Class 11 Chemistry Chapter 6 Question Answer | Thermodynamics | English Medium

Class 11 Chemistry Chapter 6 Question Answer | Thermodynamics | English Medium | ASSEB

Chapter 6 — Thermodynamics

Welcome to HSLC Guru. This English-medium study guide for ASSEB Class 11 Chemistry Chapter 6 — Thermodynamics — gives you a complete, exam-ready resource. You will find a detailed conceptual summary, all the important KaTeX formulas rendered cleanly, short and long answer questions with full solutions, numerical problems on enthalpy, entropy and Gibbs free energy, MCQs, fill in the blanks, true or false items, a glossary and a formula sheet — all aligned strictly with the ASSEB syllabus.

Summary

Thermodynamics is the branch of science that deals with the quantitative relationship between heat and other forms of energy in chemical and physical processes. The portion of the universe under study is called the system, while everything outside the system that can interact with it is called the surroundings. The boundary separating system and surroundings may be real or imaginary, rigid or flexible. Systems are classified as open (exchanges both matter and energy, e.g. boiling water in an open beaker), closed (exchanges only energy, e.g. water in a sealed conducting flask) and isolated (exchanges neither matter nor energy, e.g. a thermos flask). The macroscopic properties such as pressure, volume, temperature, internal energy, enthalpy, entropy and Gibbs energy that depend only on the present state of the system are called state functions; quantities like work and heat which depend on the path are path functions. Properties are also divided into extensive (depend on amount — mass, volume, internal energy) and intensive (independent of amount — temperature, pressure, density, molar volume).

The First Law of Thermodynamics is a statement of conservation of energy: the change in internal energy of a closed system equals the heat absorbed plus the work done on the system. For mechanical PV-work, work done on the system is w = -p_ext ΔV. For an isothermal reversible expansion of an ideal gas, w = -nRT ln(V2/V1); for an adiabatic process q = 0 so ΔU = w. Since most laboratory reactions occur at constant pressure, a more convenient state function — enthalpy H = U + PV — is defined, with ΔH = qp. For ideal gas reactions, ΔH and ΔU are related through the change in moles of gas. The molar heat capacities Cv (constant volume) and Cp (constant pressure) of an ideal gas obey Mayer’s relation Cp − Cv = R, with γ = Cp/Cv being the adiabatic index.

Thermochemistry applies the first law to chemical reactions. The standard enthalpy of reaction ΔH° is measured at 1 bar and a specified temperature (usually 298 K). Important standard enthalpies include enthalpy of formation ΔfH° (formation of one mole of compound from its elements in their reference states), combustion ΔcH° (one mole burned completely in oxygen), neutralisation (almost constant at −57.1 kJ/mol for any strong acid–strong base pair), fusion, vaporisation and sublimation. Hess’s law of constant heat summation states that the net enthalpy change of a reaction is the same whether the reaction takes place in one step or many. Bond enthalpy and lattice enthalpy let us estimate ΔH for gaseous and ionic reactions respectively (Born–Haber cycle).

The first law allows energy book-keeping but does not predict spontaneity. The Second Law introduces entropy S, a measure of molecular disorder, with dS = q_rev/T. For any spontaneous process the total entropy of system plus surroundings must increase: ΔS_total ≥ 0. To use only system properties, J. W. Gibbs defined the free energy G = H − TS, giving ΔG = ΔH − TΔS at constant T and P. A reaction is spontaneous when ΔG < 0, non-spontaneous when ΔG > 0 and at equilibrium when ΔG = 0. The standard Gibbs energy is linked to the equilibrium constant by ΔG° = −RT ln K. The Third Law states that the entropy of a perfectly crystalline substance approaches zero as the temperature approaches absolute zero, allowing absolute entropies to be tabulated.

Key Formulas (KaTeX)

$\Delta U = q + w$

$\Delta H = \Delta U + \Delta n_g RT$

$C_p – C_v = R$

$\Delta H_{rxn} = \sum \Delta H_f(\text{products}) – \sum \Delta H_f(\text{reactants})$

$\Delta G = \Delta H – T\Delta S$

$\Delta G^\circ = -RT \ln K$

$\Delta S = \frac{q_{rev}}{T}$

$w = -nRT \ln\frac{V_2}{V_1}$

Section A — One-Mark Questions

Q1. Define a thermodynamic system.

Answer: A thermodynamic system is the specific portion of the universe selected for study, separated from the surroundings by a real or imaginary boundary.

Q2. What is an isolated system? Give one example.

Answer: An isolated system can exchange neither matter nor energy with its surroundings. Example: a perfectly insulated thermos flask containing hot tea.

Q3. Name two intensive properties.

Answer: Temperature and pressure (also density, viscosity, refractive index, molar volume).

Q4. State the sign convention for work in modern IUPAC notation.

