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Thermodynamics

JEE Main 2025 (Online) 2nd April Evening ShiftMCQHard📈 Topic in 29% papers+4−1
Ques. 1JEE Main 2025, 2 April (Evening Shift)

Which of the following graphs correctly represents the variation of thermodynamic properties of Haber's process?

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]

Answer

(1) [figure]

Solution

Haber's process: N2(g) + 3H2(g) 2NH3(g)

Here rH -92 kJ mol-1 (negative, nearly independent of T) and rS -198 J K-1 mol-1 (4 mol gas 2 mol gas; negative, nearly constant).

So - H T = + 92000T is positive ( +300 J K-1 mol-1 at 300 K) and falls as T rises.

G = H - T S ,

- G T = - H T + S

- G T 300 - 200 = +100 at 300 K; it falls with T and becomes negative above T = H S 465 K.

Graph (1) shows exactly this: - H /T from about +300 falling, - G /T from about +100 falling through zero, H about −100 and S about −200, both nearly constant.

(2) plots H /T as negative and - H as a falling curve; (3) puts - G /T above - H /T, which needs S > 0; (4) shows S and H positive. All three are wrong.

Answer: (1)

Keep practising — Thermodynamics

Ques. 2JEE Main 2025, 29 January (Morning Shift)

500 J of energy is transferred as heat to 0.5 mol of Argon gas at 298 K and 1.00 atm. The final temperature and the change in internal energy respectively are: Given: R = 8.3 J K-1 mol-1

  1. 368 K and 500 J
  2. 348 K and 300 J
  3. 378 K and 300 J
  4. 378 K and 500 J
Ques. 3JEE Main 2025, 29 January (Evening Shift)

If C (diamond) C (graphite) + X kJ mol-1 C (diamond) + O2(g) CO2(g) + Y kJ mol-1 C (graphite) + O2(g) CO2(g) + Z kJ mol-1 at constant temperature. Then

  1. -X = Y + Z
  2. X = Y - Z
  3. X = -Y + Z
  4. X = Y + Z
Ques. 4JEE Main 2025, 28 January (Morning Shift)

The formation enthalpies, Hf for H(g) and O(g) are 220.0 and 250.0 kJ mol-1, respectively, at 298.15 K, and Hf for H2O(g) is -242.0 kJ mol-1 at the same temperature. The average bond enthalpy of the O-H bond in water at 298.15 K is kJ mol-1 (nearest integer).

Ques. 5JEE Main 2025, 28 January (Evening Shift)

[figure] An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path A→B→C→D→A as shown in the three cases above. Choose the correct option regarding U :

  1. U(Case-III) > U(Case-II) > U(Case-I)
  2. U(Case-I) > U(Case-II) > U(Case-III)
  3. U(Case-I) = U(Case-II) = U(Case-III)
  4. U(Case-I) > U(Case-III) > U(Case-II)
Ques. 6JEE Main 2025, 28 January (Evening Shift)

Consider the following data: Heat of formation of CO2(g) = -393.5 kJ mol-1 Heat of formation of H2O(l) = -286.0 kJ mol-1 Heat of combustion of benzene = -3267.0 kJ mol-1 The heat of formation of benzene is kJ mol-1. (Nearest integer)

Ques. 7JEE Main 2025, 24 January (Morning Shift)

Let us consider an endothermic reaction which is non-spontaneous at the freezing point of water. However, the reaction is spontaneous at boiling point of water. Choose the correct option.

  1. Both and are (-ve)
  2. is (+ve) but is (-ve)
  3. is (-ve) but is (+ve)
  4. Both and are (+ve)
Ques. 8JEE Main 2025, 24 January (Morning Shift)

Standard entropies of X2, Y2 and XY5 are 70, 50 and 110 J K-1 mol-1 respectively. The temperature in Kelvin at which the reaction 12X2 + 52Y2 XY5 H = -35 kJ mol-1 will be at equilibrium is (Nearest integer)

Ques. 9JEE Main 2025, 24 January (Evening Shift)

S(g)+ 32O2(g) SO3(g)+2x kcal SO2(g)+ 12O2(g) SO3(g)+y kcal The heat of formation of SO2(g) is given by :

  1. 2x+y kcal
  2. 2xy kcal
  3. y-2x kcal
  4. x+y kcal
Ques. 10JEE Main 2025, 24 January (Evening Shift)

Which of the following mixing of 1 M base and 1 M acid leads to the largest increase in temperature?

  1. 50 mL HCl and 20 mL NaOH
  2. 30 mL HCl and 30 mL NaOH
  3. 45 mL CH3COOH and 25 mL NaOH
  4. 30 mL CH3COOH and 30 mL NaOH
Ques. 11JEE Main 2025, 23 January (Morning Shift)

Ice at -5 C is heated to become vapor with temperature of 110 C at atmospheric pressure. The entropy change associated with this process can be obtained from

  1. 268 K273 K Cp,m dT + Hm, fusionTf + Hm, vaporisationTb + 273 K373 K Cp,m dT + 373 K383 K Cp,m dT
  2. 268 K383 K Cp dT + Hmelting273 + Hboiling373
  3. 268 K383 K Cp dT + qrevT
  4. 268 K273 K Cp,mT dT + Hm, fusionTf + Hm, vaporisationTb + 273 K373 K Cp,m dTT + 373 K383 K Cp,m dTT

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