NCERT Solutions for Class 11 Chemistry Chapter 5: Thermodynamics

Concept. A state function (state variable) is a property of a system whose value depends only on the present state of the system — fixed by quantities such as p, V, T and composition — and not on how that state was reached. Pressure, volume, temperature, internal energy U, enthalpy H, entropy S and Gibbs energy G are state functions; heat (q) and work (w) are not (they are path functions). This idea is the foundation of Hess’s law and of the first law.

1. (ii) is correct: its value is independent of path. If a system goes from state 1 to state 2, the change in a state function, ΔX = X₂ − X₁, is the same whatever route is taken. Joule showed that a given amount of work done adiabatically gives the same change of state irrespective of how the work is done, so U is a state function; NCERT’s other examples: a temperature change from 25°C to 35°C is +10°C whether done directly or by cooling first, and the volume of water in a pond is independent of whether it was filled by rain or tube well.
2. (i) used to determine heat changes — wrong. Heat q is a path-dependent quantity, not a state function. Although ΔU = q_V and ΔH = q_p allow heat changes to be found under specific conditions, that is a consequence of the conditions (constant V or p), not the definition of a state function.
3. (iii) used to determine pressure–volume work — wrong. Work w = −p_exΔV depends on the path (single step, many steps, reversible), so work is not a state function.
4. (iv) whose value depends on temperature only — wrong. State functions depend on the whole state (p, V, T, n); internal energy of a gas depends on temperature (ideal gas) but enthalpy, entropy, Gibbs energy, volume and pressure depend on other variables too. Moreover temperature is only one state function.

Extra depth. Since q + w = ΔU is a state-function change although q and w individually depend on path, the first law ties path-dependent and path-independent quantities together. Because ΔH is path independent, Hess’s law lets us add reaction enthalpies of any convenient steps; because ΔS_total and ΔG are state-function based, spontaneity can be predicted from initial and final states alone, as thermodynamics only concerns the initial and final states.

Conclusion: A thermodynamic state function is a quantity whose value is independent of the path followed, option (ii).

Concept. An adiabatic process is one in which there is no transfer of heat between the system and the surroundings; the system is enclosed in an adiabatic wall (for example a thermos flask or a well-insulated beaker). The key idea is the exchange of heat, not of work or the values of T or p.

1. (iii) q = 0 is correct. The defining condition of an adiabatic process is that no heat enters or leaves, so q = 0 and the first law becomes ΔU = q + w = w_ad: all the change of internal energy appears as work done on or by the system.
2. (i) ΔT = 0 is the condition for an isothermal process, not adiabatic. In an adiabatic process the temperature generally changes: compression heats the gas and expansion cools it (in the Joule experiment churning water in an adiabatic vessel raised its temperature from T_A to T_B).
3. (ii) Δp = 0 is an isobaric process (constant pressure), for which q_p = ΔH.
4. (iv) w = 0 describes a process in which no work is done — e.g. free expansion into vacuum or a constant-volume process (ΔV = 0 gives ΔU = q_V). It is not the definition of adiabatic.

Extra depth. In an adiabatic process the system is in effect isolated from heat exchange: a thermos flask, or a fast process (gas expansion in a cylinder so quick that heat has no time to flow) are practical examples. Adiabatic walls are drawn as thick insulating boundaries (Fig. 5.3 with q = 0 marked). An isolated system (no heat, no work, no matter) has ΔU = 0 and is a special case.

More to know. Adiabatic versus isothermal, side by side. For an ideal gas compressed reversibly, an isothermal compression removes heat to the surroundings so the temperature stays constant (ΔU = 0, w = −q), whereas an adiabatic compression keeps all the work as internal energy (ΔU = w_ad > 0) and the gas warms up — which is why a bicycle pump gets hot. Likewise in an adiabatic expansion the gas does work at the expense of its own internal energy and cools. Joule’s paddle-wheel and immersion-rod experiments used adiabatic walls: the same work, however delivered, produced the same ΔT, which is why U is a state function and why w_ad equals ΔU (Section 5.1.4).

Conclusion: The correct condition for an adiabatic process is q = 0, option (iii).

Concept. Enthalpy H = U + pV is a state function whose absolute value cannot be known; only changes (ΔH) can be measured. To tabulate and compare reaction enthalpies a reference point is chosen by convention — and this convention concerns the standard enthalpy of formation.

1. (ii) zero is correct. By convention, the standard enthalpy of formation, Δ_fH^⊖, of an element in its reference state (its most stable state of aggregation at 25°C and 1 bar) is taken as zero. Examples: H₂(g), O₂(g), N₂(g), Cl₂(g), Br₂(l), C(graphite), S(rhombic), Fe(s), Na(s) all have Δ_fH^⊖ = 0 (Table 5.2).
2. Why zero? Δ_fH^⊖ is the enthalpy change for forming a compound from its elements in their reference states; forming an element from itself involves no change, so the reference values are set to zero and all other values are measured relative to them. This lets Δ_rH^⊖ = Σa_iΔ_fH^⊖(products) − Σb_iΔ_fH^⊖(reactants) work.
3. (i) unity, (iii) < 0, (iv) different for each element — wrong: the values are neither 1 nor negative nor element-specific by definition. (A non-reference allotrope has a non-zero value, e.g. C(diamond) +1.89 and Br₂(g) +30.91 kJ mol⁻¹, and atoms such as C(g) +716.68 or H(g) +217.97 kJ mol⁻¹; only the most stable form is taken as zero.)

Extra depth. Always specify the physical state and allotrope. Because C(graphite) is the reference state of carbon, the enthalpy of combustion of graphite (−393.5 kJ mol⁻¹) is also the enthalpy of formation of CO₂; likewise the combustion of H₂(g) to H₂O(l) (−285.8) equals Δ_fH^⊖ of liquid water (see Ex 5.5).

