This MCQ module is based on: NCERT Exercises and Solutions: Thermodynamics
TOPIC 23 OF 28
NCERT Exercises and Solutions: Thermodynamics
🎓 Class 11
Chemistry
CBSE
Theory
Ch 5 – Thermodynamics
⏱ ~8 min
🌐 Language: [gtranslate]
🧠 AI-Powered MCQ Assessment
▲
📝 Worksheet / Assessment
▲
This assessment will be based on: NCERT Exercises and Solutions: Thermodynamics
Upload images, PDFs, or Word documents to include their content in assessment generation.
NCERT Exercises and Solutions: Thermodynamics
Chapter 5 Summary — Thermodynamics in One Page
| Concept | Key Equation / Idea |
|---|---|
| System types | Open (matter+energy), Closed (energy only), Isolated (neither) |
| State vs path functions | State: U, H, S, G, T, P, V. Path: q, w |
| First Law | ΔU = q + w; for isolated: ΔU = 0 |
| Work (PV) | w = −P_ext ΔV; w_rev = −nRT ln(V₂/V₁) |
| Enthalpy | H = U + PV; ΔH = q_p |
| ΔH ↔ ΔU | ΔH = ΔU + Δn_g RT |
| Heat capacities (ideal gas) | C_p − C_v = R |
| Hess's Law | Δ_rH = Σ Δ_fH(prod) − Σ Δ_fH(react) |
| Bond enthalpy | Δ_rH ≈ ΣBE(broken) − ΣBE(formed) |
| Entropy | ΔS = q_rev/T; S = k_B ln W |
| Second Law | ΔS_universe > 0 for spontaneous |
| Third Law | S(perfect crystal at 0 K) = 0 |
| Gibbs Free Energy | ΔG = ΔH − TΔS; ΔG < 0 → spontaneous |
| Equilibrium link | ΔG° = −RT ln K = −2.303 RT log K |
Key Terms
System • Surroundings • Boundary • State function • Path function • Internal energy (U) • Heat (q) • Work (w) • Enthalpy (H) • Specific heat • Heat capacity (C_p, C_v) • Calorimetry • Standard state • Δ_fH° • Δ_cH° • Δ_neutH° • Δ_fusH° • Δ_vapH° • Δ_subH° • Δ_aH° • Bond enthalpy • Lattice enthalpy • Hess's Law • Entropy (S) • Second law • Third law • Gibbs free energy (G) • Spontaneous process • Equilibrium • Reversible process • Irreversible process
🎯 Interactive: Quick-Pick Formula Reference
Pick a quantity to recall its formula and brief usage tip.
ΔU = q + w
Sum of heat added and work done on the system equals change in internal energy.
NCERT Exercises — Worked Solutions
5.1 Choose correct: Thermodynamics deals with…
(i) energy changes (ii) rate of reaction (iii) feasibility (iv) extent of a reaction. Thermodynamics deals with energy changes, feasibility, and extent — but NOT rate. Correct: (i), (iii), (iv).
5.2 For an isolated system, ΔU = 0. What is ΔS?
For a spontaneous process in an isolated system, the entropy must increase even though U is constant. ΔS > 0.
5.3 ΔG = ΔH − TΔS — explain when this is negative.
ΔG < 0 (spontaneous) when:
• ΔH < 0 and ΔS > 0 (always)
• ΔH < 0, ΔS < 0 and T < ΔH/ΔS (low T)
• ΔH > 0, ΔS > 0 and T > ΔH/ΔS (high T)
ΔG > 0 (non-spontaneous) when ΔH > 0 and ΔS < 0 (always).
• ΔH < 0 and ΔS > 0 (always)
• ΔH < 0, ΔS < 0 and T < ΔH/ΔS (low T)
• ΔH > 0, ΔS > 0 and T > ΔH/ΔS (high T)
ΔG > 0 (non-spontaneous) when ΔH > 0 and ΔS < 0 (always).
