Thermodynamics Class 11 Chemistry: First Law, Enthalpy, Entropy and Gibbs Energy

What Thermodynamics is about

Weightage: This chapter is Unit V, Chemical Thermodynamics, which the syllabus gives 9 of the 70 theory marks in the school annual exam.Source: CBSE Chemistry (043) syllabus 2026-27, Class XI course structure, Unit V Chemical Thermodynamics

Key concepts

Thermodynamic terms: system, surroundings, state functions and internal energy

NCERT section 5.1 Thermodynamic Terms (5.1.1 to 5.1.4)

• System: the part of the universe under study (for example, the reactants in a beaker). Surroundings: everything else that can exchange energy with it. System + surroundings = universe. They are separated by a boundary, real or imaginary.
• Open system: exchanges both energy and matter (reactants in an open beaker). Closed system: exchanges energy but not matter (a sealed flask). Isolated system: exchanges neither (reactants in a perfect thermos flask).
• State of a system: described by state variables such as p, V, T and amount n. A property whose value depends only on the state, not on how it was reached, is a state function (U, H, S, G, p, V, T). Heat (q) and work (w) depend on the path, so they are not state functions.
• Internal energy (U): the total energy of the system (kinetic and potential energy of its molecules, chemical, electrical and other forms). We can measure only changes, ΔU. U can change by heat flow, by work, or by matter entering or leaving.
• An adiabatic process is one with no heat exchange (q = 0): for example, a gas in a perfectly insulated container.

First law of thermodynamics: energy can be neither created nor destroyed; the energy of an isolated system is constant. For a closed system,
$\Delta U=q+w$
with the IUPAC sign convention: q is positive when heat is absorbed by the system, w is positive when work is done on the system. Heat released or work done by the system is negative.

Worked example

A gas absorbs 500 J of heat and expands, doing 200 J of work on the surroundings. (a) Find ΔU. (b) In another process the same gas is compressed adiabatically with 300 J of work done on it. Find ΔU. (c) Which of q, w and ΔU are state functions? (3 marks)

Try it first, then show the solutionSolution

1. (a) q = +500 J (absorbed); w = −200 J (done by the system). ΔU = 500 + (−200) = +300 J. (1)
2. (b) Adiabatic: q = 0; w = +300 J (done on the system). ΔU = 0 + 300 = +300 J. (1)
3. (c) Only ΔU is a state function; both processes give the same ΔU although q and w differ, which shows q and w depend on the path. (1)

Answer: (a) +300 J (b) +300 J (c) only U (ΔU) is a state function

Pressure-volume work: free, irreversible and reversible expansion

NCERT section 5.2.1 Work

When a gas expands against an external pressure $p_{ex}$, it does work on the surroundings:
$w=-p_{ex}\Delta V=-p_{ex}(V_f-V_i)$
The minus sign follows the convention: expansion (ΔV > 0) gives negative w (the system loses energy as work). Compression gives positive w. A useful conversion: 1 L bar = 100 J (1 L atm = 101.3 J).

Three kinds of expansion of an ideal gas at constant temperature (isothermal, so ΔU = 0 and q = −w):
1. Free expansion (into a vacuum, $p_{ex}$ = 0): w = 0, q = 0, ΔU = 0, whether reversible or not.
2. Irreversible expansion against a constant external pressure: $w=-p_{ex}(V_f-V_i)$.
3. Reversible expansion: carried out in infinitely small steps, with the external pressure always only slightly less than the gas pressure, so the system stays in near-equilibrium:
$w_{rev}=-2.303nRT\log \frac{V_f}{V_i}$

The reversible path gives the maximum work that the gas can do. A reversible process is an idealisation; real processes are irreversible.

On a p–V graph, the work is the area under the curve. For a reversible isothermal expansion the curve is a hyperbola (pV = constant), and the area under it is larger than for an irreversible expansion against a fixed lower pressure.

