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Thermochemistry and Thermodynamics

Thermochemistry deals mainly with the heat changes accompanying physical and chemical processes, while thermodynamics gives the broader framework for understanding energy, work, heat, spontaneity and equilibrium.

These topics are closely connected. When a chemical reaction occurs, energy may be released or absorbed. Thermodynamics helps us describe where that energy goes, how much useful work can be obtained, and whether a process is thermodynamically favorable.

Main idea: Do not memorize thermodynamic equations separately. First identify the system, determine what happens to heat and work, and then choose the equation that describes that situation.

1. Introduction to Thermodynamics

Thermodynamics is the study of energy changes and the relationships between heat, work and other forms of energy.

It does not normally describe the detailed motion of individual molecules. Instead, it deals with measurable macroscopic properties such as temperature, pressure, volume and energy.

Thermochemistry is the branch concerned particularly with heat changes in chemical reactions and physical transformations.

Important quantities

  • Temperature
  • Pressure
  • Volume
  • Internal energy
  • Enthalpy
  • Entropy
  • Gibbs free energy

2. System, Surroundings and Boundary

System

The part of the universe selected for study is called the system.

Surroundings

Everything outside the system that can interact with it is called the surroundings.

Boundary

The boundary separates the system from the surroundings. It may be real or imaginary.

For a reaction taking place inside a flask, the reacting chemicals can be considered the system, while the flask, water bath and surrounding air form the surroundings depending on how the experiment is defined.

3. Types of Thermodynamic Systems

System Mass exchange Energy exchange Example
Open Yes Yes Open beaker
Closed No Yes Sealed flask that can exchange heat
Isolated No No Ideal thermos-like system

Open system

Both matter and energy can cross the boundary.

Closed system

Matter remains inside the boundary, but energy may cross as heat or work.

Isolated system

Neither matter nor energy crosses the boundary.

4. Thermodynamic Properties

A thermodynamic property is a measurable characteristic of a system.

Intensive properties

These do not depend on the amount of substance.

  • Temperature
  • Pressure
  • Density
  • Concentration

Extensive properties

These depend on the amount of matter.

  • Mass
  • Volume
  • Number of moles
  • Internal energy
  • Enthalpy

An extensive property can often be converted to an intensive quantity by dividing by the amount of substance.

5. State Functions and Path Functions

State function

A state function depends only on the initial and final states, not on the path used between them.

Examples:

  • Pressure
  • Volume
  • Temperature
  • Internal energy
  • Enthalpy
  • Entropy
  • Gibbs free energy

Path function

A path function depends on how the process occurs.

Heat and work are path functions.

Remember: ΔU, ΔH, ΔS and ΔG describe changes in state functions. q and w describe energy transfer through a particular process.

6. Thermodynamic Processes

Process Constant quantity
Isothermal Temperature
Adiabatic q = 0
Isobaric Pressure
Isochoric Volume

Isothermal

ΔT = 0

Adiabatic

q = 0

Isobaric

P = constant

Isochoric

V = constant

7. Heat and Work

Heat

Heat is energy transferred between a system and its surroundings because of a temperature difference.

Work

Work is energy transferred when a force acts through a distance or when a system undergoes another form of organized energy transfer.

For chemical thermodynamics, expansion/compression work is especially important.

8. Sign Convention

Using the chemistry convention:

Process Sign
Heat enters the system q > 0
Heat leaves the system q < 0
Work done on system w > 0
Work done by system w < 0
Be careful: different textbooks may use different work conventions. In this chapter, the chemistry convention w = -PextΔV is used.

9. Internal Energy

Internal energy, represented by U, is the total microscopic energy associated with the system.

It includes contributions from molecular translation, rotation, vibration, electronic energy and intermolecular interactions.

The absolute value of internal energy is generally not measured directly. Thermodynamics usually deals with the change in internal energy.

ΔU = Ufinal − Uinitial

10. First Law of Thermodynamics

The first law is the law of conservation of energy applied to a thermodynamic system.

ΔU = q + w

The equation says that the change in internal energy equals heat supplied to the system plus work done on the system.

Special cases

If no work occurs:

ΔU = q

If no heat is exchanged:

ΔU = w

If neither heat nor work is exchanged:

ΔU = 0

11. Expansion Work

When a gas expands against an external pressure, it does work on the surroundings.

w = −PextΔV
where
ΔV = V2 − V1

For expansion:

V2 > V1

Therefore ΔV is positive and w is negative.

