Consider the reaction 2H2S(g) + 3O2(g) 2H2O(l) + 2SO2(g) The magnitude of enthalpy change for the reaction in kJ mol-1 is . (Nearest integer) Given : f H (H2S) = -20.1 kJ mol-1 f H (H2O) = -286.0 kJ mol-1 f H (SO2) = -297.0 kJ mol-1
Thermodynamics
Previous year questions with solutions · 2024 – 2026
Year-wise statistics questions · papers · average per paper · weightage
| Year | Questions | Papers | Avg / paper | Weightage |
|---|---|---|---|---|
| 2026 | 21 | 19 | 1.1 | 4.42% ▼ 32.3% vs last year |
| 2025 | 31 | 19 | 1.6 | 6.53% ▲ 78.3% vs last year |
| 2024 | 22 | 21 | 1 | 3.66% |
Consider the reaction X Y at 300 K. If H and K are 28.40 kJ mol-1 and 1.8 10-7 at the same temperature, then the magnitude of S for the reaction in J K-1 mol-1 is . (Nearest integer) (Given : R = 8.3 J K-1 mol-1, ln 10 = 2.3, log 3 = 0.48, log 2 = 0.30)
If 3.365 g of ethanol (l) is burnt completely in a bomb calorimeter at 298.15 K, the heat produced is 99.472 kJ. The | Hf | of ethanol at 298.15 K is 102 kJ mol-1. (Nearest integer) Given: Standard enthalpy for combustion of graphite = −393.5 kJ mol-1 Standard enthalpy of formation of water (l) = −285.8 kJ mol-1 Molar mass in g mol-1 of C, H, O are 12, 1 and 16 respectively
At the transition temperature T, A B and G = 105 - 35 T where A and B are two states of substance X. The transition temperature in °C when pressure is 1 atm is . (Nearest integer)
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)
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)
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
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)
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]
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).
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).
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)
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)
The standard enthalpy and standard entropy of decomposition of N2O4 to NO2 are 55.0 kJ mol-1 and 175.0 J/K/mol respectively. The standard free energy change for this reaction at 25 in J mol-1 is (Nearest integer)
The bond dissociation enthalpy of X2 H bond calculated from the given data is kJ mol-1. (Nearest integer) M+X-(s) M+(g)+X-(g) H lattice=800 kJ mol-1 M(s) M(g) H sub=100 kJ mol-1 M(g) M+(g)+e-(g) H i=500 kJ mol-1 X(g)+e-(g) X-(g) H eg=-300 kJ mol-1 M(s)+ 12X2(g) M+X-(s) H f=-400 kJ mol-1 [Given : M+X- is a pure ionic compound and X forms a diatomic molecule X2 in gaseous state]
Consider the following cases of standard enthalpy of reaction ( Hr in kJ mol-1) C2H6(g)+ 72O2(g) 2CO2(g)+3H2O(l) H1 =-1550 C(graphite)+O2(g) CO2(g) H2 =-393.5 H2(g)+ 12O2(g) H2O(l) H3 =-286 The magnitude of H f C2H6(g) is kJ mol-1 (Nearest integer).
When equal volume of 1 M HCl and 1 M H2SO4 are separately neutralised by excess volume of 1M NaOH solution. x and y kJ of heat is liberated respectively. The value of y/x is .
The heat of solution of anhydrous CuSO4 and CuSO4 5H2O are -70 kJ mol-1 and +12 kJ mol-1 respectively. The heat of hydration of CuSO4 to CuSO4 5H2O is -x kJ. The value of x is . (nearest integer).
When Hvap = 30 kJ/mol and Svap = 75 J mol-1 K-1, then the temperature of vapour, at one atmosphere is K.
[figure] Consider the figure provided. 1 mol of an ideal gas is kept in a cylinder, fitted with a piston, at the position A, at 18°C. If the piston is moved to position B, keeping the temperature unchanged, then 'x' L atm work is done in this reversible process. x = L atm. (nearest integer) [Given : Absolute temperature = °C + 273.15, R = 0.08206 L atm mol-1 K-1]
vapH for water is +40.79 kJ mol-1 at 1 bar and 100 C. Change in internal energy for this vapourisation under same condition is kJ mol-1. (Integer answer) (Given R = 8.3 JK-1 mol-1)
An ideal gas, Cv = 52R, is expanded adiabatically against a constant pressure of 1 atm until it doubles in volume. If the initial temperature and pressure is 298 K and 5 atm, respectively then the final temperature is K (nearest integer). [ cv is the molar heat capacity at constant volume]
For the reaction at 298 K, 2A + B C, H = 400 kJ mol-1 and S = 0.2 kJ mol-1 K-1. The reaction will become spontaneous above K.
The heat of combustion of solid benzoic acid at constant volume is -321.30 kJ at 27 C. The heat of combustion at constant pressure is (-321.30 - xR) kJ, the value of x is .
Combustion of 1 mole of benzene is expressed as C6H6(l) + 152O2(g) 6CO2(g) + 3H2O(l). The standard enthalpy of combustion of 2 mol of benzene is −'x' kJ. x = . Given : 1. Standard Enthalpy of formation of 1 mol of C6H6(l), for the reaction 6C(graphite) + 3H2(g) C6H6(l) is 48.5 kJ mol-1. 2. Standard Enthalpy of formation of 1 mol of CO2(g), for the reaction C(graphite) + O2(g) CO2(g) is -393.5 kJ mol-1. 3. Standard Enthalpy of formation of 1 mol of H2O(l), for the reaction H2(g) + 12O2(g) H2O(l) is -286 kJ mol-1.
The enthalpy of formation of ethane (C2H6) from ethylene by addition of hydrogen where the bond-energies of C-H, C-C, C=C, H-H are 414 kJ, 347 kJ, 615 kJ and 435 kJ respectively is − kJ
Three moles of an ideal gas are compressed isothermally from 60 L to 20 L using constant pressure of 5 atm. Heat exchange Q for the compression is - Lit. atm.
For a certain reaction at 300 K, K = 10, then for the same reaction is − 10-1 kJ mol-1. (Given R = 8.314 J K-1 mol-1)
Consider the following reaction at 298 K: 32O2(g) O3(g); KP = 2.47 10-29. rG for the reaction is kJ. (Given R = 8.314 JK-1 mol-1)
If 5 moles of an ideal gas expands from 10 L to a volume of 100 L at 300 K under isothermal and reversible condition then work, w, is −x J. The value of x is . (Given R = 8.314 J K-1 mol-1)