JEE Main Thermochemistry Previous Year Questions with Solutions

Thermochemistry JEE Main previous year questions with solutions are given here. Thermochemistry is the study of heat and energy related to various physical transformations and chemical reactions. It is mainly concerned with a change in energy mainly regarding an exchange of energy of a system with its surroundings. A reaction may absorb or release energy and phase change may include the same, namely boiling and melting. Regarding determination of entropy, it is used to detect whether a reaction is favourable or unfavourable, spontaneous or nonspontaneous. It is likewise used in determining product and reactant quantities through a complete reaction.

As far as the JEE exam is concerned, Thermochemistry is an important topic. The important concepts include open system, closed system, isolated system, laws of thermodynamics, etc. This article gives an idea of what type of questions can be expected from this topic. BYJU’S provides accurate solutions prepared by our subject experts. Students can easily download the questions and solutions in PDF format.

Download Thermochemistry Previous Year Solved Questions PDF

JEE Main Previous Year Solved Questions on Thermochemistry

1. The process with negative entropy change is

(1) dissolution of iodine in water

(2) sublimation of dry ice

(3) synthesis of ammonia from N₂ and H₂

(4) dissociation of CaSO_4(s) to CaO_(s) and SO_3(g).

Solution:

N_2(g) + 3H₂ → 2NH_3(g)

∆s = 2 – 4 = – 2 < 0

So entropy is negative.

Hence option (3) is the answer.

2. The standard enthalpy of formation of NH₃ is -46 kJ mol^–1. If the enthalpy of formation of H₂ from its atoms is -436 kJ mol^–1 and that of N₂ is -712 kJ mol^–1, the average bond enthalpy of N-H bond in NH₃ is

(1) -964 kJ mol^–1

(2) +352 kJ mol^–1

(3) +1056 kJ mol^–1

(4) -1102 kJ mol^–1

Solution:

½ N₂ + (3/2)H₂ → NH₃

(∆H_f) NH₃ = [½ B.E N₂ + (3/2) B.E H₂ – 3 B.E N-H]

-46 = [½ 712 + (3/2) 436 – 3 B.E N-H]

-46 = 356 + 654 – 3 B.E N-H

3 B.E N-H = 1056

B.E N-H = 1056/3 = 352 kJ mol^–1

Hence option (2) is the answer.

3. The enthalpy change for a reaction does not depend upon

(1) use of different reactants for the same product

(2) the nature of intermediate reaction steps

(3) the differences in initial or final temperature of involved substances

(4) the physical states of reactants and products

Solution:

The enthalpy change for a reaction does not depend upon the nature of intermediate reaction steps.

Hence option (2) is the answer.

4. Identify the correct statement regarding a spontaneous process

(1) For a spontaneous process in an isolated system, the change in entropy is positive

(2) Endothermic processes are never spontaneous

(3) Exothermic processes are always spontaneous

(4) Lowering of energy in the reaction process is the only criterion for spontaneity

Solution:

For a spontaneous process in an isolated sys­tem, the change in entropy is positive.

Hence option (1) is the answer.

5. For a particular reversible reaction at temperature T, ∆H and ∆S were found to be both +ve. If T_e is the temperature at equilibrium, the reaction would be spontaneous when (1) T_e > T

(2) T > T_e

(3) T_e is 5 times T

(4) T_e = T

Solution:

At equilibrium, ∆G = 0.

∆G = ∆H – T∆S

For a reaction to be spontaneous ∆G should be negative.

So T > T_e

Hence option (2) is the answer.

6. The entropy change involved in the isothermal reversible expansion of 2 moles of an ideal gas from a volume of 10 dm³ to a volume of 100 dm³ at 27°C is

(1) 32.3 J mol^-1K^-1

(2) 42.3 J mol^-1K^-1

(3) 38.3 J mol^-1K^-1

(4) 35.8 J mol^-1K^-1

Solution:

Given V₂ = 100

V₁ = 10

n = 2

∆S = 2.303nR log (V₁/V₂)

= 2.303 × 2 × 8.314 × log (100/10)

= 2.303 × 2 × 8.314 × log 10

= 38.29 J mol^-1 K^-1

Hence option (3) is the answer.

7. A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 37⁰C. As it does so, it absorbs 208J of heat. The values of q and w for the process will be (R = 8.314J/molK and In 7.5 = 2.01)

(1) q = + 208J, w = – 208 J

(2) q = – 208J, w = – 208 J

(3) q = – 208J, w = + 208 J

(4) q = + 208J, w = + 208 J

Solution:

∆E = q+w

The process is isothermal reversible expansion. So ∆E = 0

Hence q = -w

Since heat is absorbed, q = +208 J

So w = -208 J

Hence option (1) is the answer.

8. During compression of a spring, the work done is 10 kJ and 2 kJ escaped to the surrounding as heat. The change in internal energy ∆U (in kJ) is

(1) -8

(2) 12

(3) 8

(4) -12

Solution:

Given w = 10 kJ

q = –2 kJ

According to first law of thermodynamics, ∆U = q + w

= –2 + 10 = 8 kJ

Hence option (3) is the answer.