For compression, ΔV is negative and w becomes positive.

Constant external pressure

w = −Pext(V2 − V1)

12. Reversible Process

A reversible process is an idealized process that can be reversed by an infinitesimally small change in conditions, restoring both system and surroundings to their original states.

A reversible expansion occurs through a sequence of states that are very close to equilibrium.

For reversible isothermal expansion of an ideal gas:

wrev = −nRT ln(V2/V1)

Using common logarithm:

wrev = −2.303nRT log(V2/V1)

13. Irreversible Process

An irreversible process occurs through conditions that are not continuously at equilibrium.

Examples include rapid expansion and expansion against a fixed external pressure.

wirrev = −Pext(V2 − V1)

For the same initial and final states, the reversible process gives the maximum magnitude of expansion work for an idealized reversible expansion.

14. Isothermal Process

An isothermal process occurs at constant temperature.

T = constant

For an ideal gas, internal energy depends only on temperature. Therefore:

ΔU = 0

From the first law:

0 = q + w
Therefore:
q = −w

For reversible isothermal expansion:

w = −nRT ln(V2/V1)

Therefore:

q = nRT ln(V2/V1)

15. Adiabatic Process

In an adiabatic process there is no heat exchange between system and surroundings.

q = 0

From the first law:

ΔU = w

For a reversible adiabatic process involving an ideal gas:

PVγ = constant
where
γ = Cp/Cv

Other useful relationships are:

TVγ−1 = constant
TγP1−γ = constant

16. Enthalpy

Enthalpy is defined as:

H = U + PV
For a change:
ΔH = ΔU + Δ(PV)

At constant pressure, when only pressure-volume work is involved:

qp = ΔH

Therefore enthalpy is particularly useful for reactions carried out at constant atmospheric pressure.

17. Heat Capacity

Heat capacity is the amount of heat required to raise the temperature of a system by one degree.

C = q/ΔT

For a sample:

q = CΔT

For mass-specific heat capacity:

q = mcΔT

For molar heat capacity:

q = nCmΔT

18. Cp and Cv

Heat capacity at constant volume

Cv = (∂q/∂T)V

For an ideal gas:

qv = ΔU

Heat capacity at constant pressure

Cp = (∂q/∂T)P

For an ideal gas:

qp = ΔH

For an ideal gas:

Cp − Cv = R

Therefore:

Cp = Cv + R

19. Relationship Between ΔH and ΔU

Starting with:

H = U + PV
Therefore:
ΔH = ΔU + Δ(PV)

For an ideal gas:

PV = nRT
Therefore:
Δ(PV) = Δn(g)RT
Thus:
ΔH = ΔU + Δn(g)RT
where Δn(g) is:
Δn(g) = moles of gaseous products − moles of gaseous reactants
Only gaseous species are included when calculating Δn(g). Solids and liquids are not counted in this expression.

20. Thermochemistry

Thermochemistry deals with heat changes associated with chemical reactions and physical changes.

A reaction may be:

  • Exothermic
  • Endothermic

Exothermic reaction

Heat is released to the surroundings.

ΔH < 0
Example:
CH4 + 2O2 → CO2 + 2H2O

Combustion reactions are commonly exothermic.

Endothermic reaction

Heat is absorbed by the system.

ΔH > 0

21. Standard Enthalpy Changes

Standard enthalpy changes are measured under standard-state conditions. The standard state commonly involves a pressure of approximately 1 bar and specified temperature, often 298 K unless otherwise stated.

Always check the conditions specified in the question.

22. Standard Enthalpy of Formation

The standard enthalpy of formation, ΔH°f, is the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states.

For water:

H2(g) + 1/2 O2(g) → H2O(l)

The coefficient of the compound being formed must correspond to one mole.

Important point

The standard enthalpy of formation of an element in its standard state is taken as zero.

For example:

ΔH°f[O2(g)] = 0

23. Enthalpy of Combustion

The standard enthalpy of combustion is the enthalpy change when one mole of a substance undergoes complete combustion in oxygen under specified standard conditions.

For methane:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)

Complete combustion of hydrocarbons generally produces carbon dioxide and water.

24. Enthalpy of Neutralization

The enthalpy of neutralization is the enthalpy change when an acid and base react to form one mole of water under specified conditions.

For a strong acid and strong base:

H+(aq) + OH−(aq) → H2O(l)

The value is approximately −57 kJ mol−1 at ordinary conditions, with the exact experimental value depending on conditions.

