Thermodynamics JEE questions: formulas, patterns and the mistakes that cost marks

System, surroundings and sign convention

ΔU = q + w with w the work done on the system. Heat absorbed by the system is positive; work done by the system in expanding is negative. Fixing this convention once removes most of the errors in the chapter.

State functions and path functions

U, H, S and G depend only on the state, so their changes are the same by any route. q and w depend on the path — reversible and irreversible expansions between the same states give different work.

Enthalpy versus internal energy

H = U + PV, and for reactions involving gases ΔH = ΔU + Δn_g RT. Δn_g is the change in the number of moles of gas only, so a reaction with no gas-phase change has ΔH = ΔU.

Heat capacities

q_v = ΔU and q_p = ΔH, which is why bomb calorimetry measures ΔU and open-vessel work measures ΔH. For an ideal gas, C_p − C_v = R per mole.

Hess’s law and enthalpies of formation

Because H is a state function, reaction enthalpies add like the equations that produce them: ΔH_rxn = ΣΔH_f(products) − ΣΔH_f(reactants). With bond enthalpies the rule inverts: bonds broken minus bonds formed.

Entropy, Gibbs energy and spontaneity

ΔG = ΔH − TΔS at constant T and P. A negative ΔG means spontaneous. When ΔH and ΔS share a sign, temperature decides: an endothermic reaction with positive ΔS becomes spontaneous above T = ΔH/ΔS.

First law

ΔU = q + w

IUPAC convention: w is work done on the system.

Pressure–volume work

w = −p_ext ΔV

Irreversible expansion against constant external pressure.

Reversible isothermal work

w = −2.303 nRT log(V₂/V₁)

Ideal gas; ΔU = 0, so q = −w.

Enthalpy and internal energy

ΔH = ΔU + Δn_g RT

Δn_g counts gaseous moles only.

Heat at constant volume and pressure

q_v = ΔU · q_p = ΔH

Bomb calorimeter gives ΔU.

Heat capacities of an ideal gas

C_p − C_v = R

Per mole; γ = C_p/C_v.

Hess’s law

ΔH_rxn = Σ ΔH_f(products) − Σ ΔH_f(reactants)

Elements in their standard states have ΔH_f = 0.

Bond enthalpies

ΔH = Σ BE(broken) − Σ BE(formed)

Opposite order to the formation-enthalpy rule.

Entropy change

ΔS = q_rev/T · ΔS_universe > 0 for a spontaneous change

Entropy of the system alone may fall.

Gibbs energy

ΔG = ΔH − TΔS · ΔG° = −2.303RT log K

ΔG < 0 ⇒ spontaneous at that temperature.

ΔH and ΔU · 3 marks · moderate

For N₂(g) + 3H₂(g) → 2NH₃(g), ΔH° = −92.4 kJ mol⁻¹ at 298 K. Calculate ΔU°. (R = 8.314 J K⁻¹ mol⁻¹)

1. Count gaseous moles: Δn_g = 2 − 4 = −2.
2. Rearrange ΔH = ΔU + Δn_gRT to get ΔU = ΔH − Δn_gRT.
3. Δn_gRT = (−2)(8.314)(298) = −4955 J ≈ −4.96 kJ.
4. ΔU = −92.4 − (−4.96) = −87.4 kJ mol⁻¹.
5. The magnitude of ΔU is smaller because the surroundings do compression work on the shrinking gas.

Answer: ΔU° ≈ −87.4 kJ mol⁻¹.

Isothermal reversible expansion · 3 marks · moderate

Two moles of an ideal gas expand reversibly and isothermally at 300 K from 5 L to 50 L. Calculate w, ΔU and q.

1. Reversible isothermal work: w = −2.303 nRT log(V₂/V₁).
2. log(50/5) = log 10 = 1.
3. w = −2.303 × 2 × 8.314 × 300 × 1 ≈ −11488 J ≈ −11.5 kJ.
4. Isothermal, ideal gas: ΔU = 0.
5. First law: q = ΔU − w = 0 + 11.5 = +11.5 kJ, absorbed from the surroundings.

Answer: w ≈ −11.5 kJ, ΔU = 0, q ≈ +11.5 kJ.

Gibbs energy and spontaneity · 3 marks · easy

A reaction has ΔH = +30 kJ mol⁻¹ and ΔS = +75 J K⁻¹ mol⁻¹. Is it spontaneous at 298 K, and above what temperature does it become spontaneous?

1. Convert to matching units: ΔS = 0.075 kJ K⁻¹ mol⁻¹.
2. At 298 K: ΔG = 30 − (298)(0.075) = 30 − 22.35 = +7.65 kJ mol⁻¹, so it is not spontaneous.
3. Spontaneity begins when ΔG = 0: T = ΔH/ΔS = 30/0.075 = 400 K.
4. Above 400 K the TΔS term outweighs the endothermic ΔH.

Answer: Not spontaneous at 298 K (ΔG ≈ +7.7 kJ mol⁻¹); spontaneous above about 400 K.