Practice high-yield Chemistry & Applied Chemical Sciences MCQs on Chemical Thermodynamics with worked academic solutions for MDCAT, CSS, PMS, PPSC, FPSC.
Questions
21
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Exam-Level
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MDCAT, CSS, PMS...
Estimated Time
26 mins
Exam-Level MDCAT CSS PMS PPSC FPSC
Verified for Commission Syllabus • Oct 14, 2026
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Question 1 of 21
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Standard entropy of crystalline carbon monoxide (in kJ/mol) at 0 K is around
Correct Answer: Option D — 5.76
Detailed Academic Explanation:
5.76 is the correct answer to “Standard entropy of crystalline carbon monoxide (in kJ/mol) at 0 K is around”. This question is standard in competitive examination syllabi for Chemical Thermodynamics.
💡Key Conceptual Anchor:Chemical Thermodynamics
Question 2 of 21
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★ MDCAT
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For an irreversible adiabatic expansion of a perfect gas from volume V i to V f , the change in entropy of the gas is
Correct Answer: Option D — greater than zero
Detailed Academic Explanation:
greater than zero is the correct answer to “For an irreversible adiabatic expansion of a perfect gas from volume V i to V f , the change in entropy of the gas is”. This question is standard in competitive examination syllabi for Chemical Thermodynamics.
💡Key Conceptual Anchor:Chemical Thermodynamics
Question 3 of 21
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The change in entropy when two moles of argon gas are heated at constant volume from 300 K to 500 K is
Correct Answer: Option D — 12.74 J/K/mol
Detailed Academic Explanation:
12.74 J/K/mol is the correct answer to “The change in entropy when two moles of argon gas are heated at constant volume from 300 K to 500 K is”. This question is standard in competitive examination syllabi for Chemical Thermodynamics.
💡Key Conceptual Anchor:Chemical Thermodynamics
Question 4 of 21
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For the reaction, N 2 (g) + 3H 2 (g) → 2NH 3 (g) compute the entropy change (in J/K/mol) for the process and comment on the sign of the property. Data Species S°(J/K/mol) NH 3 (g) 192.3 N 2 (g) 191.5 H 2 (g) 130.6
Correct Answer: Option B — ΔS° = -198.7 J/K/mol; negative sign indicates that there is a decrease in the gaseous species during the reaction
Detailed Academic Explanation:
ΔS° = -198.7 J/K/mol; negative sign indicates that there is a decrease in the gaseous species during the reaction is the correct answer to “For the reaction, N 2 (g) + 3H 2 (g) → 2NH 3 (g) compute the entropy change (in J/K/mol) for the process and comment on the sign of the property. Data Species S°(J/K/mol) NH 3 (g) 192.3 N 2 (g) 191.5 H 2 (g) 130.6”. This question is standard in competitive examination syllabi for Chemical Thermodynamics.
💡Key Conceptual Anchor:Chemical Thermodynamics
Question 5 of 21
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★ MDCAT
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Given that standard molar enthalpies of formation of NO(g) and NO 2 (g) are respectively 90.3 kJ/mol and 33.2 kJ/mol, the enthalpy change for the reaction 2NO(g) + O 2 (g) → 2NO 2 (g) is
Correct Answer: Option C — -114.2 kJ
Detailed Academic Explanation:
-114.2 kJ is the correct answer to “Given that standard molar enthalpies of formation of NO(g) and NO 2 (g) are respectively 90.3 kJ/mol and 33.2 kJ/mol, the enthalpy change for the reaction 2NO(g) + O 2 (g) → 2NO 2 (g) is”. This question is standard in competitive examination syllabi for Chemical Thermodynamics.
💡Key Conceptual Anchor:Chemical Thermodynamics
Question 6 of 21
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The standard free energies of formation of H 2 S(g) and CdS(s) at 1000°C are -49.0 kJ/mol and -127.2 kJ/mol respectively. Use these data to predict whether H 2 (g) will reduce CdS(s) to metallic Cd at this temperature.