For weak acids or weak bases, the observed enthalpy of neutralization can differ because some energy is involved in ionization of the weak electrolyte.

25. Enthalpy of Solution and Hydration

Enthalpy of solution

This is the enthalpy change when one mole of a substance dissolves in a specified amount of solvent under specified conditions.

Enthalpy of hydration

This refers to the enthalpy change when gaseous ions become surrounded by water molecules and form hydrated ions.

For an ionic solid, the overall enthalpy of solution can be related to the energy required to separate the lattice and the energy released during hydration.

26. Hess's Law of Constant Heat Summation

Hess's law states that the total enthalpy change of a reaction is independent of the route by which the reaction occurs, provided the initial and final states are the same.

This follows because enthalpy is a state function.

Rules for using Hess's law

  1. If a reaction is reversed, change the sign of ΔH.
  2. If all coefficients are multiplied by a number, multiply ΔH by the same number.
  3. Add equations algebraically to obtain the required reaction.

Example:

Suppose:

C + O2 → CO2    ΔH1
CO + 1/2O2 → CO2    ΔH2

To obtain:

C + 1/2O2 → CO

Reverse the second equation and add it to the first.

Therefore:

ΔH = ΔH1 − ΔH2

27. Bond Enthalpy

Bond enthalpy is the energy required to break a particular type of bond, usually considered under gaseous conditions.

For an approximate reaction enthalpy:

ΔH ≈ Σ bond enthalpies of bonds broken − Σ bond enthalpies of bonds formed

The important rule is:

Breaking bonds requires energy.
Forming bonds releases energy.

Example

For a general reaction:

A–B + C–D → A–C + B–D

The calculation is:

ΔH ≈ [D(A–B) + D(C–D)] − [D(A–C) + D(B–D)]

28. Kirchhoff's Equation

The enthalpy of a reaction can change with temperature because the heat capacities of reactants and products are different.

The temperature dependence can be represented by:

d(ΔH)/dT = ΔCp

For approximately constant heat capacities:

ΔHT2 = ΔHT1 + ΔCp(T2 − T1)
where:
ΔCp = ΣCp(products) − ΣCp(reactants)

29. Second Law of Thermodynamics

The first law tells us that energy is conserved. It does not tell us the direction in which a process will naturally occur.

The second law introduces entropy and provides a criterion for the direction of spontaneous change.

One statement is that a spontaneous process is accompanied by an increase in the entropy of the universe:

ΔSuniverse > 0

For a reversible process:

ΔSuniverse = 0

30. Entropy

Entropy, S, is a thermodynamic state function associated with the dispersal of energy and the number of accessible microscopic arrangements.

For a reversible process:

dS = δqrev/T
Therefore:
ΔS = qrev/T

for an isothermal reversible process.

General tendencies

Entropy usually increases when:

  • A solid becomes a liquid.
  • A liquid becomes a gas.
  • The number of gas molecules increases.
  • A substance becomes more dispersed.

Entropy usually decreases in the reverse situations.

31. Third Law of Thermodynamics

The third law states that the entropy of a perfect crystalline substance approaches zero as the temperature approaches absolute zero.

S → 0 as T → 0 K

This provides a reference point for determining absolute entropy values.

32. Gibbs Free Energy

Gibbs free energy is especially useful for predicting spontaneity at constant temperature and pressure.

G = H − TS
Therefore:
ΔG = ΔH − TΔS

This is one of the most important equations in thermodynamics.

33. Spontaneity

ΔG Meaning at constant T and P
ΔG < 0 Thermodynamically spontaneous in the forward direction
ΔG > 0 Forward process is non-spontaneous as written
ΔG = 0 System is at equilibrium
Spontaneous does not mean fast. A process can be thermodynamically favorable but kinetically very slow.

34. Effect of ΔH and ΔS on Spontaneity

ΔH ΔS Spontaneity tendency
Negative Positive Spontaneous at all temperatures
Positive Negative Non-spontaneous at all temperatures
Negative Negative Favored at lower temperature
Positive Positive Favored at higher temperature

Why?

Use:

ΔG = ΔH − TΔS

The temperature multiplies the entropy term. Therefore the importance of ΔS increases as temperature increases.

35. Worked Numericals

Numerical 1: Heat absorbed by water

Question: Calculate the heat required to raise 100 g of water from 25°C to 35°C. Take c = 4.18 J g−1 K−1.