Answer: Work done on the system is positive (+w); work done by the system on the surroundings is negative (−w).

Q5. Write the mathematical statement of the First Law of Thermodynamics.

Answer: $\Delta U = q + w$, where ΔU is the change in internal energy, q is heat absorbed by the system and w is work done on the system.

Q6. Define enthalpy.

Answer: Enthalpy is a state function defined by H = U + PV. Its change at constant pressure equals the heat exchanged: ΔH = qp.

Q7. What is meant by a state function? Give one example.

Answer: A state function depends only on the present state of the system, not on the path taken. Example: internal energy U.

Q8. Write Mayer’s relation for an ideal gas.

Answer: $C_p – C_v = R$ (per mole of an ideal gas).

Q9. What is the standard enthalpy of formation of an element in its reference state?

Answer: Zero, by convention.

Q10. State the criterion for spontaneity in terms of Gibbs energy.

Answer: A process at constant T and P is spontaneous if ΔG < 0, at equilibrium if ΔG = 0, and non-spontaneous if ΔG > 0.

Section B — Two/Three-Mark Questions

Q11. Distinguish between extensive and intensive properties with two examples each.

Answer: Extensive properties depend on the amount of substance — when the system is divided, they are divided too. Examples: mass, volume, internal energy, enthalpy, number of moles. Intensive properties are independent of amount and are characteristic of the substance. Examples: temperature, pressure, density, viscosity, molar heat capacity. Note that the ratio of two extensive properties (e.g. molar volume = V/n) is intensive.

Q12. Derive the relation $\Delta H = \Delta U + \Delta n_g RT$ for a gaseous reaction.

Answer: By definition H = U + PV, so for a change at constant temperature and pressure ΔH = ΔU + Δ(PV). Treating gaseous reactants and products as ideal, PV = n_g RT, hence Δ(PV) = (Δn_g)RT, where Δn_g = (moles of gaseous products) − (moles of gaseous reactants). Substituting gives $\Delta H = \Delta U + \Delta n_g RT$. For reactions where Δn_g = 0, ΔH = ΔU.

Q13. State and explain Hess’s law of constant heat summation.

Answer: Hess’s law states that the enthalpy change for a chemical reaction is independent of the path followed and depends only on the initial and final states. This is a direct consequence of enthalpy being a state function. Hess’s law allows us to add, subtract or multiply thermochemical equations to calculate enthalpy changes that are difficult to measure directly — for instance, the enthalpy of formation of CO from C and ½O2.

Q14. Distinguish between isothermal and adiabatic processes.

Answer: In an isothermal process the temperature of the system is held constant (ΔT = 0); for an ideal gas ΔU = 0 and q = −w, so heat is exchanged with the surroundings. In an adiabatic process no heat is exchanged with the surroundings (q = 0); therefore ΔU = w and the temperature changes during expansion or compression.

Q15. Explain why ΔH of neutralisation of a strong acid by a strong base is almost constant at −57.1 kJ/mol.

Answer: Strong acids and strong bases are completely ionised in dilute aqueous solution. The actual reaction is therefore the same in every case: H+(aq) + OH−(aq) → H2O(l). Because the reactants and products are identical regardless of which strong acid and strong base were used, the enthalpy released — about −57.1 kJ per mole of water formed — is essentially constant.

Q16. Define entropy. Why is the entropy change of fusion positive?

Answer: Entropy S is a thermodynamic state function that measures the disorder or randomness of a system; for a reversible process dS = q_rev/T. When a solid melts, its highly ordered crystal lattice breaks down into a less ordered liquid in which molecules can translate. Disorder increases, so ΔS_fusion > 0.

Section C — Five/Seven-Mark Questions (with Numericals)

Q17. State the First Law of Thermodynamics and apply it to (i) isothermal reversible expansion and (ii) adiabatic expansion of an ideal gas. Calculate the work done when 2 mol of an ideal gas expands isothermally and reversibly from 5 L to 25 L at 300 K (R = 8.314 J K⁻¹ mol⁻¹).

Answer: The first law states that energy can neither be created nor destroyed, only converted from one form to another. Mathematically $\Delta U = q + w$.
(i) Isothermal reversible expansion of an ideal gas: ΔT = 0, so ΔU = 0 and q = −w with $w = -nRT \ln\frac{V_2}{V_1}$.
(ii) Adiabatic expansion: q = 0, so ΔU = w; the gas does work at the cost of its internal energy and its temperature falls.
Numerical: w = −nRT ln(V2/V1) = −(2)(8.314)(300) ln(25/5) = −(4988.4)(ln 5) = −(4988.4)(1.6094) ≈ −8028 J ≈ −8.03 kJ. The negative sign shows work is done by the gas on the surroundings.