Worked example

One mole of an ideal gas at 300 K expands isothermally from 5.0 L to 25.0 L. Calculate the work done (a) when it expands into a vacuum, (b) against a constant external pressure of 1.0 bar, (c) reversibly. (R = 8.314 J K⁻¹ mol⁻¹, log 5 = 0.699, 1 L bar = 100 J) (3 marks)

Try it first, then show the solutionSolution

1. (a) Free expansion: $p_{ex}$ = 0, so w = 0. (½)
2. (b) w = −1.0 × (25.0 − 5.0) = −20 L bar = −2000 J = −2.0 kJ. (1)
3. (c) w = −2.303 × 1 × 8.314 × 300 × log(25/5) = −2.303 × 8.314 × 300 × 0.699 = −4015 J ≈ −4.0 kJ. (1)
4. The reversible path gives the most work (in magnitude), as expected. Since ΔU = 0 for an isothermal ideal gas, q = −w in each case. (½)

Answer: (a) 0 (b) −2.0 kJ (c) about −4.0 kJ (maximum work)

Enthalpy, ΔH and ΔU, heat capacity

NCERT section 5.2.2 Enthalpy, H (extensive and intensive properties, heat capacity, Cp − Cv)

Most reactions are carried out at constant pressure in open vessels. The heat absorbed at constant pressure is $q_p$. Writing ΔU = q_p − pΔV and rearranging defines enthalpy, a new state function:
$H=U+pV$, so $\Delta H=\Delta U+p\Delta V=q_p$.
At constant volume, w = 0 and $\Delta U=q_V$.

For solids and liquids, ΔV is tiny, so ΔH ≈ ΔU. For reactions involving gases, using pV = nRT:
$\Delta H=\Delta U+\Delta n_{g}RT$
where $\Delta n_g$ = moles of gaseous products − moles of gaseous reactants.

• Exothermic reaction: heat released, ΔH negative. Endothermic: heat absorbed, ΔH positive.
• Extensive properties depend on the amount of matter (mass, volume, U, H, heat capacity). Intensive properties do not (temperature, pressure, density, molar heat capacity). Dividing an extensive property by the amount gives an intensive one.

Heat capacity: $q=C\Delta T$. Molar heat capacity $C_m=C/n$; specific heat capacity c is per gram, so $q=c\times m\times \Delta T$. For an ideal gas, $C_p-C_V=R$ (since ΔH − ΔU = RΔT for one mole).

Worked example

For $N_2(g)+3H_2(g)\to 2N{H}_3(g)$, ΔH = −92.4 kJ mol⁻¹ at 298 K. Calculate ΔU. (R = 8.314 J K⁻¹ mol⁻¹) (2 marks)

Try it first, then show the solutionSolution

1. $\Delta n_g$ = 2 − (1 + 3) = −2. (½)
2. ΔU = ΔH − Δn_g RT = −92.4 − (−2 × 8.314 × 298 ÷ 1000) = −92.4 + 4.96 (1)
3. ΔU = −87.4 kJ mol⁻¹. (½)

Answer: ΔU ≈ −87.4 kJ mol⁻¹

Calorimetry: measuring ΔU and ΔH

NCERT section 5.3 Measurement of ΔU and ΔH: Calorimetry

Calorimetry measures energy changes by observing the temperature change of a known body (the calorimeter, usually with water) whose heat capacity is known.

• ΔU at constant volume: bomb calorimeter. A steel vessel (the 'bomb') containing the substance and oxygen is immersed in water. The substance is ignited electrically. Because the bomb is sealed, no work is done (ΔV = 0), so the heat released equals ΔU. Heat = $C_{cal}\times \Delta T$. The value is then scaled to one mole.
• ΔH at constant pressure: an open calorimeter (for example, a coffee-cup calorimeter for reactions in solution). The heat measured is $q_p=\Delta H$.

Sign care: if the temperature of the calorimeter rises, the reaction released heat, so ΔU or ΔH of the reaction is negative.

Worked example

1.60 g of methane is burnt in a bomb calorimeter whose heat capacity is 15.0 kJ K⁻¹. The temperature rises by 5.95 K. Calculate ΔU and ΔH for the combustion of methane per mole at 298 K, with water formed as liquid. (C = 12, H = 1, R = 8.314 J K⁻¹ mol⁻¹) (3 marks)

Try it first, then show the solutionSolution

1. Heat released = 15.0 × 5.95 = 89.25 kJ for 1.60 ÷ 16 = 0.100 mol, so ΔU = −89.25 ÷ 0.100 = −892.5 kJ mol⁻¹. (1)
2. $C{H}_4(g)+2O_2(g)\to C{O}_2(g)+2H_{2}O(l)$: Δn_g = 1 − 3 = −2. (1)
3. ΔH = ΔU + Δn_g RT = −892.5 + (−2 × 8.314 × 298 ÷ 1000) = −892.5 − 4.96 = −897.5 kJ mol⁻¹. (1)

