Chemical Thermodynamics

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Chemical Thermodynamics — solved practice questions

8 JEE Chemistry questions with step-by-step solutions. Attempt each, then reveal the answer.

1. Q1easy

   For an isothermal process on an ideal gas, which of the following is always true?

   • AChange in internal energy is zero
   • BHeat exchanged is zero
   • CWork done is zero
   • DEnthalpy change is non-zero
   Show answer & solution

   Correct answer: (A) Change in internal energy is zero

   The internal energy of an ideal gas depends only on temperature. In an isothermal process temperature is constant, so the change in internal energy (delta U) is zero.

2. Q2easy

   Which one of the following is an intensive property?

   • AVolume
   • BEnthalpy
   • CTemperature
   • DInternal energy
   Show answer & solution

   Correct answer: (C) Temperature

   Temperature does not depend on the amount of substance, so it is an intensive property. Volume, enthalpy, and internal energy are extensive properties that depend on the quantity of matter.

3. Q3medium

   The standard enthalpy of formation of CO(g) using the data delta H(C + O2 -> CO2) = -393.5 kJ/mol and delta H(CO + 1/2 O2 -> CO2) = -283.0 kJ/mol is:

   • A-676.5 kJ/mol
   • B-110.5 kJ/mol
   • C+110.5 kJ/mol
   • D-283.0 kJ/mol
   Show answer & solution

   Correct answer: (B) -110.5 kJ/mol

   By Hess's law, delta Hf(CO) = delta H(C -> CO2) − delta H(CO -> CO2) = -393.5 − (-283.0) = -110.5 kJ/mol.

4. Q4medium

   For the reaction H2(g) + Cl2(g) -> 2HCl(g), the bond enthalpies are H–H = 436, Cl–Cl = 242, and H–Cl = 431 kJ/mol. The enthalpy of the reaction is:

   • A+184 kJ/mol
   • B-247 kJ/mol
   • C+247 kJ/mol
   • D-184 kJ/mol
   Show answer & solution

   Correct answer: (D) -184 kJ/mol

   delta H = (bonds broken) − (bonds formed) = (436 + 242) − 2(431) = 678 − 862 = -184 kJ/mol.

5. Q5medium

   When 2 moles of an ideal gas expand isothermally and reversibly to double their volume at 300 K, the entropy change of the gas (R = 8.314 J/K/mol) is approximately:

   • A+11.5 J/K
   • B-11.5 J/K
   • C+5.76 J/K
   • D+23.1 J/K
   Show answer & solution

   Correct answer: (A) +11.5 J/K

   For isothermal reversible expansion, delta S = nR ln(V2/V1) = 2 x 8.314 x ln 2 = 11.53 J/K. The change is positive since volume increases.

6. Q6medium

   A reaction is spontaneous at all temperatures when:

   • Adelta H > 0 and delta S > 0
   • Bdelta H < 0 and delta S > 0
   • Cdelta H < 0 and delta S < 0
   • Ddelta H > 0 and delta S < 0
   Show answer & solution

   Correct answer: (B) delta H < 0 and delta S > 0

   delta G = delta H − T delta S must be negative for spontaneity. If delta H < 0 (exothermic) and delta S > 0, then delta G is negative at every temperature.

7. Q7hard

   For an endothermic reaction with delta H = +30 kJ and delta S = +100 J/K, the temperature above which the reaction becomes spontaneous is:

   • A30 K
   • B100 K
   • C300 K
   • D3000 K
   Show answer & solution

   Correct answer: (C) 300 K

   At equilibrium delta G = 0, so T = delta H / delta S = 30000 J / 100 J/K = 300 K. Above 300 K the T delta S term dominates and delta G becomes negative, making it spontaneous.

8. Q8hard

   One mole of an ideal gas expands isothermally and reversibly at 300 K from 1 L to 10 L. The work done by the gas (R = 8.314 J/K/mol) is approximately:

   • A2.29 kJ
   • B13.2 kJ
   • C0.574 kJ
   • D5.74 kJ
   Show answer & solution

   Correct answer: (D) 5.74 kJ

   w done by gas = nRT ln(V2/V1) = 1 x 8.314 x 300 x ln 10 = 5743 J ≈ 5.74 kJ. (The work done on the gas is -5.74 kJ.)

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