What is the difference between delta H and delta U?

delta U is the change in internal energy (heat at constant volume); delta H is the change in enthalpy (heat at constant pressure). They are related by delta H = delta U + delta ng RT, where delta ng is the change in the number of moles of gas. They are equal when delta ng = 0.

NCERT Solutions for Class 11 Chemistry Chapter 5: Thermodynamics

Q: What is Class 11 Chemistry Chapter 5 Thermodynamics about?

Chapter 5 studies energy changes in chemical and physical processes: the first law (delta U = q + w), enthalpy and its measurement, Hess's law, standard enthalpies of formation and combustion, entropy, and Gibbs energy with the spontaneity criterion delta G < 0 and the relation delta rG = -RT ln K.

Q: How many exercise questions are there in Class 11 Chemistry Chapter 5?

The NCERT Exercises for Unit 5 contain 22 questions (5.1 to 5.22). Questions 5.1 to 5.6 are objective/MCQ-type and 5.7 to 5.22 are numerical and reasoning questions, all reproduced verbatim and solved step by step on this page.

These Class 11 Chemistry Chapter 5 solutions cover Thermodynamics with every NCERT exercise (5.1–5.22) reproduced verbatim and solved step by step — the six MCQ-type questions and all numericals on the first law, enthalpy, Hess’s law, entropy and Gibbs energy, with units verified. Answers are written in CBSE exam-ready style and updated for the 2026–27 session.

Class: 11 Subject: Chemistry Chapter: 5 Topic: Thermodynamics Exercises: 5.1–5.22 Session: 2026–27

On this page

Chapter overview
Key concepts & definitions
Important formulas
NCERT Exercises (5.1–5.22) — solutions
Extra practice questions
MCQs & Assertion–Reason
Common mistakes
Exam tips
FAQs

Chapter Overview

Chapter 5, Thermodynamics, studies the energy changes that accompany chemical and physical processes. We divide the universe into a system and its surroundings, and classify systems as open, closed or isolated. The first law of thermodynamics (ΔU = q + w) expresses conservation of energy. Because most reactions happen at constant pressure, we define the state function enthalpy (H = U + pV), and learn to compute reaction enthalpies using standard enthalpies of formation, bond enthalpies and Hess’s law. The chapter then asks which way a process goes: it introduces entropy (S, a measure of disorder) and Gibbs energy (G = H − TS), giving the spontaneity criterion ΔG < 0 and the link ΔrG° = −RT ln K. These ideas are the foundation for chemical equilibrium, electrochemistry and the rest of physical chemistry.

Key Concepts & Definitions

System and surroundings: the system is the part of the universe under study; everything else is the surroundings. An open system exchanges both matter and energy, a closed system only energy, and an isolated system neither.

State function: a property (such as U, H, S, G, p, V, T) whose value depends only on the state of the system, not on the path taken. Heat (q) and work (w) are not state functions.

Internal energy (U): the total energy stored in a system. Only changes ΔU can be measured, not absolute values.

Enthalpy (H): H = U + pV. At constant pressure ΔH = q_p (heat of reaction). ΔH is negative for exothermic and positive for endothermic reactions.

Heat capacity (C): q = CΔT. For an ideal gas C_p − C_V = R.

Entropy (S): a measure of randomness/disorder. ΔS = q_rev/T. For a spontaneous change the total entropy (system + surroundings) increases.

Gibbs energy (G): G = H − TS. ΔrG = ΔrH − TΔrS. A process is spontaneous when ΔG < 0, at equilibrium when ΔG = 0.

Hess’s law: the enthalpy change of a reaction is the same whether it occurs in one step or several, because H is a state function.

Important Formulas

First law: ΔU = q + w (w = −p_exΔV for expansion work)

Isothermal reversible work (ideal gas): w_rev = −2.303 nRT log(V_f/V_i)

Enthalpy: ΔH = ΔU + Δn_gRT (Δn_g = moles of gaseous products − reactants)

Heat: q = CΔT = mcΔT; C_p − C_V = R (ideal gas)

Reaction enthalpy: ΔrH° = Σ a_iΔ_fH°(products) − Σ b_iΔ_fH°(reactants)

Bond enthalpy route: ΔrH° = Σ bond enthalpies(reactants) − Σ bond enthalpies(products)

Entropy: ΔS = q_rev/T; ΔS_surr = −ΔH_sys/T (constant T, p)

Gibbs energy: ΔrG = ΔrH − TΔrS; ΔrG° = −RT ln K = −2.303 RT log K

NCERT Exercises (5.1–5.22) — Solutions

Questions reproduced verbatim from the NCERT textbook (Unit 5); every numerical is solved step by step with units and the final answer cross-checked against the NCERT answer key.

