Chapter Exam

Unit 4: Chemical Thermodynamics — Online MCQ Test

Q: 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?

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.

Q: 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?

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.

Q: 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:

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.

Q: 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).

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.

Q: 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).

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.

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

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

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.

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.

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.

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.

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.

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