Chemical Thermodynamics | ISC Class 11 Chemistry Notes

This note covers thermodynamic systems and processes, energy conservation, heat and work, internal energy and enthalpy, heat capacities, thermochemical equations, Hess’s law, reaction and phase-change enthalpies, neutralisation, entropy, spontaneity, free energies, equilibrium and the third law of thermodynamics.

What are a system, its surroundings and its boundary?

Thermodynamics studies energy transformations in macroscopic systems, meaning systems containing large numbers of particles. It does not describe the mechanism of a reaction or how quickly the reaction proceeds.

A system is the part of the universe selected for study. The surroundings comprise everything outside it. For practical purposes, attention is directed to the surroundings that can interact with the system. Usually, this is the region near the system.

The system and surroundings together constitute the thermodynamic universe. A real or imaginary boundary separates them and defines which matter and energy transfers to track.

How are systems classified?

Closing a vessel prevents matter transfer, but a conducting wall can still allow heat transfer.

An ideal system is a simplified model obeying specified ideal assumptions. An ideal gas, for example, obeys the ideal-gas equation. A real system is an actual system whose behaviour may depart from the ideal model.

Three vessels are surrounded by a green region labelled surroundings. The open vessel shows matter and energy crossing its boundary. The closed vessel shows energy crossing; the isolated vessel shows neither crossing.

See Fig. 5.2 in your NCERT textbook

How do state properties and thermodynamic processes differ?

The state of a system is described by measurable bulk properties, such as pressure, volume, temperature and composition. Let p denote pressure, V volume, T absolute temperature and n amount of substance in moles.

A state function depends on the current state, irrespective of the route used to reach it. Pressure, volume, temperature and internal energy are examples. A path function, such as heat or work, depends on the process followed between the states.

An extensive property depends on the amount of matter. Mass, volume, internal energy, enthalpy and heat capacity are extensive. An intensive property does not depend on that amount. Temperature, pressure and density are intensive.

Which conditions define a process?

A thermodynamic process changes a system from one state to another. At thermodynamic equilibrium, there is no driving tendency for thermal, mechanical or chemical change. Its macroscopic properties remain unchanged with time.

A reversible process proceeds infinitely slowly through a series of equilibrium states, with system and surroundings always in near equilibrium.

Note: Isothermal means constant temperature; adiabatic means no heat transfer. These conditions describe different restrictions and must not be substituted for each other.

How does the first law connect heat, work and internal energy?

Internal energy, U, represents the energy of the system. Its absolute value is not measured directly in this treatment, but changes can be measured. The symbol Δ means final value minus initial value, so ΔU is the internal-energy change.

Heat, q, is energy transferred because of a temperature difference. Heat is energy in transit, rather than a stored property called “heat content”. Net heat transfer driven by a temperature difference ceases when the bodies attain thermal equilibrium at the same temperature.

Work, w, is energy transferred through an organised action such as moving a piston, stirring water or passing an electric current. Energy is the capacity to do work. Mechanical and electrical work can both change internal energy.

The first law of thermodynamics expresses conservation of energy: energy cannot be created or destroyed. The energy of an isolated system is constant. For a closed system, using the chemical sign convention:

ΔU = q + w

SI means the International System of Units. The SI unit of heat is the joule, symbol J. The SI unit of work is the joule. The SI unit of internal energy is the joule. A kilojoule, symbol kJ, equals one thousand joules.

What happens in isolated and cyclic processes?

For an isolated system, q and w are both zero, so ΔU is zero. A non-isolated system can gain or lose energy, but an energy transfer between system and surroundings cancels when both are considered together as an isolated whole.

For a complete cycle, ΔU is zero because the initial state is restored. Hence q = −w for the whole cycle. Heat and work need not separately be zero: they depend on the route, while their sum restores the original internal energy.

Worked example 1. A closed system absorbs 701 J of heat and does 394 J of work. Calculate its internal-energy change.

Formula: ΔU = q + w. Substitute: q = +701 J and w = −394 J. Answer: ΔU = 701 − 394 = +307 J. The positive result means the system gains internal energy.

How is expansion work calculated and compared graphically?

