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.

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Practice Unit 4: Chemical Thermodynamics with a free chapter-wise online MCQ test for NEET NEET (All) CHEMISTRY. This chapter covers: Fundamentals of thermodynamics: System and surroundings extensive and intensive properties state functions types of processes. The first law of thermodynamics - Concept of work hea.... AI-generated questions from basic to board-exam level, with instant results and explanations.

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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.