Chapter 5: Thermodynamics Long Questions | chemistry Class 11 CBSE

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Chapter 5: Thermodynamics Long Questions | chemistry Class 11 CBSE | ByteNotes.in

Long Answer

Hard difficulty • Structured explanation

Hard

Question 1

Long Form

Analyse the concept of internal energy as a state function using Joule's experiments, and derive the mathematical statement of the first law of thermodynamics for a general process involving both heat and work.

Joule performed experiments (1840-50) showing that a given amount of work done on an adiabatic system produced the same temperature change regardless of the type of work (mechanical paddle-wheel or electrical immersion rod), proving ΔU is path-independent.
Internal energy U is defined such that ΔU = U2 - U1 = wad for adiabatic processes; this means U is characteristic of the state, making it a state function like T, p, and V.
For a non-adiabatic process, internal energy can change via heat transfer: at constant volume, ΔU = qV (since w = 0 when ΔV = 0), establishing the link between heat and internal energy.
For the general case where both heat transfer and work occur: ΔU = q + w (Equation 5.1), the mathematical statement of the first law; while q and w individually depend on the path, their sum ΔU depends only on the initial and final states.
IUPAC sign convention: q is positive when absorbed by the system (energy input); w is positive when work is done on the system (compression). For an isolated system (q = 0, w = 0), ΔU = 0, confirming energy conservation.
Unlike volume (an absolute value can be assigned), only changes in internal energy ΔU can be measured experimentally, not the absolute value of U itself.

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Long Answer

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Hard

Question 2

Long Form

Derive the expression for work done during reversible isothermal expansion of an ideal gas, and compare it with irreversible expansion against a constant external pressure.

For work done on a gas: w = -pex ΔV = -pex(Vf - Vi); for compression pex > pin, giving positive w (work done on system); for expansion pex < pin, giving negative w (work done by system).
For reversible isothermal expansion, pex = pin = p at every stage (infinitesimal changes), so w_rev = -∫pin dV from Vi to Vf; using ideal gas law p = nRT/V, this gives w_rev = -nRT ln(Vf/Vi) = -2.303 nRT log(Vf/Vi).
For irreversible isothermal expansion against constant external pressure pex: w_irrev = -pex(Vf - Vi); this is less work (in magnitude) than the reversible case because pex < pin throughout.
Reversible work represents the maximum work obtainable from a process; on a p-V diagram, it equals the area under the curve (larger area), while irreversible work equals a smaller rectangular area.
For isothermal free expansion (pex = 0): w = 0 and q = 0 (Joule's finding for ideal gas), so ΔU = 0; this applies both to reversible and irreversible free expansion of ideal gases.
For isothermal processes on ideal gas: ΔU = 0, therefore q = -w; heat absorbed equals work done by the system, showing that at constant temperature all heat input converts to work.

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Long Answer

Hard difficulty • Structured explanation

Hard

Question 3

Long Form

Explain enthalpy as a useful state function, derive the relationship ΔH = ΔU + ΔngRT, and discuss when the difference between ΔH and ΔU is significant.

Most reactions occur at constant atmospheric pressure, not constant volume. At constant pressure: ΔU = qp - pΔV, so qp = ΔU + pΔV = (U2 + pV2) - (U1 + pV1); defining H = U + pV gives qp = ΔH, making enthalpy the natural heat function at constant pressure.
Enthalpy H is a state function (depends on U, p, V — all state functions), so ΔH is independent of path, making qp also path-independent at constant pressure.
For reactions involving gases at constant p and T: pVA = nART (reactant gases) and pVB = nBRT (product gases), so p(VB - VA) = (nB - nA)RT = ΔngRT; substituting into ΔH = ΔU + pΔV gives ΔH = ΔU + ΔngRT.
When Δng = 0 (no change in gaseous moles), ΔH = ΔU. When Δng = +1 (one more mole of gas in products), ΔH > ΔU by RT. The difference amounts to ~2.5 kJ mol-1 per mole of Δng at 298 K.
For reactions with only solids and liquids, volume changes are negligible, so ΔH ≈ ΔU. The difference becomes significant only when gases are involved.
ΔH is negative for exothermic reactions (heat evolved, products at lower enthalpy) and positive for endothermic reactions (heat absorbed); this is directly measurable in constant-pressure calorimetry.

