Chemical Thermodynamics ISC Class 11: Energy, Entropy and Free Energy

What Chemical Thermodynamics is about

Weightage: Chemical Thermodynamics belongs to unit 5 of the ISC Class XI Chemistry syllabus. CISCE weights units 1-6 together (Physical Chemistry) at 41 of the 70 marks in Paper I (Theory). Class XI is examined by the school, so your school sets the paper within this weightage.Source: CISCE ISC Chemistry syllabus, examination year 2028, Class XI unit weightage

Key concepts

Systems, properties, state functions and processes

An open system exchanges both energy and matter with its surroundings (an open beaker); a closed system exchanges energy only (a sealed flask); an isolated system exchanges neither (an ideal vacuum flask). Extensive properties depend on the amount of matter (mass, volume, U, H, S, G, heat capacity); intensive properties do not (T, p, density, molar quantities). A state function depends only on the present state (U, H, S, G, p, V, T); q and w are path functions. Processes: isothermal (T constant), adiabatic (q = 0), isobaric (p constant), isochoric (V constant), cyclic (returns to the start, so ΔU = 0), reversible (carried out in infinitesimal steps, always near equilibrium) and irreversible (real, finite changes).

Worked example

Classify (i) tea in a closed vacuum flask (ii) volume (iii) density (iv) work done.

Try it first, then show the solutionSolution

1. (i) Isolated system (ideally)
2. (ii) Extensive property, and a state function
3. (iii) Intensive property (mass/volume ratio)
4. (iv) Path function

Answer: Isolated; extensive; intensive; path function

First law and sign convention

The first law of thermodynamics: energy can be neither created nor destroyed; for a closed system ΔU = q + w. Sign convention (IUPAC): q > 0 when heat is absorbed by the system; w > 0 when work is done on the system (compression); both are negative when energy leaves the system. Significance: it lets us account for every joule; it shows that a perpetual-motion machine producing energy from nothing is impossible; and for an isolated system ΔU = 0.

Worked example

A system absorbs 500 J of heat and does 200 J of work on its surroundings. Find ΔU.

Try it first, then show the solutionSolution

1. q = +500 J (absorbed); w = −200 J (done by the system)
2. ΔU = 500 + (−200) = 300 J

Answer: +300 J

Work: irreversible and reversible

For expansion or compression against a constant external pressure, w = −p_ext ΔV (1 L atm = 101.3 J). For a reversible isothermal change of an ideal gas, the external pressure is always only infinitesimally different from the gas pressure, and w = −2.303 nRT log(V₂/V₁) = −2.303 nRT log(p₁/p₂). Reversible expansion gives the maximum work obtainable. For an ideal gas at constant temperature, ΔU = 0, so q = −w. Expansion into a vacuum (p_ext = 0) does no work (free expansion).

Worked example

2.0 mol of an ideal gas at 300 K expands from 10 L to 20 L. Find the work (i) reversibly and isothermally (ii) against a constant external pressure of 1.0 atm.

Try it first, then show the solutionSolution

1. (i) w = −2.303 × 2.0 × 8.314 × 300 × log 2 = −2.303 × 4988 × 0.301 = −3458 J
2. (ii) w = −1.0 atm × 10 L = −10 L atm = −10 × 101.3 = −1013 J
3. More work is done by the gas in the reversible process

Answer: (i) −3.46 kJ (ii) −1.01 kJ

Enthalpy and the relation between ΔH and ΔU

Enthalpy H = U + pV. At constant pressure, the heat exchanged equals ΔH (q_p = ΔH), while at constant volume it equals ΔU (q_v = ΔU). For reactions involving gases, ΔH = ΔU + Δn_g RT, where Δn_g = moles of gaseous products − moles of gaseous reactants. Liquids and solids are ignored in Δn_g. ΔH < 0: exothermic; ΔH > 0: endothermic.

Worked example

For CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔU = −885.4 kJ mol⁻¹ at 298 K. Find ΔH.

