Chemical Thermodynamics

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Chemical Thermodynamics — solved practice questions

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

1. Q1easy

   For an isolated system, which of the following is always true?

   • Adelta U = 0
   • Bdelta H = 0
   • Cq = w
   • Ddelta S = 0
   Show answer & solution

   Correct answer: (A) delta U = 0

   An isolated system exchanges neither matter nor energy with its surroundings, so its internal energy remains constant (delta U = 0).

2. Q2easy

   The first law of thermodynamics is expressed as:

   • Adelta U = q - w
   • Bdelta U = q + w
   • Cdelta H = q + w
   • Ddelta U = w - q
   Show answer & solution

   Correct answer: (B) delta U = q + w

   The first law states delta U = q + w, where q is heat added to the system and w is work done on the system (IUPAC sign convention).

3. Q3medium

   For the reaction N2(g) + 3H2(g) -> 2NH3(g) at constant temperature, the relation between delta H and delta U is (delta n(gas) = -2):

   • Adelta H = delta U + 2RT
   • Bdelta H = delta U
   • Cdelta H = delta U - 2RT
   • Ddelta H = delta U - 4RT
   Show answer & solution

   Correct answer: (C) delta H = delta U - 2RT

   delta H = delta U + delta n(gas) RT. Here delta n(gas) = 2 - 4 = -2, so delta H = delta U - 2RT.

4. Q4hard

   The enthalpy of combustion of carbon to CO2 is -393.5 kJ/mol and that of carbon monoxide to CO2 is -283.0 kJ/mol. The enthalpy of formation of CO from carbon is:

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

   Correct answer: (A) -110.5 kJ/mol

   By Hess's law: C + O2 -> CO2 (-393.5) minus CO + 1/2 O2 -> CO2 (-283.0) gives C + 1/2 O2 -> CO = -393.5 - (-283.0) = -110.5 kJ/mol.

5. Q5medium

   For a spontaneous process at constant temperature and pressure, the correct condition is:

   • Adelta G > 0
   • Bdelta G < 0
   • Cdelta G = 0
   • Ddelta H < 0 only
   Show answer & solution

   Correct answer: (B) delta G < 0

   A process is spontaneous at constant T and P when the Gibbs free energy change is negative, i.e. delta G < 0.

6. Q6easy

   The relationship between Gibbs free energy change, enthalpy change and entropy change is:

   • Adelta G = delta H + T(delta S)
   • Bdelta G = T(delta S) - delta H
   • Cdelta G = delta H x T(delta S)
   • Ddelta G = delta H - T(delta S)
   Show answer & solution

   Correct answer: (D) delta G = delta H - T(delta S)

   The Gibbs–Helmholtz relation is delta G = delta H - T(delta S), valid at constant temperature and pressure.

7. Q7hard

   A reaction has delta H = +30 kJ/mol and delta S = +100 J/K/mol. Above what temperature does the reaction become spontaneous?

   • A100 K
   • B273 K
   • C300 K
   • D3000 K
   Show answer & solution

   Correct answer: (C) 300 K

   The reaction is spontaneous when delta G = delta H - T(delta S) < 0, i.e. T > delta H / delta S = 30000 J / 100 J/K = 300 K.

8. Q8medium

   For an ideal gas undergoing isothermal expansion into a vacuum (free expansion), the value of the work done is:

   • AZero
   • BPositive and maximum
   • CNegative and maximum
   • DEqual to nRT ln(V2/V1)
   Show answer & solution

   Correct answer: (A) Zero

   In free expansion the external pressure is zero, so work w = -P(ext) x delta V = 0. Since it is isothermal for an ideal gas, delta U = 0 and q = 0 as well.

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