Chemical Thermodynamics · Lesson 10 of 10

Chapter Summary and Practice

“Connect energy accounting, reaction-energy cycles, entropy, and equilibrium in a complete thermodynamics review.”

Learning Objectives

• Choose an energy or spontaneity method appropriate to the question. • Connect system boundaries, heat, work, internal energy, and enthalpy. • Combine calorimetry, formation data, bond data, and Hess cycles consistently. • Use entropy and Gibbs energy to distinguish direction from rate. • Solve mixed chapter problems while checking phases, units, signs, and assumptions.

One chapter answers three different questions

Thermodynamics first asks how much energy changes when a process occurs. It then asks which direction is thermodynamically favoured. Finally, it connects that direction criterion with the equilibrium composition. These questions use related quantities, but they are not interchangeable: calculating a heat release does not by itself establish spontaneity, speed, or reaction yield.

Begin by drawing a boundary around the system. State whether matter can cross it, whether heat can cross it, and whether the boundary can do work. An open beaker, a sealed metal vessel, and an ideal isolated vessel require different descriptions. Describe initial and final states with amount, composition, pressure, temperature, and phase before selecting an equation.

Quantity	Meaning	Key distinction
q	Energy crossing because of a temperature difference	Path dependent; positive into system
w	Energy transferred through work	Path dependent; positive on system
U	Internal energy	State function; change follows q + w
H	U + pV	State function; its change equals q_p under the usual work assumptions
S	Entropy	State function; total entropy decides spontaneous direction
G	H − TS	State function; decreases spontaneously at fixed T and p

The word “state” is the link between U, H, S, and G. Their endpoint differences do not depend on the calculation path. That is why Joule’s work comparisons, formation-enthalpy subtraction, Hess cycles, reversible routes for entropy, and Gibbs calculations can all be consistent. Heat and work are the transfers along a route and must not be treated as stored state properties.

Account for energy before predicting direction

Write the first law with signed transfers. For pressure–volume work, an outward change against positive external pressure gives negative work; compression gives positive work. Reversible isothermal ideal-gas work uses the logarithmic volume ratio, whereas a finite irreversible step uses the actual opposing pressure. A free expansion has zero external pressure and therefore zero boundary work.

Energy-accounting toolkitLaTeX

Δ U = q + w, w = - p_{ext} Δ V, w_{rev, iso} = - n R T ln \frac{V _{f}}{V _{i}}

Use the last expression only for a reversible isothermal ideal-gas route. Adiabatic means q = 0; rigid plus no other work means w = 0.

Example — The same endpoints, different transfers

Problem
One route between two states has q = +800 J and w = −300 J. Another has q = +150 J. Find the second-route work.

1.Use the first route to determine the state-function change: $Δ U = 800 - 300 = + 500 J$ .
2.The same endpoints require the same $Δ U$ on the other route.
3.Thus $500 = 150 + w_{2}$ , giving $w_{2} = + 350 J$ .
4.One route lets the system do net work; the other requires work on it. The heat and work differ, while the internal-energy change agrees.

For heat measured in a calorimeter, identify whether its constraint gives $Δ U$ or $Δ H$ . The calorimeter and reacting system have opposite heat signs. A rising temperature does not mean the reaction absorbs heat: the reaction is the energy source for the warming calorimeter. Phase changes require a separate enthalpy contribution when temperature is constant.

Heat and enthalpy toolkitLaTeX

H = U + p V, Δ_{r} H \approx Δ_{r} U + Δ n_{g} R T, q = C Δ T, q_{phase} = n Δ H_{phase}

The gas correction assumes ideal gases and negligible condensed-phase pV terms. Choose total, specific, or molar heat capacity consistently.

Example — A calorimeter result needs a gas correction

Problem
A bomb experiment gives $Δ_{r} U = - 100.0 kJ mo l^{- 1}$ for an ideal-gas reaction with $Δ n_{g} = - 1$ at 298 K. Find its enthalpy change.