Try it first, then show the solutionSolution

1. Δn_g = 1 − 3 = −2 (water is liquid)
2. Δn_g RT = −2 × 8.314 × 10⁻³ × 298 = −4.96 kJ
3. ΔH = −885.4 − 4.96 = −890.4 kJ mol⁻¹

Answer: −890.4 kJ mol⁻¹

Heat capacity, specific heat and Cp − Cv

Heat capacity C is the heat needed to raise the temperature of a body by 1 K; specific heat c is per gram (q = mcΔT); molar heat capacity is per mole. At constant volume q_v = C_vΔT = ΔU; at constant pressure q_p = C_pΔT = ΔH. For one mole of an ideal gas, ΔH − ΔU = Δ(pV) = RΔT, so C_p − C_v = R (8.314 J K⁻¹ mol⁻¹); C_p is larger because at constant pressure some heat goes into expansion work.

Worked example

How much heat is needed to warm 100 g of water from 25 °C to 35 °C (c = 4.18 J g⁻¹ K⁻¹)?

Try it first, then show the solutionSolution

1. q = mcΔT = 100 × 4.18 × 10
2. = 4180 J

Answer: 4.18 kJ

Hess's law and standard enthalpy of formation

Hess's law of constant heat summation: the enthalpy change of a reaction is the same whether it occurs in one step or several, because H is a state function. Thermochemical equations can therefore be added, subtracted and multiplied like algebraic equations (ΔH changes accordingly; reversing a reaction changes its sign). The standard enthalpy of formation Δ_fH° is the enthalpy change when one mole of a compound forms from its elements in their standard states (Δ_fH° of an element in its standard state is zero). Then Δ_rH° = ΣΔ_fH°(products) − ΣΔ_fH°(reactants).

Worked example

Use Δ_fH° (kJ mol⁻¹): CO₂(g) −393.5, H₂O(l) −285.8, CH₄(g) −74.8 to find the enthalpy of combustion of methane.

Try it first, then show the solutionSolution

1. $C{H}_4(g)+2O_2(g)\to C{O}_2(g)+2H_{2}O(l)$
2. Δ_cH° = [−393.5 + 2(−285.8)] − [−74.8 + 0]
3. = −965.1 + 74.8 = −890.3 kJ mol⁻¹

Answer: −890.3 kJ mol⁻¹

Bond enthalpy and other enthalpy changes

Bond dissociation enthalpy: energy to break one mole of a given bond in gaseous molecules; for polyatomic molecules an average value is used. For gaseous reactions, Δ_rH ≈ Σ(bond enthalpies broken) − Σ(bond enthalpies formed). Other named enthalpies: combustion (1 mol burnt completely in oxygen), atomisation (1 mol of gaseous atoms from the element), sublimation (solid → gas), phase transition (fusion, vaporisation), ionisation, solution (1 mol dissolved in a stated amount of solvent) and dilution (when a solution is diluted further).

Worked example

Estimate Δ_rH for $H_2(g)+C{l}_2(g)\to 2HCl(g)$ from bond enthalpies H–H 436, Cl–Cl 243, H–Cl 431 kJ mol⁻¹.

Try it first, then show the solutionSolution

1. Broken: 436 + 243 = 679
2. Formed: 2 × 431 = 862
3. Δ_rH = 679 − 862 = −183 kJ

Answer: −183 kJ (exothermic)

Enthalpy of neutralisation and calorific value

The enthalpy of neutralisation is the enthalpy change when one mole of water forms from an acid and a base in dilute solution. For strong acids and strong bases it is constant, about −57.1 kJ mol⁻¹, because they are fully ionised and the only reaction is $H^{+}(aq)+O{H}^{-}(aq)\to H_{2}O(l)$. With a weak acid or base it is less negative, because some energy is used to ionise the weak electrolyte. The calorific value of a fuel is the heat released when 1 g of it burns completely: calorific value = −Δ_cH/M.

Worked example

100 mL of 0.50 M HCl is mixed with 100 mL of 0.50 M NaOH. Find the heat released and the temperature rise (take 200 g of solution, c = 4.18 J g⁻¹ K⁻¹, ΔH = −57.1 kJ mol⁻¹).

Try it first, then show the solutionSolution

1. Moles of water formed = 0.100 × 0.50 = 0.050 mol
2. Heat released = 0.050 × 57.1 = 2.86 kJ
3. ΔT = 2855/(200 × 4.18) = 3.4 K

Answer: 2.86 kJ; about 3.4 K

Second law, entropy and spontaneity

A spontaneous process occurs by itself without continuous outside help (it may still be slow). The second law: in any spontaneous process the entropy of the universe increases: ΔS_total = ΔS_system + ΔS_surroundings > 0; at equilibrium ΔS_total = 0. Entropy S measures the dispersal of energy and matter (disorder); for a reversible change at temperature T, ΔS = q_rev/T. Entropy increases on melting, vaporisation, dissolving and when gas moles increase. The idea that the universe's entropy keeps increasing towards a maximum, after which no useful work could be done, is called the thermal (heat) death of the universe. Third law: the entropy of a perfectly crystalline substance at 0 K is zero (statement only).

