chemical Reactions
W = −ΔngRT
- Sublimation Relationship
ΔsubH = ΔfusH + ΔvapH
- Enthalpy of Solution
ΔsolnH = ΔLH + ΔhydH
- Enthalpy of Reaction
ΔrH = ΣH(products) − ΣH(reactants)
Final Chapter Quick Revision
- Thermodynamics deals with energy changes during physical and chemical processes.
- Heat and work are path functions, while internal energy and enthalpy are state functions.
- The First Law of Thermodynamics is expressed as ΔU = Q + W.
- PV work is calculated using W = −PextΔV.
- Maximum work is obtained in a reversible isothermal process.
- Enthalpy is defined as H = U + PV.
- At constant pressure, ΔH = Qp.
- For gaseous reactions, ΔH = ΔU + ΔngRT.
- Physical transformations include fusion, vaporization and sublimation.
- Thermochemistry studies heat changes during chemical reactions.
- Exothermic reactions release heat (ΔH < 0), while endothermic reactions absorb heat (ΔH > 0).
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