1.The rigid-bomb result concerns internal energy. Calculate the correction with matching units: $Δ n_{g} R T = (- 1) (8.314) (298) /1000 = - 2.478 kJ mo l^{- 1}$ .
2.Then $Δ_{r} H = - 100.0 - 2.478 = - 102.48 kJ mo l^{- 1}$ .
3.The enthalpy change is more negative because fewer gaseous moles are present in the products.
4.A constant-volume measurement does not erase the distinction between the state functions H and U.

Choose a consistent enthalpy route

Reaction enthalpy can be measured directly or assembled indirectly. Formation data use reference-state elements as a shared starting point. Combustion cycles use common oxidation products. Atomisation and mean bond enthalpies use separated gaseous atoms as an intermediate reference. Born–Haber cycles add gaseous-ion steps, while dissolution cycles compare lattice separation with hydration.

Available information	Appropriate route	Check before calculating
Formation enthalpies	Products minus reactants	Coefficients, reference states, and phases
Related thermochemical equations	Hess sum	Scaling, reversed signs, and cancelled species
Mean bond enthalpies	Bonds broken minus formed	Gas-phase endpoints and approximation
Ionisation, electron gain, sublimation, bond data	Born–Haber cycle	Ion charges and lattice sign convention
Lattice and hydration enthalpies	Solution cycle	Final solution composition
Two solution compositions	Dilution difference	Same reference solute and added solvent

Example — A formation value hidden in combustion

Problem
Methanol combustion is −726 kJ/mol. Formation values for carbon dioxide gas and liquid water are −393 and −286 kJ/mol. Find liquid methanol’s formation enthalpy.

1.Write $C H_{3} OH (l) + \frac{3}{2} O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)$ .
2.The product formation sum is $- 393 + 2 (- 286) = - 965 kJ mo l^{- 1}$ .
3.Since oxygen’s reference formation value is zero, $- 726 = - 965 - Δ_{f} H^{\circ} (C H_{3} OH, l)$ .
4.Therefore $Δ_{f} H^{\circ} = - 239 kJ mo l^{- 1}$ using this rounded dataset.

Example — A crystal can dissolve endothermically

Problem
A salt has lattice dissociation enthalpy +800 kJ/mol and hydration enthalpy −790 kJ/mol. What follows, and what remains undecided?

1.The solution enthalpy is $800 - 790 = + 10 kJ mo l^{- 1}$ .
2.At constant pressure the specified dissolution absorbs heat.
3.This result alone does not decide whether dissolution proceeds spontaneously: the system and surroundings entropy contributions are also needed.
4.A positive enthalpy change is compatible with favourable Gibbs energy if the entropy contribution is sufficiently positive.

Connect entropy, Gibbs energy, and equilibrium

Entropy explains why conserving energy does not fix a process direction. Gas spreading and mixing create more accessible arrangements. System entropy can decrease during an exothermic reaction, provided the surroundings increase more. For an isolated whole, spontaneous irreversible change produces positive total entropy. The reversible limit produces zero total entropy, while the third law anchors absolute entropy to an ideal ordered crystal approaching 0 K.

Direction and equilibrium toolkitLaTeX

Δ S = \int \frac{δ q _{rev}}{T}, Δ G = Δ H - T Δ S, Δ_{r} G^{\circ} = - R T ln K

The Gibbs difference shown assumes constant T, and its spontaneity criterion also requires fixed p. The entropy route is reversible even when the actual process is irreversible.

Temperature weights entropy’s influence. Negative enthalpy with positive entropy favours a negative Gibbs change; positive enthalpy with negative entropy opposes it. When both signs match, calculate the crossover rather than guessing. A minimum in G describes equilibrium composition, not the stopping of molecular events. Actual $Δ_{r} G$ is zero there, whereas its standard value generally is not.

Example — A system entropy decrease and a surrounding increase

Problem
At 300 K a constant-pressure process has $Δ H = - 12.0 kJ$ and $Δ S_{sys} = - 20.0 J K^{- 1}$ . Check direction by both entropy and Gibbs energy.

