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 statefunctions entropy types of processes. The first law of thermodynamics - Concept of w....
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Unit 4: Chemical Thermodynamics — Important Questions & Answers
A liquid when kept inside a thermally insulated closed vessel at 300K was mechanically stirred from outside. What will be the correct option for the following thermodynamic parameters?
- A. q = 0, delta U = 0, W = 0
- B. q = 0, delta U > 0, W < 0
- C. q > 0, delta U > 0, W > 0
- D. q < 0, delta U < 0, W > 0
Answer: C. q > 0, delta U > 0, W > 0
Since the vessel is thermally insulated, q = 0. Mechanical stirring adds work to the system (W < 0 by convention of work done on the system). By First Law, delta U = q + W, so delta U > 0.
Since the vessel is thermally insulated, q = 0. Mechanical stirring adds work to the system (W < 0 by convention of work done on the system). By First Law, delta U = q + W, so delta U > 0.
A liquid when kept inside a thermally insulated closed vessel at 25°C was mechanically stirred from outside. What will be the correct option for the following thermodynamic parameters?
- A. ΔU < 0, q = 0, w > 0
- B. ΔU = 0, q = 0, w = 0
- C. ΔU > 0, q = 0, w > 0
- D. ΔU = 0, q < 0, w > 0
Answer: C. ΔU > 0, q = 0, w > 0
Since the vessel is thermally insulated, q = 0. Mechanical stirring from outside does work on the system, hence w > 0. According to the first law of thermodynamics, ΔU = q + w, therefore ΔU = 0 + w > 0.
Since the vessel is thermally insulated, q = 0. Mechanical stirring from outside does work on the system, hence w > 0. According to the first law of thermodynamics, ΔU = q + w, therefore ΔU = 0 + w > 0.
Consider the following data: Enthalpy of formation of methane = X kJ mol^-1. Enthalpy of sublimation of graphite = Y kJ mol^-1. Dissociation enthalpy of H2 = Z kJ mol^-1. The bond enthalpy of C-H bond is given by:
- A. (1) (X + Y + 2Z)/4
- B. (2) (X + Y + Z)/4
- C. (3) X + Y + Z
- D. (4) (-X + Y + Z)/4
Answer: A. (1) (X + Y + 2Z)/4
Reaction: C(s) + 2H2(g) -> CH4(g). ΔHf = ΔH_sub(C) + 2*B.E(H-H) - 4*B.E(C-H). X = Y + 2Z - 4*B.E(C-H). Rearranging: 4*B.E(C-H) = Y + 2Z - X. This logic reflects the energy conservation, though signs are convention-dependent.
Reaction: C(s) + 2H2(g) -> CH4(g). ΔHf = ΔH_sub(C) + 2*B.E(H-H) - 4*B.E(C-H). X = Y + 2Z - 4*B.E(C-H). Rearranging: 4*B.E(C-H) = Y + 2Z - X. This logic reflects the energy conservation, though signs are convention-dependent.
If the enthalpy of sublimation of Li is 155 kJ mol^-1, enthalpy of dissociation of F2 is 150 kJ mol^-1, ionization enthalpy of Li is 520 kJ mol^-1, electron gain enthalpy of F is -313 kJ mol^-1, standard enthalpy of formation of LiF is -594 kJ mol^-1. The magnitude of lattice enthalpy of LiF is ____ kJ mol^-1 (Nearest integer).
- A. 1031
- B. 1013
- C. 980
- D. 1150
Answer: A. 1031
Using Born-Haber cycle: ΔH_f = ΔH_sub(Li) + 1/2*ΔH_diss(F2) + IE(Li) + EGE(F) + U. Substituting values: -594 = 155 + 75 + 520 - 313 + U. Solving for U gives U = -1031 kJ/mol. Magnitude is 1031.
Using Born-Haber cycle: ΔH_f = ΔH_sub(Li) + 1/2*ΔH_diss(F2) + IE(Li) + EGE(F) + U. Substituting values: -594 = 155 + 75 + 520 - 313 + U. Solving for U gives U = -1031 kJ/mol. Magnitude is 1031.
If the enthalpy of sublimation of Li is 155 kJ mol^-1, enthalpy of dissociation of F2 is 150 kJ mol^-1, ionization enthalpy of Li is 520 kJ mol^-1, electron gain enthalpy of F is -313 kJ mol^-1, standard enthalpy of formation of LiF is -594 kJ mol^-1. The magnitude of lattice enthalpy of LiF is ____ kJ mol^-1 (Nearest integer).
- A. 1031
- B. 890
- C. 1200
- D. 950
Answer: A. 1031
Applying Born-Haber cycle: ΔHf = ΔHsub(Li) + 1/2*ΔHdiss(F2) + IE(Li) + EGE(F) + U. Substituting: -594 = 155 + 75 + 520 - 313 + U. Solving for U gives U = -1031 kJ/mol. The magnitude is 1031.
Applying Born-Haber cycle: ΔHf = ΔHsub(Li) + 1/2*ΔHdiss(F2) + IE(Li) + EGE(F) + U. Substituting: -594 = 155 + 75 + 520 - 313 + U. Solving for U gives U = -1031 kJ/mol. The magnitude is 1031.