Unit 4: Chemical Thermodynamics — Online MCQ Test
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Practice Unit 4: Chemical Thermodynamics with a free chapter-wise online MCQ test.
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....
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Unit 4: Chemical Thermodynamics — Important Questions & Answers
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.
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.
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).
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.
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.
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 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.
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 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)
- A. T > 298 K
- B. T < 425 K
- C. T > 425 K
- D. All temperatures
Answer: C. T > 425 K
For spontaneity, ΔG < 0, so ΔH < TΔS. T > ΔH/ΔS = 35500 / 83.6 ≈ 425 K.
For spontaneity, ΔG < 0, so ΔH < TΔS. T > ΔH/ΔS = 35500 / 83.6 ≈ 425 K.
Which amongst the following options is the correct relation between change in enthalpy and change in internal energy?
- A. ΔH = ΔU + ΔngRT
- B. ΔH - ΔU = -ΔnRT
- C. ΔH + ΔU = ΔnR
- D. ΔH = ΔU - ΔngRT
Answer: A. ΔH = ΔU + ΔngRT
The standard relation is ΔH = ΔU + PΔV, which for gaseous reactions becomes ΔH = ΔU + ΔngRT.
The standard relation is ΔH = ΔU + PΔV, which for gaseous reactions becomes ΔH = ΔU + ΔngRT.