Q18. Calculate ΔH for the reaction C2H4(g) + H2(g) → C2H6(g) at 298 K using ΔfH°(C2H4) = +52.3 kJ/mol, ΔfH°(C2H6) = −84.7 kJ/mol, ΔfH°(H2) = 0. State whether the reaction is exothermic or endothermic.

Answer: Using $\Delta H_{rxn} = \sum \Delta H_f(\text{products}) – \sum \Delta H_f(\text{reactants})$:
ΔH = [ΔfH°(C2H6)] − [ΔfH°(C2H4) + ΔfH°(H2)]
= [−84.7] − [+52.3 + 0]
= −84.7 − 52.3 = −137.0 kJ/mol.
Because ΔH is negative, the hydrogenation of ethene is exothermic.

Q19. For the reaction N2(g) + 3H2(g) → 2NH3(g), ΔH° = −92.4 kJ and ΔS° = −198.3 J/K at 298 K. (a) Calculate ΔG° at 298 K. (b) Determine the temperature above which the reaction becomes non-spontaneous.

Answer: (a) Convert ΔS° to kJ/K: ΔS° = −0.1983 kJ/K. Apply $\Delta G = \Delta H – T\Delta S$:
ΔG° = (−92.4) − (298)(−0.1983) = −92.4 + 59.09 = −33.31 kJ.
Since ΔG° < 0, ammonia synthesis is spontaneous at 298 K.
(b) The reaction becomes non-spontaneous when ΔG° > 0, i.e. when T > ΔH°/ΔS°. Crossover temperature T = (−92.4)/(−0.1983) ≈ 466 K. Above ~466 K the reaction is non-spontaneous in the forward direction.

Q20. The standard Gibbs free energy change for a reaction at 298 K is −13.6 kJ/mol. Calculate the equilibrium constant K (R = 8.314 J K⁻¹ mol⁻¹).

Answer: Apply $\Delta G^\circ = -RT \ln K$, so ln K = −ΔG°/(RT).
ln K = −(−13600)/(8.314 × 298) = 13600/2477.6 = 5.489.
K = e^{5.489} ≈ 242.
Since K is much greater than unity, the products are favoured at equilibrium.

Q21. Calculate the entropy change when 1 mol of ice melts at 0°C. The molar enthalpy of fusion of ice is 6.01 kJ/mol.

Answer: Melting at the normal melting point is reversible, so $\Delta S = \frac{q_{rev}}{T}$. Here q_rev = ΔH_fus = 6010 J/mol and T = 273.15 K.
ΔS = 6010 / 273.15 = 22.0 J K⁻¹ mol⁻¹.
The positive value confirms that disorder increases on melting.

Q22. Define bond enthalpy and lattice enthalpy. Using the Born–Haber cycle, list the steps required to calculate the lattice enthalpy of NaCl(s).

Answer: The bond enthalpy (or bond dissociation enthalpy) is the average energy required to break one mole of a particular type of bond in gaseous molecules to give gaseous atoms. For diatomic molecules it equals the bond dissociation energy; for polyatomic molecules it is taken as a mean over similar bonds in different compounds.
The lattice enthalpy of an ionic solid is the enthalpy change when one mole of the solid is formed from its constituent gaseous ions: M+(g) + X−(g) → MX(s). It cannot be measured directly and is obtained using the Born–Haber cycle.
Steps for NaCl: (i) sublimation of Na(s) → Na(g), ΔH_sub; (ii) ionisation of Na(g) → Na+(g) + e−, ΔH_IE; (iii) dissociation of ½Cl2(g) → Cl(g), ½ΔH_diss; (iv) electron gain by Cl(g) + e− → Cl−(g), ΔH_EA; (v) lattice formation Na+(g) + Cl−(g) → NaCl(s), ΔH_lat. By Hess’s law: ΔfH°(NaCl) = ΔH_sub + ΔH_IE + ½ΔH_diss + ΔH_EA + ΔH_lat. Solving for ΔH_lat gives the lattice enthalpy.

Q23. Predict the sign of ΔS for each of the following processes and justify briefly:
(a) H2O(l) → H2O(g)
(b) 2H2(g) + O2(g) → 2H2O(l)
(c) Dissolution of NH4Cl(s) in water
(d) Crystallisation of sugar from a saturated solution.

Answer: (a) ΔS > 0 — vaporisation greatly increases molecular disorder (gas has many more accessible microstates than liquid).
(b) ΔS < 0 — three moles of gas combine to form two moles of liquid, sharply lowering randomness.
(c) ΔS > 0 — the rigid ionic lattice breaks up and ions diffuse through the solvent.
(d) ΔS < 0 — randomly moving solute molecules become locked into an ordered crystal.

Multiple Choice Questions

MCQ1. Which of the following is not a state function?
(a) Enthalpy (b) Internal energy (c) Work (d) Entropy
Answer: (c) Work.

