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Chemical Equilibrium revision notes

Every idea in the Chemical Equilibrium chapter for JEE Main, with the usual mistakes and the fast methods. Topic 1 is open to everyone; sign in free to read all 18 and download the PDF.

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How we count: 7× asked means 7 past JEE Main questions in the PrepWiser bank needed that concept, and "last" is the latest year one appeared. ★ marks ideas asked 8 or more times.

Dynamic and physical equilibrium

Dynamic equilibrium
5× asked, last 2025
rf=rb,Kc=kfkb  (one-step reaction)r_f = r_b,\qquad K_c = \frac{k_f}{k_b}\ \ (\text{one-step reaction})
Closed system, fixed TT. Both reactions continue at equal rates; amounts stay constant, not equal. Reached from either side.
ttconc.0.250.751[A][B]
From pure A (solid) or pure B (dashed): the same end mixture, [B]/[A]=Kc=3[B]/[A]=K_c=3
ttraterf=kf[A]r_f=k_f[A]rb=kb[B]r_b=k_b[B]rf=rbr_f=r_b
Both reactions go on; at equilibrium their rates are equal, not zero
  • Not 'all properties become equal' at equilibrium. They become constant. The two rates are equal; the concentrations stop changing but usually differ from each other.
  • Quick method. Rate written as a forward term minus a backward term: at equilibrium the two terms are equal, so Kc=kf/kbK_c=k_f/k_b. Read each rate constant off its own term.
Equilibria in physical processes
4× asked, last 2025
HX2O(s)⇌HX2O(l),HX2O(l)⇌HX2O(g),IX2(s)⇌IX2(g)\ce{H2O(s) <=> H2O(l)},\qquad \ce{H2O(l) <=> H2O(g)},\qquad \ce{I2(s) <=> I2(g)}
Closed system, fixed TT. Normal m.p./b.p. at 1.013 bar. Henry: dissolved gas ∝ its pressure. Solubility is fixed at fixed TT.
VVpplast drop gonepvapp_{\rm vap}liquid left: pp fixedall vapour: p=nRT/Vp=nRT/V
Closed vessel at fixed TT: the vapour pressure does not depend on volume while some liquid remains
  • A pure liquid's vapour pressure depends only on TT: while any liquid is left it stays fixed, whatever the volume. Once no liquid is left, p=nRT/Vp=nRT/V and falls as VV grows. A non-volatile solute lowers the vapour pressure.
  • Not 'more ice' on compressing ice and water. Ice has the larger volume, so pressure favours the liquid and lowers the melting point. At 273.15 K and above 1 atm, the ice melts.

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Practise Chemical Equilibrium

16 more concepts in 4 topics

Law of mass action and K, Reaction quotient and equilibrium calculations, Gibbs energy and temperature dependence of K and Le Chatelier's principle. Read the first part free. Sign in free with Google to read the rest and download the PDF.

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Law of mass action and K

  • Law of mass action: Kc7×
  • Kp and its relation to Kc9×
  • Units of Kc and Kp2×
  • Heterogeneous equilibria5×
  • Manipulating K10×

Reaction quotient and equilibrium calculations

  • Reaction quotient Q6×
  • Partial pressures in an ICE table12×
  • Degree of dissociation and Kp13×
  • Small α with a fractional coefficient5×

Gibbs energy and temperature dependence of K

  • ΔG° and K13×
  • Temperature dependence: van 't Hoff2×

Le Chatelier's principle

  • Adding or removing a species7×
  • Pressure or volume change4×
  • Adding an inert gas6×
  • Temperature change3×
  • Effect of a catalyst2×

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Practise Chemical Equilibrium

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