Pressure-volume work occurs when a boundary moves against external pressure. Let pₑₓ denote external pressure, Vᵢ initial volume and V𝒻 final volume. For a single expansion or compression against constant external pressure:

w = −pₑₓΔV

Here ΔV = V𝒻 − Vᵢ. Expansion gives positive ΔV and negative w, because the gas does work. Compression gives negative ΔV and positive w, because the surroundings do work on the gas.

The SI unit of pressure is the pascal, symbol Pa. The SI unit of volume is the cubic metre, symbol m³. Pressure in pascals multiplied by volume in cubic metres gives work in joules. The litre, symbol L, is also used in gas calculations.

What changes under reversible conditions?

For changing external pressure, the work is the negative integral of external pressure with respect to volume. Under reversible conditions, the external pressure differs infinitesimally from the gas pressure. For an ideal gas:

pV = nRT

Here R is the molar gas constant. Using R = 8.314 J mol⁻¹ K⁻¹ gives energy in joules; K denotes kelvin. The gas law relates the already defined pressure, volume, amount and absolute temperature.

For reversible isothermal expansion of a fixed amount of ideal gas:

w = −nRT ln(V𝒻/Vᵢ)

The symbol ln means natural logarithm. This expression follows by substituting p = nRT/V into the work integral while T remains constant. Under these conditions ΔU = 0 and q = −w.

Both plots put pressure vertically and volume horizontally. Compression runs from the larger initial volume on the right to the smaller final volume on the left. The finite-step plot has shaded rectangles; the reversible plot shades the area under a smooth curve.

See Figs. 5.5(b) and 5.5(c) in your NCERT textbook

For the same end states, finite-step compression requires more work input than reversible compression. Conversely, reversible isothermal expansion gives the maximum work output between the specified states. Compare magnitudes carefully: expansion work carries a negative sign in the chemical convention.

Worked example 2. An ideal gas expands isothermally at 25°C from 2 L to 10 L into a vacuum, where external pressure is zero. Find work and heat transfer.

Formula: w = −pₑₓΔV; q = −w. Substitute: ΔV = 10 − 2 = 8 L; pₑₓ = 0. Answer: w = 0 J and q = 0 J. This is free expansion: the gas encounters no opposing pressure.

Why is enthalpy useful at constant pressure?

Many chemical reactions take place under constant atmospheric pressure. The energy supplied as heat can both change internal energy and support expansion work. Enthalpy, H, combines internal energy with the pressure-volume term:

H = U + pV

Enthalpy is an extensive state function because U, p and V are state properties. The SI unit of enthalpy is the joule. For heat transfer at constant pressure, write qₚ; for heat transfer at constant volume, write qᵥ.

Derivation: Heat at constant pressure

1. Assume a closed system with only pressure-volume work at constant pressure. Then w = −pΔV.
2. Substitute into the first law: ΔU = qₚ − pΔV.
3. Rearrange to obtain qₚ = ΔU + pΔV. At constant pressure, the definition of enthalpy gives ΔH = ΔU + pΔV.
4. Equating these expressions identifies the measured constant-pressure heat with the enthalpy change.

ΔH = qₚ

At constant volume, pressure-volume work is zero. With no other work, ΔU = qᵥ. These conditions matter: neither heat relationship should be used without identifying the constraint and the permitted work.

An exothermic process releases heat and has negative ΔH at constant pressure. An endothermic process absorbs heat and has positive ΔH. The difference between ΔH and ΔU is not usually significant for systems containing only solids and liquids.

How are enthalpy and internal energy related for gas reactions?

For ideal gases at the same temperature and pressure, the pressure-volume change can be written in terms of the change in gaseous amount. Let Δn𝗀 mean moles of gaseous products minus moles of gaseous reactants for the specified reaction amount.

ΔH = ΔU + Δn𝗀RT

Count gaseous substances from the balanced equation; solids and liquids do not enter Δn𝗀. For molar reaction energies, use the gas-mole change per mole of reaction.

Worked example 3. Evaporate 18 g of water at 298 K. Its molar mass is 18 g mol⁻¹ and molar vaporisation enthalpy is 44.01 kJ mol⁻¹. Treat the vapour as ideal; use R = 8.314 J mol⁻¹ K⁻¹. Find heat absorbed and internal-energy change.

