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Class 11 Chemistry Notes

Thermodynamics Class 11 Chemistry Notes

Complete, exam-ready notes on chemical thermodynamics: systems and state functions, work and heat, the first law, enthalpy and calorimetry, Hess's law, bond enthalpies, entropy, Gibbs free energy and spontaneity — written for CBSE, JEE and NEET revision.

Class11SubjectChemistryCoversCBSE · JEE · NEET

Written byDeep Narayan· Science & Mathematics EducatorReviewed byPushpanjali

What is thermodynamics in one line?

Thermodynamics is the study of energy changes (heat and work) during physical and chemical processes, and the conditions (entropy and Gibbs free energy) that determine whether a process is spontaneous.

Systems, Work and Heat

  • System types: open (matter + energy exchange), closed (energy only), isolated (neither).
  • State functions (H, U, V, P, T) depend only on the current state, not the path; path functions (q, w) depend on how the change happens.
  • Work done by a gas:
  • w=−PextΔVw = -P_{\text{ext}}\Delta V
  • .
  • Sign convention: heat absorbed by the system q > 0, work done on the system w > 0.
w=−Pext ΔV=−nR ΔT (isobaric)w = -P_{\text{ext}}\,\Delta V = -nR\,\Delta T\ \text{(isobaric)}
Work done by a gas

First Law of Thermodynamics

First law

ΔU=q+w\Delta U = q + w

The change in internal energy of a system equals the heat added plus the work done on it (considering sign conventions). Equivalently, energy is conserved.

  • For an ideal gas,
  • ΔU=nCVΔT\Delta U = nC_V\Delta T
  • .
  • Adiabatic process (q = 0):
  • ΔU=w\Delta U = w
  • .
  • Isothermal process (ΔT = 0):
  • ΔU=0\Delta U = 0
  • , so
  • q=−wq = -w
  • .

Internal energy is a state function

ΔU depends only on the initial and final states, but q and w depend on the path. For the same change, different paths give different q and w, but the sum q + w is always the same ΔU.

Enthalpy and Calorimetry

Enthalpy

H=U+PV,ΔH=ΔU+PΔVH = U + PV,\quad \Delta H = \Delta U + P\Delta V

Enthalpy is the heat absorbed at constant pressure. For reactions at constant pressure, qp=ΔHq_p = \Delta H. Exothermic reactions have ΔH<0\Delta H < 0; endothermic have ΔH>0\Delta H > 0.

  • Enthalpy of formation: heat to form 1 mole of a compound from its elements.
  • Standard enthalpy of reaction is measured at 298 K and 1 bar.
  • Calorimetry measures heat using
  • q=mcΔTq = mc\Delta T
  • .
ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT
Relation between ΔH and ΔU (gas reactions)

Hess's Law of Constant Heat Summation

Hess's law

ΔHrxn=∑ΔHf(products)−∑ΔHf(reactants)\Delta H_{\text{rxn}} = \sum \Delta H_f (\text{products}) - \sum \Delta H_f (\text{reactants})

The enthalpy change of a reaction is independent of the path taken — it depends only on the initial and final states. This lets us calculate ΔH for reactions that cannot be measured directly.

Reverse and scale reactions

Reverse a reaction to reverse the sign of ΔH; multiply a reaction by a factor to scale ΔH by the same factor. Sum the steps to get the target ΔH.

Bond Enthalpy and Enthalpy of Combustion

Bond enthalpy

ΔHrxn=∑BE(broken)−∑BE(formed)\Delta H_{\text{rxn}} = \sum BE (\text{broken}) - \sum BE (\text{formed})

Bond enthalpy is the energy to break one mole of a bond. Estimate reaction ΔH by subtracting the bond enthalpies of bonds formed from those broken.

Sign care

Breaking bonds absorbs energy (positive); forming bonds releases energy (negative). So ΔH = Σ(BE broken) − Σ(BE formed).

Entropy and Spontaneity

Entropy

ΔS=qrevT\Delta S = \frac{q_{\text{rev}}}{T}

Entropy measures randomness (disorder). Systems tend toward increasing entropy. For an isolated system, ΔS≥0\Delta S \geq 0 — spontaneous processes increase entropy.

Gibbs free energy

ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S

Gibbs free energy determines spontaneity at constant T and P: spontaneous if ΔG<0\Delta G < 0, at equilibrium if ΔG=0\Delta G = 0, non-spontaneous if ΔG>0\Delta G > 0.

Predicting spontaneity

ΔH negative and ΔS positive is always spontaneous. ΔH positive and ΔS negative is never spontaneous. If ΔH and ΔS have the same sign, spontaneity depends on temperature (TΔS term).

Solved Examples

Example: A reaction is exothermic with ΔH=−20 kJ\Delta H = -20\,\text{kJ} and ΔS=+50 J/K\Delta S = +50\,\text{J/K} at 300 K. Is it spontaneous?

Solution: ΔG=−20−0.050×300=−20−15=−35 kJ\Delta G = -20 - 0.050\times300 = -20 - 15 = -35\,\text{kJ}. Since ΔG<0\Delta G < 0, the process is spontaneous.

Revision

Key formulas at a glance

Memorise these before attempting numericals — most exam questions hinge on one of them.

First law

ΔU=q+w\Delta U = q + w

Enthalpy

ΔH=ΔU+PΔV\Delta H = \Delta U + P\Delta V

Hess's law

ΔH=∑ΔHf(P)−∑ΔHf(R)\Delta H = \sum \Delta H_f (P) - \sum \Delta H_f (R)

Bond enthalpy

ΔH=∑BE(broken)−∑BE(formed)\Delta H = \sum BE(broken) - \sum BE(formed)

Gibbs free energy

ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S

Spontaneity

spontaneous if ΔG<0\text{spontaneous if } \Delta G < 0

Exam tips

How this chapter is asked

Where this topic appears in CBSE, JEE Main and NEET papers.

  • ΔU = q + w is the first law (sign conventions matter).
  • q_p = ΔH at constant pressure.
  • Hess's law: ΔH is path independent.
  • ΔG < 0 → spontaneous; ΔG = 0 → equilibrium.
  • Breaking bonds is endothermic; forming bonds is exothermic.

FAQ

Common questions

What is Gibbs free energy?

Gibbs free energy G = H − TS. The change ΔG = ΔH − TΔS decides spontaneity at constant temperature and pressure: negative means spontaneous, zero means equilibrium.

What is the first law of thermodynamics?

Energy is conserved: the change in internal energy ΔU equals heat added (q) plus work done on the system (w), i.e. ΔU = q + w.

What is Hess's law?

The enthalpy change of a reaction depends only on the initial and final states, not the path, so it is the sum of the ΔH of its steps. It is used to find ΔH for unmeasurable reactions.

What is entropy?

Entropy is a measure of disorder or randomness. Spontaneous processes in an isolated system increase entropy, with ΔS ≥ 0.

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