Conclusion: The standard enthalpy of formation of every element in its standard (reference) state is zero, option (ii).

Concept. ΔH and ΔU are related by ΔH = ΔU + Δn_gRT (eq. 5.10), where Δn_g = (moles of gaseous products) − (moles of gaseous reactants). So the difference depends on the change in the number of moles of gas in the reaction.

1. Reaction: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l). Gaseous reactants: 1 + 2 = 3 mol; gaseous products: 1 mol CO₂ (liquid water is not counted). So Δn_g = 1 − 3 = −2.
2. Calculation: ΔH = ΔU + (−2)RT = ΔU − 2RT. Since RT is positive, ΔH is less than ΔU (more negative): numerically, for ΔU = −X, ΔH = −X − 2RT = −(X + 4.96) kJ mol⁻¹ at 298 K (2RT = 2 × 8.314 × 298/1000 = 4.96 kJ mol⁻¹).
3. Answer: (iii) ΔH < ΔU. Physically, at constant pressure the surroundings compress the system as the volume falls (3 mol gas → 1 mol), doing work on the system and releasing an extra 2RT of energy as heat.
4. Why the others are wrong: (i) ΔH = ΔU only if Δn_g = 0; (ii) ΔH > ΔU needs Δn_g > 0 (gas moles increase); (iv) ΔH = 0 would mean no heat change at all.

Extra depth. If the water formed were counted as vapour, Δn_g would be 3 − 3 = 0 and ΔH = ΔU; NCERT’s standard enthalpies of combustion (−890.3 kJ mol⁻¹ for methane) refer to liquid water, so Δn_g = −2. Compare Problem 5.5 (vaporisation, Δn_g = +1, ΔU = ΔH − RT) and Problem 5.6 (graphite, Δn_g = 0).

More to know. How large is the difference? At 298 K, RT = 2.48 kJ mol⁻¹, so Δn_gRT is only a few kJ: for N₂ + 3H₂ → 2NH₃ (Δn_g = −2) ΔH = ΔU − 4.96 kJ; for CaCO₃ → CaO + CO₂ (Δn_g = +1) ΔH = ΔU + 2.48 kJ. Compared with the typical magnitude of reaction enthalpies (hundreds of kJ) the correction is a percent or less, which is why ΔH and ΔU are often loosely equated for reactions without a large change in gas moles. For reactions of only solids and liquids (Δn_g = 0) they are practically identical, because solids and liquids do not undergo significant volume change on heating.

Conclusion: For the combustion of methane Δn_g = −2, so ΔH = ΔU − 2RT and ΔH < ΔU, option (iii).

Concept. This is a typical Hess’s law problem: the standard enthalpy of formation of methane cannot be measured directly (carbon and hydrogen do not combine cleanly to CH₄), but it can be obtained from the enthalpies of combustion of methane, graphite and hydrogen by combining their thermochemical equations so that they add up to the formation reaction C(graphite) + 2H₂(g) → CH₄(g).

1. (a) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l); Δ_cH = −890.3 kJ mol⁻¹
2. (b) C(graphite) + O₂(g) → CO₂(g); Δ_cH = −393.5 kJ mol⁻¹
3. (c) H₂(g) + ½O₂(g) → H₂O(l); Δ_cH = −285.8 kJ mol⁻¹
4. Target: C(graphite) + 2H₂(g) → CH₄(g); Δ_fH = ?. Take (b) as written, 2 × (c), and reverse (a) (CO₂ + 2H₂O → CH₄ + 2O₂, ΔH = +890.3): sum = C + O₂ + 2H₂ + O₂ + CO₂ + 2H₂O → CO₂ + 2H₂O + CH₄ + 2O₂; cancelling common species gives the target equation.
5. Δ_fH(CH₄) = 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.
6. Answer: (i) −74.8 kJ mol⁻¹ (NCERT Table 5.2 gives −74.81 kJ mol⁻¹).

Extra depth. The shortcut is that Δ_fH = ΣΔ_cH(reactants) − ΣΔ_cH(products) — the combustion enthalpy of the elements/compounds consumed minus that of the compounds formed (the sign is reversed relative to the usual Δ_fH formula, because combustion is the ‘down-hill’ route to the same oxides). Check by magnitude: the answer is negative, as expected for the exothermic formation of methane, and option (iv) +52.26 is the Δ_fH of ethene (C₂H₄, Table 5.2), a distractor. Option (ii) −52.27 would result from a sign mistake.

Conclusion: Δ_fH(CH₄) = −393.5 + 2(−285.8) + 890.3 = −74.8 kJ mol⁻¹, option (i).

Concept. Spontaneity at constant temperature and pressure is decided by the sign of the Gibbs energy change, ΔG = ΔH − TΔS: the process is spontaneous (‘possible’) if ΔG < 0. We therefore need the signs of ΔH and ΔS. (In the NCERT paper the last option is numbered (v); it is the ‘any temperature’ option.)

1. Sign of ΔH. The equation is written as A + B → C + D + q, i.e. heat q is a product: the reaction releases heat and is exothermic, so ΔH < 0 (negative).
2. Sign of ΔS. It is given that the entropy change is positive: ΔS > 0, so −TΔS is negative at every temperature (T is always positive).
3. Combine. ΔG = ΔH − TΔS = (negative) − T(positive) = (negative) + (negative) = negative at all temperatures. Hence the reaction is spontaneous (possible) at any temperature.
4. Answer: option (iv) — possible at any temperature. This corresponds to the first row of Table 5.4 (Δ_rH −, Δ_rS +, Δ_rG −): spontaneous at all temperatures.