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Long Answer

Medium difficulty • Structured explanation

Medium

Question 4

Long Form

Describe the bomb calorimeter in detail, explain how ΔU is measured, and outline how ΔH is obtained from ΔU for the combustion of 1g of graphite at 298 K.

A bomb calorimeter consists of a sealed steel vessel (the bomb) containing a sample and pure dioxygen, immersed in a known volume of water in an outer jacket; the entire assembly is the calorimeter.
The steel vessel is sealed and immersed in a water bath to ensure no heat is lost to the surroundings. A combustible substance is burned, and the temperature rise of the surrounding water is monitored.
Since the bomb is sealed, its volume does not change (ΔV = 0); thus no work is done (w = 0), and by the first law ΔU = qV, measured by ΔT × heat capacity of the calorimeter.
For 1g of graphite burned (C + O2 → CO2): the heat released = -CV × ΔT = -20.7 kJ/K × 1 K = -20.7 kJ; for 1 mol (12g) of graphite, ΔU = -20.7 × 12 = -248 kJ mol-1.
Since Δng = 0 for this reaction (1 mol gas reactant → 1 mol gas product), ΔH = ΔU + ΔngRT = ΔU + 0 = -248 kJ mol-1.
The negative sign confirms the exothermic nature; bomb calorimetry provides precise ΔU values which are then converted to ΔH using the ΔngRT correction when gases are involved.

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Long Answer

Hard difficulty • Structured explanation

Hard

Question 5

Long Form

Discuss Hess's law with a suitable example. How is it applied to calculate the standard enthalpy of formation of benzene from combustion data?

Hess's law: the standard enthalpy of a reaction is the same whether it occurs in one step or multiple steps, since enthalpy is a state function independent of path. Mathematically, ΔrH = ΔrH1 + ΔrH2 + ΔrH3 ...
Application to benzene: the direct formation reaction 6C(graphite) + 3H2(g) → C6H6(l) cannot be measured directly; instead, combustion data is used.
Step 1: 6C(graphite) + 6O2(g) → 6CO2(g); ΔfH° = 6 × (-393.5) = -2361 kJ mol-1. Step 2: 3H2(g) + 3/2 O2(g) → 3H2O(l); ΔfH° = 3 × (-285.83) = -857.49 kJ mol-1.
Adding steps 1 and 2: 6C + 3H2 + 15/2 O2 → 6CO2 + 3H2O; ΔH = -3218.49 kJ mol-1.
Reversing combustion of benzene: 6CO2 + 3H2O → C6H6(l) + 15/2 O2; ΔH = +3267 kJ mol-1.
Adding to get formation of benzene: ΔfH°(C6H6) = -3218.49 + 3267 = +48.51 kJ mol-1, confirming benzene is slightly endothermic to form.

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Long Answer

Hard difficulty • Structured explanation

Hard

Question 6

Long Form

Explain the Born-Haber cycle for NaCl, listing all the steps involved, and use Hess's law to calculate the lattice enthalpy.

The Born-Haber cycle constructs an indirect enthalpy path to find lattice enthalpy, since it cannot be measured directly. The cycle for NaCl involves five steps whose enthalpies sum to zero (Hess's law).
Step 1: Sublimation of sodium — Na(s) → Na(g); ΔsubH° = +108.4 kJ mol-1. Step 2: Ionization of sodium — Na(g) → Na+(g) + e-; ΔiH° = +496 kJ mol-1.
Step 3: Dissociation of chlorine — 1/2 Cl2(g) → Cl(g); ΔH = 1/2 × 242 = +121 kJ mol-1. Step 4: Electron gain by chlorine — Cl(g) + e- → Cl-(g); ΔegH° = -348.6 kJ mol-1.
Step 5: Lattice formation — Na+(g) + Cl-(g) → NaCl(s); ΔlatticeH° = ? (this is what we calculate).
The overall reaction Na(s) + 1/2 Cl2(g) → NaCl(s) has ΔfH° = -411.2 kJ mol-1. By Hess's law: ΔfH° = ΔsubH° + ΔiH° + 1/2 ΔbondH° + ΔegH° + ΔlatticeH°.
ΔlatticeH° = -411.2 - 108.4 - 121 - 496 + 348.6 = -788 kJ mol-1; equivalently, for dissociation of NaCl(s) → Na+(g) + Cl-(g), ΔlatticeH° = +788 kJ mol-1.

