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JEE Mains Chemistry · Chemical Thermodynamics

Gibbs Energy and the Equilibrium Constant

How far a reaction goes: ΔG° = −RT ln K, the equilibrium constant from a degree of dissociation or from rate constants, and the graphs of log K against 1/T and of G against the extent of reaction.

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Why this matters

Fourteen PYQs, ten of them numerical, and five from 2026. Nine link ΔG° to K, often through a degree of dissociation, rate constants or a table of ΔH and S. Five read a graph: log K against 1/T, G against the extent of reaction, or ΔH and ΔS against temperature.

Concept 1 of 2: Standard Gibbs energy and the equilibrium constant

ΔG° compares products and reactants in their standard states; K says where the real mixture settles. The more negative ΔG°, the further the equilibrium lies towards products. A positive ΔG° does not stop the reaction — it only means K is less than 1.

Definition

  • ΔG∘=−RTln⁡K=−2.303 RTlog⁡K\Delta G^\circ = -RT\ln K = -2.303\,RT\log K.
  • ΔG∘<0⇒K>1\Delta G^\circ < 0 \Rightarrow K > 1; ΔG∘=0⇒K=1\Delta G^\circ = 0 \Rightarrow K = 1; ΔG∘>0⇒K<1\Delta G^\circ > 0 \Rightarrow K < 1.
  • Combine it with ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ to find any one of ΔH∘\Delta H^\circ, ΔS∘\Delta S^\circ or K from the other two.
  • From rate constants: K=kfkrK = \frac{k_f}{k_r}.
  • From a degree of dissociation: find the moles at equilibrium, turn them into partial pressures pi=xiPp_i = x_i P, then build KpK_p.

Gibbs energy and the equilibrium constant

ΔG∘=−RTln⁡K=−2.303 RTlog⁡K\Delta G^\circ = -RT\ln K = -2.303\,RT\log K

Worked example

A reaction has ΔH∘=−20\Delta H^\circ = -20 kJ mol⁻¹ and ΔS∘=+12\Delta S^\circ = +12 J K⁻¹ mol⁻¹. Find ΔG∘\Delta G^\circ and K at 500 K. (R=8.314R = 8.314 J K⁻¹ mol⁻¹)
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 1 · Q50Moderate

Example 1 · Chemical Thermodynamics · Gibbs Energy and the Equilibrium Constant

Consider the reaction X⇌YX \rightleftharpoons Y at 300 K . If ΔHθ\Delta H^{\theta} and K are 28.40 kJ mol−128.40\text{ }kJ{\text{ }mol}^{- 1} and 1.8×10−71.8 \times 10^{- 7} at the same temperature, then the magnitude of ΔSθ\Delta S^{\theta} for the reaction in JK−1 mol−1JK^{- 1}{\text{ }mol}^{- 1} is ____\_\_\_\_ .(Nearest Integer). (Given : R=8.3JK−1 mol−1,ln⁡10=2.3R = 8.3{JK}^{- 1}{\text{ }mol}^{- 1},\ln10 = 2.3, log⁡3=0.48,log⁡2=0.30\log3 = 0.48,\log2 = 0.30

A positive ΔG° is not 'no reaction'

Some product always forms. ΔG∘>0\Delta G^\circ > 0 only means the equilibrium lies on the reactant side, with K below 1. Statements that the reaction 'will not occur at all' are wrong.

Moles are not partial pressures

KpK_p uses partial pressures. Divide each equilibrium amount by the total moles to get a mole fraction, then multiply by the total pressure, before building KpK_p.
Drill 8 more on standard gibbs energy and the equilibrium constant

Concept 2 of 2: Graphs of log K against 1/T and of G against extent

Put ΔG° = ΔH° − TΔS° into ΔG° = −2.303RT log K, and log K becomes a straight line in 1/T. Its slope carries ΔH° and its intercept carries ΔS°. On a curve of G against how far the reaction has gone, the lowest point is equilibrium.

Definition

  • log⁡K=−ΔH∘2.303R⋅1T+ΔS∘2.303R\log K = -\frac{\Delta H^\circ}{2.303R}\cdot\frac{1}{T} + \frac{\Delta S^\circ}{2.303R}.
  • Slope =−ΔH∘2.303R= -\frac{\Delta H^\circ}{2.303R}: negative for an endothermic reaction (K rises with T), positive for an exothermic one.
  • Intercept =ΔS∘2.303R= \frac{\Delta S^\circ}{2.303R}.
  • In ln form: ln⁡K=−ΔH∘−TΔS∘RT\ln K = -\frac{\Delta H^\circ - T\Delta S^\circ}{RT}, and K=e−ΔG∘/RTK = e^{-\Delta G^\circ/RT}.
  • G against extent of reaction: left of the minimum the forward reaction is spontaneous; at the minimum, equilibrium; right of it, the reverse. ΔG∘\Delta G^\circ = G(pure products) − G(pure reactants).
  • If ΔH∘\Delta H^\circ and ΔS∘\Delta S^\circ do not change with T, their graphs against T are flat lines, while ΔG∘\Delta G^\circ is a sloping line.

log K against 1/T

log⁡K=−ΔH∘2.303R⋅1T+ΔS∘2.303R\log K = -\frac{\Delta H^\circ}{2.303R}\cdot\frac{1}{T} + \frac{\Delta S^\circ}{2.303R}

Worked example

A plot of log K against 1/T is a straight line with slope −2000 K and intercept 3.0. Find ΔH∘\Delta H^\circ and ΔS∘\Delta S^\circ. (R=8.314R = 8.314 J K⁻¹ mol⁻¹)
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 2 · Q30Moderate

Example 2 · Chemical Thermodynamics · Gibbs Energy and the Equilibrium Constant

The plot of log⁡10 K\log_{10}\text{ }K vs 1 T\frac{1}{\text{ }T} gives a straight line. The intercept and slope respectively are (where K is equilibrium constant).

Dropping the minus sign

ln⁡K=ΔH∘−TΔS∘RT\ln K = \frac{\Delta H^\circ - T\Delta S^\circ}{RT} is the planted wrong form. Since ln⁡K=−ΔG∘/RT\ln K = -\Delta G^\circ/RT, the numerator is TΔS∘−ΔH∘T\Delta S^\circ - \Delta H^\circ.

Reading the slope

The slope is −ΔH∘/2.303R-\Delta H^\circ/2.303R. A line that falls from left to right means ΔH∘>0\Delta H^\circ > 0: K grows as the temperature rises.
Drill 4 more on graphs of log k against 1/t and of g against extent

Summary — formulas & gotchas at a glance

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Formulas (2)

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