5.1 Choose the correct answer. A thermodynamic state function is a quantity (i) used to determine heat changes (ii) whose value is independent of path (iii) used to determine pressure volume work (iv) whose value depends on temperature only.

ANSWER (ii) whose value is independent of path. A state function (U, H, S, G, p, V, T) depends only on the initial and final states, not on the route. Heat and work depend on path, so options (i) and (iii) describe path functions.

5.2 For the process to occur under adiabatic conditions, the correct condition is: (i) ΔT = 0 (ii) Δp = 0 (iii) q = 0 (iv) w = 0

ANSWER (iii) q = 0. An adiabatic process is, by definition, one in which there is no transfer of heat between the system and the surroundings, so q = 0.

5.3 The enthalpies of all elements in their standard states are: (i) unity (ii) zero (iii) <0 (iv) different for each element

ANSWER (ii) zero. By convention, the standard enthalpy of formation of an element in its most stable reference state (e.g. O₂ gas, C_graphite) is taken as zero.

5.4 ΔU° of combustion of methane is − X kJ mol⁻¹. The value of ΔH° is (i) = ΔU° (ii) > ΔU° (iii) < ΔU° (iv) = 0

ANSWER (iii) < ΔU°. For CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), Δn_g = 1 − 3 = −2. Since ΔH = ΔU + Δn_gRT, the term Δn_gRT is negative, so ΔH is more negative (smaller) than ΔU.

5.5 The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, −890.3 kJ mol⁻¹, −393.5 kJ mol⁻¹, and −285.8 kJ mol⁻¹ respectively. Enthalpy of formation of CH₄(g) will be (i) −74.8 kJ mol⁻¹ (ii) −52.27 kJ mol⁻¹ (iii) +74.8 kJ mol⁻¹ (iv) +52.26 kJ mol⁻¹

ANSWER (i) −74.8 kJ mol⁻¹. Target: C(s) + 2H₂(g) → CH₄(g). Using Hess’s law, Δ_fH = Δ_cH(C) + 2Δ_cH(H₂) − Δ_cH(CH₄). = (−393.5) + 2(−285.8) − (−890.3) = −393.5 − 571.6 + 890.3 = −74.8 kJ mol⁻¹.

5.6 A reaction, A + B → C + D + q is found to have a positive entropy change. The reaction will be (i) possible at high temperature (ii) possible only at low temperature (iii) not possible at any temperature (iv) possible at any temperature

ANSWER (iv) possible at any temperature. The reaction releases heat (+q), so it is exothermic and ΔH is negative; it also has ΔS positive. Then ΔG = ΔH − TΔS is negative at all temperatures, so the reaction is spontaneous at any temperature.

5.7 In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?

ANSWER Heat absorbed by the system: q = +701 J. Work done by the system, so w is negative: w = −394 J. ΔU = q + w = 701 + (−394) = +307 J.

5.8 The reaction of cyanamide, NH₂CN (s), with dioxygen was carried out in a bomb calorimeter, and ΔU was found to be −742.7 kJ mol⁻¹ at 298 K. Calculate enthalpy change for the reaction at 298 K. NH₂CN(s) + (3/2) O₂(g) → N₂(g) + CO₂(g) + H₂O(l)

ANSWER Gaseous moles: products = 2 (N₂ + CO₂); reactants = 3/2 (O₂ only; NH₂CN is solid, H₂O is liquid). Δn_g = 2 − 3/2 = +1/2 = 0.5 mol. ΔH = ΔU + Δn_gRT = −742.7 kJ + (0.5)(8.314×10⁻³ kJ K⁻¹ mol⁻¹)(298 K) = −742.7 + 1.239 = −741.5 kJ mol⁻¹ (NCERT key: −741.5 kJ).

5.9 Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol⁻¹ K⁻¹.