Formula: n = mass/molar mass; qₚ = n × molar vaporisation enthalpy; ΔU = qₚ − Δn𝗀RT. Substitute: n = 18/18 = 1 mol; Δn𝗀 = 1 mol; Δn𝗀RT = 2.48 kJ. Answer: qₚ = 44.01 kJ and ΔU = 44.01 − 2.48 = 41.53 kJ, equivalent to 41530 J.

How do heat capacities help measure energy changes?

Heat capacity, C, is the heat needed per unit rise in temperature under specified conditions. For a range over which C can be treated as constant, the heat supplied is related to the temperature change:

q = CΔT

Here ΔT is final temperature minus initial temperature. Heat capacity has units J K⁻¹ and depends on the amount of material.

Specific heat capacity, c, is heat capacity per unit mass. If m is the sample mass, then:

q = mcΔT

The SI unit of specific heat capacity is J kg⁻¹ K⁻¹, where kg denotes kilogram. Data may instead use grams, symbol g, with J g⁻¹ K⁻¹. Keep the mass unit consistent with the quoted specific heat capacity.

Molar heat capacity is heat capacity per mole, with units J mol⁻¹ K⁻¹. In the derivation below, Cₚ and Cᵥ mean molar heat capacities at constant pressure and constant volume respectively, rather than heat capacities of an unspecified sample.

Derivation: The ideal-gas heat-capacity relation

1. For one mole of ideal gas, the equation of state gives pV = RT, so H = U + RT.
2. For a temperature interval, ΔH = ΔU + RΔT.
3. Write the changes using molar heat capacities treated as constant over the interval: ΔH = CₚΔT and ΔU = CᵥΔT.
4. Substitute and divide by the non-zero temperature interval: Cₚ = Cᵥ + R.

Cₚ − Cᵥ = R

For n moles, ΔU = nCᵥΔT. The larger constant-pressure heat capacity reflects the additional expansion work. The relation above applies to an ideal gas; it is not a general relation for every substance.

What does a calorimeter measure?

Calorimetry measures energy changes through observed temperature changes and known heat capacities. In a rigid bomb calorimeter, combustion heats the surrounding water and apparatus. Constant volume makes the reaction heat a measure of ΔU when other work is absent.

A constant-pressure calorimeter measures ΔH under the corresponding work restriction. Heat gained by the calorimeter has the opposite sign to heat lost by the reaction. Include the heat capacities of the parts that absorb the released energy.

How should thermochemical equations and standard states be read?

A thermochemical equation is a balanced chemical equation accompanied by its reaction enthalpy. Physical states matter: (s) means solid, (l) liquid and (g) gas. The enthalpy changes when the physical state of a reactant or product changes.

The symbol ΔᵣH denotes reaction enthalpy, with the subscript r identifying a reaction. It equals total product enthalpy minus total reactant enthalpy, using the amounts specified by the balanced equation.

A standard state is the pure substance at 1 bar and a specified temperature. A bar is a pressure unit. The superscript ° denotes standard-state quantities. Data are usually quoted at 298 K, but standard state does not itself fix the temperature at 298 K.

What does standard enthalpy of formation mean?

Standard molar enthalpy of formation, Δ𝒻H°, is the enthalpy change when one mole of a compound forms in its standard state from its elements in their reference states at the specified temperature. Here f identifies formation. A reference state is the element’s most stable state under the specified standard conditions.

For example, at 298 K the reference forms include dihydrogen gas, dioxygen gas and graphite. By convention, an element in its reference state has zero standard enthalpy of formation. This does not assert that its absolute internal energy or enthalpy is zero.

The formation equation for liquid water is H₂(g) + ½O₂(g) → H₂O(l). Formation must start from elements and produce one mole of the compound. Producing a compound from other compounds is a reaction, but not its formation reaction.

How are reaction equations scaled?

1. Write the balanced equation and identify the physical states of every substance.
2. Interpret its coefficients as molar amounts for the thermochemical calculation.
3. Multiply the enthalpy change by the same factor if all equation coefficients are multiplied.
4. Reverse the sign of the enthalpy change if the equation is reversed.

How does Hess’s law determine an indirect enthalpy change?

Hess’s law of constant heat summation states that the enthalpy change for a reaction equals the sum of the enthalpy changes of the steps into which it can be divided, at the same temperature. The initial and final states must match.

