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.

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## 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 N_{2} and H_{2}

(4) dissociation of CaSO_{4(s)} to CaO_{(s)} and SO_{3(g)}.

**Solution:**

N_{2(g) }+ 3H_{2} → 2NH_{3(g)}

∆s = 2 – 4 = – 2 < 0

So entropy is negative.

Hence option (4) is the answer.

**2. The standard enthalpy of formation of NH _{3} is -46 kJ mol^{–1}. If the enthalpy of formation of H_{2} from its atoms is -436 kJ mol^{–1} and that of N_{2} is -712 kJ mol^{–1}, the average bond enthalpy of N-H bond in NH_{3} is **

(1) -964 kJ mol^{–1 }

(2) +352 kJ mol^{–1 }

(3) +1056 kJ mol^{–1}

(4) -1102 kJ mol^{–1 }

**Solution:**

½ N_{2} + (3/2)H_{2} → NH_{3}

**(∆**H_{f}) NH_{3} = [½ B.E N_{2} + (3/2) B.E H_{2} – 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 dm ^{3} to a volume of 100 dm^{3} at 27°C is **

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

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

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

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

**Solution:**

Given V_{2} = 100

V_{1} = 10

n = 2

∆S = 2.303nR log (V_{1}/V_{2})

= 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 37 ^{0}C. 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

V_{1} = 1 L

V_{2} = 10 L

w = -P_{ext}(V_{2}-V_{1})

= -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 CH _{3}OH_{(l)}+ (3/2)O_{2(g)} → CO_{2(g)} + 2H_{2}O_{(l)}**

**At 298 K standard Gibb’s energies of formation for CH _{3}OH(l), H_{2}O (l) and CO_{2} (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:**

CH_{3}OH_{(l)}+ (3/2)O_{2(g)} → CO_{2(g)} + 2H_{2}O_{(l)}

Given ΔG_{f}^{0} CH_{3}OH (*l*) = -166.2 kJ mol^{-1}

ΔG_{f}^{0} H_{2}O (*l*) = -394.4 kJ mol^{-1}

ΔG = ΣΔG_{f}^{0} products – ΣΔG_{f}^{0} 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 :**

**4NO _{2} (g) + O_{2} (g) → 2N_{2}O_{5} (g). Δ_{r}H = -111kJ**

**If N _{2}O_{5} (s) is formed instead of N_{2}O_{5} (g) in the above reaction, the Δ_{r}H value will be:-**

**(given, ΔH of sublimation for N _{2}O_{5} 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.