Equilibrium
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Direct answer
Equilibrium in a reversible reaction is dynamic: forward and backward rates become equal, concentrations stop changing, and the ratio they settle into defines the equilibrium constant Kc (or Kp for gases, with Kp = Kc(RT)^Δn). The reaction quotient Q compared with K tells the direction of shift, and Le Chatelier's principle predicts how concentration, pressure or temperature changes disturb the balance — though only temperature actually changes K itself. The ionic half builds pH = −log[H+], the ionic product of water Kw = 10^-14 at 298 K, Henderson's buffer equation, and the solubility product Ksp that governs precipitation.
What you must remember
- Pure solids and pure liquids do not appear in the equilibrium expression; only gases and aqueous species do.
- Haber process: N2 + 3H2 ⇌ 2NH3 run at about 700 K and 200 atm over finely divided iron with molybdenum as promoter.
- Contact process for H2SO4: about 720 K and 2 bar over V2O5 — the compromise between yield and rate that NCERT explains.
- Adding an inert gas at constant volume changes nothing; at constant pressure it shifts equilibrium towards more gas moles.
- For conjugate acid-base pairs, pKa + pKb = pKw (14 at 298 K).
- Ka of acetic acid is 1.8 × 10^-5, so its buffer with sodium acetate sits near pH 4.74.
- A buffer of equimolar weak acid and salt has pH = pKa; buffer capacity is maximum there.
- Ksp with common ion: Mg(OH)2 dissolves less in NaOH than in water — the common ion effect that also drives qualitative analysis.
A buffer, taken apart step by step
An NCERT-style favourite: equal volumes of 0.1 M acetic acid and 0.1 M sodium acetate are mixed — what is the pH? Apply Henderson's equation, pH = pKa + log([salt]/[acid]). The concentrations are equal, so the log term is log 1 = 0 and pH = pKa = −log(1.8 × 10^-5) = 5 − log 1.8 = 5 − 0.26 = 4.74. Now push the logic one step further, because NEET usually does: add a small amount of HCl. The added H+ is consumed by acetate ions to form more acetic acid, so the ratio [salt]/[acid] falls slightly, say to 0.9/1.1 — the log term changes by about 0.09 units, and the pH barely moves. That resistance is the entire meaning of buffer action, and the arithmetic shows why the buffer works best when acid and salt are equimolar: the ratio can absorb disturbance in either direction. Dilution leaves the pH unchanged because the ratio survives; this single statement defeats a whole family of trick questions.
Traps in the question stem
The first error is claiming a catalyst increases the equilibrium yield — it accelerates both directions equally and leaves K untouched. The second is applying Le Chatelier's pressure rule when the number of gas moles is equal on both sides, as in N2 + O2 ⇌ 2NO, where pressure changes nothing. Third, temperature direction: for an exothermic reaction, K falls as T rises, which is why the Haber process is not run at the lowest possible temperature — rate would collapse — but at a compromise 700 K; the examiner's assertion-reason format targets exactly this tension between thermodynamics and kinetics. In ionic equilibrium, students forget that pH of a strong acid is fixed by concentration alone, and that Ksp problems require the proper power — for CaF2, Ksp = [Ca2+][F−]^2, and forgetting the square doubles the solubility error.
Frequently asked questions
What happens to N2 + 3H2 ⇌ 2NH3 when the pressure is increased?
Equilibrium shifts right, towards fewer gas moles, so ammonia yield rises; this is why the Haber process uses about 200 atm.
How does raising the temperature affect K for an exothermic reaction?
K decreases, because the backward (endothermic) direction is favoured; K is temperature-dependent and unaffected by catalysts or concentration changes.
What is the pH of an equimolar acetic acid and sodium acetate mixture?
4.74, since pH = pKa + log(salt/acid) and the log term vanishes for equal concentrations.
Why is the Contact process operated at only 2 bar and 720 K?
Because the conversion of SO2 to SO3 is already nearly complete at low pressure, and 720 K balances the exothermic equilibrium with a workable rate over the V2O5 catalyst.
Which species are omitted from an equilibrium constant expression?
Pure solids and pure liquids, including solvent water in dilute aqueous equilibria; only gases and dissolved species enter.