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Thermodynamics · Question 23

JEE Main 2025 (Online) 7th April Morning ShiftMCQMedium📈 Topic in 16% papers+4−1
1

Total enthalpy change for freezing of 1 mol of water at 10 C to ice at -10 C is (Given: fusH = x kJ/mol Cp[H2O(l)] = y J mol-1 K-1 Cp[H2O(s)] = z J mol-1 K-1)

  1. -x-10y-10z
  2. x-10y-10z
  3. -10(100x+y+z)
  4. 10(100x+y+z)

Answer

(3) -10(100x+y+z)

Solution

Step 1: H2O(l), 10 C H2O(l), 0 C

H1 = Cp T = y(0-10) = -10y J

Step 2: H2O(l), 0 C H2O(s), 0 C (freezing is the reverse of fusion)

H2 = - fusH = -x kJ = -1000x J

Step 3: H2O(s), 0 C H2O(s), -10 C

H3 = z(-10-0) = -10z J

Htotal = H1 + H2 + H3

= -1000x - 10y - 10z

= -10(100x + y + z) J

Answer: (3)

JEE Main 2025 (Online) 7th April Morning Shift

Keep practising — Thermodynamics

2

The hydration energies of K+ and Cl- are -x and -y kJ/mol respectively. If the lattice energy of KCl is -z kJ/mol, then the heat of solution of KCl is :

  1. x + y + z
  2. z - (x + y)
  3. -z - (x + y)
  4. x - y - z
JEE Main 2025 (Online) 7th April Evening Shift
3

The correct statement amongst the following is :

  1. The standard state of a pure gas is the pure gas at a pressure of 1 bar and temperature 273 K.
  2. f H 500 is zero for O2(g).
  3. f H 298 is zero for O(g).
  4. The term 'standard state' implies that the temperature is 0°C.
JEE Main 2025 (Online) 7th April Evening Shift
4

One mole of an ideal gas expands isothermally and reversibly from 10 dm3 to 20 dm3 at 300 K. U, q and work done in the process respectively are Given: R = 8.3 J K-1 mol-1 10 = 2.3 2 = 0.30 3 = 0.48

  1. 0, 21.84 kJ, -1.726 J
  2. 0, 21.84 kJ, 21.84 kJ
  3. 0, 1.718 kJ, -1.718 kJ
  4. 0, -17.18 kJ, 1.718 J
JEE Main 2025 (Online) 4th April Morning Shift
5

Let us consider a reversible reaction at temperature, T. In this reaction, both H and S were observed to have positive values. If the equilibrium temperature is Te, then the reaction becomes spontaneous at:

  1. Te > T
  2. T > Te
  3. T = Te
  4. Te = 5T
JEE Main 2025 (Online) 4th April Morning Shift
6

Consider the given data : (a) HCl(g) + 10H2O(l) HCl 10H2O; H = -69.01 kJ mol-1 (b) HCl(g) + 40H2O(l) HCl 40H2O; H = -72.79 kJ mol-1 Choose the correct statement :

  1. The heat of dilution for the HCl (HCl 10H2O to HCl 40H2O) is 3.78 kJ mol-1.
  2. Dissolution of gas in water is an endothermic process.
  3. The heat of solution depends on the amount of solvent.
  4. The heat of formation of HCl solution is represented by both (a) and (b).
JEE Main 2025 (Online) 4th April Evening Shift
7

Which of the following properties will change when system containing solution 1 will become solution 2 ? [figure]

  1. Molar heat capacity
  2. Concentration
  3. Gibbs free energy
  4. Density
JEE Main 2025 (Online) 3rd April Morning Shift
8

Given : sub H [C (graphite)] = 710 kJ mol-1 C-H H = 414 kJ mol-1 H-H H = 436 kJ mol-1 C=C H = 611 kJ mol-1 The f H for CH2=CH2 is kJ mol-1 (nearest integer value)

JEE Main 2025 (Online) 3rd April Morning Shift
9

Given below are two statements: In the light of the above statements, choose the correct answer from the options given below

  1. Both Statement I and Statement II are false
  2. Statement I is true but Statement II is false
  3. Both Statement I and Statement II are true
  4. Statement I is false but Statement II is true
JEE Main 2025 (Online) 3rd April Evening Shift
10

A perfect gas (0.1 mol) having Cv = 1.50 R (independent of temperature) undergoes the following transformation from point 1 to point 4. If each step is reversible, the total work done (w) while going from point 1 to point 4 is (-) J (nearest integer) [Given: R = 0.082 L atm K-1 mol-1] [figure]

JEE Main 2025 (Online) 3rd April Evening Shift
11

A sample of n-octane (1.14 g) was completely burnt in excess of oxygen in a bomb calorimeter, whose heat capacity is 5 kJ K-1. As a result of combustion reaction, the temperature of the calorimeter is increased by 5 K. The magnitude of the heat of combustion of octane at constant volume is kJ mol-1 (nearest integer).

JEE Main 2025 (Online) 3rd April Evening Shift

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