Extra depth. Options (i) and (ii) describe cases where ΔH and ΔS have the same sign (so that a temperature dependence exists); option (iii) describes the (+, −) case. Since here the enthalpy term and the entropy term work in the same direction, no temperature is needed to force the reaction. ‘Possible’ refers to thermodynamic feasibility only — the rate may still be extremely slow (a high activation energy), as for the combination of H₂ and O₂ at room temperature.

Conclusion: The reaction is exothermic (ΔH < 0) with ΔS > 0, so ΔG is negative at all temperatures: option (iv), possible at any temperature.

Concept. The first law of thermodynamics states the conservation of energy: ΔU = q + w (eq. 5.1), where q is the heat absorbed by the system and w the work done on the system, with the IUPAC sign convention that energy added to the system is positive and energy removed from it is negative. Internal energy is a state function and q and w separately depend on the path, but their sum depends only on the initial and final states.

1. Heat: 701 J of heat is absorbed by the system → energy enters the system → q = +701 J.
2. Work: 394 J of work is done by the system (the system loses energy as work) → w = −394 J.
3. Apply the first law: ΔU = q + w = (+701 J) + (−394 J) = +307 J.
4. Interpretation: the internal energy of the system increases by 307 J: it gained 701 J as heat but gave away 394 J as work, so the net gain is 307 J.

Extra depth. The common error is to write ΔU = q − w without sign care, or to take w = +394 J. NCERT’s Problem 5.1(iii) writes ΔU = q − w for the case where w is the magnitude of work done by the system — the same result: ΔU = 701 − 394 = 307 J. Always tell which convention you use. 307 J is only the net change; the individual quantities q and w would be different for a different path between the same two states, but q + w would stay 307 J.

More to know. Checking the signs with the three cases. (a) Heat in, work out (this problem): ΔU = +701 − 394 = +307 J — internal energy rises. (b) If the same system absorbed 701 J but did 800 J of work done by it, ΔU = −99 J and the internal energy would fall. (c) For an isolated system q = w = 0, ΔU = 0 (conservation of energy). The values q = 701 J and w = −394 J could correspond to, say, a gas expanding against an external pressure (work out) while being heated (heat in); the path may change q and w, but q + w = 307 J is fixed by the two end states — the essence of ‘ΔU is a state function’.

Conclusion: ΔU = q + w = 701 − 394 = +307 J; the internal energy of the system increases by 307 J.

Concept. A bomb calorimeter measures heat at constant volume, so the measured quantity is q_V = ΔU. Chemists usually need ΔH (heat at constant pressure). The two differ only through the work of the gas volume change: ΔH = ΔU + Δn_gRT (eq. 5.10), with Δn_g = (moles of gaseous products) − (moles of gaseous reactants). Note on the question: the statement calls cyanamide a solid, NH₂CN(s), while the equation shows NH₂CN(g); in the standard solution it is treated as a solid (it is a solid at room temperature), so only the dioxygen counts on the reactant side.

1. Count gaseous moles (cyanamide as solid): gaseous reactants: O₂ = 3/2 mol. Gaseous products: N₂ = 1 and CO₂ = 1, i.e. 2 mol (H₂O is liquid). Δn_g = 2 − 3/2 = +1/2 = +0.5.
2. RT: R = 8.314 × 10⁻³ kJ K⁻¹ mol⁻¹, T = 298 K → RT = 2.478 kJ mol⁻¹. Δn_gRT = 0.5 × 2.478 = +1.239 kJ mol⁻¹.
3. ΔH = ΔU + Δn_gRT = −742.7 + 1.239 = −741.5 kJ mol⁻¹.
4. Check of the other reading: if cyanamide were counted as a gas (as printed in the equation), Δn_g = 2 − 5/2 = −0.5 and ΔH = −742.7 − 1.24 = −743.9 kJ mol⁻¹; the accepted textbook answer is −741.5 kJ mol⁻¹ (solid cyanamide).

Extra depth. The small difference (1.2 kJ against 742.7 kJ) shows that for most combustion reactions ΔH and ΔU differ by only a fraction of a per cent unless Δn_g is large. Since gas is produced (Δn_g > 0), the system does expansion work against the atmosphere at constant pressure, so less heat is evolved at constant pressure: ΔH is slightly less negative than ΔU. For Problem 5.6 (graphite, Δn_g = 0) ΔH = ΔU exactly.

Conclusion: With Δn_g = +0.5 the enthalpy change is ΔH = −742.7 + 1.24 = −741.5 kJ mol⁻¹.

Concept. Heat capacity relates the heat absorbed by a substance to its temperature rise. The molar heat capacity C_m is the heat needed to raise the temperature of one mole by one kelvin (or one degree Celsius): q = n × C_m × ΔT (equivalent to q = CΔT with C = nC_m, eq. 5.11). A temperature difference in °C equals the same difference in kelvin.

1. Moles of aluminium: molar mass of Al = 27 g mol⁻¹; n = 60.0 g / 27 g mol⁻¹ = 2.22 mol (2.222).
2. Temperature change: ΔT = 55 − 35 = 20°C = 20 K.
3. Heat required: q = n × C_m × ΔT = 2.222 mol × 24 J mol⁻¹ K⁻¹ × 20 K = 1066.7 J ≈ 1.07 kJ.
4. Hence about 1.07 kJ of heat is required.

Extra depth. The equivalent calculation by specific heat: c = C_m/M = 24/27 = 0.889 J g⁻¹ K⁻¹, q = c × m × ΔT = 0.889 × 60.0 × 20 = 1067 J, the same. The result is positive because heat is absorbed (q > 0). Aluminium has a low heat capacity compared with water (75.3 J mol⁻¹ K⁻¹), so a small amount of heat raises its temperature considerably — one reason why aluminium utensils heat up quickly.