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Long Answer

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Hard

Question 7

Long Form

Analyse entropy as a thermodynamic state function. Using the concept of entropy, explain why spontaneity cannot be decided by enthalpy change alone, giving examples of both exothermic and endothermic spontaneous reactions.

Entropy (S) is a state function measuring the degree of disorder or randomness. For a reversible process, ΔS = qrev/T; since qrev/T is independent of path, ΔS is a state function. Greater disorder means higher entropy; gaseous state has highest entropy, crystalline solid has lowest.
Enthalpy alone fails as a spontaneity criterion: while many exothermic reactions are spontaneous (NH3 formation: ΔrH° = -46.1 kJ mol-1; HCl formation: ΔrH° = -92.32 kJ mol-1), some endothermic reactions are also spontaneous.
Endothermic spontaneous examples: formation of NO2 (ΔrH° = +33.2 kJ mol-1) and CS2 from graphite and sulfur (ΔrH° = +128.5 kJ mol-1). These occur because entropy increases significantly, making ΔStotal > 0.
For diffusion of two ideal gases into each other (ΔH = 0), the process is spontaneous due to a large increase in entropy (increased disorder when gases mix), demonstrating that entropy is the true driving force.
For a spontaneous process in an isolated system: ΔStotal = ΔSsystem + ΔSsurroundings > 0. For exothermic reactions, ΔSsurr = -ΔrH/T is large and positive, ensuring spontaneity even when ΔSsystem is negative.
Neither enthalpy nor entropy alone determines spontaneity; both must be considered together — which is precisely why Gibbs energy G = H - TS was introduced to provide a unified criterion.

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Long Answer

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Hard

Question 8

Long Form

Derive the Gibbs equation ΔG = ΔH - TΔS from the concept of total entropy change, and analyse all possible combinations of signs of ΔH and ΔS to predict the spontaneity of reactions at different temperatures.

Starting from: ΔStotal = ΔSsys + ΔSsurr. At constant pressure, ΔSsurr = -ΔHsys/T (since heat lost by system = heat gained by surroundings). So: T × ΔStotal = TΔSsys - ΔHsys.
For spontaneous process ΔStotal > 0, so TΔSsys - ΔHsys > 0, meaning -(ΔHsys - TΔSsys) > 0, or ΔG = ΔH - TΔS < 0. Gibbs energy G = H - TS is thus defined such that ΔG < 0 for spontaneous processes.
Case 1 (ΔH < 0, ΔS > 0): ΔG = negative - T×positive = always negative → spontaneous at all temperatures.
Case 2 (ΔH < 0, ΔS < 0): ΔG = negative - T×negative; ΔG < 0 only at low T where |ΔH| > T|ΔS| → spontaneous at low temperature only.
Case 3 (ΔH > 0, ΔS > 0): ΔG = positive - T×positive; ΔG < 0 only at high T where TΔS > ΔH → spontaneous at high temperature only.
Case 4 (ΔH > 0, ΔS < 0): ΔG = positive - T×negative = always positive → non-spontaneous at all temperatures. These four cases form Table 5.4 in the chapter.

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Long Answer

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Hard

Question 9

Long Form

Explain the relationship between standard Gibbs energy change and the equilibrium constant, and discuss how temperature affects the value of K for endothermic and exothermic reactions.