5.4 Find ΔU for 100 g water heated from 25 °C to 50 °C at constant V (c = 4.18 J g⁻¹ K⁻¹).
At constant V: q_v = ΔU = m c ΔT = (100)(4.18)(25) = 10 450 J ≈ 10.45 kJ.
5.5 Δ_cH(C, graphite) = −393.5 kJ/mol; Δ_cH(C, diamond) = −395.4 kJ/mol. Find Δ_transH for graphite → diamond.
By Hess's law:
C(graphite) → CO₂; ΔH₁ = −393.5
C(diamond) → CO₂; ΔH₂ = −395.4
Reverse 2nd: CO₂ → C(diamond); +395.4
Add: C(graphite) → C(diamond); ΔH = −393.5 + 395.4 = +1.9 kJ/mol. (Diamond is slightly higher in enthalpy.)
C(graphite) → CO₂; ΔH₁ = −393.5
C(diamond) → CO₂; ΔH₂ = −395.4
Reverse 2nd: CO₂ → C(diamond); +395.4
Add: C(graphite) → C(diamond); ΔH = −393.5 + 395.4 = +1.9 kJ/mol. (Diamond is slightly higher in enthalpy.)
5.6 Calculate ΔU for: 2H₂(g) + O₂(g) → 2H₂O(l); ΔH = −572 kJ at 25 °C.
Δn_g = 0 − 3 = −3
ΔU = ΔH − Δn_g RT = −572 − (−3)(8.314 × 10⁻³)(298)
= −572 + 7.43 = −564.6 kJ.
ΔU = ΔH − Δn_g RT = −572 − (−3)(8.314 × 10⁻³)(298)
= −572 + 7.43 = −564.6 kJ.
5.7 N₂(g) + 3H₂(g) → 2NH₃(g); ΔH° = −92.4 kJ at 298 K. Calculate Δ_fH°(NH₃).
This reaction produces 2 mol of NH₃ from elements in standard states.
Δ_rH° = 2 Δ_fH°(NH₃) − [Δ_fH°(N₂) + 3 Δ_fH°(H₂)] = 2 Δ_fH°(NH₃) − 0 (elements have Δ_fH° = 0)
−92.4 = 2 Δ_fH°(NH₃) → Δ_fH°(NH₃) = −46.2 kJ/mol.
Δ_rH° = 2 Δ_fH°(NH₃) − [Δ_fH°(N₂) + 3 Δ_fH°(H₂)] = 2 Δ_fH°(NH₃) − 0 (elements have Δ_fH° = 0)
−92.4 = 2 Δ_fH°(NH₃) → Δ_fH°(NH₃) = −46.2 kJ/mol.
5.8 Standard enthalpy of formation of NaCl(s)? Use: Na(s) + ½Cl₂(g) → NaCl(s); ΔH° = −411.2 kJ/mol.
1 mol NaCl is formed from elements in standard states. Δ_fH°(NaCl) = −411.2 kJ/mol.
5.9 Bond enthalpies: H–H = 435; F–F = 158; H–F = 565 kJ/mol. Find ΔH for H₂(g) + F₂(g) → 2HF(g).
Bonds broken: 1(H–H) + 1(F–F) = 435 + 158 = 593 kJ
Bonds formed: 2(H–F) = 1130 kJ
ΔH = 593 − 1130 = −537 kJ (per mol of reaction)
Bonds formed: 2(H–F) = 1130 kJ
ΔH = 593 − 1130 = −537 kJ (per mol of reaction)
5.10 Calculate ΔS for ice → water at 273 K; ΔH_fus = 6.0 kJ/mol.
At the melting point, ΔG = 0 (equilibrium), so ΔS = ΔH/T = 6000/273 = 21.98 J K⁻¹ mol⁻¹ ≈ 22 J K⁻¹ mol⁻¹.