MCQ2. A thermos flask is an example of —
(a) Open system (b) Closed system (c) Isolated system (d) None
Answer: (c) Isolated system.

MCQ3. For an ideal gas, Cp − Cv equals —
(a) 0 (b) R (c) 2R (d) R/2
Answer: (b) R.

MCQ4. In an adiabatic process —
(a) ΔT = 0 (b) q = 0 (c) w = 0 (d) ΔU = 0
Answer: (b) q = 0.

MCQ5. The standard enthalpy of formation of O2(g) is —
(a) +498 kJ/mol (b) −498 kJ/mol (c) Zero (d) +205 kJ/mol
Answer: (c) Zero.

MCQ6. Which sign of ΔG indicates a spontaneous reaction?
(a) ΔG > 0 (b) ΔG = 0 (c) ΔG < 0 (d) Sign immaterial
Answer: (c) ΔG < 0.

MCQ7. The unit of entropy in SI is —
(a) J K⁻¹ mol⁻¹ (b) kJ mol⁻¹ (c) J mol⁻¹ (d) K J⁻¹
Answer: (a) J K⁻¹ mol⁻¹.

MCQ8. ΔH of neutralisation of a strong acid by a strong base is approximately —
(a) −13.7 kJ/mol (b) −57.1 kJ/mol (c) −285.8 kJ/mol (d) +57.1 kJ/mol
Answer: (b) −57.1 kJ/mol.

MCQ9. Hess’s law is a consequence of —
(a) First law (b) Second law (c) Third law (d) Zeroth law
Answer: (a) First law (enthalpy is a state function).

MCQ10. ΔG° = −RT ln K. If K > 1 then ΔG° is —
(a) Positive (b) Negative (c) Zero (d) Cannot tell
Answer: (b) Negative.

Fill in the Blanks

F1. The mathematical form of the first law is ΔU = q + ______.
Answer: w.

F2. Heat absorbed at constant pressure equals the change in ______.
Answer: enthalpy (H).

F3. A reaction is at equilibrium when ΔG = ______.
Answer: 0.

F4. The entropy of a perfectly crystalline solid at 0 K is ______ (Third Law).
Answer: zero.

F5. Cp − Cv for an ideal gas equals ______.
Answer: R.

True or False

T1. Internal energy is a path function. Answer: False — it is a state function.

T2. In an isothermal expansion of an ideal gas, ΔU = 0. Answer: True.

T3. ΔH and ΔU are equal for reactions in which Δn_g = 0. Answer: True.

T4. An exothermic reaction is always spontaneous. Answer: False — spontaneity depends on ΔG, not ΔH alone.

T5. The entropy of the universe always tends to increase. Answer: True.

Glossary

Term	Meaning
System	Part of the universe under thermodynamic study.
Surroundings	Everything outside the system that can interact with it.
Open system	Exchanges both matter and energy with surroundings.
Closed system	Exchanges only energy with surroundings.
Isolated system	Exchanges neither matter nor energy.
State function	Property depending only on present state (U, H, S, G, T, P, V).
Path function	Quantity depending on the path followed (q, w).
Extensive property	Depends on amount of matter (mass, volume).
Intensive property	Independent of amount (T, P, density).
Internal energy (U)	Total energy stored in a system.
Enthalpy (H)	H = U + PV; heat at constant pressure.
Entropy (S)	Measure of disorder; dS = q_rev/T.
Gibbs energy (G)	G = H − TS; spontaneity criterion at constant T, P.
Hess’s law	Net ΔH is independent of the path of the reaction.
Bond enthalpy	Energy needed to break one mole of a bond in the gas phase.
Lattice enthalpy	Energy released when one mole of an ionic solid forms from gaseous ions.
Standard state	Pure substance at 1 bar pressure and a stated temperature (usually 298 K).

Formula Sheet

Concept	Formula
First law	ΔU = q + w
PV work (constant p_ext)	w = −p_ext ΔV
Isothermal reversible work (ideal gas)	w = −nRT ln(V2/V1)
Adiabatic process	q = 0; ΔU = w
Enthalpy definition	H = U + PV
Enthalpy–internal energy relation	ΔH = ΔU + Δn_g RT
Mayer’s relation	Cp − Cv = R
Reaction enthalpy	ΔH_rxn = Σ ΔH_f(products) − Σ ΔH_f(reactants)
Entropy (reversible)	ΔS = q_rev/T
Gibbs energy	ΔG = ΔH − TΔS
Standard Gibbs–equilibrium	ΔG° = −RT ln K
Crossover temperature	T = ΔH/ΔS

This guide is prepared exclusively for the ASSEB Class 11 Chemistry syllabus. Practice the numerical problems, memorise the formula sheet and revise the glossary regularly to score well in the board examination. For more chapter-wise notes and ASSEB resources, keep visiting HSLC Guru.