Worked example

The enthalpy of fusion of ice is 6.01 kJ mol⁻¹ at 273 K. Find the entropy change of fusion.

Try it first, then show the solutionSolution

1. At the melting point, melting is reversible: ΔS = q_rev/T
2. ΔS = 6010/273 = 22.0 J K⁻¹ mol⁻¹

Answer: +22.0 J K⁻¹ mol⁻¹

Gibbs and Helmholtz energy; ΔG and K

Gibbs energy G = H − TS and Helmholtz energy A = U − TS; they are related by G = A + pV. At constant T and p, ΔG = ΔH − TΔS and ΔG = −TΔS_total, so the criteria are: ΔG < 0 spontaneous; ΔG = 0 equilibrium; ΔG > 0 non-spontaneous (the reverse is spontaneous). Signs: ΔH < 0, ΔS > 0: spontaneous at all T; ΔH > 0, ΔS < 0: never; ΔH < 0, ΔS < 0: spontaneous below T = ΔH/ΔS; ΔH > 0, ΔS > 0: spontaneous above T = ΔH/ΔS. The standard Gibbs energy change is linked to the equilibrium constant: ΔG° = −2.303 RT log K.

Worked example

For $2N{O}_2(g)\to N_2{O}_4(g)$, ΔH° = −57.2 kJ mol⁻¹ and ΔS° = −175.8 J K⁻¹ mol⁻¹. Over what range of temperature is the reaction spontaneous?

Try it first, then show the solutionSolution

1. Both ΔH and ΔS are negative, so ΔG < 0 only while |ΔH| > T|ΔS|
2. T < 57 200/175.8 = 325.4 K

Answer: Below about 325 K (52 °C); above this, N₂O₄ tends to dissociate

Chemistry in real life: Chemical 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 Bond Enthalpy
• Entropy, Gibbs Energy and Spontaneity

Chemical Thermodynamics formula sheet

First law

$\Delta U=q+w$

q absorbed, w done on the system: positive.

Work

$w=-p_{\text{ext}}\Delta V;w_{\text{rev}}=-2.303nRT\log \frac{V_2}{V_1}$

1 L atm = 101.3 J.

Enthalpy

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

Count gaseous moles only.

Heat and heat capacity

$q=mc\Delta T;C_p-C_v=R$

R = 8.314 J K⁻¹ mol⁻¹.

From enthalpies of formation

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

Elements in standard states: zero.

From bond enthalpies

${\Delta }_{r}H=\sum BE(\text{broken})-\sum BE(\text{formed})$

Gaseous reactions; approximate.

Entropy

$\Delta S=\frac{q_{\text{rev}}}{T};\Delta S_{\text{total}}=\Delta S_{\text{sys}}+\Delta S_{\text{surr}}$

Spontaneous: ΔS_total > 0.

Gibbs energy

$\Delta G=\Delta H-T\Delta S;\Delta G^{\circ }=-2.303RT\log K$

G = A + pV; A = U − TS.

Chemical Thermodynamics: diagrams to practise

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

How the school exam asks Chemical Thermodynamics

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

Tricks to remember Chemical Thermodynamics

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

• Work on is plus

  Energy into the system (heat absorbed, work done on it) is positive.

  Use it when: First-law signs.

• Only gases count

  Δn_g ignores solids and liquids.

  Use it when: ΔH = ΔU + Δn_gRT.

• Products minus reactants; broken minus made

  Formation data: products − reactants. Bond data: broken − formed.

  Use it when: Not mixing the two orders.

• Kilo meets joule

  ΔH in kJ but ΔS in J: convert before using ΔH − TΔS.

  Use it when: Gibbs energy numericals.

• Strong + strong = 57

  Strong acid + strong base: −57.1 kJ per mole of water, always.

  Use it when: Neutralisation.

Chemical 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 D style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 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 Chemical Thermodynamics

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

• ✗Mistake: Sign errors in ΔU = q + w

  Students treat work done by the gas as positive.

