Chemical Thermodynamics — JEE Main Chemistry Notes & Formulas

1. Chapter at a Glance

Chemical thermodynamics studies energy changes in physical and chemical processes and predicts whether a process can happen by itself (spontaneity). It rests on a few laws: energy is conserved (first law), the entropy of the universe increases in a spontaneous process (second law), and a perfect crystal has zero entropy at 0 K (third law). JEE Main asks one or two questions — numericals on ΔH and ΔU, work, Hess's law and Gibbs energy.

What JEE asks most

First law: ΔU = q + w, sign conventions and work in expansions.
ΔH − ΔU = Δn_gRT for gaseous reactions.
Enthalpy of reaction using Hess's law or bond enthalpies.
Spontaneity: sign of ΔG = ΔH − TΔS and ΔG° = −RT ln K.

2. Basic Terms

System: the part under study; surroundings: everything else. Systems are open (exchange matter and energy), closed (energy only) or isolated (neither).
State functions depend only on the initial and final states: U, H, S, G, P, V, T. Path functions depend on the path: q and w.
Extensive properties depend on the amount (mass, volume, U, H); intensive properties do not (T, P, density, molar quantities).
Processes: isothermal (constant T), adiabatic (q = 0), isobaric (constant P), isochoric (constant V), cyclic (returns to the start, so ΔU = 0).

3. First Law and Work

ΔU = q + w

Sign convention (IUPAC, used in NCERT): heat absorbed by the system and work done on the system are positive.

Work of expansion against constant external pressure: w = −P_extΔV. Free expansion (P_ext = 0): w = 0.
Isothermal reversible expansion of an ideal gas: w = −2.303 nRT log(V₂/V₁) = −2.303 nRT log(P₁/P₂). For an ideal gas at constant T, ΔU = 0, so q = −w.
Reversible work is the maximum work obtainable.

On a P–V diagram, an adiabatic curve is steeper than an isothermal curve; the area under a curve is the work done.

4. Enthalpy and Thermochemistry

Enthalpy H = U + PV. At constant pressure, the heat exchanged equals ΔH; at constant volume, it equals ΔU.

ΔH = ΔU + Δn_gRT

Δn_g = (moles of gaseous products) − (moles of gaseous reactants). Negative ΔH means exothermic; positive means endothermic.

Hess's law: the enthalpy change of a reaction is the same whether it occurs in one step or several. Thermochemical equations can be added, subtracted and multiplied like algebraic equations.
Standard enthalpy of formation ΔH_f°: for 1 mol of compound from its elements in their standard states (ΔH_f° of elements = 0). ΔH_r° = ΣΔH_f°(products) − ΣΔH_f°(reactants).
Bond enthalpy method: ΔH_r = Σ(bond enthalpies of bonds broken) − Σ(bond enthalpies of bonds formed) — used for gaseous reactions.
Enthalpy of neutralisation of a strong acid by a strong base ≈ −57.1 kJ mol⁻¹; less negative for weak acids or bases (energy used for ionisation).
Heat capacity: q = C ΔT; C_p − C_v = R for one mole of ideal gas.

5. Entropy, Gibbs Energy and Spontaneity

Entropy (S) measures the randomness or disorder. ΔS = q_rev/T. Entropy increases on melting, vaporisation, dissolving and when Δn_g > 0.
Second law: for a spontaneous process ΔS_total = ΔS_system + ΔS_surroundings > 0.
Gibbs energy: G = H − TS; at constant T and P,

ΔG = ΔH − TΔS; ΔG < 0 spontaneous, ΔG = 0 equilibrium, ΔG > 0 non-spontaneous

ΔH	ΔS	Spontaneous?
−	+	At all temperatures
+	−	Never
−	−	At low temperatures (T < ΔH/ΔS)
+	+	At high temperatures (T > ΔH/ΔS)

ΔG° = −RT ln K = −2.303 RT log K; ΔG = ΔG° + RT ln Q

If K > 1, ΔG° is negative (products favoured); if K = 1, ΔG° = 0.

6. Formula Master Table

Quantity	Formula
First law	ΔU = q + w
Irreversible work	w = −P_extΔV
Isothermal reversible work	w = −2.303 nRT log(V₂/V₁)
ΔH and ΔU	ΔH = ΔU + Δn_gRT
Reaction enthalpy (formation)	ΣΔH_f°(P) − ΣΔH_f°(R)
Reaction enthalpy (bonds)	ΣBE(broken) − ΣBE(formed)
Entropy change	ΔS = q_rev/T
Phase change	ΔS = ΔH/T (at the transition temperature)
Gibbs energy	ΔG = ΔH − TΔS
Equilibrium constant	ΔG° = −2.303 RT log K
Ideal gas heat capacities	C_p − C_v = R

7. Mind Map

Thermodynamics

Basics

System types
State vs path functions
Extensive vs intensive

First law

ΔU = q + w
w = −PΔV
Isothermal reversible work

Enthalpy

ΔH = ΔU + Δn_gRT
Hess's law
Bond enthalpies

Spontaneity

ΔS, second law
ΔG = ΔH − TΔS
ΔG° = −RT ln K

8. Tips and Tricks

Shortcuts

Count only gaseous species for Δn_g; ignore liquids and solids.
Convert units before combining: ΔH is usually in kJ and ΔS in J K⁻¹ — change one of them.
For the temperature at which a reaction becomes spontaneous, set ΔG = 0: T = ΔH/ΔS.
Reverse a thermochemical equation → change the sign of ΔH; multiply the equation → multiply ΔH.