Given:

m = 100 g

c = 4.18 J g−1 K−1

ΔT = 35 − 25 = 10 K

Formula:

q = mcΔT

Substitution:

q = 100 × 4.18 × 10

Answer:

q = 4180 J = 4.18 kJ

Numerical 2: Internal energy

A system absorbs 500 J of heat and does 200 J of work on the surroundings. Calculate ΔU.

Given:

q = +500 J

w = −200 J

Formula:

ΔU = q + w

Substitution:

ΔU = 500 + (−200)

Answer:

ΔU = +300 J

Numerical 3: Expansion work

A gas expands from 2.0 L to 5.0 L against an external pressure of 1 atm. Calculate the work done.

Given:

Pext = 1 atm

V1 = 2.0 L

V2 = 5.0 L

ΔV = 3.0 L

Formula:

w = −PextΔV

Substitution:

w = −1 × 3.0 = −3.0 L atm

Using:

1 L atm = 101.325 J

Answer:

w = −303.98 J ≈ −304 J

Numerical 4: ΔH from ΔU

For a gaseous reaction, ΔU = −100 kJ mol−1 and Δn(g) = 2. Calculate ΔH at 298 K.

Given:

ΔU = −100 kJ mol−1

Δn(g) = 2

T = 298 K

R = 8.314 J mol−1 K−1

Formula:

ΔH = ΔU + Δn(g)RT

Convert R term to kJ:

R = 0.008314 kJ mol−1 K−1

Substitution:

ΔH = −100 + (2)(0.008314)(298)
ΔH ≈ −95.05 kJ mol−1

Answer:

ΔH ≈ −95.1 kJ mol−1

Numerical 5: Gibbs free energy

For a reaction, ΔH = −50 kJ mol−1 and ΔS = −100 J mol−1 K−1 at 298 K. Calculate ΔG.

Given:

ΔH = −50 kJ mol−1

ΔS = −100 J mol−1 K−1

T = 298 K

Convert ΔS to kJ:

ΔS = −0.100 kJ mol−1 K−1

Formula:

ΔG = ΔH − TΔS

Substitution:

ΔG = −50 − (298)(−0.100)
ΔG = −50 + 29.8

Answer:

ΔG = −20.2 kJ mol−1

The reaction is thermodynamically spontaneous under these specified conditions.

Numerical 6: Enthalpy from bond energies

Suppose the total bond energy of bonds broken is 800 kJ mol−1 and the total bond energy of bonds formed is 950 kJ mol−1. Calculate the approximate reaction enthalpy.

Formula:

ΔH ≈ bonds broken − bonds formed

Substitution:

ΔH ≈ 800 − 950

Answer:

ΔH ≈ −150 kJ mol−1

The negative value indicates an exothermic reaction.

36. Understand It: Reconstruct the Equations

The following reasoning method helps when you forget which thermodynamic equation to use.

Situation 1: Heat and work are both involved

Start from conservation of energy:

ΔU = q + w

This is the first equation to consider.

Situation 2: Gas expands against external pressure

Expansion means the system pushes the surroundings, so work is done by the system.

w = −PextΔV

Situation 3: Constant pressure

At constant pressure with only pressure-volume work:

qp = ΔH

Situation 4: Constant volume

At constant volume, ΔV = 0. Therefore pressure-volume work is zero.

qv = ΔU

Situation 5: Need to judge spontaneity

Ask whether the question is at constant temperature and pressure. If so, Gibbs free energy is usually the most convenient criterion.

ΔG = ΔH − TΔS

Situation 6: Temperature is changing

If heat capacity is given:

q = mcΔT

or for molar heat capacity:

q = nCmΔT

37. Common Wrong Thinking

Wrong idea 1: Heat and temperature are the same thing.

They are not. Temperature describes the thermal state of a system, whereas heat is energy transferred because of a temperature difference.

Wrong idea 2: ΔU is always equal to q.

Only under conditions where work is zero does ΔU = q.

Wrong idea 3: Expansion work is positive.

Using the chemistry sign convention:

w = −PextΔV

Expansion gives positive ΔV and therefore negative w.

Wrong idea 4: Negative ΔH automatically means spontaneous.

Not necessarily. Spontaneity depends on ΔG:

ΔG = ΔH − TΔS

Wrong idea 5: Spontaneous means fast.

Thermodynamics describes feasibility, not reaction rate. Kinetics determines how rapidly a reaction occurs.

Wrong idea 6: Entropy always increases inside the system.

The entropy of the system can decrease during a spontaneous process as long as the total entropy change of the universe is positive.