Answer: ΔU = −892.5 kJ mol⁻¹; ΔH ≈ −897.5 kJ mol⁻¹

Reaction enthalpy, standard enthalpy of formation and Hess's law

NCERT section 5.4 Enthalpy Change, ΔrH of a Reaction (standard enthalpy, phase transformations, formation, thermochemical equations, Hess's law)

Reaction enthalpy (${\Delta }_{r}H$) is the enthalpy change for a reaction as written: Σ(H of products) − Σ(H of reactants). Its standard value (${\Delta }_r{H}^{\ominus }$) refers to all substances in their standard states: the pure form at 1 bar and the stated temperature (usually 298 K).

Standard enthalpy of formation (${\Delta }_f{H}^{\ominus }$): the enthalpy change when one mole of a compound is formed from its elements in their most stable states (reference states). By convention ${\Delta }_f{H}^{\ominus }$ of an element in its reference state (H₂ gas, O₂ gas, C graphite, Br₂ liquid) is zero.
Then ${\Delta }_r{H}^{\ominus }=\sum a_i{\Delta }_f{H}^{\ominus }(\text{products})-\sum b_i{\Delta }_f{H}^{\ominus }(\text{reactants})$.

Thermochemical equations give the physical states and ΔH. Rules: ΔH is for the moles shown; multiplying the equation multiplies ΔH; reversing the equation changes the sign of ΔH.

Hess's law of constant heat summation: if a reaction takes place in several steps, its standard enthalpy is the sum of the standard enthalpies of the steps, at the same temperature. It follows from H being a state function. It lets us find enthalpies that cannot be measured directly, such as the formation of CO from carbon (which always also gives some CO₂).

Worked example

The standard enthalpy of combustion of ethane, C₂H₆(g), is −1560 kJ mol⁻¹. The standard enthalpies of formation of CO₂(g) and H₂O(l) are −393.5 and −285.8 kJ mol⁻¹. Calculate the standard enthalpy of formation of ethane. (3 marks)

Try it first, then show the solutionSolution

1. Combustion: $2C_2{H}_6+7O_2\to 4C{O}_2+6H_{2}O$ (or per mole, C₂H₆ + 3½O₂ → 2CO₂ + 3H₂O), ΔcH = −1560 kJ per mole of ethane. (1)
2. ΔcH = [2ΔfH(CO₂) + 3ΔfH(H₂O)] − [ΔfH(C₂H₆) + 0], so −1560 = [2(−393.5) + 3(−285.8)] − ΔfH(C₂H₆). (1)
3. −1560 = −787.0 − 857.4 − ΔfH, so ΔfH(C₂H₆) = −1644.4 + 1560 = −84.4 kJ mol⁻¹. (1)

Answer: ΔfH⦵(C₂H₆) ≈ −84.4 kJ mol⁻¹

Enthalpies of different types of reactions; bond and lattice enthalpy

NCERT section 5.5 Enthalpies for Different Types of Reactions (combustion, atomisation, bond, lattice, solution, dilution)

• Standard enthalpy of combustion (${\Delta }_c{H}^{\ominus }$): enthalpy change when one mole of a substance burns completely in oxygen, all reactants and products in standard states. Always negative. Useful for fuels and foods.
• Enthalpy of atomisation (${\Delta }_a{H}^{\ominus }$): enthalpy change to break one mole of bonds completely into gaseous atoms (H₂ → 2H: 435.0 kJ mol⁻¹; for a metal, it equals the enthalpy of sublimation).
• Bond enthalpy: for a diatomic molecule, the bond dissociation enthalpy (H–H 435.0, Cl–Cl 242, O=O 498 kJ mol⁻¹). For polyatomic molecules, successive bonds break with different energies, so we use the mean bond enthalpy (C–H in CH₄: 1665 ÷ 4 = 416 kJ mol⁻¹). For gas-phase reactions:

${\Delta }_r{H}^{\ominus }=\sum$ bond enthalpies of reactants − $\sum$ bond enthalpies of products.