1.The surroundings receive 12000 J, giving $Δ S_{surr} = 12000/300 = + 40.0 J K^{- 1}$ .
2.Total entropy change is $- 20.0 + 40.0 = + 20.0 J K^{- 1}$ , so the specified direction is favourable.
3.Convert the system entropy to $- 0.0200 kJ K^{- 1}$ . Then $Δ G = - 12.0 - 300 (- 0.0200) = - 6.00 kJ$ .
4.The results agree because $- T Δ S_{total} = - (300) (20.0) J = - 6.00 kJ$ .

Example — From equilibrium to standard Gibbs energy

Problem
A reaction has dimensionless K = 10 at 300 K. Determine its standard reaction Gibbs energy and explain what happens to actual reaction Gibbs energy at equilibrium.

1.Use $Δ_{r} G^{\circ} = - (8.314) (300) ln 10 = - 5743 J mo l^{- 1} \approx - 5.74 kJ mo l^{- 1}$ .
2.K exceeds one, so the negative standard value is consistent with product-favoured equilibrium in its activity expression.
3.The actual equilibrium mixture nevertheless has $Δ_{r} G = 0$ : its composition contribution cancels the standard term.
4.Neither K nor Gibbs energy alone tells us how rapidly that equilibrium is reached.

Common mistake

Check five recurring traps: “adiabatic” is not “isothermal”; q and w are not state functions; elemental formation enthalpy zero is not elemental entropy zero; breaking bonds costs energy; and equilibrium sets actual reaction Gibbs energy to zero, not every standard Gibbs value. State phases before comparing numbers.

Test your connections

Use the following questions to choose a principle, not just recognise a symbol. Before calculating, identify the system and process conditions. After calculating, interpret the sign and check whether the result answers an energy, direction, or equilibrium question.

Quick check

Which pair contains only state functions?

Quick check

An ideal gas expands isothermally against a constant external pressure. What is fixed by ideal-gas isothermality?

Quick check

A bomb calorimeter warms when fuel burns. What is the reaction heat sign?

Quick check

Which reference-state convention is correct?

Quick check

Why can a Hess cycle use an imagined intermediate route?

Quick check

Which sum gives a solution enthalpy under the positive lattice-dissociation convention?

Quick check

Which condition is required for spontaneous change of an isolated whole?

Quick check

A reaction has positive enthalpy and positive entropy changes. How can increasing temperature affect its driving force?

Quick check

At chemical equilibrium, which statement is correct?

Quick check

What must be used inside the logarithm in the thermodynamic K relation?