How should component equations be combined?

1. Write the target equation, including physical states and required molar coefficients.
2. Reverse supplied equations when necessary, reversing their enthalpy signs too.
3. Multiply equations and their enthalpies by matching factors to obtain the required amounts.
4. Add the equations and cancel identical species on opposite sides, then add their enthalpy changes.

If ΔH₁ and ΔH₂ denote enthalpy changes of two component steps, their sum gives the overall change:

ΔH = ΔH₁ + ΔH₂

Worked example 4. At the same temperature, C(graphite, s) + O₂(g) → CO₂(g) has enthalpy −393.5 kJ mol⁻¹, while CO(g) + ½O₂(g) → CO₂(g) has enthalpy −283.0 kJ mol⁻¹. Find the enthalpy for C(graphite, s) + ½O₂(g) → CO(g).

Formula: ΔH = ΔH₁ − ΔH₂. Reverse the carbon monoxide combustion equation, making its enthalpy +283.0 kJ mol⁻¹. Substitute: ΔH = −393.5 + 283.0. Answer: ΔH = −110.5 kJ mol⁻¹. Carbon dioxide cancels when the equations are added.

What do the named enthalpy changes describe?

Which enthalpies accompany phase changes?

Fusion, vaporisation and sublimation absorb heat. Reversing a phase change reverses its enthalpy sign. During a phase change at fixed pressure, heat can change the phase while the temperature remains constant; it need not cause a temperature rise.

Calorific value is the heat released by complete combustion of a unit mass of fuel, commonly expressed in kJ g⁻¹ or kJ kg⁻¹. It is a positive heat-output magnitude; the combustion enthalpy of the reacting system is negative.

How do bond dissociation and mean bond enthalpy differ?

Bond dissociation enthalpy is the enthalpy needed to break one mole of specified covalent bonds in gaseous species, forming gaseous products. For a diatomic molecule, separating its two atoms corresponds to breaking its bond.

In a polyatomic molecule, successive bond-breaking steps can require different energies. Mean bond enthalpy averages those values. Methane’s total atomisation enthalpy is 1665 kJ mol⁻¹, and its mean carbon-hydrogen bond enthalpy is quoted as 416 kJ mol⁻¹.

Mean carbon-hydrogen bond enthalpies differ slightly from compound to compound. Estimates based on average bond enthalpies are therefore approximate. For gaseous reactants and products, subtract the sum of bond enthalpies for bonds formed from the sum for bonds broken.

How do solution, dilution and neutralisation enthalpies differ?

Enthalpy of solution is the enthalpy change when one mole of solute dissolves in a specified amount of solvent. The solute is the dissolved substance; the solvent is the dissolving medium. Dilute means low solute concentration, or little solute relative to the solution amount.

In an ionic solid, ions occupy an ordered crystal structure called a lattice. Dissolution separates ions and surrounds them with solvent molecules, a process called solvation. When the solvent is water, this is called hydration.

The energy needed to separate the lattice competes with energy released during hydration. For sodium chloride, the stated contributions are +788 and −784 kJ mol⁻¹, giving a small positive solution enthalpy of +4 kJ mol⁻¹.

What is enthalpy of dilution?

Enthalpy of dilution is the enthalpy change when additional solvent is added to a solution. It depends on the starting concentration and the amount added. At infinite dilution, interactions between solute particles become negligible and solution enthalpy approaches a limiting value.

Worked example 5. At the same temperature and pressure, dissolving one mole of hydrogen chloride gas in 25 mol of water gives −72.03 kJ mol⁻¹; dissolving it in 40 mol of water gives −72.79 kJ mol⁻¹. Find the enthalpy when the first solution receives 15 mol more water.

Formula: dilution enthalpy = final solution enthalpy − initial solution enthalpy. Substitute: −72.79 − (−72.03). Answer: −0.76 kJ mol⁻¹, equivalent to −760 J mol⁻¹. Dilution releases heat for this particular change.

Why is strong-acid and strong-base neutralisation nearly constant?

Enthalpy of neutralisation is the enthalpy change per mole of water formed when an acid and a base react in dilute solution. Strong acids and strong bases are essentially completely ionised in dilute aqueous solution.