• Lattice enthalpy: enthalpy change when one mole of an ionic solid separates into gaseous ions (NaCl: +788 kJ mol⁻¹). It cannot be measured directly; it is found with a Born-Haber cycle (Hess's law around a cycle: sublimation of the metal, ionisation, atomisation of the non-metal, electron gain, and formation of the solid).
• Enthalpy of solution: enthalpy change when one mole dissolves in a large amount of solvent. For an ionic solid, ${\Delta }_{sol}{H}^{\ominus }={\Delta }_{lattice}{H}^{\ominus }+{\Delta }_{hyd}{H}^{\ominus }$; if the lattice enthalpy is very large compared with the hydration enthalpy, the solid dissolves poorly or not at all.
• Enthalpy of dilution: the heat change when more solvent is added to a solution; it depends on the original concentration and amount of solvent added.

Worked example

Use the Born-Haber cycle to calculate the lattice enthalpy of KCl from: ΔfH(KCl) = −437 kJ mol⁻¹; sublimation of K = +89; ionisation enthalpy of K = +419; ½ × bond enthalpy of Cl₂ = +121; electron gain enthalpy of Cl = −349 kJ mol⁻¹. (3 marks)

Try it first, then show the solutionSolution

1. Hess's law: ΔfH = ΔsubH + ΔiH + ½ΔbondH + ΔegH − ΔlatticeH (the last step forms the solid from gaseous ions, so its enthalpy is −ΔlatticeH). (1)
2. −437 = 89 + 419 + 121 − 349 − ΔlatticeH = 280 − ΔlatticeH. (1)
3. ΔlatticeH = 280 + 437 = +717 kJ mol⁻¹. (1)

Answer: Lattice enthalpy of KCl ≈ +717 kJ mol⁻¹

Spontaneity: entropy, the second law and the third law

NCERT section 5.6 Spontaneity (entropy, second law, third law)

A spontaneous process happens on its own without continuous outside help (it may still be slow). Many spontaneous reactions are exothermic, but not all: ice melts and NH₄Cl dissolves spontaneously though they absorb heat. So a decrease in enthalpy is a contributing factor but not the whole story.

Entropy (S) measures the degree of randomness or disorder of a system, or the dispersal of its energy. It is a state function.

• S increases from solid to liquid to gas.
• S increases with temperature.
• S increases when the number of gaseous molecules increases in a reaction.
• For a reversible change at temperature T: $\Delta S=\frac{q_{rev}}{T}$. The same heat causes a larger entropy change at a low temperature than at a high one.

Second law of thermodynamics: the total entropy change ($\Delta S_{total}=\Delta S_{system}+\Delta S_{surr}$) is positive for every spontaneous process. At equilibrium, $\Delta S_{total}$ = 0. For an isolated system, ΔU = 0 and spontaneity requires ΔS > 0.

The surroundings gain entropy when the system releases heat: $\Delta S_{surr}=\frac{-\Delta H_{system}}{T}$. This explains why an exothermic reaction with ΔS_system < 0 (such as rusting) can still be spontaneous.

Third law: the entropy of a perfectly crystalline substance is zero at absolute zero (0 K). This lets us find absolute entropies of substances from heat-capacity data.

Worked example

The enthalpy of vaporisation of water at 373 K is 40.7 kJ mol⁻¹. (a) Calculate the entropy change for vaporising 1 mol of water at 373 K. (b) Find ΔS of the surroundings and ΔS_total for this change at 373 K, and comment. (3 marks)

Try it first, then show the solutionSolution

1. (a) Boiling at 373 K is a reversible change: ΔS_sys = q_rev/T = 40 700 ÷ 373 = +109 J K⁻¹ mol⁻¹. (1)
2. (b) The surroundings lose 40.7 kJ: ΔS_surr = −40 700 ÷ 373 = −109 J K⁻¹ mol⁻¹. (1)
3. ΔS_total = 109 − 109 = 0, as expected for a system at equilibrium (water and steam at the boiling point). (1)

Answer: (a) +109 J K⁻¹ mol⁻¹ (b) ΔS_surr = −109 J K⁻¹ mol⁻¹; ΔS_total = 0 (equilibrium)

Gibbs energy, spontaneity and equilibrium

NCERT section 5.6 Spontaneity (Gibbs energy); 5.7 Gibbs Energy Change and Equilibrium

Working with ΔS_total is awkward, so Gibbs energy is defined: $G=H-TS$. At constant temperature and pressure,
$\Delta G=\Delta H-T\Delta S$
and $\Delta G=-T\Delta S_{total}$. So:

• ΔG < 0: spontaneous.
• ΔG > 0: non-spontaneous (the reverse reaction is spontaneous).
• ΔG = 0: equilibrium.

