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MHT-CET Chemistry · Chemical Thermodynamics and Energetics

Gibbs Free Energy and Spontaneity

ΔG = ΔH − TΔS packs the second law into system-only quantities: negative means spontaneous, zero means equilibrium, and the temperature at which it crosses zero is ΔH/ΔS.

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Why this matters

23 PYQs, none HARD. Half are the sign table — which combination of ΔH and ΔS is spontaneous at all, high or no temperatures — and half are arithmetic: ΔG from ΔH, T and ΔS (units!), the boiling point or equilibrium temperature as ΔH/ΔS, and ΔG° = −2.303RT log K.

Concept 1 of 2: ΔG = ΔH − TΔS and the Sign Table

Two things drive a process: releasing heat (ΔH < 0) and increasing disorder (ΔS > 0). ΔG weighs them at the given temperature. When both favour, it is spontaneous at every T; when both oppose, at none; when they disagree, temperature decides — entropy wins when T is high.

Definition

  • ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S; ΔG < 0 spontaneous, ΔG > 0 non-spontaneous, ΔG = 0 equilibrium. Keep ΔS in kJ K⁻¹ when ΔH is in kJ.
  • ΔH < 0, ΔS > 0: spontaneous at ALL temperatures. ΔH > 0, ΔS < 0: at NONE.
  • ΔH > 0, ΔS > 0: spontaneous at HIGH T only (T>ΔH/ΔST > \Delta H/\Delta S). ΔH < 0, ΔS < 0: at LOW T only (T<ΔH/ΔST < \Delta H/\Delta S) — A + B with −84.2 kJ and −200 J K⁻¹ is spontaneous below 421 K.
  • Numbers: ΔH = 31400 J, ΔS = 32 J K⁻¹, 1273 K: 31400−40736=−933631400 - 40736 = -9336 J. Ice melting, 7 kJ, 24.8 J K⁻¹, 300 K: 7−7.44=−0.447 - 7.44 = -0.44 kJ. ΔG° = 28000 J, ΔS° = 120 J K⁻¹, 298 K: ΔH∘=28000+35760=63.76\Delta H^\circ = 28000 + 35760 = 63.76 kJ.
  • ΔG=−TΔStotal\Delta G = -T\Delta S_{\text{total}}, which is why a negative ΔG is the same statement as the second law.

Gibbs energy

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

Worked example

A reaction has ΔH = +40 kJ and ΔS = +125 J K⁻¹. Is it spontaneous at 300 K? Above what temperature does it become so?
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 2 · Q94Moderate

Example 1 · Chemical Thermodynamics and Energetics · Gibbs Free Energy and Spontaneity

For a reaction A + B →\rightarrow products ΔH=−84.2 kJ\Delta H = -84.2\,\text{kJ} and ΔS=−200 J K−1\Delta S = -200\,\text{J K}^{-1}. Calculate the highest value of temperature for forward spontaneity.

Subtracting joules from kilojoules

7 − 300 × 24.8 = −7433 'kJ' is nonsense; 24.8 J K⁻¹ is 0.0248 kJ K⁻¹, giving −0.44 kJ. The mismatched-unit answer is on the option list.
Drill 6 more on δg = δh − tδs and the sign table

Concept 2 of 2: ΔG = 0: Boiling Points, Transition Temperatures and K

At equilibrium ΔG = 0, so ΔH = TΔS and T = ΔH/ΔS — that is how a boiling point follows from the enthalpy and entropy of vaporisation. For a reaction the standard Gibbs energy sets the equilibrium constant: ΔG° = −2.303RT log K, a large K meaning a strongly negative ΔG°.

Definition

  • At equilibrium: ΔH∘=TΔS∘\Delta H^\circ = T\Delta S^\circ, T=ΔHΔST = \dfrac{\Delta H}{\Delta S}. Boiling point: 30 kJ / 75 J K⁻¹ = 400 K. ΔH = −210 kJ, ΔS = −150 J K⁻¹: 1400 K.
  • ΔG∘=−RTln⁡K=−2.303 RTlog⁡10K\Delta G^\circ = -RT\ln K = -2.303\,RT\log_{10} K, R = 8.314 × 10⁻³ kJ. Kp=3.356×1017K_p = 3.356 \times 10^{17} at 298 K: log⁡K=17.53\log K = 17.53, ΔG∘=−2.303×8.314×10−3×298×17.53=−100\Delta G^\circ = -2.303 \times 8.314 \times 10^{-3} \times 298 \times 17.53 = -100 kJ.
  • K > 1 ⇔ ΔG° < 0 (products favoured); K < 1 ⇔ ΔG° > 0.
  • Non-standard: ΔG=ΔG∘+2.303 RTlog⁡Q\Delta G = \Delta G^\circ + 2.303\,RT\log Q.

Equilibrium conditions

Teq=ΔHΔS,ΔG∘=−2.303 RTlog⁡10KT_{\text{eq}} = \frac{\Delta H}{\Delta S},\qquad \Delta G^\circ = -2.303\,RT\log_{10} K

Worked example

Water has ΔH_vap = 40.7 kJ mol⁻¹ and ΔS_vap = 109 J K⁻¹ mol⁻¹. Predict its boiling point.
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 1 · Q67Easy

Example 2 · Chemical Thermodynamics and Energetics · Gibbs Free Energy and Spontaneity

The enthalpy of vaporisation of a liquid is 30 kJ mol−130\,\text{kJ mol}^{-1} and entropy of vaporisation is 75 J K−1mol−175\,\text{J K}^{-1}\text{mol}^{-1}. Calculate boiling point of liquid at 1 atm.

Inverting to T = ΔS/ΔH

75/30000 = 0.0025 is not a temperature. ΔH (J) over ΔS (J K⁻¹) gives kelvin; the inverted ratio gives K⁻¹, and the wrong option is built from it.
Drill 4 more on δg = 0: boiling points, transition temperatures and k

Summary — formulas & gotchas at a glance

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