More to know. Why a molar heat capacity of 24 J mol⁻¹ K⁻¹ for a metal? For many metals the molar heat capacity is close to 3R = 24.9 J mol⁻¹ K⁻¹ (the Dulong–Petit rule), so equal amounts (moles) of different metals need about the same heat for the same temperature rise. In contrast liquid water has C_m = 75.3 J mol⁻¹ K⁻¹ (about 4.18 J g⁻¹ K⁻¹), so 60 g of water would need 60 × 4.18 × 20 = 5016 J — nearly five times more than 60 g of aluminium. This is why water is used as a coolant and why coastal regions have milder climates.

Conclusion: The heat required is q = (60.0/27) mol × 24 J mol⁻¹ K⁻¹ × 20 K = 1.07 kJ.

Concept. Enthalpy is a state function, so the enthalpy change between the initial state (liquid water at 10°C) and the final state (ice at −10°C) can be computed along any convenient path — Hess’s law applied to a physical change. The path used is: (1) cool the liquid from 10°C to 0°C, (2) freeze it at 0°C, (3) cool the ice from 0°C to −10°C. At constant pressure, heat changes equal enthalpy changes: ΔH = C_pΔT for heating or cooling, and ΔH = −Δ_fusH for freezing (the reverse of fusion).

1. Step 1 — cool water 10°C → 0°C: ΔH₁ = C_p(l) × ΔT = 75.3 J mol⁻¹ K⁻¹ × (0 − 10) K = −753 J mol⁻¹ = −0.753 kJ mol⁻¹.
2. Step 2 — freeze water at 0°C: H₂O(l) → H₂O(s) is the reverse of fusion, so ΔH₂ = −Δ_fusH = −6.03 kJ mol⁻¹.
3. Step 3 — cool ice 0°C → −10°C: ΔH₃ = C_p(s) × ΔT = 36.8 × (−10 − 0) = −368 J mol⁻¹ = −0.368 kJ mol⁻¹.
4. Total: ΔH = ΔH₁ + ΔH₂ + ΔH₃ = −0.753 − 6.03 − 0.368 = −7.151 kJ mol⁻¹.
5. The negative sign shows that 7.151 kJ of heat is evolved per mole of water in this process (exothermic).

Extra depth. Notice that the freezing step contributes most of the heat; the sensible-heat steps are small because C_p values are only tens of J mol⁻¹ K⁻¹. Care: the phase change takes place at the melting point (0°C), where the given Δ_fusH applies — we cannot freeze supercooled water directly at 10°C with this value. This is also an example of the Δ_fusH being positive for melting and negative for freezing (Section 5.4(b)). Since Δn_g = 0 for liquid → solid, ΔU ≈ ΔH.

Conclusion: The enthalpy change is ΔH = −0.753 − 6.03 − 0.368 = −7.151 kJ mol⁻¹.

Concept. The standard enthalpy of combustion is the enthalpy change per mole of substance burnt. For carbon (graphite) burning to CO₂ it is also the standard enthalpy of formation of CO₂, because C(graphite) and O₂(g) are the elements in their reference states: C(graphite) + O₂(g) → CO₂(g); Δ_cH^⊖ = Δ_fH^⊖ = −393.5 kJ mol⁻¹. The negative sign means that 393.5 kJ of heat is released for each mole of CO₂ formed.

1. Molar mass of CO₂: 12 + 2 × 16 = 44 g mol⁻¹.
2. Moles of CO₂ formed: n = 35.2 g / 44 g mol⁻¹ = 0.8 mol.
3. Heat released: 1 mol CO₂ formed releases 393.5 kJ; so 0.8 mol releases 0.8 × 393.5 = 314.8 kJ (ΔH = −314.8 kJ).
4. Thus 314.8 kJ of heat is released when 35.2 g of CO₂ is formed from carbon and dioxygen gas.

Extra depth. This is the proportionality rule of thermochemistry: because enthalpy is an extensive property, the heat change scales with the amount of reactant converted. Equivalent mass-based reasoning: 44 g of CO₂ ↔ 393.5 kJ → 35.2 g ↔ 393.5 × 35.2/44 = 314.8 kJ. Note also that this assumes complete combustion to CO₂ (not CO, whose formation releases only 110.5 kJ mol⁻¹). The same method gives the heat from burning other fuels: e.g. 1 mol butane releases 2658 kJ, 1 mol glucose 2802 kJ.

More to know. Heat per gram of fuel. From the same enthalpy data we can compare fuels per unit mass: carbon 393.5/12 = 32.8 kJ g⁻¹; hydrogen 285.8/2.016 = 142 kJ g⁻¹ (to liquid water); butane 2658/58 = 45.8 kJ g⁻¹; glucose 2802/180 = 15.6 kJ g⁻¹. This is why hydrogen is considered for rocket fuel, why cooking gas gives more heat per kilogram than coal, and why food energy (about 16 kJ g⁻¹ for carbohydrate) is lower. The problem assumed complete combustion to CO₂ and that heat measured is the enthalpy change at constant pressure for the standard-state reaction.

Conclusion: Heat released = (35.2/44) mol × 393.5 kJ mol⁻¹ = 314.8 kJ.

Concept. When standard enthalpies of formation are available, the enthalpy change of any reaction is the difference between the total enthalpy of formation of the products and that of the reactants: Δ_rH = Σa_iΔ_fH(products) − Σb_iΔ_fH(reactants) (eq. 5.15), where a_i and b_i are the stoichiometric coefficients of the balanced equation. This is Hess’s law in compact form: every compound is imagined to be made from the elements (whose Δ_fH is zero) and the elements are the common starting point of both sides.

1. Products: 1 mol N₂O(g) + 3 mol CO₂(g): Σ = 1(81) + 3(−393) = 81 − 1179 = −1098 kJ.
2. Reactants: 1 mol N₂O₄(g) + 3 mol CO(g): Σ = 1(9.7) + 3(−110) = 9.7 − 330 = −320.3 kJ.
3. Δ_rH = (−1098) − (−320.3) = −1098 + 320.3 = −777.7 kJ (per mole of the reaction as written).
4. The reaction is strongly exothermic: Δ_rH = −777.7 kJ mol⁻¹.