At equilibrium, the free energy of the system is minimum and ΔrG = 0. The standard Gibbs energy is related to K by: ΔrG° = -RT ln K = -2.303 RT log K, derived from the condition that at equilibrium the sum of free energies of products equals that of reactants.
Also, ΔrG° = ΔrH° - TΔrS°; combining: -RT ln K = ΔrH° - TΔrS°, or ln K = -ΔrH°/RT + ΔrS°/R. This shows K depends exponentially on ΔrH° and linearly on ΔrS°.
For exothermic reactions (ΔrH° large and negative): ΔrG° is large and negative, making K >> 1. These reactions go nearly to completion and are favored.
For endothermic reactions (ΔrH° large and positive): K << 1 at low temperatures, products are not favored. However, increasing T increases K if ΔrS° > 0 (since ΔrG° = ΔrH° - TΔrS° becomes less positive or negative at high T).
Practical use: K can be estimated from ΔrH° and ΔrS° data without direct measurement, allowing prediction of product yields at any temperature.
If K is measured in the laboratory, ΔrG° at any temperature can be back-calculated using ΔrG° = -RT ln K, enabling thermodynamic characterization of the reaction.

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Long Answer

Medium difficulty • Structured explanation

Medium

Question 10

Long Form

Compare and contrast standard enthalpy of fusion, vaporization, and sublimation. Explain why the magnitude of these quantities depends on intermolecular interactions, using water and acetone as examples.

Standard enthalpy of fusion (ΔfusH°): heat absorbed when 1 mol of a solid melts at its melting point and standard pressure. Always positive (endothermic). Example: H2O(s) → H2O(l); ΔfusH° = 6.00 kJ mol-1.
Standard enthalpy of vaporization (ΔvapH°): heat absorbed when 1 mol of liquid vaporizes at its boiling point and standard pressure. Always positive. Example: H2O(l) → H2O(g); ΔvapH° = 40.79 kJ mol-1.
Standard enthalpy of sublimation (ΔsubH°): heat absorbed when 1 mol of solid converts directly to vapour at standard pressure. Always positive. Example: CO2(s) → CO2(g); ΔsubH° = 25.2 kJ mol-1. By Hess's law: ΔsubH° = ΔfusH° + ΔvapH°.
The magnitude depends on strength of intermolecular interactions: stronger interactions mean more energy needed to separate molecules. Water has strong hydrogen bonds, requiring 40.79 kJ mol-1 to vaporize.
Acetone has weaker dipole-dipole interactions, so ΔvapH°(acetone) = 29.1 kJ mol-1, much less than water. This difference reflects the greater energy needed to disrupt water's hydrogen bond network.
Among the three processes, vaporization requires the most energy (largest separation of molecules from liquid phase to gas), fusion requires the least (solid to liquid, partial disruption), and sublimation = fusion + vaporization combined.

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Chapter 5: Thermodynamics | Long Questions

Analyse the concept of internal energy as a state function using Joule's experiments, and derive the mathematical statement of the first law of thermodynamics for a general process involving both heat and work.

Derive the expression for work done during reversible isothermal expansion of an ideal gas, and compare it with irreversible expansion against a constant external pressure.

Explain enthalpy as a useful state function, derive the relationship ΔH = ΔU + ΔngRT, and discuss when the difference between ΔH and ΔU is significant.

Describe the bomb calorimeter in detail, explain how ΔU is measured, and outline how ΔH is obtained from ΔU for the combustion of 1g of graphite at 298 K.

Discuss Hess's law with a suitable example. How is it applied to calculate the standard enthalpy of formation of benzene from combustion data?

Explain the Born-Haber cycle for NaCl, listing all the steps involved, and use Hess's law to calculate the lattice enthalpy.

Analyse entropy as a thermodynamic state function. Using the concept of entropy, explain why spontaneity cannot be decided by enthalpy change alone, giving examples of both exothermic and endothermic spontaneous reactions.

Derive the Gibbs equation ΔG = ΔH - TΔS from the concept of total entropy change, and analyse all possible combinations of signs of ΔH and ΔS to predict the spontaneity of reactions at different temperatures.

Explain the relationship between standard Gibbs energy change and the equilibrium constant, and discuss how temperature affects the value of K for endothermic and exothermic reactions.

Compare and contrast standard enthalpy of fusion, vaporization, and sublimation. Explain why the magnitude of these quantities depends on intermolecular interactions, using water and acetone as examples.

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