Previous year questions: Thermodynamics

Q1 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

Use the following data to find the enthalpy of formation of anhydrous solid Al2Cl6. (i) 2Al(s) + 6HCl(aq) giving Al2Cl6(aq) + 3H2(g), delta H = -1200 kJ mol^-1. (ii) H2(g) + Cl2(g) giving 2HCl(g), delta H = -164 kJ mol^-1. (iii) HCl(g) + aq giving HCl(aq), delta H = -83 kJ mol^-1. (iv) Al2Cl6(s) + aq giving Al2Cl6(aq), delta H = -663 kJ mol^-1.

Answer: The formation equation is 2Al(s) + 3Cl2(g) giving Al2Cl6(s). Start with reaction (i) as written, and note that its enthalpy is -1200 kJ. Multiply equation (iii) by six to cancel the six HCl(aq) on the left, adding 6 x (-83) = -498 kJ. Reverse equation (iv) to convert Al2Cl6(aq) back to the solid, adding +663 kJ. Multiply equation (ii) by three to produce the six HCl(g) the previous step needs, adding 3 x (-164) = -492 kJ. Adding the four steps: -1200 - 498 + 663 - 492 = -1527 kJ mol^-1. The enthalpy of formation of anhydrous aluminium chloride is -1527 kJ mol^-1, so the correct option is -1527.

Q2 (JEE Main 2026, 2 Apr Shift 2, 4 marks)

A gas changes from state X to state Y by absorbing 10 J of heat and doing 18 J of work. It is then brought back to state X by a process in which 6 J of heat is evolved. In the reverse process from Y to X, how much work is done and on which side?

Answer: For the path X to Y, q = +10 J and w = -18 J, so the internal energy change is 10 - 18 = -8 J. Internal energy is a state function, so the change from Y back to X must be +8 J. The return path evolves 6 J of heat, so q = -6 J. Using the first law, +8 = -6 + w, which gives w = +14 J. A positive work means work is done on the gas by the surroundings, so 14 J of work is done on the gas in the reverse process. The correct option is 14 J of work done on the gas by the surroundings.

Q3 (JEE Main 2026, 4 Apr Shift 1, 4 marks)

Statement I: for an ideal gas, the heat capacity at constant volume is always greater than the heat capacity at constant pressure. Statement II: in a constant volume process, no work is produced and all the heat supplied goes into the chaotic motion of the gas, which shows up as a temperature rise. Which statement is correct?

Answer: Statement I is false. For an ideal gas the heat capacity at constant pressure is larger than at constant volume, because at constant pressure part of the heat supplied is used as pressure-volume work and only the rest raises the temperature. The difference is R, so C_p = C_V + R and C_p is always greater. Statement II is true. At constant volume the work term is zero, so the first law gives delta U = q, and the internal energy of an ideal gas depends only on temperature, so all the heat appears as a temperature rise. The correct option is that statement I is false and statement II is true.

Q4 (JEE Main 2026, 6 Apr Shift 1, 4 marks)

Arrange these isothermal processes between the same two states in order of the magnitude of pressure-volume work: A, expansion in a single stage; B, expansion in multiple stages; C, compression in a single stage; D, compression in multiple stages.

Answer: The area under the p-V curve measures the magnitude of the work, and the reversible path encloses the largest area. For expansion, a single-stage expansion against a constant final pressure gives the smallest magnitude, and carrying the expansion out in several stages gives a larger magnitude because the gas pushes against a pressure closer to its own for more of the path. For compression, the single-stage compression against the high final pressure gives the largest magnitude of work, and multi-stage compression gives less because the pressure rises gradually. The order of decreasing magnitude is therefore single-stage compression, multi-stage compression, multi-stage expansion, single-stage expansion. The correct option is |WC| greater than |WD| greater than |WB| greater than |WA|.

Q5 (JEE Main 2026, 6 Apr Shift 1, 4 marks)

For the reaction X giving Y at 300 K, the standard enthalpy change is 28.40 kJ mol^-1 and the equilibrium constant is 1.8 x 10^-7. Find the magnitude of the standard entropy change of the reaction in J K^-1 mol^-1. (R = 8.3 J K^-1 mol^-1, ln 10 = 2.3)

Answer: First find the standard Gibbs energy change from the equilibrium constant: delta G standard = -RT ln K. Here R is 8.3 J K^-1 mol^-1, T is 300 K and K is 1.8 x 10^-7, so ln K is about -15.5. That gives delta G standard = +3.86 x 10^4 J mol^-1, about +38.6 kJ mol^-1. The reaction is non-spontaneous under standard conditions, as expected from K much smaller than one. Now use delta G standard = delta H standard - T delta S standard. Rearranging, delta S standard = (28.4 - 38.6) x 1000 / 300, which is about -34 J K^-1 mol^-1. The magnitude is 34.