Wrong idea 7: ΔH and ΔU are always identical.

They can differ when the number of gaseous moles changes.

ΔH = ΔU + Δn(g)RT

38. What the Examiner Actually Wants

For 2 marks

Give the definition followed by the essential equation or one distinguishing point.

Example: Define enthalpy.

H = U + PV

Then state that enthalpy is a state function.

For 5 marks

A five-mark answer should contain enough explanation to show the relationship between the concepts, not just a list of equations.

For example, for the first law:

  1. Statement
  2. Energy conservation explanation
  3. Mathematical expression
  4. Meaning of q and w
  5. One special case

For 10 marks

For a long question such as “Explain the first law of thermodynamics and its applications,” organize the answer as:

  1. Introduction
  2. Statement of first law
  3. Mathematical expression
  4. Sign convention
  5. Expansion work
  6. Reversible and irreversible work
  7. Constant-volume application
  8. Constant-pressure application
  9. Enthalpy relationship
  10. Relevant equations
  11. Worked example
  12. Conclusion

For “Discuss Gibbs free energy and spontaneity,” include the definition, equation, meaning of positive/negative/zero ΔG, effect of ΔH and ΔS, and at least one numerical example.

39. Thermochemistry Calculator

This small calculator allows you to practice three common calculations: heat using q = mcΔT, Gibbs free energy, and ΔH from ΔU.

Heat Calculator

Enter values and calculate.

Gibbs Free Energy Calculator

Enter values and calculate.

ΔH from ΔU Calculator

Enter values and calculate.

40. Practice Questions

1. Why is ΔU a state function while q is a path function?

ΔU depends only on the initial and final states, whereas the heat transferred depends on the path by which the process occurs.

2. A gas expands against constant external pressure. What is the sign of w?

Using w = −PextΔV, expansion gives ΔV > 0, so w is negative.

3. What happens to ΔU for an ideal gas in an isothermal process?

For an ideal gas, internal energy depends only on temperature. Since temperature is constant, ΔU = 0.

4. Why is SN1/SN2 not related to thermodynamics?

SN1/SN2 describes organic reaction mechanisms, while thermodynamics deals with energy, heat, work and spontaneity. They can be connected through reaction energetics, but they answer different questions.

5. A reaction has ΔH < 0 and ΔS > 0. What can you say about ΔG?

ΔG = ΔH − TΔS. Both terms favor a negative ΔG, so the reaction is thermodynamically spontaneous at all temperatures under the stated assumptions.

6. Why is Hess's law possible?

Because enthalpy is a state function. Therefore the total enthalpy change depends only on the initial and final states, not on the reaction pathway.

41. Quick Revision

  • Thermodynamics deals with energy changes.
  • Thermochemistry deals mainly with heat changes in chemical and physical processes.
  • System = part under study.
  • Surroundings = everything outside the system.
  • Open system exchanges matter and energy.
  • Closed system exchanges energy but not matter.
  • Isolated system exchanges neither matter nor energy.
  • U, H, S and G are state functions.
  • q and w are path functions.
  • First law: ΔU = q + w.
  • Expansion work: w = −PextΔV.
  • At constant volume: qv = ΔU.
  • At constant pressure: qp = ΔH.
  • H = U + PV.
  • ΔH = ΔU + Δn(g)RT.
  • q = mcΔT.
  • Breaking bonds requires energy.
  • Forming bonds releases energy.
  • Hess's law follows from the state-function nature of enthalpy.
  • ΔS = qrev/T for an isothermal reversible process.
  • For a spontaneous process at constant T and P, ΔG < 0.
  • ΔG = ΔH − TΔS.
  • At equilibrium, ΔG = 0.
  • Spontaneous does not mean fast.

42. Quick Check

1. Which is a state function?

2. What is the first law equation?

3. During expansion, using the chemistry convention, w is:

4. At constant pressure, q is equal to:

5. Which equation is used to predict spontaneity at constant temperature and pressure?

6. If ΔG = 0, the system is:

7. Which process has q = 0?

8. Hess's law is based mainly on the fact that enthalpy is:

Connect the chapters:

Thermochemistry connects naturally with chemical kinetics, equilibrium, electrochemistry, phase changes and solution chemistry. It also connects with molecular bonding because bond breaking and bond formation are directly related to reaction enthalpy.

For example, bond energies can help estimate ΔH, while ΔG can tell whether a process is thermodynamically favorable. Kinetics then asks a different question: how quickly does that process occur?