  Wrong working

  Gas absorbs 500 J and does 200 J of work: ΔU = 700 J.

  Right working

  w = −200 J: ΔU = 300 J.

  ✓Do this: Energy leaving the system is negative.

  Usually costs: 1-2 marks.

• ✗Mistake: Including liquids in Δn_g

  Students count all moles.

  Wrong working

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

  Right working

  Δn_g = 1 − 3 = −2.

  ✓Do this: Write states and cross out (l) and (s).

  Usually costs: 1 mark.

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

  ΔS is usually in J K⁻¹ mol⁻¹.

  Wrong working

  ΔG = 178 − 298 × 160 = −47 502.

  Right working

  ΔG = 178 − 298 × 0.160 = +130.3 kJ.

  ✓Do this: Convert ΔS to kJ K⁻¹ first.

  Usually costs: Full marks for the part.

• ✗Mistake: Bond enthalpy in the wrong order

  Students copy 'products − reactants' from formation data.

  Wrong working

  Δ_rH = Σ(formed) − Σ(broken) = +183 kJ for H₂ + Cl₂.

  Right working

  Σ(broken) − Σ(formed) = −183 kJ.

  ✓Do this: Breaking costs energy (+), making releases it (−).

  Usually costs: 1-2 marks.

• ✗Mistake: Thinking exothermic means spontaneous

  The entropy term is forgotten.

  Wrong working

  NH₄NO₃ dissolving is endothermic, so it cannot be spontaneous.

  Right working

  ΔS > 0 makes ΔG < 0 at room temperature.

  ✓Do this: Always check the sign of ΔG, not just ΔH.

  Usually costs: 1 mark.

Practice questions on Chemical 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

   Which is an intensive property?
   (a) mass
   (b) volume
   (c) temperature
   (d) enthalpy

   Hint

   Independent of amount.

   Answer

   (c)

2. Q2 · 1 mark · MCQExam level

   An isolated system exchanges:
   (a) matter only
   (b) energy only
   (c) both
   (d) neither

   Hint

   Ideal vacuum flask.

   Answer

   (d)

3. Q3 · 1 mark · Very short answerExam level

   Write the relation between Cp and Cv for one mole of an ideal gas.

   Hint

   Difference.

   Answer

   Cp − Cv = R

4. Q4 · 1 mark · Very short answerExam level

   What is the sign of ΔS when water freezes?

   Hint

   Order increases.

   Answer

   Negative

5. Q5 · 2 marks · Short answerExam level

   How much heat is needed to raise the temperature of 50 g of aluminium (c = 0.90 J g⁻¹ K⁻¹) by 20 K?

   Hint

   q = mcΔT.

   Answer

   50 × 0.90 × 20 = 900 J

6. Q6 · 2 marks · Short answerHarder exam level

   A gas expands from 2.0 L to 7.0 L against a constant external pressure of 2.0 atm. Find the work done in joules.

   Hint

   1 L atm = 101.3 J.

   Answer

   w = −2.0 × 5.0 = −10 L atm = −1013 J

7. Q7 · 2 marks · Short answerHarder exam level

   A reaction has ΔH = −50 kJ mol⁻¹ and ΔS = +20 J K⁻¹ mol⁻¹. Is it spontaneous at all temperatures? Explain.

   Hint

   Both terms favour.

   Answer

   Yes: ΔH < 0 and −TΔS < 0, so ΔG is negative at every temperature.

8. Q8 · 3 marks · Short answerHarder exam level

   Estimate Δ_rH for C₂H₄(g) + H₂(g) → C₂H₆(g) from bond enthalpies C=C 614, H–H 436, C–C 348, C–H 414 kJ mol⁻¹.

   Hint

   Only the changed bonds matter.

   Answer

   Broken: 614 + 436 = 1050; formed: 348 + 2(414) = 1176; Δ_rH = −126 kJ

9. Q9 · 2 marks · Short answerHarder exam level

   Find Δ_rH for the thermite reaction $F{e}_2{O}_3(s)+2Al(s)\to A{l}_2{O}_3(s)+2Fe(s)$ given Δ_fH (kJ mol⁻¹): Fe₂O₃ −824.2, Al₂O₃ −1675.7.

   Hint

   Elements have zero.

   Answer

   −1675.7 − (−824.2) = −851.5 kJ

10. Q10 · 2 marks · Short answerHarder exam level

    A reaction has ΔH = +30 kJ mol⁻¹ and ΔS = +105 J K⁻¹ mol⁻¹. Above what temperature is it spontaneous?