9. An ideal gas undergoes isothermal expansion at constant pressure. During the process

(1) enthalpy increases but entropy decreases

(2) enthalpy remains constant but entropy increases

(3) enthalpy decreases but entropy increases

(4) both enthalpy and entropy remain constant.

Solution:

During isothermal expansion at constant pressure, ∆H = nC_p ∆T = 0

∆S = nRln(V_f/V_i) > 0

Hence option (2) is the answer.

10. Assuming that water vapour is an ideal gas, the internal energy change (U) when 1 mol of water is vaporized at 1 bar pressure and 100°C, (given: molar enthalpy of vapourisation of water at 1 bar and 373 K = 41 kJ mol^–1 and R = 8.3 J mol^–1 K^–1) will be

(1) 41.00 kJ mol^-1

(2) 4.100 kJ mol^–1

(3) 3.7904 kJ mol^–1

(4) 37.904 kJ mol^–1

Solution:

Given ∆H = 41000

T = 373 K

∆U = ∆H-∆nRT

= 41000 – 1× 8.314× 373

= 41000 – 3101.122

= 37898.878 J mol^-1

= 37.9 kJ mol^-1

Hence option (4) is the answer.

11. A gas undergoes change from state A to state B. In this process, the heat absorbed and work done by the gas is 5 J and 8 J, respectively. Now gas is brought back to A by another process during which 3 J of heat is evolved. In this reverse process of B to A

(1) 10 J of the work will be done by the surrounding on gas

(2) 10 J of the work will be done by the gas

(3) 6 J of the work will be done by the surrounding on gas

(4) 6 J of the work will be done by the gas.

Solution:

From A to B:

Given q = + 5 J

w = – 8 J (work done by the system)

According to first law of thermodynamics,

U = q + w

= 5 -8

= -3 J

From B to A:

U = 3 J

(As internal energy is state function and does not depend on path)

q = -3 J (heat evolved),

w = U-q

= + 3-(-3)

= + 6 J

Therefore 6 J work will be done by the surrounding on gas.

Hence option (3) is the answer.

12. An ideal gas is allowed to expand from 1 L to 10 L against a constant external pressure of 1 bar. The work done in kJ is

(1) +10.0

(2) -0.9

(3) -2.0

(4) -9.0

Solution:

Given P_ext = 1 bar

V₁ = 1 L

V₂ = 10 L

w = -P_ext(V₂-V₁)

= -1(10-1)

= -9 bar L

= -900 J

= -0.9 kJ

Hence option (2) is the answer.

13. Among the following, the set of parameters that represents path function, is

(A) q + w

(B) q

(C) w

(D) H – TS

(1) (B) and (C)

(2) (A) and (D)

(3) (B), (C) and (D)

(4) (A), (B) and (C)

Solution:

∆U = q+w

q ⇒ path function

w ⇒ path function

∆U ⇒ state function

H -TS = G ⇒ state function

Hence option (1) is the answer.

14. In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is CH₃OH_(l)+ (3/2)O_2(g) → CO_2(g) + 2H₂O_(l)

At 298 K standard Gibb’s energies of formation for CH₃OH(l), H₂O (l) and CO₂ (g) are

-166.2, -237.2 and -394.4 kJ mol^-1 respectively. If the standard enthalpy of combustion of methanol is –726 kJmol^-1, the efficiency of the fuel cell will be

(1) 90%

(2) 97%

(3) 80%

(4) 87%

Solution:

CH₃OH_(l)+ (3/2)O_2(g) → CO_2(g) + 2H₂O_(l)

Given ΔG_f⁰ CH₃OH (l) = -166.2 kJ mol^-1

ΔG_f⁰ H₂O (l) = -394.4 kJ mol^-1

ΔG = ΣΔG_f⁰ products – ΣΔG_f⁰ reactants

= -394.4 -2(237.2) + 166.2

= -702.6 kJ mol^-1

The efficiency of fuel cell = ( ΔG/ ΔH)× 100

= (702.6/726)× 100

= 97%

Hence option (2) is the answer.

15. Consider the reaction :

4NO₂ (g) + O₂ (g) → 2N₂O₅ (g). Δ_rH = -111kJ

If N₂O₅ (s) is formed instead of N₂O₅ (g) in the above reaction, the Δ_rH value will be:-

(given, ΔH of sublimation for N₂O₅ is 54 kJ mol^-1)

(1) –165 kJ

(2) +54 kJ

(3) +219 kJ

(4) –219 kJ

Solution:

Given ΔH_reaction = -111 kJ

ΔH_sub = 54 kJ

ΔH_f+ΔH_sub=ΔH_reaction [Hess’s law]

ΔH_f = ΔH_reaction-ΔH_sub

= -111 -54

= -165 kJ

Hence option (1) is the answer.

Thermochemistry Revision – Video Lesson