Mixed Chapter Practice

State what makes a quantity a thermodynamic state function. Explain why heat and work do not meet that definition.
For an adiabatic process, which condition must hold: zero temperature change, zero pressure change, zero heat transfer, or zero work? Explain why the others need not hold.
State the zero convention for standard formation enthalpies of reference-state elements. Explain why the statement “the absolute enthalpy of every element is zero” is not equivalent.
Methane combustion forms carbon dioxide gas and liquid water at 298 K. If its molar internal-energy change is −X kJ/mol, determine whether its enthalpy change is greater than, equal to, or less than that value by counting gaseous moles.
Use combustion enthalpies of methane, graphite, and hydrogen of −890.3, −393.5, and −285.8 kJ/mol to calculate methane’s standard formation enthalpy.
An exothermic reaction also has positive entropy change. Discuss its standard Gibbs sign at positive temperatures when both values are treated as constant.
A system absorbs 701 J of heat and does 394 J of work. Calculate its internal-energy change and interpret the sign.
For $N H_{2} CN (s) + \frac{3}{2} O_{2} (g) \to N_{2} (g) + C O_{2} (g) + H_{2} O (l)$ , use $Δ U = - 742.7 kJ mo l^{- 1}$ at 298 K to calculate $Δ H$ . Treat the gases as ideal and retain the stated solid phase for cyanamide.
Calculate the heat required to warm 60.0 g of aluminium from 35 to 55 degrees Celsius. Use molar mass 27.0 g/mol and molar heat capacity 24 J mol−1 K−1.
Calculate the enthalpy change when one mole of liquid water at 10 degrees Celsius becomes ice at −10 degrees Celsius. Use liquid and solid molar heat capacities 75.3 and 36.8 J mol−1 K−1 and fusion enthalpy 6.03 kJ/mol at 0 degrees Celsius.
Find the heat released when 35.2 g of carbon dioxide forms from graphite and oxygen. Use molar mass 44.0 g/mol and carbon combustion enthalpy −393.5 kJ/mol.
Calculate $Δ_{r} H$ for $N_{2} O_{4} (g) + 3CO (g) \to N_{2} O (g) + 3C O_{2} (g)$ using formation enthalpies of 9.7, −110, 81, and −393 kJ/mol respectively.
Given $N_{2} (g) + 3 H_{2} (g) \to 2N H_{3} (g)$ has standard reaction enthalpy −92.4 kJ/mol, find the standard formation enthalpy of ammonia gas.
Find the formation enthalpy of liquid methanol from its combustion enthalpy −726 kJ/mol and formation values of −393 and −286 kJ/mol for carbon dioxide gas and liquid water.
Find the atomisation enthalpy for $CC l_{4} (g) \to C (g) + 4Cl (g)$ and the mean C–Cl bond enthalpy. Use liquid CCl4 formation enthalpy −135.5 kJ/mol, its vaporisation enthalpy +30.5 kJ/mol, graphite atomisation +715.0 kJ/mol, and $C l_{2} (g) \to 2Cl (g)$ enthalpy +242 kJ/mol. Construct a phase-consistent Hess route.
For an isolated system with $Δ U = 0$ , state the permitted entropy change for a spontaneous irreversible process and for equilibrium. Explain why energy constancy alone cannot distinguish them.
A reaction has $Δ H = + 400 kJ mo l^{- 1}$ and $Δ S = + 0.200 kJ K^{- 1} mo l^{- 1}$ . Find its crossover temperature and the temperature range for a negative Gibbs change if both values remain constant.
For $2Cl (g) \to C l_{2} (g)$ , predict the signs of enthalpy and entropy changes and explain them separately.
At 298 K, $2A (g) + B (g) \to 2D (g)$ has $Δ U^{\circ} = - 10.5 kJ$ and $Δ S^{\circ} = - 44.1 J K^{- 1}$ per written reaction amount. Calculate $Δ H^{\circ}$ and $Δ G^{\circ}$ and interpret the standard driving force.
For a reaction with dimensionless K = 10 at 300 K, calculate $Δ_{r} G^{\circ}$ using R = 8.314 J mol−1 K−1.
Discuss the enthalpic tendency of nitric oxide given $\frac{1}{2} N_{2} + \frac{1}{2} O_{2} \to NO$ has $Δ H^{\circ} = + 90 kJ mo l^{- 1}$ , while $NO + \frac{1}{2} O_{2} \to N O_{2}$ has $Δ H^{\circ} = - 74 kJ mo l^{- 1}$ . Explain why full thermodynamic stability and persistence also require entropy, conditions, and kinetic information.
Calculate the entropy change of a 298 K surrounding reservoir when one mole of liquid water forms under standard conditions with $Δ_{f} H^{\circ} = - 286 kJ mo l^{- 1}$ .

Key Takeaways

Energy accounting begins with a system boundary and the first law. Calorimetry and enthalpy cycles measure or reconstruct the same endpoint changes using different information. Entropy connects microscopic dispersal with the preferred macroscopic direction. Gibbs energy combines enthalpy and entropy at fixed temperature and pressure, while standard Gibbs energy links to equilibrium through K. Correct signs, phases, reaction amounts, units, and assumptions are part of the explanation, not optional finishing details.

Previous · Lesson 9

Gibbs Energy, Temperature, and Equilibrium

End of chapter

One chapter answers three different questions

Account for energy before predicting direction

Choose a consistent enthalpy route

Connect entropy, Gibbs energy, and equilibrium

Test your connections

Quiz

Practice Problems

Key Takeaways