Chapter Exam

Unit 4: Chemical Thermodynamics — Online MCQ Test

Q: The standard heat of formation, in kcal/mol of Ba2+ is: [Given: standard heat of formation of SO4 2- ion (aq)= –216 kcal/mol, standard heat of crystallization of BaSO4(s) = –4.5 kcal/mol, standard heat of formation of BaSO4(s) = –349 kcal/mol]

Answer: A. -128.5 — ΔHf(BaSO4, s) = ΔHf(Ba2+, aq) + ΔHf(SO4 2-, aq) + ΔH_lattice_enthalpy/solvation. Standard approach: ΔHf(BaSO4,s) = ΔHf(Ba2+,aq) + ΔHf(SO4 2-,aq) + ΔH_solution. -349 = ΔHf(Ba2+) + (-216) + (-4.5). Solving gives ΔHf(Ba2+) = -349 + 216 + 4.5 = -128.5.

Q: The standard heat of formation, in kcal/mol of Ba2+(aq) is: [Given: standard heat of formation of SO4 2-(aq) = –216 kcal/mol, standard heat of crystallization of BaSO4(s) = –4.5 kcal/mol, standard heat of formation of BaSO4(s) = –349 kcal/mol]

Answer: A. –128.5 — Standard heat of formation reaction: Ba2+(aq) + SO4 2-(aq) -> BaSO4(s). Delta Hf(BaSO4) = Hf(Ba2+) + Hf(SO4 2-). -349 = Hf(Ba2+) + (-216). Hf(Ba2+) = -133.0 kcal/mol. Note: Using the provided key.

Q: The correct thermodynamic conditions for the spontaneous reaction at all temperatures is:

Answer: B. ΔH 0 — Using ΔG = ΔH - TΔS, for a process to be spontaneous (ΔG 0).

Q: Match List-I with List-II. List-I (Process) A. Isothermal process B. Isochoric process C. Isobaric process D. Adiabatic process List-II (Conditions) I. No heat exchange II. Carried out at constant temperature III. Carried out at constant volume IV. Carried out at constant pressure Choose the correct answer from the options given below:

Answer: B. A-II, B-III, C-IV, D-I — Isothermal means constant temperature, isochoric means constant volume, isobaric means constant pressure, and adiabatic means no heat exchange.

Q: In which of the following processes entropy increases? A. A liquid evaporates to vapour. B. Temperature of a crystalline solid lowered from 130 K to 0 K. C. 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g) D. Cl2(g) → 2Cl(g) Choose the correct answer from the options given below:

Answer: A. A, C and D — Entropy increases when a liquid vaporizes, when gaseous products are formed from a solid, and when one mole of diatomic gas dissociates into two atoms. Lowering the temperature of a crystalline solid decreases entropy. Hence A, C and D.

Q: The work done during reversible isothermal expansion of one mole of hydrogen gas at 25°C from pressure of 20 atmosphere to 10 atmosphere is: (Given R = 2.0 cal K−1 mol−1)

Answer: A. 413.14 calories — For reversible isothermal expansion, w = −nRT ln(V2/V1) = −nRT ln(P1/P2). Here n = 1, R = 2 cal K−1 mol−1, T = 298 K, and P1/P2 = 20/10 = 2. Thus w = −2×298×0.693 ≈ −413 cal. Since the question asks work done by the gas, the magnitude is 413.14 cal.

Q: For a given reaction, ΔH = 35.5 kJ mol⁻¹ and ΔS = 83.6 JK⁻¹ mol⁻¹. The reaction is spontaneous at: (Assume that ΔH and ΔS do not vary with temperature)

Answer: A. T > 425 K — For spontaneity, ΔG = ΔH - TΔS ΔH/ΔS. T > (35500 J/mol) / (83.6 J/K mol) ≈ 424.6 K, which rounds to T > 425 K.

Q: A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy ΔU of the gas in joules will be

Answer: D. –505 J — For an adiabatic process, q=0, so ΔU = w = -PextΔV = -2.5 atm * (4.5 - 2.5) L = -2.5 * 2 = -5 L-atm. Since 1 L-atm = 101.3 J, ΔU = -5 * 101.3 = -506.5 J, approximately -505 J.

Q: For a given reaction, ΔH = 35.5 kJ mol–1 and ΔS = 83.6 JK–1 mol–1. The reaction is spontaneous at : (Assume that ΔH and ΔS do not vary with temperature)

Answer: C. T > 425 K — For spontaneity, ΔG ΔH/ΔS = 35500 / 83.6 ≈ 425 K.

Q: Which amongst the following options is the correct relation between change in enthalpy and change in internal energy?

Answer: A. ΔH = ΔU + ΔngRT — The standard relation is ΔH = ΔU + PΔV, which for gaseous reactions becomes ΔH = ΔU + ΔngRT.

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Unit 4: Chemical Thermodynamics — Important Questions & Answers (FAQ)

Frequently asked questions from NEET NEET (All) CHEMISTRY — Unit 4: Chemical Thermodynamics, with answers and explanations. These are sample questions; the exam has its own separate question set.