Extra depth. The main contribution comes from CO₂, which is highly stable (large negative Δ_fH) — converting three CO to three CO₂ alone releases 3(−393 + 110) = −849 kJ, partly offset by the +71.3 kJ needed to convert N₂O₄ into N₂O (81 − 9.7). Take care with the arithmetic: multiply each Δ_fH by its coefficient, and subtract the reactant sum as a whole (the sign of −(−320.3) becomes +320.3).

Conclusion: Δ_rH = [81 + 3(−393)] − [9.7 + 3(−110)] = −777.7 kJ.

Concept. The standard enthalpy of formation, Δ_fH^⊖, is the enthalpy change when one mole of a compound is formed from its elements in their standard (reference) states. It is therefore a special case of Δ_rH^⊖, with the equation written so that exactly one mole of the compound forms. The given reaction forms two moles of NH₃, so the value must be divided by two. (NCERT illustrates this with HBr: H₂ + Br₂ → 2HBr has Δ_rH^⊖ = 2Δ_fH^⊖.)

1. Given: N₂(g) + 3H₂(g) → 2NH₃(g); Δ_rH^⊖ = −92.4 kJ per mole of reaction (2 mol NH₃ formed).
2. Divide the equation by 2 to get one mole of NH₃: ½N₂(g) + 3/2 H₂(g) → NH₃(g).
3. Divide the enthalpy change by 2 (enthalpy is extensive): Δ_fH^⊖(NH₃) = −92.4/2 = −46.2 kJ mol⁻¹.
4. Check by the formation formula: Δ_rH^⊖ = 2Δ_fH^⊖(NH₃) − [Δ_fH^⊖(N₂) + 3Δ_fH^⊖(H₂)] = 2x − 0 = −92.4, so x = −46.2 (the elements N₂ and H₂ are zero by convention). Table 5.2 gives −46.11 kJ mol⁻¹.

Extra depth. This is Convention 2 and 3 of thermochemical equations: Δ_rH^⊖ refers to the number of moles written, and multiplying or dividing an equation by a factor multiplies or divides ΔH by the same factor (compare Fe₂O₃ reduction: −33.3 vs −16.6 kJ mol⁻¹). The negative Δ_fH^⊖ shows NH₃ is thermodynamically stable with respect to its elements at 298 K; the reverse reaction, decomposition, would have ΔH = +46.2 kJ mol⁻¹.

Conclusion: The standard enthalpy of formation of NH₃(g) is −92.4/2 = −46.2 kJ mol⁻¹.

Concept. Methanol cannot be made directly from its elements, so its enthalpy of formation is found from the enthalpies of other reactions using Hess’s law: the enthalpy change of the overall reaction is the sum of the enthalpy changes of the steps into which it can be divided. Target: C(graphite) + 2H₂(g) + ½O₂(g) → CH₃OH(l), Δ_fH^⊖ = ?

1. (i) CH₃OH(l) + 3/2 O₂ → CO₂ + 2H₂O(l); −726   (ii) C + O₂ → CO₂; −393   (iii) H₂ + ½O₂ → H₂O(l); −286.
2. Adjust the equations: methanol is a product in the target, so reverse (i): CO₂ + 2H₂O(l) → CH₃OH(l) + 3/2 O₂; ΔH = +726. Carbon is a reactant, so keep (ii) as written: −393. Two moles of H₂ are needed, so multiply (iii) by 2: 2H₂ + O₂ → 2H₂O(l); ΔH = −572.
3. Add: C + O₂ + 2H₂ + O₂ + CO₂ + 2H₂O → CO₂ + 2H₂O + CH₃OH + 3/2 O₂. Cancel CO₂, 2H₂O and 3/2 O₂ (of the 2 O₂ on the left, 3/2 cancel, leaving ½O₂): C(graphite) + 2H₂(g) + ½O₂(g) → CH₃OH(l).
4. Δ_fH^⊖ = (−393) + 2(−286) + (+726) = −393 − 572 + 726 = −239 kJ mol⁻¹.

Extra depth. Using the shortcut for combustion data: Δ_fH(CH₃OH) = ΣΔ_cH(C, 2H₂) − Δ_cH(CH₃OH) = (−393 − 572) − (−726) = −239 kJ mol⁻¹. The value agrees with Table 5.2 (−238.86 kJ mol⁻¹). The −239 kJ means that formation of liquid methanol from its elements is exothermic. Note that the Δ_cH of graphite and the Δ_fH of water are both formation enthalpies, which is why they can be used directly.

Conclusion: Δ_fH^⊖(CH₃OH, l) = −393 + 2(−286) + 726 = −239 kJ mol⁻¹.

Concept. The enthalpy of the process CCl₄(g) → C(g) + 4Cl(g) is the enthalpy of atomisation of CCl₄: all bonds of one mole of gaseous CCl₄ are broken to give gaseous atoms. Since four identical C–Cl bonds are broken, the mean C–Cl bond enthalpy is one-fourth of this value (just as the mean C–H bond enthalpy in CH₄ is 1665/4 = 416 kJ mol⁻¹). The given Δ_fH^⊖ = −135.5 kJ mol⁻¹ belongs to liquid CCl₄ (a Δ_vapH is given to convert it to the gas).