The common net ionic change is H⁺(aq) + OH⁻(aq) → H₂O(l), where H⁺ is the hydrogen ion, OH⁻ the hydroxide ion and (aq) means dissolved in water. Other ions are spectators, meaning they do not participate in this net change.

For dilute strong-acid and strong-base solutions under comparable conditions, the molar neutralisation enthalpy is approximately constant because the net ionic reaction is the same. A weak acid or weak base can require energy for further ionisation, altering the observed heat.

To verify this experimentally, mix measured amounts initially at the same temperature in an insulated calorimeter. Measure the temperature rise, calculate the heat gained by solution and apparatus, reverse its sign for the reaction, and divide by moles of water formed. Compare different strong-acid and strong-base pairs under matched conditions.

Why are entropy and the second law needed?

The first law balances energy but does not determine the direction of change. Heat flows spontaneously from hotter to colder bodies, although conservation of energy alone would not prohibit the reverse transfer. This limitation requires a further criterion.

A spontaneous process has the potential to proceed without assistance from an external agency. It need not be rapid. A non-spontaneous process does not proceed unaided in the specified direction under the given conditions.

Entropy, S, is a state function associated with the dispersal of energy and, as a useful qualitative picture, the degree of disorder. Greater disorder in an isolated system corresponds to higher entropy. The SI unit of entropy is J K⁻¹.

How is entropy change calculated?

For a reversible isothermal process, let qᵣₑᵥ denote reversible heat transfer. At absolute temperature T:

ΔS = qᵣₑᵥ/T

Entropy change depends on the endpoints. To evaluate an irreversible change, one can consider a reversible path between the same states; inserting irreversible heat directly into this expression is not generally valid.

The second law of thermodynamics states that the entropy of an isolated system increases in a spontaneous process. For the combined system and surroundings, let ΔSₜₒₜ denote total entropy change and ΔSₛᵤᵣᵣ the surroundings’ entropy change:

ΔSₜₒₜ = ΔS + ΔSₛᵤᵣᵣ

Total entropy change is positive for a spontaneous irreversible process and zero for a reversible process. The system’s entropy alone can decrease while the surroundings’ entropy increases by a larger amount.

The upper drawing has black and pale dots on opposite sides of a partition. The lower drawing shows the two kinds of dots mixed after the partition is removed, illustrating the increase in disorder during diffusion.

See Fig. 5.11 in your NCERT textbook

The second law also limits heat-to-work conversion: a cyclic engine cannot turn heat drawn from a single reservoir entirely into work with no other effect. A reservoir is a body that supplies or receives heat without appreciable temperature change.

Thermal death describes the hypothetical final equilibrium condition of the universe, with no available temperature differences to drive heat engines. It means loss of the ability to obtain useful work from such differences, rather than disappearance of energy.

What does the third law state?

The third law of thermodynamics states that the entropy of a pure, perfectly crystalline substance approaches zero as its temperature approaches absolute zero, 0 K. The restriction to a pure, perfectly ordered crystal is part of the statement.

How do Gibbs and Helmholtz energies describe spontaneity?

Gibbs energy, G, combines enthalpy and entropy. It is an extensive state function defined by:

G = H − TS

At constant temperature, its change is:

ΔG = ΔH − TΔS

Express ΔH and TΔS in matching energy units. If entropy change is in J K⁻¹ mol⁻¹ and enthalpy change in kJ mol⁻¹, convert one before subtracting. Temperature must be in kelvin.

What is the constant-temperature, constant-pressure criterion?

A favourable enthalpy change and a favourable entropy change reinforce each other. If ΔH is negative and ΔS positive, ΔG is negative. If ΔH is positive and ΔS negative, ΔG is positive.

When both are positive, increasing temperature can make TΔS exceed ΔH. When both are negative, lower temperature favours negative ΔG. “High” and “low” temperature are relative to the particular reaction; high temperature can even mean room temperature.

Worked example 6. For 2A + B → C, reaction enthalpy is 400 kJ mol⁻¹ and reaction entropy is 0.2 kJ K⁻¹ mol⁻¹. Assume both remain constant over the temperature range. Find the temperature above which the forward reaction is spontaneous at constant pressure.