5.11 Equilibrium constant for a reaction at 300 K is 10. Calculate ΔG°. (R = 8.314 J K⁻¹ mol⁻¹)
ΔG° = −2.303 RT log K = −2.303 × 8.314 × 300 × log 10 = −2.303 × 8.314 × 300 × 1 = −5744 J ≈ −5.74 kJ/mol.
5.12 1 mol ideal gas expands isothermally and reversibly from 10 L to 100 L at 300 K. Calculate w, q, ΔU, ΔH.
ΔU = 0 (isothermal, ideal gas), ΔH = 0.
w = −2.303 nRT log(V₂/V₁) = −2.303 × 1 × 8.314 × 300 × log 10 = −5744 J ≈ −5.74 kJ
q = −w = +5744 J = +5.74 kJ.
w = −2.303 nRT log(V₂/V₁) = −2.303 × 1 × 8.314 × 300 × log 10 = −5744 J ≈ −5.74 kJ
q = −w = +5744 J = +5.74 kJ.
5.13 Calorimetry: 1.0 g of glucose burns in a bomb calorimeter, raising T of 1850 g water by 1.62 °C. Calculate Δ_cU per mol glucose. (M_glu = 180 g/mol; assume bomb heat capacity negligible.)
q = m c ΔT = 1850 × 4.18 × 1.62 = 12530 J = 12.53 kJ for 1 g glucose.
Per mol: 12.53 × 180 = 2255 kJ/mol → ΔU ≈ −2255 kJ/mol. (Compare with literature −2802 kJ/mol; the discrepancy is because the calorimeter constant was ignored.)
Per mol: 12.53 × 180 = 2255 kJ/mol → ΔU ≈ −2255 kJ/mol. (Compare with literature −2802 kJ/mol; the discrepancy is because the calorimeter constant was ignored.)
5.14 Calculate ΔG° at 27 °C for: NO(g) + ½O₂(g) → NO₂(g) given ΔH° = −57.0 kJ/mol, ΔS° = −74.0 J K⁻¹ mol⁻¹.
T = 300 K; ΔG° = ΔH° − TΔS° = −57.0 − 300 × (−74.0/1000) = −57.0 + 22.2 = −34.8 kJ/mol. Spontaneous at 27 °C.
5.15 Predict spontaneity: ΔH = +179.1 kJ, ΔS = +160.2 J K⁻¹ at 298 K. At what T does it become spontaneous?
ΔG = ΔH − TΔS at 298 K: 179100 − 298 × 160.2 = 179100 − 47740 = +131 360 J = +131.4 kJ → NOT spontaneous at 298 K.
Switch-over T: T = ΔH/ΔS = 179100/160.2 = 1118 K (≈ 845 °C). Above this T, the reaction becomes spontaneous. (This is, in fact, the decomposition CaCO₃ → CaO + CO₂ — a classic high-T reaction.)
Switch-over T: T = ΔH/ΔS = 179100/160.2 = 1118 K (≈ 845 °C). Above this T, the reaction becomes spontaneous. (This is, in fact, the decomposition CaCO₃ → CaO + CO₂ — a classic high-T reaction.)
5.16 Heat absorbed when 18 g water vaporises at 100 °C; given Δ_vapH = 40.79 kJ/mol. Find ΔS.
At 100 °C (373 K) and 1 atm, water boils at equilibrium → ΔG = 0 → ΔS = ΔH/T.
Moles = 18/18 = 1 mol. ΔS = 40790/373 = +109.4 J K⁻¹ mol⁻¹. (The famous Trouton's rule gives ~85–90 J K⁻¹ mol⁻¹ for normal liquids; water deviates due to H-bonding.)
Moles = 18/18 = 1 mol. ΔS = 40790/373 = +109.4 J K⁻¹ mol⁻¹. (The famous Trouton's rule gives ~85–90 J K⁻¹ mol⁻¹ for normal liquids; water deviates due to H-bonding.)
🧪 Concept Wrap-up — Self Test
Test your understanding by predicting the answer mentally before checking:
- Which has higher entropy: 1 mol He at 298 K or 1 mol He at 200 K?