Correct Answer: Option D — ΔG = 78.2 kJ/mol and the reaction is not feasible
Detailed Academic Explanation:
ΔG = 78.2 kJ/mol and the reaction is not feasible is the correct answer to “The standard free energies of formation of H 2 S(g) and CdS(s) at 1000°C are -49.0 kJ/mol and -127.2 kJ/mol respectively. Use these data to predict whether H 2 (g) will reduce CdS(s) to metallic Cd at this temperature.”. This question is standard in competitive examination syllabi for Chemical Thermodynamics.
💡Key Conceptual Anchor:Chemical Thermodynamics
Question 7 of 21
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For the reaction, 2Cl(g) → Cl 2 (g) the thermodynamic properties
Correct Answer: Option B — ΔG, ΔH and ΔS are negative
Detailed Academic Explanation:
ΔG, ΔH and ΔS are negative is the correct answer to “For the reaction, 2Cl(g) → Cl 2 (g) the thermodynamic properties”. This question is standard in competitive examination syllabi for Chemical Thermodynamics.
💡Key Conceptual Anchor:Chemical Thermodynamics
Question 8 of 21
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According to conventional transition state theory, for elementary bimolecular reactions, the molar entropy of activation ΔS ± is
Correct Answer: Option C — negative
Detailed Academic Explanation:
negative is the correct answer to “According to conventional transition state theory, for elementary bimolecular reactions, the molar entropy of activation ΔS ± is”. This question is standard in competitive examination syllabi for Chemical Thermodynamics.
💡Key Conceptual Anchor:Chemical Thermodynamics
Question 9 of 21
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The enthalpy of formation of AgCl is obtained from the enthalpy change from which one of the following process?
Correct Answer: Option B — Ag(s) + $$\frac{1}{2}$$Cl2(g) → AgCl(s)
Detailed Academic Explanation:
Ag(s) + $$\frac{1}{2}$$Cl2(g) → AgCl(s) is the correct answer to “The enthalpy of formation of AgCl is obtained from the enthalpy change from which one of the following process”. This question is standard in competitive examination syllabi for Chemical Thermodynamics.
💡Key Conceptual Anchor:Chemical Thermodynamics
Question 10 of 21
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For the process, 1Ar(300 K, 1 bar) → 1Ar(200 K, 10 bar) assuming ideal gas behaviour, the change in molar entropy is
Correct Answer: Option A — -27.57 J/K/mol
Detailed Academic Explanation:
-27.57 J/K/mol is the correct answer to “For the process, 1Ar(300 K, 1 bar) → 1Ar(200 K, 10 bar) assuming ideal gas behaviour, the change in molar entropy is”. This question is standard in competitive examination syllabi for Chemical Thermodynamics.
💡Key Conceptual Anchor:Chemical Thermodynamics
Question 11 of 21
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The rate of evaporation of a liquid is always faster at a higher temperature because
Correct Answer: Option A — the enthalpy of vaporization is always endothermic
Detailed Academic Explanation:
the enthalpy of vaporization is always endothermic is the correct answer to “The rate of evaporation of a liquid is always faster at a higher temperature because”. This question is standard in competitive examination syllabi for Chemical Thermodynamics.
💡Key Conceptual Anchor:Chemical Thermodynamics
Question 12 of 21
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Choose the correct criterion of spontaneity in terms of the properties of the system alone.
Correct Answer: Option A — (dS)U,V > 0
Detailed Academic Explanation:
(dS)U,V > 0 is the correct answer to “Choose the correct criterion of spontaneity in terms of the properties of the system alone.”. This question is standard in competitive examination syllabi for Chemical Thermodynamics.
💡Key Conceptual Anchor:Chemical Thermodynamics
Question 13 of 21
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The criterion for the spontaneity of a process is
Correct Answer: Option C — ΔSsys + ΔSsurr > 0
Detailed Academic Explanation:
ΔSsys + ΔSsurr > 0 is the correct answer to “The criterion for the spontaneity of a process is”. This question is standard in competitive examination syllabi for Chemical Thermodynamics.