ΔG is the free energy available to do useful (non-expansion) work.

Effect of temperature (signs of ΔH and ΔS):

• ΔH −, ΔS +: spontaneous at all temperatures.
• ΔH +, ΔS −: non-spontaneous at all temperatures.
• ΔH −, ΔS −: spontaneous at low temperatures (|ΔH| > |TΔS|).
• ΔH +, ΔS +: spontaneous at high temperatures, above $T=\frac{\Delta H}{\Delta S}$.

Gibbs energy and equilibrium: for a reaction at equilibrium, ΔG = 0 and
${\Delta }_r{G}^{\ominus }=-RT\ln K=-2.303RT\log K$
A large negative ${\Delta }_r{G}^{\ominus }$ means K ≫ 1, and the reaction goes nearly to completion; a large positive value means K ≪ 1, and little product forms.

Thermodynamics says only whether a reaction can happen, not how fast. Diamond turning into graphite is spontaneous but immeasurably slow.

Worked example

For $CaC{O}_3(s)\to CaO(s)+C{O}_2(g)$, ΔH⦵ = +178 kJ mol⁻¹ and ΔS⦵ = +160 J K⁻¹ mol⁻¹. (a) Is the reaction spontaneous at 298 K? (b) Above what temperature does it become spontaneous? (3 marks)

Try it first, then show the solutionSolution

1. (a) ΔG = ΔH − TΔS = 178 − 298 × 0.160 = 178 − 47.7 = +130.3 kJ mol⁻¹. Positive, so not spontaneous at 298 K. (1)
2. (b) It becomes spontaneous when TΔS > ΔH, i.e. T > ΔH/ΔS. (1)
3. T = 178 000 ÷ 160 = 1112.5 K ≈ 1113 K. (Lime kilns run at about this temperature or above.) (1)

Answer: (a) No, ΔG ≈ +130 kJ mol⁻¹ (b) above about 1113 K

Chemistry in real life: Thermodynamics

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Topics in this chapter

• Enthalpy Changes, Hess's Law and Calorimetry
• Entropy, Gibbs Energy and Spontaneity

Thermodynamics formula sheet

First law

$\Delta U=q+w$

q, w positive when energy enters the system (IUPAC).

Work against constant external pressure

$w=-p_{ex}(V_f-V_i)$

1 L bar = 100 J; 1 L atm = 101.3 J.

Reversible isothermal work (ideal gas)

$w_{rev}=-2.303nRT\log \frac{V_f}{V_i}=-2.303nRT\log \frac{p_i}{p_f}$

Maximum work; ΔU = 0 so q = −w.

Enthalpy

$H=U+pV,\Delta H=\Delta U+p\Delta V=q_p$

At constant volume ΔU = q_V.

ΔH and ΔU for gas reactions

$\Delta H=\Delta U+\Delta n_{g}RT$

Δn_g counts gaseous moles only. Use R = 8.314 × 10⁻³ kJ K⁻¹ mol⁻¹ when ΔH is in kJ.

Heat capacity

$q=C\Delta T=cm\Delta T,C_p-C_V=R$

c is specific heat (per gram).

Reaction enthalpy from formation enthalpies

${\Delta }_r{H}^{\ominus }=\sum a_i{\Delta }_f{H}^{\ominus }(\text{products})-\sum b_i{\Delta }_f{H}^{\ominus }(\text{reactants})$

ΔfH⦵ of elements in reference states = 0.

Reaction enthalpy from bond enthalpies

${\Delta }_r{H}^{\ominus }=\sum \text{B.E.(reactants)}-\sum \text{B.E.(products)}$

Gas-phase reactions only; the order is reversed compared with formation enthalpies.

Entropy change

$\Delta S=\frac{q_{rev}}{T},\Delta S_{surr}=-\frac{\Delta H_{sys}}{T}$

Second law: ΔS_total > 0 for spontaneous change.