Formula: ΔG = ΔH − TΔS; at the boundary, T = ΔH/ΔS. Substitute: T = 400/0.2. Answer: the boundary is 2000 K. At 2000 K, ΔG = 0; above 2000 K, ΔG is negative and the forward reaction is spontaneous.

How is Helmholtz energy related to Gibbs energy?

Helmholtz energy, A, is defined by A = U − TS. It is also a state function with energy units. Substituting H = U + pV into the Gibbs definition gives:

G = A + pV

For a closed system at constant temperature and volume with no imposed non-expansion work, Helmholtz energy decreases during spontaneous change and is minimum at equilibrium. Gibbs energy supplies the corresponding criterion at constant temperature and pressure.

How is standard Gibbs energy related to equilibrium?

At chemical equilibrium, the Gibbs energy is minimum under constant temperature and pressure, and the reaction Gibbs-energy change is zero. This is the change for the actual equilibrium composition, rather than an assertion that standard reaction Gibbs energy must be zero.

Let ΔᵣG° denote standard reaction Gibbs-energy change and K the thermodynamic equilibrium constant, the dimensionless equilibrium product-to-reactant activity ratio with stoichiometric powers. Activity expresses effective concentration relative to its standard state.

ΔᵣG° = −RT ln K

In this equation, K means the equilibrium constant; when written after a numerical temperature, K is the unit kelvin. R is the molar gas constant and T is the absolute temperature at which the equilibrium constant applies.

What can the sign predict?

If K exceeds one, ln K is positive and ΔᵣG° is negative. If K is less than one, ln K is negative and ΔᵣG° is positive. When K equals one, standard reaction Gibbs energy is zero.

Worked example 7. For a reaction at 300 K, the dimensionless equilibrium constant is 10. Calculate standard reaction Gibbs energy using R = 8.314 J mol⁻¹ K⁻¹ and ln 10 = 2.303.

Formula: ΔᵣG° = −RT ln K. Substitute: −8.314 × 300 × 2.303 J mol⁻¹. Answer: ΔᵣG° ≈ −5744 J mol⁻¹ = −5.744 kJ mol⁻¹. The negative value agrees with an equilibrium constant greater than one.

Glossary

• System — The selected part of the universe whose properties and energy changes are being studied.
• Surroundings — Everything outside the system, especially the region able to exchange energy or matter with it.
• State function — A property determined by the current state, independently of the path used to reach it.
• Path function — A quantity whose value for a process depends on the route between initial and final states.
• Internal energy — Energy of a system, whose change can result from heat transfer or work.
• Enthalpy — The state function equal to internal energy plus the product of pressure and volume.
• Heat capacity — Heat supplied per unit temperature rise under specified conditions for the amount of material considered.
• Adiabatic process — A process in which no heat is transferred between the system and its surroundings.
• Standard state — The pure form of a substance at a pressure of one bar and a specified temperature.
• Hess’s law — Reaction enthalpy equals the sum of component enthalpy changes connecting the same initial and final states.
• Entropy — A state function related to energy dispersal and qualitatively pictured through the degree of disorder.
• Spontaneous process — A process with the potential to proceed without external assistance under the stated conditions.
• Gibbs energy — Enthalpy minus temperature times entropy, supplying a spontaneity criterion at constant temperature and pressure.
• Helmholtz energy — Internal energy minus temperature times entropy, useful for processes at constant temperature and volume.

Common errors and misconceptions

• Misconception: Work done by the gas is positive in the chemical convention. Correct: Expansion work is negative; work done on the gas is positive.
• Misconception: A closed vessel is necessarily isolated. Correct: It prevents matter transfer but may exchange heat through conducting walls.
• Misconception: Adiabatic means constant temperature. Correct: It means zero heat transfer; temperature can change as work changes internal energy.
• Misconception: Every exothermic reaction is spontaneous under every condition. Correct: At constant temperature and pressure, evaluate ΔG using both enthalpy and entropy changes.
• Misconception: Spontaneous reactions must be fast. Correct: Spontaneity concerns thermodynamic direction and does not determine the rate.
• Misconception: A spontaneous reaction must increase the system’s entropy. Correct: The total entropy change of system and surroundings must be positive.
• Misconception: Standard formation enthalpy being zero means an element contains no energy. Correct: Zero formation enthalpy is a reference convention, not an absolute-energy measurement.
• Misconception: Equilibrium requires ΔᵣG° = 0. Correct: Actual reaction Gibbs-energy change is zero at equilibrium; ΔᵣG° depends on the equilibrium constant.