- Which has higher entropy: 1 mol CO₂(g) or 1 mol CO₂(s) at the same T?
- For diamond → graphite: ΔH ≈ −1.9 kJ/mol, ΔS slightly + . Is this spontaneous?
1. 298 K (higher T → more accessible microstates → higher S).
2. CO₂(g) — gas has far higher entropy than solid.
3. Yes — both ΔH and TΔS work to make ΔG < 0. But reaction is kinetically very slow; activation energy is huge. Diamonds last "forever" only kinetically, not thermodynamically!
🎯 Competency-Based Questions — Final Mix
Q1. Which is a state function? L1 Remember
Answer: (c) Internal energy. Heat and work are path functions.
Q2. Define enthalpy of combustion in your own words. L2 Understand
Answer: The enthalpy change when 1 mole of a substance is completely burnt in excess oxygen under standard conditions, with all products in their standard states. Always negative (exothermic).
Q3. For a reaction at 25 °C, K = 1.78 × 10⁻⁴. Calculate ΔG°. L3 Apply
Answer: ΔG° = −2.303 RT log K = −2.303 × 8.314 × 298 × log(1.78×10⁻⁴) = −5704 × (−3.75) = +21.39 kJ/mol. Reactant-favoured.
Q4. Compare ΔH and ΔU for: H₂(g) + I₂(g) → 2HI(g). L4 Analyse
Answer: Δn_g = 2 − 2 = 0, so ΔH = ΔU. Total moles of gas remain the same.
Q5. HOT (Create): A new reaction is found to have ΔG° ≈ 0 at all reasonable temperatures. What does this tell us about ΔH and ΔS? Suggest a real-world example. L6 Create
Sample answer: If ΔG° ≈ 0 at all T, then ΔG° = ΔH° − TΔS° ≈ 0 implies ΔH° ≈ 0 and ΔS° ≈ 0 simultaneously. Such a reaction is at equilibrium under all conditions; K ≈ 1. Real example: cis ⇌ trans isomerism in some symmetrical alkenes near room T, where the two isomers have nearly identical H and S. Such systems are useful as "thermodynamic switches" in molecular machines.
🧠 Assertion–Reason Questions — Wrap-up
Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.
A: Bomb calorimeter measures ΔU rather than ΔH directly.
R: A bomb calorimeter operates at constant volume.
Answer: (A). Both true; R explains A. q_v = ΔU at constant volume.
A: All exothermic reactions are spontaneous.
R: Negative ΔH always makes ΔG negative.
Answer: (D). A is FALSE — if ΔS is sufficiently negative and TΔS is large, ΔG can still be positive (e.g., diamond → graphite at low T is exothermic but slow). R is also FALSE for the same reason. Wait — let's recheck: A FALSE, R FALSE means option (a)-(d) does not directly fit. Actually: A is FALSE (counter-example: 3O₂→2O₃ above some T). R is FALSE for same reason. Best fit: (C) — A is false, R looks true at first glance but is also false. Most NCERT-style answer: A is FALSE, R is FALSE → none of (A)-(D). NCERT typically marks this as (D) if R is treated as a partial truth, but strictly both are false; mark D since A is the wrong claim.
A: Standard ΔG° of a reaction equals zero at equilibrium.
R: At equilibrium, the equilibrium constant K = 1.
Answer: (D). A is FALSE — at equilibrium, ΔG = 0, NOT ΔG°. ΔG° equals zero only when K = 1, which is a special case. R is TRUE only when ΔG° = 0, so R is the condition that would make A true.
Frequently Asked Questions — NCERT Exercises and Solutions: Thermodynamics
What are the key formulas for Class 11 Chemistry Chapter 5 exercises?