Gibbs energy

$\Delta G=\Delta H-T\Delta S,T_{switch}=\frac{\Delta H}{\Delta S}$

Convert ΔS to kJ when ΔH is in kJ.

Gibbs energy and K

${\Delta }_r{G}^{\ominus }=-RT\ln K=-2.303RT\log K$

K > 1 when ΔrG⦵ < 0.

Thermodynamics: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Thermodynamics

The question patterns to expect, and the method that works for each.

Tricks to remember Thermodynamics

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Into the system is plus

  Think of the system's bank account: heat absorbed or work done on it is a deposit (+); heat released or work done by it is a withdrawal (−).

  Use it when: Every sign in ΔU = q + w.

• Bonds: Reactants minus Products; Formation: Products minus Reactants

  'B-R-P' for bonds (broken minus made) and 'F-P-R' for formation.

  Use it when: Choosing the correct order in ΔrH calculations.

• Sign table for ΔG

  −/+ always yes; +/− always no; −/− cold is good; +/+ hot is good (ΔH/ΔS).

  Use it when: Predicting the effect of temperature.

• Bomb = constant volume = ΔU

  A sealed bomb cannot expand, so no work: the heat gives ΔU directly.

  Use it when: Calorimetry questions.

• Gases count, liquids don't

  In ΔH = ΔU + Δn_gRT, count only gas moles; liquid water is ignored.

  Use it when: Δn_g errors.

Thermodynamics: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Thermodynamics

The slips examiners see most often, and what to write instead.

• ✗Mistake: Using the old sign convention

  Physics books write ΔU = q − W with W positive for work done by the gas.

  Wrong working

  Gas absorbs 500 J and does 200 J: ΔU = 500 − (−200) = 700 J

  Right working

  ΔU = q + w = 500 + (−200) = 300 J

  ✓Do this: In Chemistry use ΔU = q + w, and give w a negative sign when the system does work.

  Usually costs: The whole answer.

• ✗Mistake: Counting liquid water in Δn_g

  Students count all product moles.

  Wrong working

  CH₄ + 2O₂ → CO₂ + 2H₂O(l): Δn_g = 3 − 3 = 0

  Right working

  Δn_g = 1 − 3 = −2

  ✓Do this: Look at the state symbols and count gases only.

  Usually costs: 1 mark.

• ✗Mistake: Mixing kJ and J in ΔG = ΔH − TΔS

  ΔH is usually given in kJ and ΔS in J K⁻¹.

  Wrong working

  ΔG = 178 − 298 × 160 = −47 502

  Right working

  ΔG = 178 − 298 × 0.160 = +130 kJ mol⁻¹

  ✓Do this: Convert ΔS to kJ K⁻¹ mol⁻¹ before substituting.

  Usually costs: The answer and the conclusion.

• ✗Mistake: Wrong order in bond-enthalpy calculations

  Students use the formation-enthalpy order (products − reactants).

  ✓Do this: ΔrH = bonds broken (reactants) − bonds formed (products).

  Usually costs: Sign of the answer.

• ✗Mistake: Forgetting to reverse the sign when reversing an equation in Hess's law

  Equations are flipped but ΔH is copied unchanged.

  ✓Do this: Write the new ΔH next to every rearranged equation before adding.

  Usually costs: 1-2 marks.

• ✗Mistake: Saying a spontaneous reaction is fast

  Spontaneous is confused with quick.

  ✓Do this: ΔG says whether a reaction can happen; rate depends on kinetics. Diamond → graphite is spontaneous but very slow.

  Usually costs: 1 mark in reasoning items.

NCERT exercise guide: Thermodynamics

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Thermodynamics

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   For an isolated system, which is true?
   (a) ΔU = 0, q = 0 (b) ΔU > 0 (c) q > 0, w = 0 (d) ΔS < 0 for a spontaneous change

   Hint

   No exchange of energy or matter.

   Answer

   (a)

2. Q2 · 1 mark · MCQExam level

   The enthalpy of formation of O₂(g) at 298 K is:
   (a) −249 kJ mol⁻¹ (b) zero (c) +498 kJ mol⁻¹ (d) cannot be defined

   Hint

   Reference state.