1. Formation of gaseous CCl₄: C(s) + 2Cl₂(g) → CCl₄(l), −135.5; CCl₄(l) → CCl₄(g), +30.5. Hence Δ_fH(CCl₄, g) = −135.5 + 30.5 = −105.0 kJ mol⁻¹.
2. Build a cycle (Hess’s law): C(s) + 2Cl₂(g) → C(g) + 4Cl(g) can be done in two ways: (a) atomise the elements: Δ_aH(C) + 2Δ_aH(Cl₂) = 715.0 + 2(242) = 1199 kJ; (b) form CCl₄(g) (−105.0) and then atomise it (ΔH = x). So 1199 = −105.0 + x.
3. Enthalpy of atomisation of CCl₄: x = 1199 + 105.0 = +1304 kJ mol⁻¹.
4. Mean C–Cl bond enthalpy = 1304/4 = 326 kJ mol⁻¹.

Extra depth. The result (326 kJ mol⁻¹) is close to the tabulated mean C–Cl bond enthalpy of 330 kJ mol⁻¹ (Table 5.3(a)), which confirms the method. Because Δ_fH^⊖ for CCl₄ is given for the liquid, forgetting Δ_vapH (taking −135.5 for the gas) would give 1273.5 and 318 kJ mol⁻¹, an error of more than 2 %. This is the standard method for finding a mean bond enthalpy from thermochemical data: atomisation enthalpy ÷ number of bonds broken.

Conclusion: The enthalpy change for CCl₄(g) → C(g) + 4Cl(g) is +1304 kJ mol⁻¹, so the mean C–Cl bond enthalpy is 1304/4 = 326 kJ mol⁻¹.

Concept. An isolated system exchanges neither energy nor matter with its surroundings, so its internal energy is constant: ΔU = 0 (this is the first law for an isolated system, w = 0 and q = 0). The first law, however, tells nothing about the direction of a spontaneous change. This is where the second law and the state function entropy (S), a measure of disorder or randomness, come in.

1. Second law. For an isolated system, increase in entropy is the natural direction of a spontaneous change: the total entropy change ΔS_total = ΔS_system + ΔS_surroundings > 0 for a spontaneous process (eq. 5.19). For an isolated system the surroundings do not interact, so the entropy of the system itself must increase: ΔS > 0.
2. At equilibrium the entropy of the isolated system is a maximum and ΔS = 0. A decrease in entropy (ΔS < 0) would mean a non-spontaneous (impossible) change in an isolated system.
3. Hence: ΔS > 0 (positive) for any spontaneous process in an isolated system; ΔU = 0 does not discriminate between spontaneous and non-spontaneous changes, whereas ΔS does.
4. Illustration: two gases mixing (Fig. 5.11): ΔH = 0, ΔU = 0 in the isolated container, but the gases mix spontaneously because the disorder, and so S, increases.

Extra depth. The statement of the second law used here is NCERT’s: ‘for an isolated system the increase in entropy is the natural direction of a spontaneous change’. The universe itself is an isolated system, so its entropy keeps increasing: ΔS_universe = ΔS_sys + ΔS_surr > 0 for all natural processes. Even when a process makes the system more ordered (as in freezing, ΔS_sys < 0), the heat given to the surroundings raises ΔS_surr by more.

More to know. Entropy of the universe. NCERT notes that increase in entropy is the natural direction of spontaneous change for isolated systems, and the universe is the ultimate isolated system. For a reversible process in an isolated system ΔS = 0 (e.g. an ideal reversible expansion where ΔS_sys + ΔS_surr = 0), while every irreversible (spontaneous) process increases the total entropy. Examples: heat flowing from a hot body (T₁) to a cold body (T₂): ΔS = q/T₂ − q/T₁ > 0 since T₂ < T₁; a gas expanding into vacuum (ΔU = 0, w = 0, q = 0) has ΔS > 0 because the molecules can occupy a larger volume. This is the sense in which ΔS, not ΔU, gives the direction of change.

Conclusion: For an isolated system with ΔU = 0 a spontaneous change has ΔS > 0 (and ΔS = 0 at equilibrium).

Concept. A reaction at constant temperature and pressure is spontaneous when the Gibbs energy change is negative: ΔG = ΔH − TΔS < 0 (eq. 5.21 and 5.22). Here both ΔH and ΔS are positive (endothermic, entropy increases) — the (+, +) case of Table 5.4: ΔG is positive at low temperature but becomes negative at high temperature, when the TΔS term outweighs ΔH. The question asks the threshold temperature, at which ΔG = 0 (equilibrium).

1. At the threshold: ΔG = 0 ⇒ ΔH = TΔS ⇒ T = ΔH/ΔS.
2. Substitute (keep both in kJ): T = 400 kJ mol⁻¹ / 0.2 kJ K⁻¹ mol⁻¹ = 2000 K.
3. At 298 K: ΔG = 400 − 298 × 0.2 = 400 − 59.6 = +340.4 kJ mol⁻¹ (positive, non-spontaneous).
4. Conclusion: for T > 2000 K, TΔS > ΔH so ΔG < 0; the reaction becomes spontaneous above 2000 K (at exactly 2000 K, ΔG = 0 and the system is at equilibrium).

Extra depth. Units matter: ΔH is in kJ, ΔS in kJ K⁻¹ mol⁻¹, so the answer is in K directly (if ΔS were 200 J K⁻¹ mol⁻¹, convert before dividing). For endothermic, entropy-increasing reactions the ‘turn-on’ temperature T = ΔH/ΔS is the reason why reactions such as CaCO₃ → CaO + CO₂ are run at high temperature in industry. The assumption of constant ΔH and ΔS is an approximation, since both vary slightly with temperature.

More to know. A real example of the same idea. The decomposition CaCO₃(s) → CaO(s) + CO₂(g) has Δ_rH^⊖ = +178.3 kJ mol⁻¹ (from Table 5.2) and a positive ΔS (a gas is produced, about +160 J K⁻¹ mol⁻¹, a standard value not given in NCERT). The threshold T = ΔH/ΔS ≈ 178.3/0.160 ≈ 1100 K, which is why limestone must be heated to about 1100 K (≈ 800–900°C) to make lime. In the exercise the ‘turn-on’ temperature of 2000 K is higher because ΔH is large relative to ΔS. The phrase ‘high temperature’ in Table 5.4 is therefore relative to this threshold.