💡Key Conceptual Anchor:Chemical Thermodynamics
Question 14 of 21
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The molar entropy of crystalline CO at absolute zero is
Correct Answer: Option C — R In2
Detailed Academic Explanation:
R In2 is the correct answer to “The molar entropy of crystalline CO at absolute zero is”. This question is standard in competitive examination syllabi for Chemical Thermodynamics.
V is the correct answer to “$${\left( {\frac{{\partial G}}{{\partial p}}} \right)_T} = $$”. This question is standard in competitive examination syllabi for Chemical Thermodynamics.
💡Key Conceptual Anchor:Chemical Thermodynamics
Question 16 of 21
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Among the following, the system that would require the least amount of thermal energy to bring its temperature to 80°C is
Correct Answer: Option C — 150 g of water at 50°C
Detailed Academic Explanation:
150 g of water at 50°C is the correct answer to “Among the following, the system that would require the least amount of thermal energy to bring its temperature to 80°C is”. This question is standard in competitive examination syllabi for Chemical Thermodynamics.
💡Key Conceptual Anchor:Chemical Thermodynamics
Question 17 of 21
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Among the following, the reaction that is accompanied by a decrease in the entropy is
Correct Answer: Option A — N2(g) + 3H2(g) → 2NH3(g)
Detailed Academic Explanation:
N2(g) + 3H2(g) → 2NH3(g) is the correct answer to “Among the following, the reaction that is accompanied by a decrease in the entropy is”. This question is standard in competitive examination syllabi for Chemical Thermodynamics.
💡Key Conceptual Anchor:Chemical Thermodynamics
Question 18 of 21
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The criterion for spontaneous change in terms of the state functions is
Correct Answer: Option D — dGp,V ≤ 0
Detailed Academic Explanation:
dGp,V ≤ 0 is the correct answer to “The criterion for spontaneous change in terms of the state functions is”. This question is standard in competitive examination syllabi for Chemical Thermodynamics.
💡Key Conceptual Anchor:Chemical Thermodynamics
Question 19 of 21
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The free energy change (ΔG) of 1 mole of an ideal gas that is compressed isothermally from 1 atm to 2 atm is
Correct Answer: Option A — RT In2
Detailed Academic Explanation:
RT In2 is the correct answer to “The free energy change (ΔG) of 1 mole of an ideal gas that is compressed isothermally from 1 atm to 2 atm is”. This question is standard in competitive examination syllabi for Chemical Thermodynamics.
💡Key Conceptual Anchor:Chemical Thermodynamics
Question 20 of 21
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Among W (work), Q (heat), U (internal energy) and S (entropy)
Correct Answer: Option D — W and Q are path functions but U and S are state functions
Detailed Academic Explanation:
W and Q are path functions but U and S are state functions is the correct answer to “Among W (work), Q (heat), U (internal energy) and S (entropy)”. This question is standard in competitive examination syllabi for Chemical Thermodynamics.
💡Key Conceptual Anchor:Chemical Thermodynamics
Question 21 of 21
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ΔH and ΔE for the reaction, Fe 2 O 3 (s) + 3H 2 (g) → 2Fe(s) + H 2 O($$l$$) at constant temperature are related as
Correct Answer: Option D — ΔH = ΔE - 3RT
Detailed Academic Explanation:
ΔH = ΔE - 3RT is the correct answer to “ΔH and ΔE for the reaction, Fe 2 O 3 (s) + 3H 2 (g) → 2Fe(s) + H 2 O($$l$$) at constant temperature are related as”. This question is standard in competitive examination syllabi for Chemical Thermodynamics.
💡Key Conceptual Anchor:Chemical Thermodynamics
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💡Examiner's Advisory for Chemistry & Applied Chemical Sciences
In competitive commission exams (FPSC, PPSC, KPPSC, SPSC, CSS, PMS), questions are frequently phrased with distractor options designed to catch rote learners. Whenever you answer an MCQ, read the Detailed Academic Explanation and Conceptual Anchor to understand why the remaining options are incorrect.