    Hint

    T = ΔH/ΔS.

    Answer

    T > 30 000/105 = 286 K

11. Q11 · 2 marks · Short answerHarder exam level

    The equilibrium constant of a reaction at 300 K is 10. Find ΔG°.

    Hint

    log 10 = 1.

    Answer

    ΔG° = −2.303 × 8.314 × 300 × 1 = −5744 J mol⁻¹ ≈ −5.74 kJ mol⁻¹

12. Q12 · 3 marks · Short answerHarder exam level

    Find ΔH − ΔU at 298 K for (i) C(graphite) + O₂(g) → CO₂(g) (ii) $2H_2(g)+O_2(g)\to 2H_{2}O(l)$.

    Hint

    Δn_g RT.

    Answer

    (i) Δn_g = 0, so ΔH = ΔU. (ii) Δn_g = −3: ΔH − ΔU = −3 × 8.314 × 298 = −7.43 kJ

13. Q13 · 5 marks · Long answerHarder exam level

    0.32 g of methane is burnt in a bomb calorimeter of heat capacity 8.0 kJ K⁻¹, and the temperature rises by 2.21 K.
    (a) Find ΔU of combustion per mole. [2]
    (b) Find ΔH at 298 K (water formed as liquid). [2]
    (c) Find the calorific value of methane. [1]

    Hint

    Bomb calorimeter: constant volume.

    Answer

    (a) q = 8.0 × 2.21 = 17.68 kJ for 0.020 mol: ΔU = −884 kJ mol⁻¹. (b) Δn_g = −2: ΔH = −884 − 4.96 = −889 kJ mol⁻¹. (c) 889/16 = 55.6 kJ g⁻¹.

14. Q14 · 3 marks · Short answerHOTS / competency

    Find Δ_fH of benzene, C₆H₆(l), given Δ_cH(C₆H₆) = −3268 kJ mol⁻¹ and Δ_fH: CO₂ −393.5, H₂O(l) −285.8 kJ mol⁻¹.

    Hint

    Write the combustion equation with Δ_fH(C₆H₆) as the unknown.

    Answer

    C₆H₆ + 7½O₂ → 6CO₂ + 3H₂O: −3268 = 6(−393.5) + 3(−285.8) − Δ_fH ⇒ Δ_fH = −2361 − 857.4 + 3268 = +49.6 kJ mol⁻¹

15. Q15 · 3 marks · Short answerHOTS / competency

    1.0 mol of an ideal gas at 300 K expands isothermally from 1.0 bar to 0.50 bar (i) reversibly (ii) against a constant 0.50 bar. Compare the work done.

    Hint

    V₁ = RT/p₁ = 24.94 L.

    Answer

    (i) w = −2.303RT log(p₁/p₂) = −8.314 × 300 × 0.693 = −1729 J. (ii) V doubles to 49.88 L: w = −0.50 bar × 24.94 L = −12.47 L bar = −1247 J. Reversible expansion does more work.

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.First law in symbols?Show answer

   ΔU = q + w.

2. 2.Work against constant pressure?Show answer

   w = −p_ext ΔV.

3. 3.ΔH and ΔU relation?Show answer

   ΔH = ΔU + Δn_gRT.

4. 4.Hess's law?Show answer

   ΔH of a reaction is independent of the route.

5. 5.Enthalpy of neutralisation, strong acid and base?Show answer

   About −57.1 kJ mol⁻¹.

6. 6.Second law criterion?Show answer

   ΔS_universe > 0 for a spontaneous change.

7. 7.Gibbs equation?Show answer

   ΔG = ΔH − TΔS.

8. 8.Third law?Show answer

   Entropy of a perfect crystal at 0 K is zero.

Quick revision checklist

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

• ☐Can you classify properties as extensive or intensive?
• ☐Can you find ΔU, q or w with correct signs?
• ☐Can you compare reversible and irreversible work?
• ☐Can you convert between ΔH and ΔU?
• ☐Can you combine thermochemical equations to get a target ΔH?
• ☐Can you estimate ΔH from bond enthalpies?
• ☐Can you find the temperature at which a reaction becomes spontaneous?
• ☐Can you find K from ΔG° and vice versa?

Chemical Thermodynamics: frequently asked questions