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Chemical Thermodynamics: concepts, formulas & solved PYQ-style examples

Thermodynamics questions in chemistry centre on sign conventions and on the Gibbs equation. Once signs are right, most numericals are one or two lines.

In chemistry, work done on the system is positive: ΔU = q + w.

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Key concepts

First law

ΔU = q + w with the IUPAC sign convention.

Work of expansion

Irreversible against constant pressure vs reversible isothermal (log formula).

Enthalpy

ΔH = ΔU + Δn_gRT; Hess’s law for combining reactions.

Entropy

Disorder increases with gas formation, melting and dissolving.

Gibbs energy

ΔG < 0 means spontaneous; the sign depends on temperature when ΔH and ΔS have the same sign.

Formula sheet

First law

ΔU=q+w\Delta U = q + w

Reversible isothermal work

w=−nRTln⁡V2V1w = -nRT\ln\dfrac{V_2}{V_1}

Enthalpy and internal energy

ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT

Gibbs equation

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

Gibbs energy and K

ΔG∘=−RTln⁡K\Delta G^\circ = -RT\ln K

What PYQs keep asking

Patterns that repeat across previous-year JEE and NEET papers for this chapter.

  • Temperature above or below which a reaction becomes spontaneous.
  • Work, heat and ΔU in isothermal reversible vs irreversible expansion.
  • Converting between ΔH and ΔU using Δn_g.
  • Predicting the sign of ΔS for a process.

Solved examples

Example 1

For a reaction, ΔH = +40 kJ mol⁻¹ and ΔS = +100 J K⁻¹ mol⁻¹. Above what temperature is it spontaneous?

  1. 1Spontaneous when ΔG=ΔH−TΔS<0\Delta G = \Delta H - T\Delta S < 0.
  2. 2Boundary: T=ΔHΔS=40000100=400T = \dfrac{\Delta H}{\Delta S} = \dfrac{40000}{100} = 400 K (convert kJ to J first).
  3. 3Both ΔH and ΔS are positive, so the reaction is spontaneous above this temperature.
Answer: Above 400 K
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Example 2

1 mol of an ideal gas expands isothermally and reversibly at 300 K from 10 L to 20 L. Find w, q and ΔU.

  1. 1Isothermal ideal gas: ΔU=0\Delta U = 0.
  2. 2w=−nRTln⁡V2V1=−(1)(8.314)(300)ln⁡2≈−1729w = -nRT\ln\dfrac{V_2}{V_1} = -(1)(8.314)(300)\ln 2 \approx -1729 J.
  3. 3From the first law: q=−w≈+1729q = -w \approx +1729 J.
Answer: w ≈ −1.73 kJ, q ≈ +1.73 kJ, ΔU = 0
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Common traps

  • Using ΔH in kJ and ΔS in J without converting.
  • Mixing the physics sign convention (W by gas) with the chemistry one (w on system).
  • Forgetting that ΔU = 0 for isothermal changes of an ideal gas.

Practise with

Stoichiometry SolverElectrochemistryChemical BondingGeneral Organic Chemistry (GOC)3 Thermodynamics practice questionsVideo solutions

FAQ

When is a reaction spontaneous at all temperatures?

When ΔH is negative and ΔS is positive — then ΔG is negative at every temperature.

What is Δn_g in ΔH = ΔU + Δn_gRT?

Moles of gaseous products minus moles of gaseous reactants in the balanced equation.

Is chemical thermodynamics in NEET?

Yes, mainly first law sign conventions, enthalpy relations and Gibbs energy spontaneity.

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