   Answer

   (b) zero

3. Q3 · 2 marks · Very short answerExam level

   State Hess's law and explain why it follows from H being a state function. (2 marks)

   Hint

   Path independence.

   Answer

   The standard enthalpy change of a reaction is the same whether it happens in one step or several, being the sum of the step enthalpies (1). Because H depends only on the initial and final states, any route between them gives the same ΔH (1).

4. Q4 · 2 marks · Very short answerHarder exam level

   Calculate ΔU for the combustion of 1 mol of propane at 298 K: $C_3{H}_8(g)+5O_2(g)\to 3C{O}_2(g)+4H_{2}O(l)$, ΔH = −2220 kJ mol⁻¹. (R = 8.314 J K⁻¹ mol⁻¹) (2 marks)

   Hint

   Δn_g = 3 − 6.

   Answer

   Δn_g = 3 − 6 = −3 (½). ΔU = ΔH − Δn_gRT = −2220 + 3 × 8.314 × 298 ÷ 1000 = −2220 + 7.43 (1) = −2212.6 kJ mol⁻¹ (½).

5. Q5 · 3 marks · Short answerHarder exam level

   Calculate the enthalpy change for $C{H}_4(g)\to C(g)+4H(g)$ and the mean C–H bond enthalpy, given: ΔfH⦵(CH₄) = −74.8, ΔaH⦵(C, graphite) = +716.7 and ΔaH⦵ of hydrogen = +218.0 kJ per mol of H atoms. (3 marks)

   Hint

   Break CH₄ into elements, then atomise them.

   Answer

   CH₄ → C(graphite) + 2H₂: +74.8 (1). C(graphite) → C(g): +716.7; 2H₂ → 4H: 4 × 218.0 = +872.0 (1). Total = 74.8 + 716.7 + 872.0 = 1663.5 kJ mol⁻¹; mean C–H = 1663.5 ÷ 4 = 415.9 kJ mol⁻¹ (1).

6. Q6 · 3 marks · Short answerHarder exam level

   For a reaction, ΔH = +40 kJ mol⁻¹ and ΔS = +100 J K⁻¹ mol⁻¹. At what temperature is it at equilibrium? Is it spontaneous at 300 K and at 500 K? (3 marks)

   Hint

   T = ΔH/ΔS.

   Answer

   T = 40 000 ÷ 100 = 400 K (1). At 300 K: ΔG = 40 − 300 × 0.100 = +10 kJ, not spontaneous (1). At 500 K: ΔG = 40 − 50 = −10 kJ, spontaneous (1).

7. Q7 · 3 marks · Short answerHarder exam level

   Suppose 2.0 mol of an ideal gas is compressed isothermally and reversibly at 300 K from 20 L to 2.0 L. Find w, q and ΔU. (R = 8.314 J K⁻¹ mol⁻¹) (3 marks)

   Hint

   log(2/20) = −1.

   Answer

   w = −2.303 × 2.0 × 8.314 × 300 × log(2.0/20) = −2.303 × 4988 × (−1) = +11 488 J ≈ +11.5 kJ (work done on the gas) (1½). ΔU = 0 (isothermal, ideal) (½); q = −w = −11.5 kJ (heat released) (1).

8. Q8 · 3 marks · Short answerHarder exam level

   Predict the sign of ΔS for: (a) $N{H}_{4}Cl(s)\to N{H}_3(g)+HCl(g)$ (b) $H_2(g)\to 2H(g)$ (c) cooling a gas from 400 K to 300 K. (3 marks)

   Hint

   Count gas moles; temperature effect.

   Answer

   (a) Positive: a solid gives two gases (1). (b) Positive: one gas molecule gives two atoms (1). (c) Negative: lower temperature means less molecular motion (1).

9. Q9 · 3 marks · Short answerHarder exam level

   Suppose a student mixes 50 mL of 1.0 M HCl with 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, both at 25.0 °C, and the temperature rises to 31.8 °C. Assume the solution has density 1.0 g mL⁻¹ and specific heat 4.18 J g⁻¹ K⁻¹. Calculate the enthalpy of neutralisation per mole of water formed. (3 marks)

   Hint

   q = m c ΔT, then divide by moles.

   Answer

   q = 100 × 4.18 × 6.8 = 2842 J (1). Moles of water = 0.050 (1). ΔH = −2842 ÷ 0.050 = −56 840 J ≈ −56.8 kJ mol⁻¹ (negative because heat was released) (1).