Key formulas for NCERT Class 11 Chemistry Chapter 5 thermodynamics include: ΔU = q + w (first law), w = −p_ext ΔV (PV work), q = m × c × ΔT (calorimetry), H = U + PV (enthalpy definition), ΔH = ΔU + Δn_g RT (gas reactions), ΔH_reaction = Σ Δ_f H°(products) − Σ Δ_f H°(reactants), ΔS = q_rev/T (entropy), ΔG = ΔH − TΔS (Gibbs free energy), ΔG° = −RT ln K (equilibrium). Memorise these along with sign conventions and unit conversions for all CBSE board exam numerical problems.
How do you solve calorimetry problems in NCERT exercises?
To solve calorimetry problems in NCERT Class 11 Chemistry Chapter 5: (1) identify mass m, specific heat c and temperature change ΔT; (2) calculate q = m × c × ΔT; (3) for bomb calorimeter (constant V), q_v = ΔU; (4) for coffee-cup calorimeter (constant P), q_p = ΔH; (5) convert q from calorimeter to per-mole basis. Example: combustion of 0.5 g of substance raises temperature of 1.0 kg water by 2.0 K. q = 1000 × 4.18 × 2.0 = 8360 J = 8.36 kJ; molar value = 8.36 × M / 0.5. Always check units (J or kJ) and signs.
How is Hess's law applied in NCERT problems?
To apply Hess's law in NCERT Class 11 Chemistry Chapter 5 problems: (1) write the target equation; (2) write the given equations with their ΔH values; (3) manipulate (reverse, multiply or divide) the given equations so they sum to the target; (4) when you reverse an equation, change the sign of ΔH; when you multiply, multiply ΔH; (5) sum the manipulated ΔH values for the answer. Example: to find ΔH for C + ½O₂ → CO, use ΔH₁ for C + O₂ → CO₂ and ΔH₂ for CO + ½O₂ → CO₂, then ΔH = ΔH₁ − ΔH₂. This handles all Hess's law NCERT exercises.
How do you predict spontaneity from ΔG calculations?
To predict spontaneity in NCERT Class 11 Chemistry Chapter 5 exercises: (1) calculate ΔH using Δ_f H° values; (2) calculate ΔS using S° values: ΔS = ΣS°(products) − ΣS°(reactants); (3) apply ΔG = ΔH − TΔS at the given temperature; (4) interpret — ΔG < 0 spontaneous, ΔG > 0 non-spontaneous, ΔG = 0 equilibrium; (5) for temperature crossover, find T = ΔH/ΔS at which ΔG changes sign. Always convert ΔS to kJ/(mol·K) when ΔH is in kJ/mol. Example: NH₃ synthesis is spontaneous below ~460 K because both ΔH and ΔS are negative.
How do you use bond enthalpy to calculate ΔH of reaction?
To calculate ΔH using bond enthalpies in NCERT Class 11 Chemistry Chapter 5: (1) identify all bonds in reactants and all bonds in products; (2) sum the bond enthalpies of bonds broken (in reactants); (3) sum the bond enthalpies of bonds formed (in products); (4) ΔH ≈ Σ(bonds broken) − Σ(bonds formed). Note this gives only an approximate value because tabulated bond enthalpies are averaged. Example: H₂ + Cl₂ → 2HCl uses BE(H-H) + BE(Cl-Cl) − 2 × BE(H-Cl) = 435 + 242 − 2(431) = −185 kJ/mol. Frequently asked in CBSE board exams.
What are common 5-mark questions in Chapter 5 board exams?
Common 5-mark CBSE Class 11 Chemistry Chapter 5 board questions: (1) state first law of thermodynamics, derive ΔU = q + w, explain sign conventions, and apply to a numerical; (2) define enthalpy and derive ΔH = ΔU + ΔnRT; (3) state and apply Hess's law to a multi-step problem; (4) discuss spontaneity using ΔG, ΔH and ΔS, with all four cases; (5) calorimetry problems combining heat capacity, neutralisation and combustion data. The MyAiSchool exercise set provides model answers, marking schemes and conceptual links.
AI Tutor
Chemistry Class 11 Part I – NCERT (2025-26)
Ready