ANSWER Moles of Al, n = 60.0 g ÷ 27 g mol⁻¹ = 2.222 mol. ΔT = 55 − 35 = 20 K (= 20 °C rise). q = n C_m ΔT = (2.222)(24)(20) = 1066.7 J ≈ 1.067 kJ.

5.10 Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at −10.0°C. Δ_fusH = 6.03 kJ mol⁻¹ at 0°C. C_p[H₂O(l)] = 75.3 J mol⁻¹ K⁻¹; C_p[H₂O(s)] = 36.8 J mol⁻¹ K⁻¹.

ANSWER The change is done in three steps for 1 mol: Step 1 — cool water 10°C → 0°C: ΔH₁ = C_p(l)ΔT = 75.3 × (0 − 10) = −753 J = −0.753 kJ. Step 2 — freeze water at 0°C: ΔH₂ = −Δ_fusH = −6.03 kJ (freezing releases heat). Step 3 — cool ice 0°C → −10°C: ΔH₃ = C_p(s)ΔT = 36.8 × (−10 − 0) = −368 J = −0.368 kJ. Total ΔH = −0.753 − 6.03 − 0.368 = −7.151 kJ mol⁻¹.

5.11 Enthalpy of combustion of carbon to CO₂ is −393.5 kJ mol⁻¹. Calculate the heat released upon formation of 35.2 g of CO₂ from carbon and dioxygen gas.

ANSWER C(s) + O₂(g) → CO₂(g); ΔH = −393.5 kJ per mole of CO₂. Moles of CO₂ = 35.2 g ÷ 44 g mol⁻¹ = 0.80 mol. Heat released = 0.80 × (−393.5) = −314.8 kJ (i.e. 314.8 kJ of heat is evolved).

5.12 Enthalpies of formation of CO(g), CO₂(g), N₂O(g) and N₂O₄(g) are −110, −393, 81 and 9.7 kJ mol⁻¹ respectively. Find the value of ΔrH for the reaction: N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g)

ANSWER ΔrH = [Δ_fH(N₂O) + 3Δ_fH(CO₂)] − [Δ_fH(N₂O₄) + 3Δ_fH(CO)]. = [81 + 3(−393)] − [9.7 + 3(−110)] = [81 − 1179] − [9.7 − 330] = (−1098) − (−320.3) = −777.7 kJ ≈ −778 kJ.

5.13 Given N₂(g) + 3H₂(g) → 2NH₃(g); ΔrH° = −92.4 kJ mol⁻¹. What is the standard enthalpy of formation of NH₃ gas?

ANSWER Δ_fH is the enthalpy change to form one mole of NH₃. The given reaction forms 2 mol, so divide by 2: Δ_fH(NH₃) = (−92.4) ÷ 2 = −46.2 kJ mol⁻¹.

5.14 Calculate the standard enthalpy of formation of CH₃OH(l) from the following data: CH₃OH(l) + (3/2) O₂(g) → CO₂(g) + 2H₂O(l); ΔrH° = −726 kJ mol⁻¹ C(graphite) + O₂(g) → CO₂(g); Δ_cH° = −393 kJ mol⁻¹ H₂(g) + (1/2) O₂(g) → H₂O(l); Δ_fH° = −286 kJ mol⁻¹

ANSWER Target: C(graphite) + 2H₂(g) + (1/2)O₂(g) → CH₃OH(l). By Hess’s law: Δ_fH(CH₃OH) = Δ_cH(C) + 2Δ_fH(H₂O) − ΔrH(combustion of CH₃OH) = (−393) + 2(−286) − (−726) = −393 − 572 + 726 = −239 kJ mol⁻¹.

5.15 Calculate the enthalpy change for the process CCl₄(g) → C(g) + 4 Cl(g) and calculate bond enthalpy of C–Cl in CCl₄(g). Δ_vapH°(CCl₄) = 30.5 kJ mol⁻¹; Δ_fH°(CCl₄) = −135.5 kJ mol⁻¹; Δ_aH°(C) = 715.0 kJ mol⁻¹ (enthalpy of atomisation); Δ_aH°(Cl₂) = 242 kJ mol⁻¹.

