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Thermodynamics · Question 36

JEE Main 2025 (Online) 2nd April Evening ShiftMCQHard📈 Topic in 23% papers+4−1
1

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)

rH -92 kJ mol-1 (negative, nearly independent of T)

rS -198 J K-1 mol-1 (4 mol gas 2 mol gas; negative, nearly constant)

- H T = + 92000T: 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)

JEE Main 2025 (Online) 2nd April Evening Shift

Keep practising — Thermodynamics

2

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
JEE Main 2025 (Online) 29th January Morning Shift
3

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).

JEE Main 2025 (Online) 28th January Morning Shift
4

[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)
JEE Main 2025 (Online) 28th January Evening Shift
5

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)

JEE Main 2025 (Online) 28th January Evening Shift

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