Class 11 — Thermodynamics

Chapter Overview & Weightage

Class 11 Chemistry Thermodynamics is the conceptual backbone for all later energy-related topics — equilibrium, electrochemistry, kinetics. The chapter is heavy on definitions, sign conventions, and a small set of state functions. Master these and the math is straightforward.

Typical CBSE weightage: $7-9$ marks per year. Usually one $3$ -mark numerical on enthalpy/internal energy and one $5$ -mark question combining first law with calorimetry or Hess’s law.

Year	Marks	Question type
2019	5	Hess’s law application
2020	3	First law numerical
2021	7	Spontaneity (Gibbs energy) + first law
2022	5	Bond enthalpy calculation
2023	3	Sign convention reasoning
2024	5	Enthalpy of formation

Key Concepts You Must Know

System, surroundings, types of systems (open, closed, isolated)
State functions vs path functions; key state functions: $U, H, S, G$
First law: $\Delta U = q + w$ (and the sign convention)
Work done in different processes: isothermal, adiabatic, isochoric, isobaric
Enthalpy: $H = U + PV$ and $\Delta H = \Delta U + \Delta n_g RT$
Hess’s law and standard enthalpies (formation, combustion, neutralisation, atomisation)
Spontaneity: $\Delta G = \Delta H - T\Delta S$
Bond enthalpy and average bond enthalpy

Important Formulas

$\Delta U = q + w$

with sign convention: $w > 0$ if work is done ON the system, $q > 0$ if heat is absorbed BY the system.

Isothermal reversible: $w = -nRT\ln(V_2/V_1)$

Isothermal irreversible (against const $P_{ext}$ ): $w = -P_{ext}(V_2 - V_1)$

Adiabatic: $w = nC_V\Delta T = (P_2V_2 - P_1V_1)/(\gamma - 1)$

$\Delta H = \Delta U + \Delta n_g RT$

$\Delta G = \Delta H - T\Delta S$

Use $\Delta G < 0$ for spontaneity at constant $T, P$ .

Solved Previous Year Questions

PYQ 1 (CBSE 2020, 3 marks)

$2$ moles of an ideal gas expand isothermally and reversibly at $300$ K from $5$ L to $20$ L. Calculate the work done.

$w = -nRT\ln(V_2/V_1) = -2 \times 8.314 \times 300 \times \ln(20/5)$ .

$w = -2 \times 8.314 \times 300 \times \ln 4 = -2 \times 8.314 \times 300 \times 1.386 \approx -6915$ J.

Negative sign: work done BY the system.

PYQ 2 (CBSE 2022, 5 marks)

Use Hess’s law to calculate $\Delta H_f$ of $CH_4$ given: $C(s) + O_2 \to CO_2$ , $\Delta H_1 = -393.5$ kJ $H_2 + \tfrac{1}{2}O_2 \to H_2O$ , $\Delta H_2 = -285.8$ kJ $CH_4 + 2O_2 \to CO_2 + 2H_2O$ , $\Delta H_3 = -890.4$ kJ

Target: $C + 2H_2 \to CH_4$ .

Use Eq1 + 2(Eq2) - Eq3.

$\Delta H_f = -393.5 + 2(-285.8) - (-890.4) = -393.5 - 571.6 + 890.4 = -74.7$ kJ/mol.

PYQ 3 (CBSE 2024, 5 marks)

For a reaction, $\Delta H = -50$ kJ/mol, $\Delta S = -100$ J/(K·mol). Find the temperature above which the reaction becomes non-spontaneous.

Reaction is spontaneous when $\Delta G = \Delta H - T\Delta S < 0$ .

$T < \Delta H/\Delta S = -50000/(-100) = 500$ K.

So reaction is non-spontaneous above $500$ K.

Difficulty Distribution

Difficulty	% of CBSE Qs	Typical type
Easy	$35\%$	Definitions, sign convention
Medium	$50\%$	First-law numericals, Hess’s law
Hard	$15\%$	Adiabatic process, Gibbs spontaneity edge cases

Expert Strategy

Memorise the IUPAC sign convention: heat absorbed positive, work done ON the system positive. Some old textbooks use the reverse — stick to NCERT.

For Hess’s law, set up the target equation first, then work backward to figure out which reactions to add/subtract/multiply.

For “spontaneity above what temperature” questions, write $\Delta G = 0$ at the threshold and solve $T = \Delta H/\Delta S$ . Direction of inequality follows from the signs.

Common Traps

Forgetting to convert $\Delta S$ from J/(K·mol) to kJ/(K·mol) when $\Delta H$ is in kJ. Mixing units gives answers off by a factor of $1000$ .

Using $\Delta H = \Delta U$ for a gas reaction. They differ by $\Delta n_g RT$ . Only when $\Delta n_g = 0$ are they equal.

Treating $w = -P\Delta V$ for any process. This is the irreversible-against-constant- $P$ formula. For reversible isothermal expansion, the formula is $w = -nRT\ln(V_2/V_1)$ .