ANSWER Build the gas-phase atomisation of CCl₄ using a Hess cycle from its formation: ΔH = −Δ_fH(CCl₄) + Δ_aH(C) + 2Δ_aH(Cl₂) − Δ_vapH(CCl₄) [Form CCl₄(l) reversed: +135.5; atomise C: +715.0; atomise 2 mol Cl₂ → 4 Cl: 2×242 = +484; vaporise CCl₄(l)→(g): subtract 30.5 because we start from gas.] ΔH = 135.5 + 715.0 + 484 − 30.5 = 1304 kJ mol⁻¹ for CCl₄(g) → C(g) + 4Cl(g). Bond enthalpy of one C–Cl bond = 1304 ÷ 4 = 326 kJ mol⁻¹.

5.16 For an isolated system, ΔU = 0, what will be ΔS?

ANSWER For an isolated system every spontaneous (natural) change must increase the total disorder, so ΔS > 0 (entropy increases / is positive). It cannot decrease for a spontaneous process.

5.17 For the reaction at 298 K, 2A + B → C, ΔH = 400 kJ mol⁻¹ and ΔS = 0.2 kJ K⁻¹ mol⁻¹. At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range.

ANSWER At the changeover (equilibrium) point, ΔG = 0, so ΔH = TΔS. T = ΔH ÷ ΔS = 400 ÷ 0.2 = 2000 K. Since ΔH and ΔS are both positive, ΔG = ΔH − TΔS becomes negative only when T > 2000 K. Hence the reaction is spontaneous above 2000 K.

5.18 For the reaction, 2 Cl(g) → Cl₂(g), what are the signs of ΔH and ΔS?

ANSWER A Cl–Cl bond is formed, so energy is released → ΔH is negative. Two moles of gaseous atoms combine into one mole of gas, reducing the number of particles and the disorder → ΔS is negative.

5.19 For the reaction 2 A(g) + B(g) → 2D(g), ΔU° = −10.5 kJ and ΔS° = −44.1 JK⁻¹. Calculate ΔG° for the reaction, and predict whether the reaction may occur spontaneously.

ANSWER Δn_g = 2 − 3 = −1. First convert ΔU to ΔH: ΔH = ΔU + Δn_gRT = −10.5 + (−1)(8.314×10⁻³)(298) = −10.5 − 2.478 = −12.978 kJ. ΔG = ΔH − TΔS = −12.978 − (298)(−44.1×10⁻³) = −12.978 + 13.142 = +0.164 kJ. Since ΔG is positive, the reaction is not spontaneous.

5.20 The equilibrium constant for a reaction is 10. What will be the value of ΔG°? R = 8.314 JK⁻¹ mol⁻¹, T = 300 K.

ANSWER ΔG° = −2.303 RT log K. = −2.303 × 8.314 × 300 × log 10 = −2.303 × 8.314 × 300 × 1 = −5744 J = −5.744 kJ mol⁻¹.

5.21 Comment on the thermodynamic stability of NO(g), given (1/2) N₂(g) + (1/2) O₂(g) → NO(g); ΔrH° = 90 kJ mol⁻¹ NO(g) + (1/2) O₂(g) → NO₂(g); ΔrH° = −74 kJ mol⁻¹

ANSWER Formation of NO from its elements is endothermic (ΔrH = +90 kJ mol⁻¹). Positive formation enthalpy means NO(g) has higher energy than its elements, so it is thermodynamically unstable and tends to decompose. The conversion of NO to NO₂ is exothermic (ΔrH = −74 kJ mol⁻¹), releasing energy. So NO readily reacts further with O₂ to form the more stable NO₂(g). In short, NO(g) is unstable, whereas NO₂(g) is comparatively stable and is formed.

5.22 Calculate the entropy change in surroundings when 1.00 mol of H₂O(l) is formed under standard conditions. Δ_fH° = −286 kJ mol⁻¹.

ANSWER Heat released by the system to the surroundings: q_surr = +286 kJ mol⁻¹ = +286000 J mol⁻¹ (formation is exothermic, so the surroundings gain this heat). ΔS_surr = q_surr / T = 286000 J ÷ 298 K = +959.73 J K⁻¹ mol⁻¹.

Extra Practice Questions

Short Answer Type Questions

Q1. Distinguish between an open, a closed and an isolated system with one example each.

ANSWERAn open system exchanges both matter and energy (reactants in an open beaker); a closed system exchanges only energy, not matter (reactants in a sealed copper vessel); an isolated system exchanges neither (hot tea in a closed thermos flask).

Q2. Why are heat and work not state functions while internal energy is?

