Class 11 Chemistry Chapter 5: Thermodynamics worksheet with answer key

Answer: (c) ΔS of the gas

Expansion into vacuum is free expansion: p_ex = 0, so w = 0; Joule found q = 0, so ΔU = q + w = 0. The expansion is spontaneous and irreversible, so ΔS_total > 0; the surroundings receive no heat (ΔS_surr = 0), so the whole increase is in the gas, which has spread into a larger volume: ΔS of the gas > 0.

Answer: (b) S(rhombic)

Δ_fH⊖ is zero only for an element in its reference state, its most stable form at 25 °C and 1 bar. For sulphur that is rhombic sulphur; C(diamond) is +1.89 and Br₂(g) is +30.91 kJ mol⁻¹ (bromine's reference state is the liquid), and oxygen's reference state is O₂(g), not O₃.

Answer: (b) a closed system

Glass conducts heat, so energy passes between the bath and the mixture, but the seal stops any matter leaving or entering: a closed system.

Answer: (a) a pure crystalline solid

The third law is confined to pure crystalline solids, which are perfectly ordered at 0 K; the entropy of solutions and supercooled liquids is not zero at absolute zero.

Answer: (c) 0.90 J g⁻¹ K⁻¹

q = c × m × ΔT, so c = 3600 J / (200 g × 20 K) = 0.90 J g⁻¹ K⁻¹.

ΔG = ΔH − TΔS. At low T the positive ΔH outweighs TΔS and ΔG > 0. As T rises, TΔS grows until it exceeds ΔH and ΔG < 0. When ΔS is small, T must be large for TΔS to outweigh ΔH, which is one reason reactions are run at high temperature.

• low T: ΔH > TΔS, ΔG > 0; high T: TΔS > ΔH, ΔG < 0 (1)
• small ΔS needs large T: hence high-temperature operation (1)

(i) Volume halves and (iv) enthalpy halves: both are extensive and depend on the amount of matter. (ii) Temperature stays 300 K and (iii) molar volume V/n stays the same: they are intensive and do not depend on the amount.

• volume and enthalpy halve (extensive) (1)
• temperature and molar volume unchanged (intensive) (1)

Solid → gas directly and solid → liquid → gas start and end at the same states. Since H is a state function, Δ_subH⊖ = Δ_fusH⊖ + Δ_vapH⊖, so Δ_vapH⊖ = 62.0 − 8.0 = 54.0 kJ mol⁻¹.

• Δ_vapH⊖ = 54.0 kJ mol⁻¹ (1)
• enthalpy is a state function (Hess's law): the two routes share their ends (1)

(i) Isolated: the closed insulated flask allows neither matter nor energy to pass. (ii) Closed: no matter leaves the sealed can, but heat flows out through the conducting steel. (iii) Open: steam escapes and heat enters, so both matter and energy are exchanged. Only the thermos flask (i) has an adiabatic wall, allowing no heat exchange.

• (i) isolated, no exchange of matter or energy (1)
• (ii) closed, energy but no matter; (iii) open, both (1)
• thermos flask is adiabatic (1)

(a) ΔS_surr = −Δ_rH/T = −30000/300 = −100 J K⁻¹ mol⁻¹. ΔS_sys = ΔS_total − ΔS_surr = −20 − (−100) = +80 J K⁻¹ mol⁻¹. (b) ΔS_total < 0: not spontaneous. (c) ΔG = ΔH − TΔS_sys = 30.0 − 300 × 0.080 = +6.0 kJ mol⁻¹; −TΔS_total = −300 × (−0.020) = +6.0 kJ mol⁻¹, the same, since TΔS_total = TΔS_sys − ΔH_sys.

• ΔS_surr = −100 J K⁻¹ mol⁻¹ and ΔS_sys = +80 J K⁻¹ mol⁻¹ (1)
• ΔS_total < 0, not spontaneous (1)
• ΔG = +6.0 kJ mol⁻¹ both ways (1)

Isothermal change of an ideal gas: ΔU = 0, so w = −q = −3.46 kJ. For the reversible path q = nRT ln(V_f/V_i): ln(V_f/V_i) = 3460 / (1.00 × 8.314 × 300) = 3460/2494.2 = 1.387, so V_f/V_i = 4.0. The volume increased four times.

• ΔU = 0 for isothermal change of an ideal gas (1)
• w = −3.46 kJ (1)
• ln(V_f/V_i) = 1.387, factor 4 (1)