Chemical Thermodynamics is an important subtopic of JEE Main Chemistry Syllabus which falls under Physical Chemistry. There are a total of 75 questions asked in the Chemistry section, out of which the average number of questions asked from Chemical Thermodynamics are usually 0-2 of 8 marks or below carrying a weightage of 0-4% in the entire portion of Chemistry. Check JEE Main Chemistry Syllabus

Sub-topics covered under JEE Main Chemical Thermodynamics section include:

• Fundamentals of thermodynamics: System and surroundings, extensive and intensive properties, state functions, types of processes;
• First law of thermodynamics: Concept of work, heat internal energy, and enthalpy, heat capacity, molar heat capacity;
• Hess’s law of constant heat summation;
• Enthalpies of bond dissociation, combustion, formation, atomization, sublimation, phase transition, hydration, ionization, and solution;
• The second law of thermodynamics: Spontaneity of processes; Delta S of the universe and Delta G of the system as criteria for spontaneity, Delta Go (Standard Gibbs energy change) and equilibrium constant.

Although Chemistry is considered to be the easiest section in JEE Main Question Paper, over the years, it has been observed that the questions asked from the Chemical Thermodynamics section are not very difficult but definitely tricky. Chemical Thermodynamics is among the toughest topics from Chemistry and so this subtopic requires in-depth knowledge, as well as daily practise of the numerical. Check JEE Main Sample Papers

JEE Main Chemical Thermodynamics Questions and Solution

Question. The process with negative entropy change is called? (JEE 2016)

1. Dissolution of iodine in water
2. Synthesis of Ammonia from N2 and H2
3. Sublimation of dry ice
4. Dissociation of CaSO4(s) to CaO(s) and SO3(g)

Solution: N2(g) +3H2= 2NH3(g)

▲s= 2-4 = -2 <0

Therefore, entropy is negative and the answer is d.)

Question. Math the following- (JEE 2011)

Solution- A matching with (p, r, s)

B matching with (r, s)

C matching with t

D matching with (p, q, t)

Question. The heats of combustion of carbon and carbon monoxide are -393.5 and -285.5 kJ

mol-1 , respectively. The heat of formation (in kJ) of Carbon Monoxide per mole is-

1. -110
2. 110.5
3. 676.5
4. -676.5 (JEE 2016)

Solution- C(s) + ½ O2(g) ------ CO(g) ; ▲Hr = ▲Hf (CO)

or, ▲Hf = ▲Hc (C) - ▲HC (CO)

or, -393.5 + 283.5

or, -110 kJ

Question. The enthalpy change for a reaction does not depend upon (JEE 2018)

1. The nature of intermediate reaction steps
2. Use of different reactants for the same product
3. The physical states of reactants and products
4. The differences in initial or final temperature of involved substances

Solution- The enthalpy change for a reaction does not depend upon the nature of intermediate reaction steps so; a.) is the answer.

Question. 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. 42.3 J mol-1K-1
2. 35.8 J mol-1K-1
3. 32.3 J mol-1K-1
4. 38.3 J mol-1K-1 (JEE 2013)

Solution- Given V2 = 100

V1 = 10

n = 2

∆S = 2.303nR log (V1/V2)

or, 2.303 × 2 × 8.314 × log (100/10)

or, 2.303 × 2 × 8.314 × log 10

or, 38.29 J mol-1 K-1

Therefore, d.) is the answer.

Question. The combustion of benzene (1) gives CO2(g) and H2O(1). Given that the heat of combustion of benzene at constant volume is -3263.9 kJ mol-1 at 25 degrees Celsius; heat of combustion of benzene at constant pressure will be – (R= 8.314 JK-1 mol-1)

1. 3260
2. -3267.6
3. 4152.6
4. -452.46 (JEE 2018)

Solution- C(6)H(6) (l) + 15/2 O2(g) ----6CO2 + 3H2O(l)

▲ng = 6 – 7.5 = -1.5 (change in gaseous mode)

▲U or ▲E= -3263.9 kJ

▲H = ▲U + ▲ngRT

▲ng = -1.5

R= 8.314 JK-1 mol-1

T= 298 K

So ▲H = -3263.9 + (-1.5) 8.314 *10-3 * 298

= -3267.6 kJ

▲H= Heat at constant pressure

▲U/▲E= Heat at constant volume

R= gas constant

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

1. enthalpy remains constant but entropy increases
2. both enthalpy and entropy remain constant.
3. enthalpy increases but entropy decreases
4. enthalpy decreases but entropy increases (JEE 2019)

Solution- During isothermal expansion at constant pressure, ∆H = nCp ∆T = 0

∆S = nRln(Vf/Vi) > 0

So, a.) is the answer

Question. 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 will be which of the following?

1. 6 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. 10 J of the work will be done by the surrounding on gas
4. 6 J of the work will be done by the gas. (JEE 2019)

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.

So, a.) will be the answer.

Question. For an ideal solution of two components A and B, which of the following is true?