ANSWERHeat (q) and work (w) depend on the path along which a change is carried out, so their values differ for different routes between the same two states. Internal energy U depends only on the state of the system; its change ΔU = U₂ − U₁ is the same for every path, which makes U a state function.

Q3. State and explain Hess’s law of constant heat summation.

ANSWERHess’s law states that the enthalpy change of a reaction is the same whether it takes place in one step or in several steps. Because enthalpy is a state function, ΔrH for an overall reaction equals the sum of the ΔrH values of the intermediate steps. It lets us calculate enthalpies that cannot be measured directly.

Q4. Calculate Δn_g and state whether ΔH is greater or smaller than ΔU for the combustion of 1 mol of graphite: C(s) + O₂(g) → CO₂(g).

ANSWERGaseous products = 1 (CO₂), gaseous reactants = 1 (O₂), so Δn_g = 0. Therefore ΔH = ΔU + Δn_gRT = ΔU; the two are equal.

Q5. Why is the entropy of a perfectly crystalline solid taken as zero at absolute zero?

ANSWERAt 0 K a perfect crystal has its atoms/ions arranged in a single, completely ordered way with no thermal motion. There is only one possible arrangement, so the disorder (and hence the entropy) is zero. This is the third law of thermodynamics.

Long Answer Type Questions

Q1. Derive the relation ΔH = ΔU + Δn_gRT and explain each term.

ANSWERBy definition H = U + pV, so for a change at constant pressure ΔH = ΔU + pΔV. For a reaction involving ideal gases, applying pV = nRT to the gaseous reactants and products at constant T and p gives pΔV = (n_products − n_reactants)RT = Δn_gRT, where Δn_g is the change in the number of moles of gas. Substituting, ΔH = ΔU + Δn_gRT. Here ΔU is the internal-energy change (measured at constant volume in a bomb calorimeter), R is the gas constant and T the absolute temperature. When Δn_g = 0, ΔH = ΔU; when gas moles increase, ΔH > ΔU, and when they decrease, ΔH < ΔU.

Q2. Explain how the sign of ΔG decides spontaneity, and how temperature can change it for a reaction with ΔH > 0 and ΔS > 0.

ANSWERThe Gibbs energy change is ΔG = ΔH − TΔS. A process is spontaneous when ΔG < 0, non-spontaneous when ΔG > 0, and at equilibrium when ΔG = 0. For an endothermic reaction with positive entropy (ΔH > 0, ΔS > 0), the term −TΔS is negative and grows with temperature. At low T the positive ΔH dominates and ΔG is positive (non-spontaneous); as T rises, −TΔS eventually outweighs ΔH, making ΔG negative. The reaction therefore becomes spontaneous above the changeover temperature T = ΔH/ΔS. This is why many decompositions (e.g. of CaCO₃) occur only on heating.

Q3. Describe how ΔU and ΔH of a reaction are measured using a calorimeter.

ANSWERΔU is measured in a bomb calorimeter: the substance is burnt in excess oxygen inside a sealed steel bomb immersed in a water bath. Since the volume is fixed (ΔV = 0), no work is done, so the heat evolved equals q_V = ΔU. The temperature rise of the known mass of water and the calorimeter, together with their heat capacity, gives q_V from q = CΔT. ΔH is measured at constant (atmospheric) pressure in an open calorimeter, where the heat of reaction q_p equals ΔH. The two are linked by ΔH = ΔU + Δn_gRT, so a value obtained at constant volume can be converted to constant pressure and vice versa.

