Skip to content
NCERT Exemplar · Q61

Q.ΔG is net energy available to do useful work and is thus a measure of "free energy". Show mathematically that ΔG is a measure of free energy. Find the unit of ΔG. If a reaction has positive enthalpy change and positive entropy change, under what condition will the reaction be spontaneous?

Punjab PsebLong· 5mImportance★★★★★est
99% · 97/98 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The Gibbs free energy change ΔG\Delta G equals the maximum useful work a system can perform, making it a measure of "free" energy. Its SI unit is the joule (J). For a reaction with ΔH>0\Delta H > 0 and ΔS>0\Delta S > 0, spontaneity requires T>ΔHΔST > \frac{\Delta H}{\Delta S} — the reaction is spontaneous only at sufficiently high temperatures.

Why ΔG\Delta G is a measure of free energy

The term "free energy" refers to energy that is available to do useful work — as opposed to energy that is inevitably lost as heat or tied up in entropy. The second law of thermodynamics tells us that for any process, the total entropy of the universe (system + surroundings) must increase. But we want a criterion that depends only on the system itself, not on the surroundings. That is exactly what Gibbs free energy provides.

The derivation connects the system's properties to the maximum work it can deliver.

  1. Start with the second law for the universe. For a spontaneous process at constant temperature and pressure, the total entropy change is:

ΔSuniverse=ΔSsystem+ΔSsurroundings>0\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}} > 0

  1. Express ΔSsurroundings\Delta S_{\text{surroundings}} in terms of the system's heat. The surroundings exchange heat with the system at constant pressure. The heat absorbed by the surroundings is −ΔHsystem-\Delta H_{\text{system}} (since heat lost by the system is gained by the surroundings). At constant temperature:

ΔSsurroundings=−ΔHsystemT\Delta S_{\text{surroundings}} = \frac{-\Delta H_{\text{system}}}{T}

  1. Combine to get a system-only condition. Substituting into the inequality:

ΔSsystem−ΔHsystemT>0\Delta S_{\text{system}} - \frac{\Delta H_{\text{system}}}{T} > 0

Multiply through by TT (positive):

TΔSsystem−ΔHsystem>0T\Delta S_{\text{system}} - \Delta H_{\text{system}} > 0

Rearranging:

−(ΔHsystem−TΔSsystem)>0-(\Delta H_{\text{system}} - T\Delta S_{\text{system}}) > 0

The quantity in parentheses is defined as the Gibbs free energy change:

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

So spontaneity means ΔG<0\Delta G < 0.

  1. Connect ΔG\Delta G to useful work. From the first law, the total work done by a system is w=wuseful+wexpansionw = w_{\text{useful}} + w_{\text{expansion}}, where wexpansion=−PΔVw_{\text{expansion}} = -P\Delta V (work done against constant pressure). At constant pressure, ΔH=qP\Delta H = q_P (heat at constant pressure). The change in internal energy is ΔU=q+w=qP+wuseful−PΔV\Delta U = q + w = q_P + w_{\text{useful}} - P\Delta V. But ΔH=ΔU+PΔV\Delta H = \Delta U + P\Delta V, so:

ΔH=qP+wuseful\Delta H = q_P + w_{\text{useful}}

The maximum work occurs for a reversible process, where qP=TΔSq_P = T\Delta S. Thus:

ΔH=TΔS+wmax, useful\Delta H = T\Delta S + w_{\text{max, useful}}

Rearranging:

wmax, useful=ΔH−TΔS=ΔGw_{\text{max, useful}} = \Delta H - T\Delta S = \Delta G

Therefore, ΔG\Delta G equals the maximum useful work (other than expansion work) that the system can perform. This is why it is called "free" energy — it is the energy free to do useful work.

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

At constant TT and PP, ΔG\Delta G is the maximum non-expansion work obtainable from the system.

Unit of ΔG\Delta G …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.