Q6 (JEE Main 2026, 8 Apr Shift 2, 4 marks)

For the reaction 2H2S(g) + 3O2(g) giving 2H2O(l) + 2SO2(g), find the magnitude of the enthalpy change in kJ mol^-1, using standard formation enthalpies of H2S(g) = -20.1, H2O(l) = -286.0 and SO2(g) = -297.0 kJ mol^-1.

Answer: Apply the formation enthalpy formula: sum the formation enthalpies of the products multiplied by their coefficients and subtract the same for the reactants. The products give 2(-286.0) + 2(-297.0) = -572.0 - 594.0 = -1166.0 kJ. The reactants give 2(-20.1) + 3(0) = -40.2 kJ, because oxygen in its standard state has zero formation enthalpy. The reaction enthalpy is -1166.0 - (-40.2) = -1125.8 kJ mol^-1, so the magnitude is 1126 kJ mol^-1.

Q7 (NCERT Exemplar, 3 marks)

Which item below lies outside the scope of thermodynamics: the energy change of a reaction, how far a reaction proceeds, how fast a reaction proceeds, or whether a reaction is feasible?

Answer: Thermodynamics deals with energy changes, with the direction and extent of change, and with the feasibility of a process under the given conditions. It says nothing about how fast the change happens. The rate of a reaction is the subject of chemical kinetics, and a reaction that is thermodynamically favourable can still be immeasurably slow, as the decomposition of diamond into graphite shows. The correct answer is the rate at which a reaction proceeds.

Source: NCERT Exemplar Class 11 Chemistry, Unit 6, MCQ 1.

Q8 (NCERT Exemplar, 3 marks)

Which statement about systems is correct: a covered beaker holding reactants is an open system, a closed system exchanges both energy and matter with the surroundings, a closed copper vessel holding reactants is a closed system, or a thermos flask holding reactants is a closed system?

Answer: A closed system exchanges energy but not matter with the surroundings. A copper vessel with a tight lid allows heat to pass through the metal but does not let matter escape, so it is a closed system. A covered beaker can still exchange matter through the imperfect cover and is better treated as open. A closed system does not exchange matter, so the second statement is wrong. A thermos flask with insulated walls allows neither heat nor matter to pass easily, so it is an isolated system rather than a closed one. The correct option is the closed copper vessel.

Source: NCERT Exemplar Class 11 Chemistry, Unit 6, MCQ 2.

Q9 (NCERT Exemplar, 3 marks)

The standard internal energy of formation of CH4(g) at a given temperature is -393 kJ mol^-1. What is the relation of the standard enthalpy of formation to this value?

Answer: The formation reaction is C(s) + 2H2(g) giving CH4(g). The change in the number of moles of gas is one minus two, which is -1. Using delta H = delta U + delta n_g RT with delta n_g negative, the enthalpy of formation is smaller (more negative) than the internal energy of formation. The standard enthalpy of formation is therefore less than the standard internal energy of formation, and the correct option is that delta H is less than delta U.

Source: NCERT Exemplar Class 11 Chemistry, Unit 6, MCQ 6.

Q10 (NCERT Exemplar, 3 marks)

For the free expansion of an ideal gas under adiabatic conditions, which set of conditions is correct: q = 0, delta T not equal to 0, w = 0; q not equal to 0, delta T = 0, w = 0; q = 0, delta T = 0, w = 0; or q = 0, delta T less than 0, w not equal to 0?

Answer: Adiabatic means no heat exchange, so q = 0. Free expansion means the gas expands into vacuum against zero external pressure, so w = 0 regardless of how much the volume changes. With both q and w zero, the first law gives delta U = 0. The internal energy of an ideal gas depends only on temperature, so delta T is also zero. The correct set is q = 0, delta T = 0, w = 0.

Source: NCERT Exemplar Class 11 Chemistry, Unit 6, MCQ 7.