ChemistryThermodynamicsClass 11

Thermodynamics: ΔG, entropy and Hess’s law: interactive simulation

A reaction is spontaneous when ΔG is negative. ΔG = ΔH − TΔS, so the sign of ΔH and ΔS together decide whether temperature can flip the verdict.

Useful for: JEE Main, JEE Advanced, NEET, CBSE Boards · 3 modes in one lab

What you can explore

  1. 1

    ΔG and spontaneity

    Live ΔG = ΔH − TΔS and the temperature where the sign flips.
  2. 2

    Hess's law

    C → CO₂ directly, or through CO. Both paths sum to −393.5 kJ/mol.
  3. 3

    Entropy of the universe

    Reversible expansion keeps ΔS(universe) at zero. Free expansion does not.

Try this

Hess’s law says the enthalpy change depends only on the initial and final states. The direct combustion of graphite and the path through carbon monoxide give the same −393.5 kJ/mol.

  1. 1

    Set ΔH = −40 kJ/mol and ΔS = −80 J/K·mol, then raise T.

    What you should see

    ΔG starts negative and becomes positive once T passes 500 K. The reaction is spontaneous only when cold.

  2. 2

    Switch entropy from reversible expansion to free expansion.

    What you should see

    ΔS(system) stays the same, but ΔS(universe) is no longer zero.

Formulas behind it

Gibbs equation

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

Temperature where ΔG = 0

T=ΔHΔST = \frac{\Delta H}{\Delta S}

Entropy of isothermal expansion

ΔS=nRln⁡V2V1\Delta S = nR \ln\frac{V_2}{V_1}

Common mistakes

  • Adding TΔS when ΔS is in J/K and ΔH is in kJ without converting units.
  • Saying an exothermic reaction is always spontaneous. If ΔS is negative enough, high temperature makes ΔG positive.
Stuck on a thermodynamics: δg, entropy and hess’s law question?Snap it and get a whiteboard video solution.

FAQ

When is a reaction spontaneous at every temperature?

When ΔH is negative and ΔS is positive, so both terms make ΔG negative.

Why do the two Hess paths match?

Enthalpy is a state function. C + O₂ and CO₂ are the same start and end either way.

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