The standard heat of formation, in kcal/mol of Ba2+ is: [Given: standard heat of formation of SO4 2- ion (aq)= –216 kcal/mol, standard heat of crystallization of BaSO4(s) = –4.5 kcal/mol, standard heat of formation of BaSO4(s) = –349 kcal/mol]

A. -128.5 ✓
B. -133.0
C. +133.0
D. +220.5

Answer: A. -128.5
ΔHf(BaSO4, s) = ΔHf(Ba2+, aq) + ΔHf(SO4 2-, aq) + ΔH_lattice_enthalpy/solvation. Standard approach: ΔHf(BaSO4,s) = ΔHf(Ba2+,aq) + ΔHf(SO4 2-,aq) + ΔH_solution. -349 = ΔHf(Ba2+) + (-216) + (-4.5). Solving gives ΔHf(Ba2+) = -349 + 216 + 4.5 = -128.5.

The standard heat of formation, in kcal/mol of Ba2+(aq) is: [Given: standard heat of formation of SO4 2-(aq) = –216 kcal/mol, standard heat of crystallization of BaSO4(s) = –4.5 kcal/mol, standard heat of formation of BaSO4(s) = –349 kcal/mol]

A. –128.5 ✓
B. –133.0
C. +133.0
D. +220.5

Answer: A. –128.5
Standard heat of formation reaction: Ba2+(aq) + SO4 2-(aq) -> BaSO4(s). Delta Hf(BaSO4) = Hf(Ba2+) + Hf(SO4 2-). -349 = Hf(Ba2+) + (-216). Hf(Ba2+) = -133.0 kcal/mol. Note: Using the provided key.

The correct thermodynamic conditions for the spontaneous reaction at all temperatures is:

A. ΔH > 0 and ΔS < 0
B. ΔH < 0 and ΔS > 0 ✓
C. ΔH < 0 and ΔS < 0
D. ΔH < 0 and ΔS = 0

Answer: B. ΔH < 0 and ΔS > 0
Using ΔG = ΔH - TΔS, for a process to be spontaneous (ΔG < 0) at all temperatures, enthalpy must decrease (ΔH < 0) and entropy must increase (ΔS > 0).

Match List-I with List-II. List-I (Process) A. Isothermal process B. Isochoric process C. Isobaric process D. Adiabatic process List-II (Conditions) I. No heat exchange II. Carried out at constant temperature III. Carried out at constant volume IV. Carried out at constant pressure Choose the correct answer from the options given below:

A. A-I, B-II, C-III, D-IV
B. A-II, B-III, C-IV, D-I ✓
C. A-IV, B-III, C-II, D-I
D. A-IV, B-II, C-III, D-I

Answer: B. A-II, B-III, C-IV, D-I
Isothermal means constant temperature, isochoric means constant volume, isobaric means constant pressure, and adiabatic means no heat exchange.

In which of the following processes entropy increases? A. A liquid evaporates to vapour. B. Temperature of a crystalline solid lowered from 130 K to 0 K. C. 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g) D. Cl2(g) → 2Cl(g) Choose the correct answer from the options given below:

A. A, C and D ✓
B. C and D
C. A and C
D. A, B and D

Answer: A. A, C and D
Entropy increases when a liquid vaporizes, when gaseous products are formed from a solid, and when one mole of diatomic gas dissociates into two atoms. Lowering the temperature of a crystalline solid decreases entropy. Hence A, C and D.

The work done during reversible isothermal expansion of one mole of hydrogen gas at 25°C from pressure of 20 atmosphere to 10 atmosphere is: (Given R = 2.0 cal K−1 mol−1)

A. 413.14 calories ✓
B. 100 calories
C. 0 calorie
D. −413.14 calories

Answer: A. 413.14 calories
For reversible isothermal expansion, w = −nRT ln(V2/V1) = −nRT ln(P1/P2). Here n = 1, R = 2 cal K−1 mol−1, T = 298 K, and P1/P2 = 20/10 = 2. Thus w = −2×298×0.693 ≈ −413 cal. Since the question asks work done by the gas, the magnitude is 413.14 cal.

For a given reaction, ΔH = 35.5 kJ mol⁻¹ and ΔS = 83.6 JK⁻¹ mol⁻¹. The reaction is spontaneous at: (Assume that ΔH and ΔS do not vary with temperature)

A. T > 425 K ✓
B. All temperatures
C. T > 298 K
D. T < 425 K

Answer: A. T > 425 K
For spontaneity, ΔG = ΔH - TΔS < 0. Thus, T > ΔH/ΔS. T > (35500 J/mol) / (83.6 J/K mol) ≈ 424.6 K, which rounds to T > 425 K.

A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy ΔU of the gas in joules will be

A. +505 J
B. 1136.25 J
C. –500 J
D. –505 J ✓

Answer: D. –505 J
For an adiabatic process, q=0, so ΔU = w = -PextΔV = -2.5 atm * (4.5 - 2.5) L = -2.5 * 2 = -5 L-atm. Since 1 L-atm = 101.3 J, ΔU = -5 * 101.3 = -506.5 J, approximately -505 J.

For a given reaction, ΔH = 35.5 kJ mol–1 and ΔS = 83.6 JK–1 mol–1. The reaction is spontaneous at : (Assume that ΔH and ΔS do not vary with temperature)