Exam-style questions with model answers

Q1. Distinguish a closed system from an isolated system in terms of matter and energy exchange. [2 marks]

1. A closed system does not exchange matter with its surroundings, but it can exchange energy, including heat.
2. An isolated system exchanges neither matter nor energy with its surroundings, so its total energy remains constant.

Q2. A closed system absorbs 701 J of heat and does 394 J of work on its surroundings. Calculate the change in internal energy, showing the chemical sign convention. [3 marks]

1. Heat enters the system, so q = +701 J. Work is done by the system on its surroundings, so w = −394 J in the chemical sign convention.
2. Apply the first law to the closed system: ΔU = q + w = 701 J + (−394 J).
3. Therefore ΔU = +307 J. The positive sign means that the internal energy increases because the absorbed heat exceeds the energy transferred out as work.

Q3. Derive the relation between molar heat capacities at constant pressure and constant volume for one mole of ideal gas. Define the symbols and state the heat-capacity assumption. [4 marks]

1. Let H be enthalpy, U internal energy, p pressure, V volume, T absolute temperature and R the molar gas constant. For one mole of ideal gas, H = U + pV = U + RT.
2. Over a temperature change ΔT, the relation becomes ΔH = ΔU + RΔT, where Δ denotes final minus initial value.
3. Cₚ and Cᵥ are the molar heat capacities at constant pressure and constant volume, respectively. Treating them as constant over the interval, substitute ΔH = CₚΔT and ΔU = CᵥΔT.
4. Dividing by non-zero ΔT gives Cₚ − Cᵥ = R. These heat capacities are per mole; the relation assumes ideal-gas behaviour.

Q4. At the same temperature, C(graphite, s) + O₂(g) → CO₂(g) has reaction enthalpy −393.5 kJ mol⁻¹. CO(g) + ½O₂(g) → CO₂(g) has reaction enthalpy −283.0 kJ mol⁻¹. Use Hess’s law to obtain the enthalpy for C(graphite, s) + ½O₂(g) → CO(g), explaining each step. [5 marks]

1. Hess’s law applies because enthalpy is a state function: the overall change is independent of the route when initial and final states agree.
2. Keep the graphite combustion equation unchanged, with its given enthalpy of −393.5 kJ mol⁻¹, because graphite must remain a reactant.
3. Reverse the carbon monoxide combustion equation to obtain CO₂(g) → CO(g) + ½O₂(g). Its enthalpy becomes +283.0 kJ mol⁻¹.
4. Add the two equations. Cancel carbon dioxide from opposite sides and cancel half a mole of oxygen, giving the required target equation.
5. Add the corresponding enthalpies: −393.5 + 283.0 = −110.5 kJ mol⁻¹. The negative result means that forming carbon monoxide by this reaction is exothermic.

Q5. Evaporate 18 g of water at 298 K at constant pressure. The molar mass is 18 g mol⁻¹ and the molar vaporisation enthalpy is 44.01 kJ mol⁻¹. Treat the vapour as ideal, neglect liquid volume, assume only pressure-volume work, and use R = 8.314 J mol⁻¹ K⁻¹. Find heat absorbed and internal-energy change, showing the steps. [5 marks]

1. The amount of water is mass divided by molar mass: 18 g ÷ 18 g mol⁻¹ = 1 mol. This is the amount undergoing vaporisation.
2. At constant pressure with only pressure-volume work, heat absorbed equals enthalpy change. Thus qₚ = 1 mol × 44.01 kJ mol⁻¹ = 44.01 kJ.
3. The process is H₂O(l) → H₂O(g). Neglecting liquid volume, formation of one mole of ideal vapour gives the expansion correction Δn𝗀RT with Δn𝗀 = 1 mol.
4. Calculate the correction: 1 × 8.314 × 298 = 2477.572 J, or approximately 2.48 kJ, using the same energy unit as the enthalpy.
5. Use ΔU = ΔH − Δn𝗀RT. Therefore ΔU = 44.01 − 2.48 = 41.53 kJ; some absorbed heat supplies expansion work.