1. A-A, B-B and A-B interactions are identical
2. A-B interaction is stronger than A-A interaction and B-B interaction
3. ▲Hmixing <0
4. ▲Hmixing >0 (JEE 2014)

Solution- c.)

Question. 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 vaporisation of water at 1 bar and 373 K = 41 kJ mol–1 and R = 8.3 J mol–1 K–1) will be?

1. 4.100 kJ mol–1
2. 37.904 kJ mol–1
3. 41.00 kJ mol-1
4. 3.7904 kJ mol–1 (JEE 2017)

Solution- Given that ∆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

Therefore, b.) is the answer.

Question. The reaction X →Y is an exothermic reaction. Activation energy for the same is 150 KJ-1 mol-1. Enthalpy of reaction is 135 kJmol-1. The activation energy for the reverse reaction will be which of the following-

1. 285 kJmol-1
2. 270 kJmol-1
3. 15 kJmol-1
4. 280 kJ (JEE 2015)

Solution- b.)

Question. Identify the correct statement regarding a spontaneous process from the following-

1. Exothermic processes are always spontaneous
2. Lowering of energy in the reaction process is the only criterion for spontaneity
3. Endothermic processes are never spontaneous
4. For a spontaneous process in an isolated system, the change in entropy is positive (JEE 2012)

Solution- When it comes to a spontaneous process in an isolated system, the change in entropy is positive.

So, d.) is the answer.

Question. The standard enthalpy of formation for methane, CH4 is -74.9 kJmol-1. In order to calculate the average energy given out in the formation of a C-H bond from it is necessary to know which of the following?

1. The dissociation of H2 and enthalpy of sublimation of carbon (graphite)
2. The dissociation energy of Hydrogen molecule, H2
3. The first four ionisation energies of carbon
4. The first four ionisation energies of carbon and electron affinity of hydrogen (JEE 2014)

Solution- For the formation of the methane C-H bond enthalpy can be calculated as – C(s) +2H2(g) →CH4.

▲H◦f (CH4 (g))= ▲H◦[H◦a C (s)] + 2(H-H) – 4(C-H)(C-H) bond shall require the data of H-H bond enthalpy and sublimation of enthalpy of carbonic solid.

Therefore, a.) is the answer.

Question. For a particular reversible reaction at temperature T, ∆H and ∆S were found to be both +ve. If Te is the temperature at equilibrium, the reaction would be spontaneous when?

1. Te is 5 times T
2. Te = T
3. Te > T
4. T > Te (JEE 2016)

Solution-It is known that at equilibrium, ∆G = 0.

∆G = ∆H – T∆S

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

So T > Te

Therefore, d.) is the answer.

Question. 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 which of the following? (JEE 2011)

1. +1056 kJ mol–1
2. -964 kJ mol–1
3. -1102 kJ mol–1
4. +352 kJ mol–1

Solution- ½ N2 + (3/2)H2 → NH3

(∆Hf) 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

Therefore, the answer is d.)

JEE Main Chemical Thermodynamics Quick Formulas

Thermodynamic Processes-

• Isochoric process: V, being constant

d V= 0

▲V= 0

• Isothermal process: T, being constant

d T =0

▲T= 0

• Isobaric process: P, being constant

d P= 0

▲P= 0

• AdiabaticProcess:

q=0

or heat exchange with the surrounding =0 (zero)

Thermal expansion -

• Linear Expansion : L = L0(1+αΔT)
• Area Expansion: A = A0(1+βΔT)
• Volume Expansion: V = V0(1+yΔT)

1st Law of Thermodynamics-

▲U= (U2- U1)= q + w

2nd Law of Thermodynamics-

▲Sunverse =▲Ssystem+ ▲Ssurrounding > 0 for a spontaneous surrounding

3rd Law of Thermodynamics-

The entropy of perfect crystals of all pure elements \& compounds is zero at the absolute zero of temperature.

Therefore, E/U=0

Reaction formulas-

If Hproducts, Reaction will be exothermic in the following cases-

• ▲Hreaction = Hproducts – Hreactants
• ▲H◦reactions= H◦products - H◦reactants
• ▲H◦reactions = positive – endothermic
• ▲H◦reactions= negative - exothermic

Entropy (S)-

▲Ssystem= {BA dqrev/T}

Entropy Calculation for an ideal gas undergoing a process-

State A = irr/▲sirr

State B = P1, V1, T1 P2, V2, T2

▲Ssystem= ncv ln T1/T2 + nRln V2/V1 ; only for an ideal gas.

Conduction-

Rate of flow of heat in conduction is given by-

dQdt=−KAdTdx\frac{dQ}{dt}=-KA\frac{dT}{dx}dtdQ=−KAdxdT

K is the thermal conductivity

A is the area of cross-section

dx is the thickness

dT is the Temperature difference

Emissivity

e = (Emissive power of a body at temperature T) / Emissive power of a black body at the same temperature