Conclusion: The reaction becomes spontaneous above T = ΔH/ΔS = 400/0.2 = 2000 K.

Concept. The reaction is the reverse of atomisation: two gaseous chlorine atoms combine to form one Cl–Cl bond in a Cl₂ molecule. The signs of ΔH and ΔS can be predicted from two principles: bond formation releases energy (breaking a bond requires energy; making one releases it), and entropy measures disorder, which depends largely on the number of gaseous particles.

1. Sign of ΔH: a covalent bond is formed. Since the dissociation Cl₂(g) → 2Cl(g) requires +242 kJ mol⁻¹ (bond dissociation enthalpy, Section 5.5), the reverse reaction releases 242 kJ per mole of Cl₂: ΔH is negative (about −242 kJ mol⁻¹); the process is exothermic.
2. Sign of ΔS: two moles of gaseous atoms (more particles, more independent translational freedom) change into one mole of gaseous molecules. The number of gas particles falls from 2 to 1, so the disorder decreases: ΔS is negative. (In the opposite process H₂ → 2H, one molecule gives two atoms and entropy increases — NCERT Problem 5.10(iv).)
3. Conclusion: ΔH < 0 and ΔS < 0.
4. Spontaneity: this corresponds to the (−, −) row of Table 5.4: ΔG = ΔH − TΔS is negative at low temperature (the enthalpy term dominates) and positive at high temperature — Cl atoms combine into molecules at ordinary temperatures but Cl₂ dissociates at very high temperature.

Extra depth. Atoms in the gas phase have higher entropy than the corresponding molecules at the same temperature because 2 mol of particles have more translational motion and more possible arrangements than 1 mol. In general: bond-forming reactions are exothermic with ΔS < 0 when the number of gaseous particles decreases (e.g. N₂ + 3H₂ → 2NH₃, ΔH = −92.4 kJ); bond-breaking reactions are endothermic with ΔS > 0 when gas moles increase (e.g. dissociation of N₂O₄).

Conclusion: For 2Cl(g) → Cl₂(g) both ΔH and ΔS are negative: bond formation releases energy and the number of gas particles decreases.

Concept. To decide spontaneity we need the Gibbs energy change ΔG^⊖ = ΔH^⊖ − TΔS^⊖. The data give ΔU, not ΔH, so we first convert ΔU to ΔH using ΔH = ΔU + Δn_gRT. (The temperature is not given in the question; as for all NCERT problems of this kind it is taken as the standard temperature, 298 K.)

1. Δn_g: gaseous reactants: 2 + 1 = 3 mol; gaseous products: 2 mol; Δn_g = 2 − 3 = −1.
2. ΔH^⊖: ΔH = ΔU + Δn_gRT = −10.5 kJ + (−1)(8.314 × 10⁻³ kJ K⁻¹ mol⁻¹)(298 K) = −10.5 − 2.48 = −12.98 kJ.
3. TΔS^⊖: TΔS = 298 K × (−44.1 × 10⁻³ kJ K⁻¹) = −13.14 kJ.
4. ΔG^⊖ = ΔH^⊖ − TΔS^⊖ = −12.98 − (−13.14) = +0.16 kJ (about +160 J).
5. Prediction: ΔG^⊖ is positive (although small), so the reaction is non-spontaneous under standard conditions at 298 K — the reverse reaction is spontaneous. Because ΔG is very close to zero, the system is almost at equilibrium (K slightly less than 1: ln K = −ΔG^⊖/RT = −160/2478 = −0.065, K ≈ 0.94).

Extra depth. Here ΔH is negative (favouring the reaction) but ΔS is also negative (the number of gas moles decreases from 3 to 2), so the entropy term opposes it: this is the (−, −) case of Table 5.4. Below a certain temperature (T = ΔH/ΔS = −12980/−44.1 = 294 K) ΔG would be negative; just above 294 K it becomes positive. At 298 K we are at the border, so a very small change in temperature changes the verdict — with the printed data the reaction is only marginally non-spontaneous. If Δn_g were neglected (using ΔH ≈ ΔU) the answer would be ΔG = −10.5 + 13.14 = +2.64 kJ, which shows why the Δn_gRT correction should be included.

Conclusion: ΔG^⊖ = ΔH^⊖ − TΔS^⊖ = −12.98 + 13.14 = +0.16 kJ, so the reaction is non-spontaneous under the given conditions.

Concept. For a reaction in which all substances are in their standard states, the standard Gibbs energy change is linked to the equilibrium constant by Δ_rG^⊖ = −RT ln K = −2.303 RT log K (eq. 5.23). This follows from the criterion for equilibrium, Δ_rG = 0 = Δ_rG^⊖ + RT ln K. If K > 1 (products favoured) ΔG^⊖ is negative; if K < 1 it is positive.

1. Data: K = 10 ⇒ log K = log 10 = 1; R = 8.314 J K⁻¹ mol⁻¹; T = 300 K.
2. Substitute: ΔG^⊖ = −2.303 × 8.314 J K⁻¹ mol⁻¹ × 300 K × log 10.
3. 2.303 × 8.314 = 19.15; 19.15 × 300 = 5744.1; × 1 = 5744 J. Hence ΔG^⊖ = −5744 J mol⁻¹ = −5.744 kJ mol⁻¹.
4. Interpretation: the negative sign means the forward reaction is spontaneous under standard conditions, consistent with K > 1.