10. Q10 · 2 marks · Very short answerHOTS / competency

    Explain, using ΔG = ΔH − TΔS, why water freezes spontaneously below 0 °C but melts above 0 °C. (2 marks)

    Hint

    Signs for freezing.

    Answer

    For freezing, ΔH < 0 and ΔS < 0 (1). Below 273 K, |ΔH| > |TΔS| so ΔG < 0 and freezing is spontaneous; above 273 K, |TΔS| > |ΔH| so ΔG > 0 and the reverse process (melting) is spontaneous (1).

11. Q11 · 5 marks · Long answerHOTS / competency

    (a) Using bond enthalpies (C–H 414, C=C 611, H–H 436, C–C 347 kJ mol⁻¹), estimate ΔH for the hydrogenation of ethene: $C_2{H}_4(g)+H_2(g)\to C_2{H}_6(g)$. (b) The value from formation enthalpies is −137 kJ mol⁻¹. Suggest why the two differ. (c) ΔS for this reaction is −121 J K⁻¹ mol⁻¹. Using ΔH = −137 kJ mol⁻¹, find the temperature above which hydrogenation is not spontaneous. (d) Why do industries still run it at about 400-450 K with a nickel catalyst? (5 marks)

    Hint

    Only C=C and H–H break; C–C and two C–H form.

    Answer

    (a) Broken: C=C 611 + H–H 436 = 1047 kJ; formed: C–C 347 + 2 C–H 828 = 1175 kJ; ΔH = 1047 − 1175 = −128 kJ mol⁻¹ (2). (b) Mean bond enthalpies are averages over many compounds, so estimates differ slightly from exact values (1). (c) T = 137 000 ÷ 121 = 1132 K (1). (d) Below about 1132 K ΔG is negative, but the reaction is slow; a catalyst and moderate heating give a fast rate while ΔG stays favourable (1).

12. Q12 · 1 mark · Assertion-reasonHOTS / competency

    Assertion (A): A reaction with ΔG⦵ = 0 has an equilibrium constant equal to 1.
    Reason (R): ΔG⦵ = −2.303RT log K.
    (a) Both A and R are true and R is the correct explanation of A.
    (b) Both A and R are true but R is not the correct explanation of A.
    (c) A is true but R is false.
    (d) A is false but R is true.

    Hint

    log K = 0 means K = ?

    Answer

    (a) If ΔG⦵ = 0, log K = 0 and K = 1.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.Give one example each of an open, a closed and an isolated system.Show answer

   Open: reactants in an open beaker; closed: a sealed flask; isolated: a perfect thermos flask.

2. 2.State the first law of thermodynamics.Show answer

   The energy of an isolated system is constant; ΔU = q + w.

3. 3.What is the work done in free expansion of an ideal gas?Show answer

   Zero.

4. 4.Write the relation between ΔH and ΔU.Show answer

   ΔH = ΔU + Δn_g RT.

5. 5.What is measured by a bomb calorimeter?Show answer

   ΔU (heat at constant volume).

6. 6.State Hess's law.Show answer

   The enthalpy change of a reaction is the same whether it occurs in one step or several; it is the sum of the enthalpy changes of the steps.

7. 7.What is the standard enthalpy of formation of an element in its reference state?Show answer

   Zero.

8. 8.State the third law of thermodynamics.Show answer

   The entropy of a perfectly crystalline substance is zero at 0 K.

9. 9.What is the sign of ΔG at equilibrium?Show answer

   Zero.

10. 10.Write the relation between ΔrG⦵ and K.Show answer

    ΔrG⦵ = −2.303 RT log K.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you state whether q, w, U, H, S and G are state functions?
• ☐Can you find ΔU from q and w with correct signs?
• ☐Can you calculate reversible and irreversible isothermal work?
• ☐Can you convert ΔU to ΔH for a reaction with gases?
• ☐Can you use Hess's law to get a formation enthalpy from combustion data?
• ☐Can you estimate ΔrH from bond enthalpies?
• ☐Can you find a lattice enthalpy from a Born-Haber cycle?
• ☐Can you predict spontaneity from ΔH and ΔS at different temperatures, and find K from ΔG⦵?

Thermodynamics: frequently asked questions