Mistakes students make

Mixing sign conventions: in the IUPAC convention, work done by the gas in expansion is negative.
Using ln instead of log (or forgetting the 2.303).
Assuming every exothermic reaction is spontaneous — entropy also matters.

9. Important One-liners

For an ideal gas, internal energy depends only on temperature.
In a cyclic process ΔU = 0 and ΔH = 0.
In an adiabatic process q = 0, so ΔU = w.
Entropy of a perfectly crystalline substance at absolute zero is zero (third law).
ΔH_f° of an element in its standard state is zero (e.g. graphite, O₂(g)).
Free expansion of an ideal gas into vacuum involves no work and, for an ideal gas, no change in temperature.
At equilibrium ΔG = 0 but ΔG° is not necessarily zero.

10. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main. (R = 8.314 J K⁻¹ mol⁻¹.)

Q1Numerical

A system absorbs 50 J of heat and does 20 J of work on the surroundings. Find ΔU (in J).

Show solution

q = +50 J (absorbed); w = −20 J (work done by the system). ΔU = q + w = 50 − 20 = 30 J.

Answer: 30

Q2Medium

For N₂(g) + 3H₂(g) → 2NH₃(g) at 298 K, the value of ΔH − ΔU is approximately:

+4.96 kJ
−4.96 kJ
−2.48 kJ
0

Show solution

Δn_g = 2 − 4 = −2. ΔH − ΔU = Δn_gRT = −2 × 8.314 × 298 ≈ −4955 J ≈ −4.96 kJ.

Answer: (B)

Q3Numerical

For a reaction, ΔH = +30 kJ mol⁻¹ and ΔS = +100 J K⁻¹ mol⁻¹. Above what temperature (in K) does it become spontaneous? (Assume ΔH and ΔS do not change with temperature.)

Show solution

ΔG = ΔH − TΔS < 0 ⇒ T > ΔH/ΔS = 30000/100 = 300 K.

Answer: 300

Q4Medium

Given C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ and CO(g) + ½O₂(g) → CO₂(g), ΔH = −283.0 kJ. The enthalpy of formation of CO(g) is:

−676.5 kJ mol⁻¹
−110.5 kJ mol⁻¹
+110.5 kJ mol⁻¹
−283.0 kJ mol⁻¹

Show solution

Target: C(s) + ½O₂(g) → CO(g). Take the first equation and subtract the second (i.e. add the reverse of the second):

ΔH = −393.5 − (−283.0) = −110.5 kJ mol⁻¹.

Answer: (B)

Q5Medium

One mole of an ideal gas expands isothermally and reversibly at 300 K from 10 L to 100 L. The work done on the gas is approximately:

+5.74 kJ
−5.74 kJ
−2.49 kJ
0

Show solution

w = −2.303nRT log(V₂/V₁) = −2.303 × 1 × 8.314 × 300 × log 10 ≈ −5744 J ≈ −5.74 kJ.

The negative sign shows that the gas does work on the surroundings during expansion.

Answer: (B)

Q6Easy

For which process is ΔS negative?

Melting of ice
Evaporation of water
2H₂(g) + O₂(g) → 2H₂O(l)
Dissolving NaCl in water

Show solution

Three moles of gas become liquid water, so disorder decreases sharply: ΔS < 0. The other processes increase disorder.

Answer: (C)

Q7Numerical

Using bond enthalpies H–H = 435, Cl–Cl = 243 and H–Cl = 431 kJ mol⁻¹, find ΔH (in kJ) for H₂(g) + Cl₂(g) → 2HCl(g).

Show solution

Bonds broken: 435 + 243 = 678 kJ. Bonds formed: 2 × 431 = 862 kJ.

ΔH = 678 − 862 = −184 kJ.

Answer: −184

Q8Hard

For a reaction at 300 K, the equilibrium constant K = 1. Which statement is correct?

ΔG° = 0
ΔG° < 0
ΔG° > 0
ΔH = 0

Show solution

ΔG° = −RT ln K = −RT ln 1 = 0. This says nothing about ΔH by itself; ΔH and TΔS° simply balance.

Answer: (A)

11. Quick Revision Summary

Remember before the exam

ΔU = q + w (heat absorbed and work done on the system are positive).
w = −P_extΔV; isothermal reversible w = −2.303nRT log(V₂/V₁).
ΔH = ΔU + Δn_gRT; Hess's law; bonds broken − bonds formed.
ΔG = ΔH − TΔS; spontaneous if ΔG < 0; T = ΔH/ΔS at the switch-over.
ΔG° = −2.303RT log K.

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