Extra depth. Each tenfold increase in K lowers ΔG^⊖ by 2.303RT = 5.74 kJ mol⁻¹ at 300 K. Compare Problem 5.13: Δ_rG^⊖ = −13.6 kJ mol⁻¹ gave K = 2.4 × 10² (the reverse calculation). Note that ΔG^⊖ is the value for standard states, while ΔG = ΔG^⊖ + RT ln Q depends on the actual composition; at equilibrium ΔG = 0 although ΔG^⊖ is not zero.

More to know. Related relations. The full expression ΔG = ΔG^⊖ + RT ln Q describes a reaction mixture away from equilibrium (Q = reaction quotient); at equilibrium Q = K and ΔG = 0, giving ΔG^⊖ = −RT ln K. Since ΔG^⊖ = ΔH^⊖ − TΔS^⊖, one can also write ln K = −ΔH^⊖/RT + ΔS^⊖/R, which explains how K changes with temperature (K rises with T for endothermic reactions). Using the result ΔG^⊖ = −5.74 kJ mol⁻¹, we can say that at 300 K the products dominate (K = 10) but the reaction is far from going to completion (that would require a very large K, e.g. 10⁶ corresponds to ΔG^⊖ = −34.5 kJ mol⁻¹).

Conclusion: ΔG^⊖ = −2.303 × 8.314 × 300 × log 10 = −5744 J mol⁻¹ (−5.74 kJ mol⁻¹).

Concept. Thermodynamic stability of a compound is judged by its enthalpy of formation (and ultimately its Gibbs energy of formation): a compound with a large negative Δ_fH^⊖ is stable with respect to its elements, whereas a compound with a positive Δ_fH^⊖ lies at a higher energy than its elements and is unstable (thermodynamically). The two reactions given let us comment on (i) the stability of NO relative to its elements and (ii) its stability relative to NO₂.

1. Reaction 1 — formation of NO: ½N₂ + ½O₂ → NO has Δ_rH^⊖ = +90 kJ mol⁻¹ (it is the Δ_fH^⊖ of NO, Table 5.2 gives +90.25). The positive value means 90 kJ is absorbed to make one mole of NO from the elements: NO is at a higher enthalpy than N₂ and O₂, so it is endothermic and thermodynamically unstable with respect to its elements (the energy released on forming NO bonds does not compensate for breaking the strong N≡N and O=O bonds).
2. Reaction 2 — oxidation of NO: NO + ½O₂ → NO₂ has Δ_rH^⊖ = −74 kJ mol⁻¹: exothermic, favourable. So NO readily reacts with oxygen to give NO₂, which lies at a lower enthalpy relative to NO + ½O₂. (By Hess’s law the enthalpy of formation of NO₂ from the elements is 90 + (−74) = +16 kJ mol⁻¹, much less positive than that of NO.)
3. Comment: NO(g) is thermodynamically unstable (Δ_fH^⊖ = +90 kJ mol⁻¹) and, in the presence of O₂, it gets converted to the more stable NO₂ (ΔH = −74 kJ mol⁻¹) — which is why colourless NO turns brown (NO₂) on exposure to air. NO is thus stable only in the sense of persisting in the absence of oxygen; it is kinetically slow to decompose into N₂ and O₂, but not thermodynamically stable.

Extra depth. Thermodynamic stability (energy) must be distinguished from kinetic stability (rate): NO does not decompose rapidly into N₂ and O₂ at ordinary temperature because the activation energy is high, even though Δ_fH is positive; the reaction N₂ + O₂ → 2NO occurs only at very high temperatures (such as in lightning or internal-combustion engines), where the positive ΔS term (which is small) and kinetics allow it. (The positive-ΔH and small ΔS means ΔG_f of NO is also positive at all common temperatures.) NCERT Table 5.2 data: NO +90.25, NO₂ +33.18 kJ mol⁻¹.

Conclusion: NO(g) is thermodynamically unstable (Δ_fH^⊖ = +90 kJ mol⁻¹) and in the presence of O₂ it is converted to the more stable NO₂ (ΔH = −74 kJ mol⁻¹).

Concept. The entropy change of the surroundings is determined by the heat they absorb: the surroundings act as a huge reservoir at constant temperature and pressure, so the heat they receive is q_surr = −q_sys = −ΔH_sys, and the entropy change is ΔS_surr = q_surr/T = −ΔH_sys/T (Section 5.6(c)). An exothermic reaction (ΔH < 0) therefore increases the entropy of the surroundings: heat released raises their molecular randomness.

1. Reaction: H₂(g) + ½O₂(g) → H₂O(l); Δ_fH^⊖ = −286 kJ mol⁻¹. This is exothermic: the system loses 286 kJ and the surroundings gain 286 kJ per mole of water formed.
2. Heat gained by surroundings: q_surr = −ΔH_sys = +286 kJ = +286 × 10³ J.
3. ΔS_surr = q_surr/T with the standard temperature T = 298 K: ΔS_surr = 286 × 10³ J mol⁻¹ / 298 K = +959.7 J K⁻¹ mol⁻¹ (≈ 960 J K⁻¹ mol⁻¹).
4. Interpretation: the large positive entropy gain of the surroundings (+960 J K⁻¹ mol⁻¹) outweighs the decrease in the entropy of the system (H₂ and O₂ gases become liquid water), so ΔS_total > 0 and the formation of water is spontaneous.

Extra depth. This is the same method as Problem 5.11 (iron oxidation: ΔS_surr = 1648 × 10³/298 = 5530 J K⁻¹ mol⁻¹). The question does not state T explicitly; ‘standard conditions’ implies 298 K. At a different temperature ΔS_surr would change (it is inversely proportional to T: the same heat produces less extra disorder in a hotter surroundings). Together with ΔS_sys (negative here) the value gives ΔS_total, the true criterion of spontaneity: ΔS_total = ΔS_sys + ΔS_surr > 0.

Conclusion: ΔS_surr = −ΔH/T = 286 000 J mol⁻¹ / 298 K = +959.7 J K⁻¹ mol⁻¹.