MCQs & Assertion–Reason

1. Which of the following is an intensive property?

(a) enthalpy (b) volume (c) temperature (d) heat capacity

2. For an isothermal expansion of an ideal gas into vacuum:

(a) q > 0 (b) w > 0 (c) ΔU = 0 and q = w = 0 (d) ΔU < 0

3. The mathematical statement of the first law of thermodynamics is:

(a) ΔU = q − w (b) ΔU = q + w (c) ΔH = q_V (d) ΔG = ΔH − TΔS

4. For an ideal gas, C_p − C_V equals:

(a) 0 (b) R (c) 2R (d) R/2

5. The standard enthalpy of formation of O₂(g) at 298 K is:

(a) +393.5 kJ mol⁻¹ (b) zero (c) −285.8 kJ mol⁻¹ (d) unity

6. Heat absorbed by a system at constant pressure is equal to:

(a) ΔU (b) ΔS (c) ΔH (d) ΔG

7. A reaction is spontaneous at all temperatures when:

(a) ΔH > 0, ΔS > 0 (b) ΔH < 0, ΔS < 0 (c) ΔH < 0, ΔS > 0 (d) ΔH > 0, ΔS < 0

8. At equilibrium, the value of ΔG is:

(a) positive (b) negative (c) zero (d) infinite

9. Which relation connects ΔrG° and the equilibrium constant K?

(a) ΔrG° = RT ln K (b) ΔrG° = −RT ln K (c) ΔrG° = −RT/K (d) ΔrG° = K/RT

10. The entropy of a system generally increases when:

(a) a gas condenses to a liquid (b) a solid dissolves to give a solution (c) a liquid freezes (d) moles of gas decrease in a reaction

Answer key: 1-(c), 2-(c), 3-(b), 4-(b), 5-(b), 6-(c), 7-(c), 8-(c), 9-(b), 10-(b).

For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.

A-R 1. Assertion: Enthalpy is a state function.

Reason: H = U + pV, and U, p and V are all state functions.

A-R 2. Assertion: For an adiabatic process q = 0.

Reason: In an adiabatic process there is no exchange of heat between the system and the surroundings.

A-R 3. Assertion: The enthalpy of formation of an element in its standard state is taken as zero.

Reason: An element cannot release or absorb energy under any condition.

A-R 4. Assertion: For the reaction 2Cl(g) → Cl₂(g), entropy decreases.

Reason: The number of gaseous particles decreases, lowering the disorder of the system.

A-R 5. Assertion: An endothermic reaction with positive entropy change becomes spontaneous only at high temperature.

Reason: ΔG = ΔH − TΔS becomes negative when TΔS exceeds the positive ΔH.

Answer key: 1-(A), 2-(A), 3-(C), 4-(A), 5-(A).

Common Mistakes to Avoid

Watch out for these

Using the old physics sign convention for work. In chemistry (IUPAC), w is positive when work is done on the system, so ΔU = q + w (not q − w).
Forgetting to count only gaseous moles in Δn_g. Solids and liquids do not contribute.
Mixing units: keep R as 8.314 J K⁻¹ mol⁻¹ and convert J ↔ kJ before adding (e.g. Δn_gRT in Q5.8 and Q5.19).
Confusing Δ_fH (one mole of product) with ΔrH for a balanced equation that forms several moles — remember to divide (Q5.13).
Dropping the sign when freezing or condensing — these release heat, so ΔH is negative (Q5.10).
Thinking a negative ΔH alone guarantees spontaneity. Spontaneity is decided by ΔG = ΔH − TΔS, not by ΔH alone.

Exam tips to score full marks

Always write the working: state the formula, substitute values with units, then give the boxed final answer. For numericals on ΔH/ΔU, write the balanced equation, count gaseous moles to find Δn_g, and convert RT to kJ before adding. For Hess’s-law and bond-enthalpy problems, set up the target equation first, then add/subtract the given reactions (reverse a reaction → reverse its sign; multiply a reaction → multiply its ΔH). For spontaneity questions, compute ΔG = ΔH − TΔS and remember the temperature condition T = ΔH/ΔS for the changeover point.

Frequently Asked Questions

What is Class 11 Chemistry Chapter 5 Thermodynamics about?

Chapter 5 studies energy changes in chemical and physical processes: the first law (ΔU = q + w), enthalpy and its measurement, Hess’s law, standard enthalpies of formation and combustion, entropy, and Gibbs energy with the spontaneity criterion ΔG < 0 and the relation ΔrG° = −RT ln K.

How many exercise questions are there in Class 11 Chemistry Chapter 5?

The NCERT Exercises for Unit 5 contain 22 questions (5.1 to 5.22). Questions 5.1–5.6 are objective/MCQ-type and 5.7–5.22 are numerical and reasoning questions — all are reproduced verbatim and solved step by step on this page.

What is the difference between ΔH and ΔU?

ΔU is the change in internal energy (heat at constant volume); ΔH is the change in enthalpy (heat at constant pressure). They are related by ΔH = ΔU + Δn_gRT, where Δn_g is the change in the number of moles of gas. They are equal when Δn_g = 0.

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