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Thermodynamics · Question 13

JEE Main 2026 (Online) 28th January Evening ShiftMCQEasy📈 Topic in 23% papers+4−1
1

The plot of 10 K vs 1T gives a straight line. The intercept and slope respectively are (where K is the equilibrium constant).

  1. - S R2.303, H R2.303
  2. - H 2.303R, S 2.303R
  3. S 2.303R, - H 2.303R
  4. 2.303R H , 2.303R S

Answer

(3) S 2.303R, - H 2.303R

Solution

G = -2.303RT 10 K and G = H - T S

-2.303RT 10 K = H - T S

10 K = - H 2.303R ( 1T ) + S 2.303R

Comparing with y = mx + c (y = 10 K, x = 1T):

Intercept = S 2.303R and slope = - H 2.303R

Answer: (3)

JEE Main 2026 (Online) 28th January Evening Shift

Keep practising — Thermodynamics

2

Match the LIST-I with LIST-II Choose the correct answer from the options given below:

  1. A-IV, B-I, C-III, D-II
  2. A-I, B-III, C-II, D-IV
  3. A-II, B-I, C-III, D-IV
  4. A-IV, B-II, C-III, D-I
JEE Main 2026 (Online) 24th January Morning Shift
3

The heats of atomisation of methane and ethane are 'x' kJ mol-1 and 'y' kJ mol-1 respectively. The longest wavelength ( ) of light capable of breaking the C-C bond can be expressed in SI unit as :

  1. NA hc250(y - 6x)
  2. NA hc (y - 6x4 )-1
  3. hc1000 ( y - 6x4 )-1
  4. NA hc250(4y - 6x)
JEE Main 2026 (Online) 24th January Evening Shift
4

A cup of water at 5 C (system) is placed in a microwave oven and the oven is turned on for one minute during which the water begins to boil. Which of the following options is true?

  1. q = +ve, w = -ve, U = +ve
  2. q = +ve, w = 0, U = -ve
  3. q = +ve, w = -ve, U = -ve
  4. q = -ve, w = -ve, U = -ve
JEE Main 2026 (Online) 23rd January Morning Shift
5

Match the LIST-I with LIST-II Choose the correct answer from the options given below :

  1. A-I, B-II, C-III, D-IV
  2. A-III, B-II, C-IV, D-I
  3. A-II, B-I, C-III, D-IV
  4. A-II, B-III, C-I, D-IV
JEE Main 2026 (Online) 22nd January Morning Shift
6

If the enthalpy of sublimation of Li is 155 kJ mol-1, enthalpy of dissociation of F2 is 150 kJ mol-1, ionization enthalpy of Li is 520 kJ mol-1, electron gain enthalpy of F is -313 kJ mol-1, standard enthalpy of formation of LiF is -594 kJ mol-1. The magnitude of lattice enthalpy of LiF is kJ mol-1. (Nearest Integer)

JEE Main 2026 (Online) 22nd January Evening Shift
7

Which of the following graphs between pressure 'p' versus volume 'V' represents the maximum work done?

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
JEE Main 2026 (Online) 21st January Morning Shift
8

Use the following data : array|c|c|c| Substance & f H (500 K) kJ mol-1 & S (500 K) J K-1 mol-1 \ AB(g) & 32 & 222 \ A2(g) & 6 & 146 \ B2(g) & x & 280 \ array One mole each of A2(g) and B2(g) are taken in a 1 L closed flask and allowed to establish the equilibrium at 500 K. A2(g) + B2(g) 2AB(g) The value of x (in kJ mol-1) is . (Nearest integer) (Given : K = 2.2, R = 8.3 J K-1 mol-1)

JEE Main 2026 (Online) 21st January Morning Shift
9

Consider the following data : f H (methane, g) = -X kJ mol-1 Enthalpy of sublimation of graphite = Y kJ mol-1 Dissociation enthalpy of H2 = Z kJ mol-1 The bond enthalpy of C–H bond is given by :

  1. X+Y+2Z4
  2. X+Y+4Z2
  3. -X+Y+Z4
  4. X+Y+Z
JEE Main 2026 (Online) 21st January Evening Shift
10

Resonance in X2Y can be represented as [figure] The enthalpy of formation of X2Y ( X X(g) + 12Y = Y(g) X2Y(g) ) is 80 kJ mol-1. The magnitude of resonance energy of X2Y is kJ mol-1 (nearest integer value). Given: Bond energies of X X, X = X, Y = Y and X = Y are 940, 410, 500, and 602 kJ mol-1 respectively. Valence X: 3, Y: 2

JEE Main 2025 (Online) 8th April Evening Shift
11

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)
JEE Main 2025 (Online) 7th April Morning Shift

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