Everything from the 5 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.
In a closed system a reversible reaction reaches dynamic equilibrium: the forward and reverse reactions continue at equal rates, so the concentrations stay constant. Le Chatelier's principle: if a condition of a system at equilibrium is changed, the position of equilibrium shifts to minimise the change. Adding a reactant shifts it right; raising the pressure shifts it towards fewer gas molecules; raising the temperature shifts it in the endothermic direction. A catalyst speeds up both reactions equally, so it does not move the position; it only gets there faster.
A catalyst does not change the position of equilibrium or the yield; it increases the rate of the forward and reverse reactions equally. And when explaining a pressure change, always count the moles of gas on each side of the equation; if they are equal, pressure has no effect on the position.
For aA + bB ⇌ cC + dD, Kc = [C]c[D]d ÷ [A]a[B]b, using equilibrium concentrations in mol dm−3. Products go on top, and each concentration is raised to the power of its coefficient. The units come from cancelling the concentration units, and may be none at all. A large Kc means the position lies to the right. Kc changes only with temperature; concentration, pressure and catalysts leave it unchanged.
Use equilibrium concentrations, not initial amounts, and convert moles to concentrations by dividing by the volume unless the volumes cancel. Products always go on top. Work out the units separately each time by cancelling, rather than guessing.
The partial pressure of a gas is the pressure it would exert alone in the container: p = mole fraction × total pressure, where mole fraction = moles of that gas ÷ total moles of gas. The partial pressures add up to the total pressure. Kp is written like Kc but with partial pressures: for N2O4 ⇌ 2NO2, Kp = p(NO2)2 ÷ p(N2O4). Units come from the pressure units (Pa, kPa or atm). Like Kc, Kp changes only with temperature.
Partial pressure uses the mole fraction of the equilibrium mixture, so first find the equilibrium moles and the new total. A common slip is dividing by the initial number of moles. Also remember that increasing the total pressure changes the position of equilibrium but not the value of Kp.
A Brønsted–Lowry acid is a proton (H+) donor; a base is a proton acceptor. When an acid donates a proton it becomes its conjugate base; when a base accepts one it becomes its conjugate acid. Each pair differs by one H+. A strong acid dissociates fully in water (HCl, HNO3, H2SO4); a weak acid dissociates only partially, setting up an equilibrium (CH3COOH). Strength is about the extent of dissociation, not concentration.
Strong and concentrated mean different things. Strong means fully dissociated; concentrated means a lot of acid per dm3. A dilute solution of HCl is still a strong acid. Also, in a conjugate pair the formulae differ by exactly one H+: the conjugate base of H2SO4 is HSO4−, not SO42−.
pH = −log10[H+]. For a strong monobasic acid, [H+] equals the acid concentration, so 0.0100 mol dm−3 HCl has pH 2.00; each tenfold dilution raises the pH by one. A titration curve plots pH against volume added. Near the equivalence point the pH changes sharply, and the vertical section sits around pH 7 for strong acid/strong base, below 7 for strong acid/weak base, and above 7 for weak acid/strong base. A suitable indicator must change colour entirely within that vertical section: methyl orange (3.1 to 4.4) for strong acid/weak base, phenolphthalein (8.3 to 10.0) for weak acid/strong base, either for strong/strong, and none for weak/weak.
The equivalence point is where the acid and base have reacted in exactly the ratio of the equation; it is not always pH 7. And the end point of an indicator is where it changes colour, which is only a good estimate of equivalence if the indicator's range lies inside the steep part of the curve.