Q6. A reaction has ΔH = +400 kJ mol⁻¹ and ΔS = +0.2 kJ K⁻¹ mol⁻¹. Assuming both remain constant with temperature, determine the boundary temperature and the temperature range for spontaneity at constant pressure. [3 marks]

1. At constant temperature and pressure, use ΔG = ΔH − TΔS, where ΔG is Gibbs-energy change and T is absolute temperature. A negative value indicates spontaneity.
2. The boundary occurs at ΔG = 0, giving T = ΔH/ΔS = 400/0.2 = 2000 K. Both given energy quantities already use kilojoules.
3. Above 2000 K, TΔS exceeds ΔH and the reaction is spontaneous. At 2000 K it is at the boundary; below this temperature the specified forward change is non-spontaneous.

Q7. For a reaction at 300 K, the dimensionless equilibrium constant is 10. Use R = 8.314 J mol⁻¹ K⁻¹ and ln 10 = 2.303 to calculate standard reaction Gibbs energy and interpret its sign. [3 marks]

1. Use ΔᵣG° = −RT ln K, where ΔᵣG° is standard reaction Gibbs energy, R the gas constant, T the absolute temperature and K the equilibrium constant.
2. Substitute the supplied data: ΔᵣG° = −8.314 × 300 × 2.303 J mol⁻¹ ≈ −5744 J mol⁻¹, or −5.744 kJ mol⁻¹.
3. The negative sign is consistent with K greater than one and a product-favoured equilibrium. It gives no information about how rapidly the reaction approaches that equilibrium.

Q8. Explain why the neutralisation enthalpy of dilute strong acids and strong bases is approximately constant, and describe how this can be checked calorimetrically. [5 marks]

1. Neutralisation enthalpy is the enthalpy change per mole of water formed by reaction between an acid and a base in dilute solution.
2. Strong acids and strong bases are essentially fully ionised under these conditions, so the same reactive ions are available in different pairs.
3. The common net ionic equation is H⁺(aq) + OH⁻(aq) → H₂O(l). Spectator ions do not change, explaining the approximately constant enthalpy under comparable conditions.
4. Mix measured amounts initially at the same temperature in an insulated calorimeter. Measure the temperature rise and calculate heat absorbed using the heat capacities of solution and apparatus.
5. Reverse the heat sign to obtain reaction heat and divide by moles of water formed. Repeat with different strong-acid and strong-base pairs under matched dilute conditions.

Key takeaways

• Choose the system boundary first, then classify the system by its permitted transfers of matter and energy.
• The first law conserves energy; use positive heat for absorption and positive work for work done on the system.
• Internal energy and enthalpy are state functions, while heat and work depend on the process connecting the states.
• Constant-volume heat measures internal-energy change, and constant-pressure heat measures enthalpy change, under the stated work restrictions.
• Hess’s law requires matching physical states and stoichiometric amounts; reversing an equation reverses its enthalpy sign.
• Spontaneous changes increase total entropy of system and surroundings, even when the system’s own entropy decreases.
• At constant temperature and pressure, Gibbs-energy change combines enthalpy and entropy to determine the thermodynamic direction.
• Standard reaction Gibbs energy determines the equilibrium constant, while thermodynamics does not predict the rate of approach to equilibrium.

Test yourself

Why can a closed system exchange heat?

Its boundary prevents matter transfer but can conduct energy between the system and its surroundings.

What is the defining condition for an adiabatic process?

No heat crosses the boundary, so q = 0; the temperature need not remain constant.

Why is internal-energy change zero over a complete cycle?

Internal energy is a state function, and the final state of a complete cycle equals its initial state.

When does constant-pressure heat equal enthalpy change?

For a closed system at constant pressure, when pressure-volume work is the only work involved.

What happens to reaction enthalpy when the equation is reversed?

Its magnitude stays the same but its sign reverses, provided the physical states and amounts match.

Can a spontaneous reaction decrease the entropy of its system?

Yes. The surroundings must gain enough entropy for the combined entropy change to remain positive.

How are Gibbs and Helmholtz energies related?

G = A + pV: Gibbs energy equals Helmholtz energy plus the pressure-volume term.

What restriction belongs in the third-law statement?

The substance must be a pure, perfectly ordered crystal as its temperature approaches absolute zero.